tag:blogger.com,1999:blog-88830342016-09-20T14:22:15.888+02:00atdotdeRobert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.comBlogger177125tag:blogger.com,1999:blog-8883034.post-39458583830332274852016-09-19T21:43:00.000+02:002016-09-19T21:43:06.273+02:00Brute forcing Crazy Game PuzzlesIn the 1980s, as a kid I loved my Crazy Turtles Puzzle ("Das verrückte Schildkrötenspiel"). For a number of variations, see <a href="http://www.geekyhobbies.com/the-crazy-game-puzzles-puzzled/">here</a> or <a href="http://www.penguin.com/static/packages/us/yr-microsites/crazygamesolution/index.php">here</a>.<br /><br />I had completely forgotten about those, but a few days ago, I saw a self-made reincarnation when staying at a friends' house:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-3EsTQ25UG2c/V-A8U8PBycI/AAAAAAAALK0/d7Wo4YqA95wxdatwCa6ZuR9_f69jA0ntgCLcB/s1600/IMG_0350%2B%25281%2529.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://4.bp.blogspot.com/-3EsTQ25UG2c/V-A8U8PBycI/AAAAAAAALK0/d7Wo4YqA95wxdatwCa6ZuR9_f69jA0ntgCLcB/s320/IMG_0350%2B%25281%2529.jpg" width="319" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div>I tried a few minutes to solve it, unsuccessfully (in case it is not clear: you are supposed to arrange the nine tiles in a square such that they form color matching arrows wherever they meet).<br /><br />So I took the picture above with the plan to either try a bit more at home or write a program to solve it. Yesterday, I had about an hour and did the latter. I am a bit proud of the implementation I came up with and in particular the fact that I essentially came up with a correct program: It came up with the unique solution the first time I executed it. So, here I share it:<br /><br /><pre style="background-color: #001800; color: #55cc66;"><span style="color: #b96969;">#!</span><span style="color: #007997;">/usr/bin/perl</span><br /><br /><span style="color: #b96969;"># 1 rot 8</span><br /><span style="color: #b96969;"># 2 gelb 7</span><br /><span style="color: #b96969;"># 3 gruen 6</span><br /><span style="color: #b96969;"># 4 blau 5</span><br /><br />@karten <span style="color: #808030;">=</span> <span style="color: #808030;">(</span><span style="color: #778c77;">7151</span><span style="color: #808030;">,</span> <span style="color: #778c77;">6754</span><span style="color: #808030;">,</span> <span style="color: #778c77;">4382</span><span style="color: #808030;">,</span> <span style="color: #778c77;">2835</span><span style="color: #808030;">,</span> <span style="color: #778c77;">5216</span><span style="color: #808030;">,</span> <span style="color: #778c77;">2615</span><span style="color: #808030;">,</span> <span style="color: #778c77;">2348</span><span style="color: #808030;">,</span> <span style="color: #778c77;">8253</span><span style="color: #808030;">,</span> <span style="color: #778c77;">4786</span><span style="color: #808030;">)</span><span style="color: purple;">;</span><br /><br /><span style="color: #508050; font-weight: bold;">foreach</span> $karte<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> $farbe<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #808030;">[</span><span style="color: #808030;">split</span><span style="color: #cc5555;"> </span><span style="color: maroon;">/</span><span style="color: maroon;">/</span><span style="color: #808030;">,</span>$karten<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: purple;">;</span><br /><span style="color: purple;">}</span><br /><span style="color: #808030;">&</span>ausprobieren<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">)</span><span style="color: purple;">;</span><br /><br /><span style="color: #508050; font-weight: bold;">sub</span> ausprobieren <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">my</span> $pos <span style="color: #808030;">=</span> <span style="color: #400000;">shift</span><span style="color: purple;">;</span><br /><br /> <span style="color: #508050; font-weight: bold;">foreach</span> <span style="color: #508050; font-weight: bold;">my</span> $karte<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">next</span> <span style="color: #508050; font-weight: bold;">if</span> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: purple;">;</span><br /> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #778c77;">1</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">foreach</span> <span style="color: #508050; font-weight: bold;">my</span> $dreh<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">3</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">if</span> <span style="color: #808030;">(</span>$pos <span style="color: #808030;">%</span> <span style="color: #778c77;">3</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #b96969;"># Nicht linke Spalte</span><br /> $suche <span style="color: #808030;">=</span> <span style="color: #778c77;">9</span> <span style="color: #808030;">-</span> $farbe<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">1</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">1</span> <span style="color: #808030;">-</span> $drehung<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">1</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">next</span> <span style="color: #508050; font-weight: bold;">if</span> $farbe<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">3</span> <span style="color: #808030;">-</span> $dreh<span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span> <span style="color: #808030;">!</span><span style="color: #808030;">=</span> $suche<span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /> <span style="color: #508050; font-weight: bold;">if</span> <span style="color: #808030;">(</span>$pos <span style="color: #808030;">></span><span style="color: #808030;">=</span> <span style="color: #778c77;">3</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #b96969;"># Nicht oberste Zeile</span><br /> $suche <span style="color: #808030;">=</span> <span style="color: #778c77;">9</span> <span style="color: #808030;">-</span> $farbe<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">3</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">2</span> <span style="color: #808030;">-</span> $drehung<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">3</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">next</span> <span style="color: #508050; font-weight: bold;">if</span> $farbe<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">4</span> <span style="color: #808030;">-</span> $dreh<span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span> <span style="color: #808030;">!</span><span style="color: #808030;">=</span> $suche<span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /><br /> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #778c77;">1</span><span style="color: purple;">;</span><br /> $gelegt<span style="color: #808030;">[</span>$pos<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> $karte<span style="color: purple;">;</span><br /> $drehung<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> $dreh<span style="color: purple;">;</span><br /> <span style="color: #b96969;">#print @gelegt[0..$pos]," ",@drehung[0..$pos]," ", 9 - $farbe[$gelegt[$pos - 1]]->[(1 - $drehung[$gelegt[$pos - 1]]) % 4],"\n";</span><br /> <br /> <span style="color: #508050; font-weight: bold;">if</span> <span style="color: #808030;">(</span>$pos <span style="color: #808030;">=</span><span style="color: #808030;">=</span> <span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">print</span> <span style="color: #cc5555;">"Fertig!</span><span style="color: #aa3333; font-weight: bold;">\n</span><span style="color: #cc5555;">"</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">for</span> $l<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">print</span> <span style="color: #cc5555;">"</span><span style="color: #cc5555;">$gelegt</span><span style="color: #808030;">[</span><span style="color: #cc5555;">$l</span><span style="color: #808030;">]</span><span style="color: #cc5555;"> </span><span style="color: #cc5555;">$drehung</span><span style="color: #808030;">[</span><span style="color: #cc5555;">$gelegt</span><span style="color: #808030;">[</span><span style="color: #cc5555;">$l</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #aa3333; font-weight: bold;">\n</span><span style="color: #cc5555;">"</span><span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /> <span style="color: purple;">}</span> <span style="color: #508050; font-weight: bold;">else</span> <span style="color: purple;">{</span><br /> <span style="color: #808030;">&</span>ausprobieren<span style="color: #808030;">(</span>$pos <span style="color: #808030;">+</span> <span style="color: #778c77;">1</span><span style="color: #808030;">)</span><span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /> <span style="color: purple;">}</span><br /> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #778c77;">0</span><span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /><span style="color: purple;">}</span></pre><br />Sorry for variable names in German, but the idea should be clear. Regarding the implementation: red, yellow, green and blue backs of arrows get numbers 1,2,3,4 respectively and pointy sides of arrows 8,7,6,5 (so matching combinations sum to 9).<br /><br />It implements depth first tree search where tile positions (numbered 0 to 8) are tried left to write top to bottom. So tile $n$ shares a vertical edge with tile $n-1$ unless it's number is 0 mod 3 (leftist column) and it shares a horizontal edge with tile $n-3$ unless $n$ is less than 3, which means it is in the first row.<br /><br />It tries rotating tiles by 0 to 3 times 90 degrees clock-wise, so finding which arrow to match with a neighboring tile can also be computed with mod 4 arithmetic.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-39113532392132865222016-06-20T22:12:00.001+02:002016-06-20T22:12:33.408+02:00Restoring deleted /etc from TimeMachineYesterday, I managed to empty the /etc directory on my macbook (don't ask how I did it. I was working on <a href="http://subsurface-divelog.org/">subsurface</a> and had written a perl script to move system files around that had to be run with sudo. And I was still debugging...).<br /><br />Anyway, once I realized what the problem was I did some googling but did not find the answer. So here, as a service to fellow humans googling for help is how to fix this.<br /><br />The problem is that in /etc all kinds of system configuration files are stored and without it the system does not know anymore how to do a lot of things. For example it contains /etc/passwd which contains a list of all users, their home directories and similar things. Or /etc/shadow which contains (hashed) passwords or, and this was most relevant in my case, /etc/sudoers which contains a list of users who are allowed to run commands with <a href="https://xkcd.com/149/">sudo</a>, i.e. execute commands with administrator privileges (in the GUI this shows as as a modal dialog asking you to type in your password to proceed).<br /><br />In my case, all was gone. But, luckily enough, I had a time machine backup. So I could go 30 minutes back in time and restore the directory contents.<br /><br />The problem was that after restoring it, it ended up as a symlink to /private/etc and user helling wasn't allowed to access its contents. And I could not sudo the access since the system could not determine I am allowed to sudo since it could not read /etc/sudoers.<br /><br />I tried a couple of things including a reboot (as a last resort I figured I could always boot in target disk mode and somehow fix the directory) but it remained in /private/etc and I could not access it.<br /><br />Finally I found the solution (so here it is): I could look at the folder in Finder (it had a red no entry sign on it meaning that I could not open it). But I could right click and select Information and there I could open the lock by tying in my password (no idea why that worked) and give myself read (and for that matter write) permissions and then everything was fine again.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-24954675567200031522016-05-24T12:03:00.001+02:002016-05-25T09:25:38.082+02:00Holographic operator ordering?Believe it or not, at the end of this week I will speak at <a href="https://www-m5.ma.tum.de/Allgemeines/LQPWorkshop">a workshop on algebraic and constructive quantum field theory</a>. And (I don't know which of these two facts is more surprising) I will advocate holography.<br /><br />More specifically, I will argue that it seems that holography can be a successful approach to formulate effective low energy theories (similar to other methods like perturbation theory of weakly coupled quasi particles or minimal models). And I will present this as a challenge to the community at the workshop to show that the correlators computed with holographic methods indeed encode a QFT (according to your favorite set of rules, e.g. Whiteman or Osterwalder-Schrader). My [kudos to an anonymous reader for pointing out a typo] guess would be that this has a non-zero chance of being a possible approach to the construction of (new) models in that sense or alternatively to show that the axioms are violated (which would be even more interesting for holography).<br /><br />In any case, I am currently preparing my slides (I will not be able to post those as I have stolen far too many pictures from the interwebs including the holographic doctor from Star Trek Voyager) and came up with the following question:<br /><br /><blockquote class="tr_bq">In a QFT, the order of insertions in a correlator matters (unless we fix an ordering like time ordering). How is that represented on the bulk side?</blockquote><br />Does anybody have any insight about this?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://www.giantfreakinrobot.com/wp-content/uploads/2014/06/helpmeobiwankenobi.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://www.giantfreakinrobot.com/wp-content/uploads/2014/06/helpmeobiwankenobi.jpg" height="136" width="320" /></a></div>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com3tag:blogger.com,1999:blog-8883034.post-82510413377926125202016-04-21T14:55:00.000+02:002016-04-21T14:55:23.079+02:00The Quantum in Quantum ComputingI am sure, by now, all of you have seen Canada's prime minister <a href="https://www.youtube.com/watch?v=rRmv4uD2RQ4">"explain" quantum computers</a> at Perimeter. It's really great that politicians care about these things and he managed to say what is the standard explanation for the speed up of quantum computers compared to their classical cousins: It is because you can have superpositions of initial states and therefore "perform many operations in parallel".<br /><br />Except of course, that this is bullshit. This is not the reason for the speed up, you can do the same with a classical computer, at least with a probabilistic one: You can also as step one perform a random process (throw a coin, turn a Roulette wheel, whatever) to determine the initial state you start your computer with. Then looking at it from the outside, the state of the classical computer is mixed and the further time evolution also "does all the computations in parallel". That just follows from the formalism of (classical) statistical mechanics.<br /><br />Of course, that does not help much since the outcome is likely also probabilistic. But it has the same parallelism. And as the state space of a qubit is all of a Bloch sphere, the state space of a classical bit (allowing mixed states) is also an interval allowing a continuum of intermediate states.<br /><br />The difference between quantum and classical is elsewhere. And it has to do with non-commuting operators (as those are essential for quantum properties) and those allow for entanglement.<br /><br />To be more specific, let us consider one of the most famous quantum algorithms, <a href="https://en.wikipedia.org/wiki/Grover%27s_algorithm">Grover's database lookup</a>, There the problem (at least in its original form) is to figure out which of $N$ possible "boxes" contains the hidden coin. Classically, you cannot do better than opening one after the other (or possibly in a random pattern), which takes $O(N)$ steps (on average).<br /><br />For the quantum version, you first have to say how to encode the problem. The lore is, that you start with an $N$-dimensional Hilbert space with a basis $|1\rangle\cdots|N\rangle$. The secret is that one of these basis vectors is picked. Let's call it $|\omega\rangle$ and it is given to you in terms of a projection operator $P=|\omega\rangle\langle\omega|$.<br /><br />Furthermore, you have at your disposal a way to create the flat superposition $|s\rangle = \frac1{\sqrt N}\sum_{i=1}^N |i\rangle$ and a number operator $K$ that act like $K|k\rangle= k|k\rangle$, i.e. is diagonal in the above basis and is able to distinguish the basis elements in terms of its eigenvalues.<br /><br />Then, what you are supposed to do is the following: You form two unitary operators $U_\omega = 1 - 2P$ (this multiplies $|\omega\rangle$ by -1 while being the identity on the orthogonal subspace, i.e. is a reflection on the plane orthogonal to $|\omega\rangle$) and $U_s = 2|s\rangle\langle s| - 1$ which reflects the vectors orthogonal to $|s\rangle$.<br /><br />It is not hard to see that both $U_s$ and $U_\omega$ map the two dimensional place spanned by $|s\rangle$ and $|\omega\rangle$ into itself. They are both reflections and thus their product is a rotation by twice the angle between the two planes which is given in terms of the scalar product $\langle s|\omega\rangle =1/\sqrt{N}$ as $\phi =\sin^{-1}\langle s|\omega\rangle$.<br /><br />But obviously, using a rotation by $\cos^{-1}\langle s|\omega\rangle$, one can rotate $|s\rangle$ onto $\omega$. So all we have to do is to apply the product $(U_sU\omega)^k$ where $k$ is the ratio between these two angles which is $O(\sqrt{N})$. (No need to worry that this is not an integer, the error is $O(1/N)$ and has no influence). Then you have turned your initial state $|s\rangle$ into $|omega\rangle$ and by measuring the observable $K$ above you know which box contained the coin.<br /><br />Since this took only $O(\sqrt{N})$ steps this is a quadratic speed up compared to the classical case.<br /><br />So how did we get this? As I said, it's not the superposition. Classically we could prepare the probabilistic state that opens each box with probability $1/N$. But we have to expect that we have to do that $O(N)$ times, so this is essential as fast as systematically opening one box after the other.<br /><br />To have a better unified classical-quantum language, let us say that we have a state space spanned by $N$ pure states $1,\ldots,N$. What we can do in the quantum case is to turn an initial state which had probability $1/N$ to be in each of these pure states into one that is deterministically in the sought after state.<br /><br />Classically, this is impossible since no time evolution can turn a mixed state into a pure state. One way to see this is that the entropy of the probabilistic state is $\log(N)$ while it is 0 for the sought after state. If you like classically, we only have the observables given by C*-algebra generated by $K$, i.e. we can only observe which box we are dealing with. Both $P$ and $U_\omega$ are also in this classical algebra (they are diagonal in the special basis) and the strict classical analogue would be that we are given a rank one projector in that algebra and we have to figure out which one.<br /><br />But quantum mechanically, we have more, we also have $U_s$ which does not commute with $K$ and is thus not in the classical algebra. The trick really is that in this bigger quantum algebra generated by both $K$ and $U_s$, we can form a pure state that becomes the probabilistic state when restricted to the classical algebra. And as a pure state, we can come up with a time evolution that turns it into the pure state $|\omega\rangle$.<br /><br />So, this is really where the non-commutativity and thus the quantumness comes in. And we shouldn't really expect Trudeau to be able to explain this in a two sentence statement.<br /><br />PS: The actual speed up in the end comes of course from the fact that probabilities are amplitudes squared and the normalization in $|s\rangle$ is $1/\sqrt{N}$ which makes the angle to be rotated by proportional to $1/\sqrt{N}$. Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-86367287727361332082016-04-21T13:56:00.000+02:002016-04-21T13:56:13.874+02:00One more resuscitationThis blog has been silent for almost two years for a number of reasons. First, I myself stopped reading blogs on a daily basis as in open Google Reader right after the arXiv an checking what's new. I had already stopped doing that due to time constraints before Reader was shut down by Google and I must say I don't miss anything. My focus shifted much more to Twitter and Facebook and from there, I am directed to the occasional blog post, but as I said, I don't check them systematically anymore. And I assume others do the same.<br /><br />But from time to time I run into things that I would like to discuss on a blog. Where (as my old readers probably know) I am mainly interested in discussions. I don't write here to educate (others) but only myself. I write about something I found interesting and would like to have further input on.<br /><br />Plus, this should be more permanent than a Facebook post (which is gone once scrolled out of the bottom of the screen) and more than the occasional 160 character remark on Twitter.<br /><br />Assuming that others have adopted their reading habits in a similar way to the year 2016, I have set up <a href="https://ifttt.com/recipes">If This Than That</a> to announce new posts to FB and Twitter so others might have a chance to find them.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-20727835141606217732014-05-23T16:15:00.000+02:002014-05-23T16:15:00.681+02:00Conference Fees for SpeakersListening to a podcast on open access I had an idea: Many conferences waive conference fees (which can be substantial) for invited speakers. But those are often enough the most senior people who would have the least difficulty in paying the fee from their budget or grant money. So wouldn't it be a good idea for conferences to offer to their invited speakers to instead waive the fee for a graduate student or junior post-doc of the speakers choice and make the speaker pay the fee from their grant (or reduce the fee by 50% for both)?<br /><br />Discuss!Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-74991781516766068172014-01-29T21:01:00.000+01:002014-01-29T21:01:01.382+01:00Questions to the inter webs: classical 't-Hooft-limit and path integral entanglementHey blog, long time no see!<br /><br />I am coming back to you with a new format: Questions. Let me start with two questions I have been thinking about recently but that I don't know a good answer to.<br /><br /><h4>'t Hooft limit of classical field equations</h4><div>The 't Hooft limit leads to important simplifications in perturbative QFT and is used for many discoveries around AdS/CFT, N=4 super YM, amplitudes etc etc. You can take it in its original form for SU(N) gauge theory where its inventor realized you can treat N as a parameter of the theory and when you do perturbation theory you can do so in terms of ribbon Feynman diagrams. Then a standard analysis in terms of Euler's polyhedron theorem (discrete version of the Gauss-Bonnet-theorem) shows that genus g diagrams come with a factor 1/N^g such that at leading order for large N only the planar diagrams survive.</div><div><br /></div><div>The argument generalizes to all kinds of theories with matrix valued fields where the action can be written as a single trace. In a similar vain, it also has a version for non-commutative theories on the Moyal plane.</div><div><br /></div><div>My question is now if there is a classical analogue of this simplification. Let's talk the classical equations of motion for SU(N) YM or any of the other theories, maybe something as simple as</div><div>d^2/dt^2 M = M^3 for NxN matrices M. Can we say anything about simplifications of taking the large N limit? Of course you can use tree level Feynman diagrams to solve those equations perturbatively (as for example I described <a href="http://homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf">here</a>), but is there a non-perturbative version of "planar"?</div><div>Can I say anything about the structure of solutions to these equations that is approached for N->infinity?</div><div><br /></div><h4>Path Integral Entanglement</h4><div>Entanglement is <b>the</b> distinguishing feature of quantum theory as compared to classical physics. It is closely tied to the non-comutativity of the observable algebra and is responsible for things like the violation of Bell's inequality.</div><div><br /></div><div>On the other hand, we know that the path integral gives us an equivalent description of quantum physics, surprisingly in terms of configurations/paths of the classical variables (that we then have to take the weighted integral over) which are intrinsically commuting objects. </div><div><br /></div><div>Properties of non-commuting operators can appear in subtle ways, like the operator ordering ambiguity how to quantize the classical observable x^2p^2, should it be xp^2x or px^2p or for example (x^2p^2 + p^2x^2)/2? This is a true quantization ambiguity and the path integral has to know about it as well. It turns out, it does: When you show the equivalence of Schroedinger's equation and the path integral, you do that by considering infinitesimal paths and you have to evaluate potentials etc on some point of those paths to compute things like V(x) in the action. Turns out, the operator ambiguity is equivalent to choosing where to evaluate V(x), at the start of the path, the end, the middle or somewhere else.</div><div><br /></div><div>So far so good. The question that I don't know the answer to is how the path integral encodes entanglement. For example can you discuss a version of Bell's inequality (or similar like GHZ) in the path integral language? Of course you would have to translate the spin operators to positions .</div>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com7tag:blogger.com,1999:blog-8883034.post-90425870797226114552012-11-06T11:57:00.000+01:002012-11-06T11:57:26.848+01:00A Few Comments On FirewallsI was stupid enough to agree to talk about <a href="http://inspirehep.net/record/1122534">Firewalls</a> in our strings lunch seminar this Wednesday without having read the paper (or <a href="http://inspirehep.net/search?ln=en&p=refersto%3Arecid%3A1122534">what other people say about them</a>) except for talking to Raphael Busso at the Strings 2012 conference and reading <a href="http://blogs.discovermagazine.com/cosmicvariance/2012/09/27/guest-post-joe-polchinski-on-black-holes-complementarity-and-firewalls/#more-8862">Joe Polichinski's guest post</a> over at the Cosmic Variance blog.<br /><br />Now, of course I had to read (some of) the papers and I have to say that I am confused. I admit, I did not get the point. Even more, I cannot understand a large part of the discussion. There is a lot of prose and very little formulas and I have failed to translate the prose to formulas or hard facts for myself. Many of the statements taken at face value do not make sense to me but on the other hand, I know the authors to be extremely clever people and thus the problem is most likely on my end.<br /><br />In this post, I would like to share some of my thoughts in my endeavor to decode these papers but probably they are to you even more confusing than the original papers to me. But maybe you can spot my mistakes and correct me in the comment section.<br /><br />I had a long discussion with Cristiano Germani on these matters for which I am extremely grateful. If this post contains any insight it is his while all errors are for course mine.<br /><br /><h3>What is the problem?</h3><div>I have a very hard time not to believe in "no drama", i.e. that anything special can happen at an event horizon. First of all, the event horizon is a global concept and its location now does in general depend on what happens in the future (e.g. how much further stuff is thrown in the black hole). So who can it be that the location of a anything like a firewall can depend on future events?</div><div><br /></div><div>Furthermore, I have never seen such a firewall so far. But I might have already passed an event horizon (who knows what happens at cosmological scales?). Even more, I cannot see a local difference between a true event horizon like that of a black hole and the horizon of an accelerated observer in the case of the Unruh-effect. That the later I am pretty sure I have crossed already many times and I have never seen a firewall.</div><div><br /></div><div>So I was trying to understand why there should be one. And whenever I tried to flesh out the argument for one they way I understood it it fell apart. So, here are some of my thoughts;</div><br /><h3>The classical situation</h3><div>No question, Hawking radiation is a quantum effect (even though it happens at tree level in QFT on curved space-time and is usually derived in a free theory or, equivalently, by studying the propagator). But apart from that not much of the discussion (besides possibly the monogamy of entanglement, see below) seems to be particular quantum. Thus we might gain some mileage by studying classical field theory on the space time of a forming and decaying black hole as given by the causal diagram:</div><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://prime-spot.de/Bilder/BR/bhevap_l.jpg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://prime-spot.de/Bilder/BR/bhevap_l.jpg" width="230" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><span style="font-size: small; text-align: -webkit-auto;">A decaying black hole, image stolen from</span><a href="http://prime-spot.de/" style="font-size: medium; text-align: -webkit-auto;"> Sabine Hossenfelder</a><span style="font-size: small; text-align: -webkit-auto;">.</span></td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">Issues of causality a determined by the characteristics of the PDE in question (take for example the wave equation) and those are invariant under conformal transformations even if the field equation is not. So, it is enough to consider the free wave equation on the causal diagram (rather than the space-time related to it by a conformal transformation). </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">For example we can give initial data on I- (and have good boundary conditions at the r=0 vertical lines). At the dashed horizontal line, the location of the singularity, we just stop evolving (free boundary conditions) and then we can read off outgoing radiation at I+. The only problematic point is the right end of the singularity: This is the end of the black hole evaporation and to me it is not clear how we can here start to impose again some boundary condition at the new r=0 line without affecting what we did earlier. But anyway, this is in a region of strong curvature, where quantum gravity becomes essential and thus what we conclude should better not depend too much on what's going on there as we don't have a good understanding of that regime.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">The firewall paper, when it explains the assumptions of complementarity mentions an S-matrix where it tries to formalize the notion of unitary time evolution. But it seems to me, this might be the wrong formalization as the S-matrix is only about asymptotic states and even fails in much simpler situations when there are bound states and the asymptotic Hilbert spaces are not complete. Furthermore, strictly speaking, this (in the sense of LSZ reduction) is not what we can observe: Our detectors are never at spatial infinity, even if CMS is huge, so we should better come up with a more local concept. <table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: 1em; margin-right: 1em; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-SdrmGhWvECs/UJePShS6UMI/AAAAAAAAKqo/A-qCDo0N_ss/s1600/causal_shadows.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="175" src="http://4.bp.blogspot.com/-SdrmGhWvECs/UJePShS6UMI/AAAAAAAAKqo/A-qCDo0N_ss/s400/causal_shadows.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Two regions M and N on a Cauchy surface C with their causal shadows</td></tr></tbody></table></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">In the case of the wave equation, this can be encoded in terms of domains of dependence: By giving initial data on a region of a Cauchy surface I determine the solution on its causal shadow (in the full quantum theory maybe plus/minus an epsilon for quantum uncertainties). In more detail: If I have two sets of initial data on one Cauchy surface that agree on a local region. Than the two solutions have to agree on the causal shadow of this region no matter what the initial data looks like elsewhere. This encodes that "my time-evolution is good and I do not lose information on the way" in a local fashion.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><h3>States</h3><div>Some of my confusion comes from talking about states in a way that at least when taken at face value is in conflict with how we understand states both in classical and in better understood quantum (both quantum mechanics and quantum field theory) circumstances.<br /><br />First of all (and quite trivially), a state is always at one instant of time, that is it lives on a Cauchy surface (or at least a space-like hyper surface, as our space-time might not be globally hyperbolic), not in a region of space-time. Hilbert space, as the space of (pure) states thus also lives on a Cauchy surface (and not for example in the region behind the horizon). If one event is after another (i.e. in its forward light-cone) it does not make sense to say they belong to different tensor factors of the Hilbert (or different Hilbert spaces for that matter).<br /><br />Furthermore, a state is always a global concept, it is everywhere (in space, but not in time!). There is nothing like "the space of this observer". What you can do of course is restrict a state to a subset of observables (possibly those that are accessible to one observer) by tracing out a tensor factor of the Hilbert space. But in general, the total state cannot be obtained by merging all these restricted states as those lack information about correlations and possible entanglement.<br /><br />This brings me to the next confusion: There is nothing wrong with states containing correlations of space-like separated observables. This is not even a distinguishing property of quantum physics, as this happens all the time even in classical situations: In the morning, I pick a pair of socks from my drawer without turning on the light and put it on my feet. Thus I do not know which socks I am wearing, in particular, I don't know their color. But as I combined matching socks when they came from the washing machine (as far as this is possible given the tendency of socks going missing) I know by looking at the sock on my right foot what the color of the sock on my left foot is, even when my two feet are spatially separated. Before looking, the state of the color of the socks was a statistical mixture but with non-local correlations. And of course there is nothing quantum about my socks (even if in German "Quanten" is not only "quantum" but also a pejorative word for feet). This would even be true (and still completely trivial) if I had put one of my feet through an event horizon while the other one is still outside. This example shows that locality is not a property that I should demand of states in order to be sure my theory is free of time travel. The important locality property is not in the states, it is in the observables: The measurement of an observable here must not depend of whether or not I apply an operator at a space-like distance. Otherwise that would imply I could send signals faster than the speed of light. But it is the operators, not the states that have to be local (i.e. commute for spatial separation).<br /><br />If two operators, however, are time-like separated (i.e. one is after the other in its forward light cone), I can of course influence one's measurement by applying the other. But this is not about correlations, this is about influence. In particular, if I write something in my notebook and then throw it across the horizon of a black hole, there is no point in saying that there is a correlation (or even entanglement) between the notebook's state now and after crossing the horizon. It's just the former influencing the later.</div><br />Which brings us to entanglement. This must not be confused with correlation, the former being a strict quantum property whereas the other can be both quantum or classical. Unfortunately, you can often see this in popular talks about quantum information where many speakers claim to explain entanglement but in fact only explain correlations. As a hint: For entanglement, one must discuss non-commuting observables (like different components of a the same spin) as otherwise (by the GNS reconstruction theorem) one deals with a commutative operator algebra which always has a classical interpretation (functions on a classical space). And of course, it is entanglement which violates Bell's inequality or shows up in the GHZ experiment. But you need something of this complexity (i.e. involving non-commuting observables) to make use of the quantumness of the situation. And it is only this entanglement (and not correlation) that is "monogamous": You cannot have three systems that are fully entangled for all pairs. You can have three spins that are entangled, but once you only look at two they are no longer entangles (which makes quantum cryptography work as the eavesdropper cannot clone the entanglement that is used for coding).<br /><br />And once more, entanglement is a property of a state when it is split according to a tensor product decomposition of the Hilbert space. And thus lives on a Cauchy surface. You can say that a state contains entanglement of two regions on a Cauchy surface but it makes no sense to say to regions that are time-like to each other to be entangled (like the notebook before and after crossing the horizon). And therefore monogamy cannot be invoked with respect to also taking the outgoing radiation in as the third player.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com6tag:blogger.com,1999:blog-8883034.post-88463238880568453532012-09-24T21:46:00.000+02:002012-09-24T21:47:13.873+02:00The future of blogging (for me) and in particular twitterAs you might have noticed, breaks between two posts here get bigger and bigger. This is mainly due to lack of ideas on my side but also as I am busy with other things (now that with Ella H. kid number two has joined the family but there is also a lot of <a href="http://www.theorie.physik.uni-muenchen.de/TMP">TMP</a> admin stuff to do).<br /><br />This is not only true for me writing blog posts but also about reading: Until about a year ago, I was using <a href="http://reader.google.com/">google reader</a> not to miss a single blog post of a list of about 50 blogs. I have completely stopped this and systematically read blogs only very occasionally (that is other than being directed to a specific post by a link from somewhere else).<br /><br />What I still do (and more than ever) is use facebook (mainly to stay in contact with not so computer affine friends) and of course twitter (you will know that I am <a href="http://www.twitter.com/atdotde">@atdotde</a> there). Twitter seems to be the ideal way to stay current on a lot of matters you are interested in (internet politics for example) while not wasting too much time given the 140 character limit.<br /><br />Twitter's only problem is that they don't make (a lot of) money. This is no problem for the original inventors of the site (they have sold their shares to investors) but the current owners now seem desperate to change this. From what they say they want to move twitter more to a many to one (marketing) communication platform and force users to see ads they mix among the genuine tweets.<br /><br />One of the key aspects of the success of twitter was its open API (application programmers interface): Everybody could write programs (and for example I did) that interacted with twitter so for example everybody can choose their favourite client program on any OS to read and write tweets. Since the recent twitter API policy changes this is no longer the case: A client can now have only 100,000 users (or if they already have more can double the number of users), a small number given the allegedly about 4,000,000 million twitter accounts. And there are severe restrictions how you may display tweets to your users (e.g. you are not allowed to use them in any kind of cloud service or mix them with other social media sites, i.e. blend them with Facebook updates). The message that this sends is clearly: "developers go away" (the idea seems to be to force users to use the twitter website and their own clients) and anybody who still invests in twitter developing is betting on a dead horse. But it is not hard to guess that in the long run this will also make the while twitter unattractive to a lot of (if not eventually all) their users.<br /><br />People (often addicted to twitter feeds) are currently evaluating alternatives (like <a href="http://app.net/">app.net</a>) but this morning I realized that maybe the twitter managers are not so stupid as they seem to be (or maybe they just want to cash in what they have and don't care if this ruins the service), there is still an alternative that would make twitter profitable and would secure the service in the long run: They could offer to developers to allow them to use the old API guidelines but for a fee (say a few $/Euros per user per month): This would bring in the cash they are apparently looking for while still keeping the healthy ecosystem of many clients and other programs. twitter.com would only be dealing with developers while those would forward the costs to their users and recollect the money by selling their apps (so twitter would not have to collect money from millions of users). <br /><br />But maybe that's too optimistic and they just want to earn advertising money NOW.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-48443041147952075922012-02-07T18:31:00.004+01:002012-02-07T18:31:46.182+01:00AdS/cond-matLast week, Subir Sachdev came to Munich to give three Arnold Sommerfeld Lectures. I want to take this opportunity to write about a subject that has attracted a lot of attention in recent years, namely applying AdS/CFT techniques to condensed matter systems like trying to write gravity duals for D-wave superconducturs or <a href="http://en.wikipedia.org/wiki/Pseudogap">strange metals</a> (it's surprisingly hard to find a good link for this keyword).<br /><br />My attitude towards this attempt has somewhat changed from "this will never work" to "it's probably as good as anything else" and in this post I will explain why I think this. I should mention as well that Sean Hartnoll has been essential in this phase transition of my mind.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://upload.wikimedia.org/wikipedia/commons/a/a0/Bi2212_Unit_Cell.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="320" src="http://upload.wikimedia.org/wikipedia/commons/a/a0/Bi2212_Unit_Cell.png" width="129" /></a></div>Let me start by sketching (actually: caricaturing) what I am talking about. You want to understand some material, typically the electrons in a horribly complicated lattice like bismuth strontium calcium copper oxide, or <a href="http://en.wikipedia.org/wiki/BSCCO">BSCCO</a>. To this end, you come up with a five dimensional theory of gravity coupled to your favorite list of other fields (gauge fields, scalars with potentials, you name it) and place that in an anti-de-Sitter background (or better, for finite temperature, in an asymptotically anti-de-Sitter black hole). Now, you compute solutions with prescribed behavior at infinity and interpret these via Witten's prescription as correlators in your condensed matter theory. For example you can read off Green functions and (frequency dependent) conductivities, densities of state.<br /><br />How can this ever work, how are you supposed to guess the correct field content (there is no D-brane/string description anywhere near that could help you out) and how can you ever be sure you got it right?<br /><br />The answer is you cannot but it does not matter. It does not matter as it does not matter elsewhere in condensed matter physics. To clarify this, we have to be clear about what it means for a condensed matter theorist to "understand" a system. Expressed in our high energy lingo, most of the time, the "microscopic theory" is obvious: It is given by the Schrödinger equation for $10^23$ electrons plus as similar number of noclei feeling the Coulomb potential of the nuclei and interacting themselves with Coulomb repulsion. There is nothing more to be known about this. Except that this is obviously not what we want. These are far too many particles to worry about and, what is more important, we are interested in the behavior at much much lower energy scales and longer wave lengths, at which all the details of the lattice structure are smoothed out and we see only the effect of a few electrons close to the Fermi surface. As an estimate, one should compare the typical energy scale of the Coulomb interactions, the binding energies of the electrons to the nucleus (Z times 13.6 eV) or in terms of temperature (where putting in the constants equates 1eV to about 10,000K) to the milli-eV binding energy of Cooper pairs or the typical temperature where superconductivity plays a role.<br /><br />In the language of the renormalization group, the Coulomb interactions are the UV theory but we want to understand the effective theory that this flows to in the IR. The convenient thing about such effective theories is that they do not have to be unique: All we want is a simple to understand theory (in which we can compute many quantities that we would like to know) that is in the same universality class as the system we started from. Differences in relevant operators do not matter (at least to leading order).<br /><br />Surprisingly often, one can find free theories or weakly (and thus almost free) theories that can act as the effective theory we are looking for. BCS is a famous example, but Landau's Fermi Liquid Theory is another: There the idea is that you can almost pretend that your fermions are free (and thus you can just add up energies taking into account the Pauli exclusion principle giving you Fermi-surfaces etc) even though your electrons are interacting (remember, there is always the Coulomb interaction around). The only effect the interactions have, is to renormalize the mass, to deform the Fermi surface away from a ball and to change the hight of the jump in the T=0 occupation number. Experience shows that this is an excellent description in more than one dimension (that has the exception of the Luttinger liquid) and can probably traced back to the fact that a four-Fermi-interaction is non-renormalizable and thus invisible in the IR.<br /><br />Only, it is important to remember that the fields/particles in that effective theories are not really the electrons you started with but just quasi-particles that are build in complicated ways out of the microscopic particles carrying around clouds of other particles and deforming the lattice they move in. But these details don't matter and that is the point.<br /><br />It is only important to guess the effective theory in the same universality class. You never derive this (or: hardly ever). Following an exact renormalization group flow is just way beyond what is possible. You make a hopefully educated guess (based on symmetries etc) and then check that you get good descriptions. But only the fact, that there are not too many universality classes makes this process of guessing worthwhile.<br /><br />Free or weakly coupled theories are not the only possible guesses for effective field theories in which one can calculate. 2d conformal field theories are others. And now, AdS-technology gives us another way of writing down correlation functions just as Feynman-rules give us correlation functions for weakly coupled theories. And that is all one needs: Correlation functions of effective field theory candidates. Once you have those you can check if you are lucky and get evidence that you are in the correct universality class. You don't have to derive the IR theory from the UV. You never do this. You always just guess. And often enough this is good enough to work. And strictly speaking, you never know if your next measurement shows deviations from what you thought would be an effective theory for your system.<br /><br />In a sense, it is like the mystery that chemistry works: The periodic table somehow pretends that the electrons in atoms are arranged in states that group together like for the hydrogen atom, you get the same n,l,m,s quantum numbers and the shells are roughly the same (although with some overlap encoded in the <a href="http://en.wikipedia.org/wiki/Aufbau_principle">Aufbau principle</a>) as for hydrogen. This pretends that the only effect of the electron-electron Coulomb potential is to shield the charge of the nucleus and every electron sees effectively a hydrogen like atom (although not necessarily with integer charge Z) and Pauli's exclusion principle regulates that no state is filled more than once. One could have thought that the effect of n-1 electrons on the last is much bigger, after all, they have a total charge that is almost the same of the nucleous, but it seems, the last electron only sees the nucleus with a 1/r potential although with reduced charge.<br /><br />If you like, the only thing one should might worry about is that the Witten prescription to obtain boundary correlators from bulk configurations really gives you valid n-point functions of<i><b> </b>a<b> </b></i>quantum theory (if you feel sufficient mathematical masochism for example in the sense of Wightman) but you don't want to show that it is <i>the</i> quantum field theory corresponding to the material you started with.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-66553343911503286262012-02-03T16:02:00.001+01:002012-02-03T16:02:51.466+01:00Write-upsNot much to say, but I would like to mention that, finally, we have been able two finalize two write-ups that I have announced here in the past:<br /><br />First, there are the notes of a block course that I have in the summer on how to fix some mathematicla lose ends in QFT (notes written by our students Mario Flory and Constantin Sluka):<br /><br /><br /><h1 class="title" style="background-color: white; font-family: 'Lucida Grande', helvetica, arial, verdana, sans-serif; font-size: x-large; line-height: 28px; margin-bottom: 0.5em; margin-left: 20px; margin-right: 0px; margin-top: 0.5em;"><a href="http://arxiv.org/abs/1201.2714">How I Learned to Stop Worrying and Love QFT</a></h1><div class="authors" style="background-color: white; font-family: 'Lucida Grande', helvetica, arial, verdana, sans-serif; line-height: 24px; margin-bottom: 0.5em; margin-left: 20px; margin-right: 0px; margin-top: 0.5em;"><a href="http://arxiv.org/find/math-ph,math/1/au:+Flory_M/0/1/0/all/0/1" style="text-decoration: none;">Mario Flory</a>, <a href="http://arxiv.org/find/math-ph,math/1/au:+Helling_R/0/1/0/all/0/1" style="text-decoration: none;">Robert C. Helling</a>, <a href="http://arxiv.org/find/math-ph,math/1/au:+Sluka_C/0/1/0/all/0/1" style="text-decoration: none;">Constantin Sluka</a></div><blockquote class="abstract" style="background-color: white; font-family: 'Lucida Grande', helvetica, arial, verdana, sans-serif; font-size: 14px; line-height: 19px; margin-bottom: 1.5em;">Lecture notes of a block course explaining why quantum field theory might be in a better mathematical state than one gets the impression from the typical introduction to the topic. It is explained how to make sense of a perturbative expansion that fails to converge and how to express Feynman loop integrals and their renormalization using the language of distribtions rather than divergent, ill-defined integrals.</blockquote><br />Then there are the contributions to a seminar on "<a href="https://wiki.physik.uni-muenchen.de/TMP/images/1/1b/Foundations.pdf">Foundations of Quantum Mechanics</a>" (including an introduction by your's truly) that I taught a year ago. From the contents:<br /><br /><br /><ol><li>C*-algebras, GNS-construction, states, (Sebastian)</li><li>Stone-von-Neumann Theorem (Dennis)</li><li>Pure Operations, POVMs (Mario)</li><li>Measurement Problem (Anupam, David)</li><li>EPR and Entanglement, Bell's Theorem, Kochen–Specker theorem (Isabel, Matthias)</li><li>Decoherence (Kostas, Cosmas)</li><li>Pointer Basis (Greeks again)</li><li>Consistent Histories (Hao)</li><li>Many Worlds (Max)</li><li>Bohmian Interpretation (Henry, Franz)</li></ol><div>See also the seminar's <a href="https://wiki.physik.uni-muenchen.de/TMP/index.php/Foundations_of_Quantum_Mechanics_Seminar_WS_10/11">wiki page</a>.</div><div><br /></div><div>Have fun!</div><h1 style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0.17em; padding-top: 0.5em;"><br /><span class="mw-headline" id="Possible_Literature" style="font-weight: bolder; line-height: 24px;"></span></h1><br />Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-62223074441715210722011-11-30T09:31:00.001+01:002011-11-30T10:22:47.475+01:00More than one nature for natural unitsHey blog, long time no see! Bee has put together <a href="http://backreaction.blogspot.com/2011/11/what-are-natural-units.html">a nice video</a> on natural units. There are one or two aspects that I would put slightly differently and rather than writing a comment I thought it might better be to write a post myself. <p>The first thing is that strictly speaking, there is not <i>the</i> natural unit system, it depends on the problem you are interested in. For example, if you are interested in atoms, the typical mass is that of the electron, so you will likely be interested in masses as multiples of $m_e$. Then, interactions are Coulomb and you will want to express charges as multiples of the electron charge $e$. Finally, quantum mechanics is your relevant framework, so it is natural to express actions in multiples of $\hbar$. Then a quick calculation shows that this unit system of setting $m_e=e=\hbar=1$ implies that distances are dimensionless and the distance $r=1$ happens to be the Bohr radius that sets the natural scale for the size of atoms. Naturalness here lets you guess the size of an atom from just identifying the electron mass, the electric charge and quantum mechanics to be the relevant ingredients. <p>When you are doing high energy particle physics quantum physics and special relativity are relevant and thus it is convenient to use units in which $\hbar=c=1$ which is Bee's example. In this unit system, masses and energy have inverse units of length. <p>If you are a classical relativist contemplating solutions of Einstein's equations, then quantum mechanics (and thus $\hbar$) does not concern you but Newton's constant $G$ does. These people thus use units with $c=G=1$. Confusingly, in this unit system, masses have units of length (and not inverse length as above). In particular, the length scale of a black hole with mass M, the Schwarzschild radius is $R=2M$ (the 2 being there to spice up life a bit). So you have to be a bit careful when you convert energies to lengths, you have to identify if you are in a quantum field theory or in a classical gravity situation. <p>My other remark is that it is conventional how many independent units you have. Many people think, that in mechanics you need three (e.g. length, mass and time, meters, kilograms and seconds in the SI system) and a fourth if you include thermodynamics (like temperature measured in Kelvins) and a fifth if there is electromagnetism (like charge or alternatively current, Amperes in SI). But these numbers are just what we are used to. This number can change when we change our understanding of a relation from "physical law" to "conversion factor". The price is a dimensionful constant: In the SI system, it is a law that in equipartition of energy $E=\frac 12k_bT$ and Coulombs law equates a mechanical force to an electrostatic expression via $F=\frac{qQ} 1{4\pi\epsilon_0r}$ and it is a law that light moves at a speed $c=s/t$. <p>But alternatively, we could use these laws to define what we actually <i>mean</i> by Temperature (then measured in units of energy), charge (effectively setting $4\pi\epsilon_0$ to unity and thereby expressing charge in mechanical units) and length (expressing a distance by the time light need to traverse it). This eliminates a law and a unit. What remains of the law is only the fact that one can do that without reference to circumstances, that a distance from here to Paris does not depend for example on the time of the year (and thus on the direction of the velocity of the earth on its orbit around the sun and thus potentially relative to the ether). If the speed of light would not be constant and we would try to measure distances by the time it takes light to traverse them then distances would suddenly vary when we would say that the speed of light varies. <p>There is even an example that you can increase the number of units to more than what we are used to (although a bit artificial): It is not god given what kinds of things we consider 'of the same type' and thus possible to be measured in the same units. We are used to measuring all distances in the same unit (like for example meters) or derived units like kilometers or feet (with a fixed numerical conversion factor). But in nautical situations it is common to treat horizontal distance to be entirely different from vertical distances. Horizontal distances like the way to the next island you would measure in nautical miles while vertical distances (like the depth of water) you measure in fathoms. It is then a natural law that the ratio between a given depth and a given horizontal distance is constant over time and there is dimensionful constant (fathoms per mile) of nature that allows to compute a horizontal distance from a depth.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com3tag:blogger.com,1999:blog-8883034.post-6005843386694821862011-06-03T15:53:00.000+02:002011-06-03T15:53:29.429+02:00Bitcoin explainedAs me, you might have recently heared about <a href=http://www.bitcoin.org/>"Bitcoin"</a>, the internet currency that tries to be safe without a central authority like a bank or a credit card company that say which transactions are legitimate. So far, all mentions in blogs, podcasts or the press that I have seen had in common that they did not say how it works, what are the mechanisms that make sure Bitcoins operate like money. So I <a href=http://www.bitcoin.org/bitcoin.pdf>looked it up</a> and this is what I found:<br /><br />Bitcoin uses to cryptographic primitives: hashes and public key encryption. I case you don't know what these are: A hash is a function that reads in a string (or file or number, those are technically all the same) and produces some sort of checksum. The important properties are that everybody can do this computation (with some small amount of effort) and produce the same checksum. On the other hand, it is "random" in the sense that you cannot work backwards, i.e. if you only know the checksum you effectively have no idea about the original string. It is computationally hard to find a string for a given checksum (more or less the best you can do is guess random strings, compute their checksums until you succeed). A related hard problem is to find such a string with prescribed first $N$ characters.<br /><br />This can be used as a proof of effort: You can pose the problem to find a string (possibly with prescribed first characters) such that the first $M$ digits of the checksum have a prescribed value. In binary notation you could for example you could ask for $M$ zeros. Then on the average you have to make $2^M$ guesses for the string until you succeed. Presenting such a string then proves you have invested an effort of $O(2^M)$. The nice thing is that this effort is additive: You can start your string with the characters "The message '....' has checksum 000000xxxxxxxxxxx" and continue it such that the checksum of the total string starts with many zeros. That proves that in addition to the zeros your new string has, somebody has already spent some work on the string I wrote as dots. Common hash functions are <a href=http://en.wikipedia.org/wiki/Sha-1>SHA-1</a> (and older and not as reliable: <a href=http://en.wikipedia.org/wiki/MD5>MD5</a>).<br /><br />The second cryptographic primitive is public key encryption. Here you have two keys $A$, the public key which you tell everybody about and $B$ your secret key (you tell nobody about). These have the properties that you can use one of the keys to "encrypt" a string and then the other key can be used to recover the original string. In particular, you need to know the private key to produce a message that can be decrypted with the public key. This is called a "signature": You have a message $M$ and encrypt it using $B$. Let us call the result $B(M)$. Then you can show $A$ and $M$ and $B(M)$ to somebody to prove that you are in possession of $B$ without revealing $B$ since that person can verify that $B(M)$ can be decrypted using $A$. Here, an example is the <a href=http://en.wikipedia.org/wiki/RSA>RSA algorithm</a>.<br /><br />Now to Bitcoin. Let's go through the list of features that you want your money to have. The first is that you want to be able to prove that your coins belong to you. This is done by making coins files that contain the public key $A$ of their owner. Then, as explained in the previous paragraph you can prove that you are the legitimate owner of the private key belonging to that coin and thus you are its owner. Note that you can have as many public-private key pairs as you like possibly one for every coin. It is just there to equate knowing of a secret (key) to owning the coin.<br /><br />Second you want to be able to transfer ownership of the coin. Let us assume that the recipient has the public key $A'$. Then you transfer the coin (which already contains your public key $A$) by appending the string "This coin is transfered to the owner of the secrete key to the public key $A'$". Then you sign the whole thing with your private key $B$. The recipient can now prove that the coin was transferred to him as the coin contains both your public key (from before) and your statement of the transfer (which only you, knowing $B$ can have authorized. This can be checked by everybody by checking the signature). So the recipient can prove you owned the coin and agreed to transfer it to him.<br /><br />The last property is that once you transfered the coin to somebody else you cannot give it to a third person as you do not own it anymore. Or put differently: If you try to transfer a coin a second time that should not work and the recipient should not accept it or at least it should be illegitimate.<br /><br />But what happens if two people claim they own the same coin, how can we resolve this conflict? This is done via a public time-line that is kept collaboratively between all participants. Once you receive a coin you want to be able to prove later that you already owned it at a specific time (in particular at the time when somebody else claims he received it).<br /><br />This is done as follows: You compute the hash function of the transfer (or the coin after transfer, see a,bove including the signature of the previous owner of the coin that he has given it to you) and add it to the time line. This means you take the hash value of the time line so far, at the hash of the transfer and compute new hash. This whole package you then send to your network peers and ask them to also include your transfer in their version of the time line.<br /><br />So the time line is a record of all the transfers that have happened in the past and each participant in the network keeps his own copy of it.<br /><br />There could still be a conflict when two incompatible time lines are around. Which is the correct one that should be trusted? One could have a majority vote amongst the participants but (as everybody knows from internet discussions) nothing is easier than to come up with a large number of sock puppets that swing any poll. Here comes the proof of work that I mentioned above in relation to hash functions: There is a field in the time line that can be filled with anything in the attempt to construct something that has a hash with as many zeros as possible. Remember, producing $N$ leading zeros amounts to $O(2^N)$ work. Having a time line with many zeros demonstrates that were willing to put a lot of effort into this time line. But as explained above, this proof of effort is additive and all the participants in the network continuously try to add zeros to their time line hashes. But if they share and combine their time lines often enough such that they stay coherent they are (due to additivity) all working on fining zeros on the same time line. So rather than everybody working for themselves everybody works together as long as their time lines stay coherent. And going back through a time line it is easy to see how much zero finding work has been but in. Thus in the case of conflicting time lines one simply takes that that contains more zero finding work. If you wanted to establish an alternative time line (possibly one where at some point in time you did not transfer a coin but rather kept it to yourself so you could give it to somebody else later) to establish it you would have to outperform all other computers in the network that are all busy working on computing zeros for the other, correct, time line.<br /><br />Of course, if you want to receive a bitcoin you should make sure that in the generally accepted time line that same coin has not already been given to somebody else. This is why the transfers take some time: You want to wait for a bit that the information that the coin has been transferred to you has been significantly spread on the network and included in the collective time line that it cannot be reversed anymore.<br /><br />There are some finer points like how subdividing coins (currently worth about 13 dollars) is done and how new coins can be created (again with a lot CPU work) but I think they are not as essential in case you want to understand the technical basis of bitcoin before you but real money in.<br /><br />BTW, if you liked this exposition (or some other here) feel free to transfer me some bitcoins (or fractions of it). My receiving address is <pre>19cFYVExc2ZS4p7ZARGyENFijnV43y6ts1</pre>.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com12tag:blogger.com,1999:blog-8883034.post-26691276564774643292011-03-24T16:23:00.000+01:002011-03-24T16:23:06.288+01:00Mixed superrationality does not beat pure in prisoner's dilemmaThe <a href=http://en.wikipedia.org/wiki/Prisoner%27s_dilemma>prisoner's dilemma</a> is probably one of the most famous toy games of game theorists. It amounts to two criminals that being caught by the police are interrogated individually are offered the following deal: If both remain silent ("cooperate" with each other) both go to prison for $S$ ('short') years for small crimes that the police can prove. But if one prisoner admits the big crime ("defects") he goes free and the other spends $L$ ('long') years in prison. But if both admit the crime they both face a $M$ ('middle') year sentence. To be a dilemma the sentences should obey $0<S<M<L$ and by picking an appropriate normalisation of the unit of time, we can set $S=1$. <br /><br />The standard (economist) analysis of the game goes as follows: I assume that the other prisoner has already made his decision. Then, no matter what he decided I am better off by defecting: If he cooperates, my choice is between going free and $S$ years while if he is defecting I can choose between $M$ and $L$. So I defect and he comes to the same conclusion, so we end up spending $M$ years in prison. Both defecting is in fact a <a href=http://en.wikipedia.org/wiki/Nash_equilibrium>Nash equilibrium</a>.<br /><br />That's not too exciting, as we could do better by both cooperating and serving only $S$ years, which is <a href=http://en.wikipedia.org/wiki/Pareto_optimum>Pareto optimal</a> but unstable because there is the temptation for each player to defect and then go free. So much for the classic analysis of this game (not iterated) which is a model for many decision problems where one has to decide between a personal advantage or the global optimum.<br /><br />I first learned about this game many many years ago when still attending high school from a Douglas Hofstaedter column in the Scientific American. He makes the following observation: When defecting, I am counting on the fact that the other prisoner is not as clever as me. It only pays if the situation is asymmetric. But since the other prisoner is faced with the same problem, he will come up with the same solution so the asymmetric case of one player cooperating and the other defecting will not occur. Thus the only real possibilities are both cooperating (yielding $S$ years) and both defecting (yielding $M$ years) of which the obvious better choice is to cooperate. Hofstadter calls this argument "superrational". It is the realization that in the analysis of the Nash equilibrium the idea that my decision is independent of the other prisoner's decision might be wrong.<br /><br />Then Hofstadter points out another version of this game: You receive a letter from a very rich person stating that she is studying human intelligence and she figured that you are one of the top ten intelligent people in the world. She offers you (and also the other nine top-brainers) the following game: On the bottom of the letter is a coupon. You can either ignore the letter (in which case nothing more will happen) or you write your name on the coupon and send it back. If out of the ten possible coupons she receives exactly one she gives the person who returned the coupon 100 Million dollars. If any other number of coupons arrive until the end of this year nobody will receive any money. And as a warning: You are watched over by a number of private investigators. If they notice you trying to find out who the other nine people are the whole thing is called off and again nobody will get any money. So don't even think about it.<br /><br />This does not look very promising: Obviously, if you don't send in the coupon you won't get any money. So you have to send the coupon but so will the other nine and again you will receive nil. Too bad.<br /><br />Well, unless you widen your strategy space and besides 'pure', deterministic strategies you also allow for 'mixed', i.e. probabilistic strategies. You could for example come up with the following strategy: You roll dice and then send the coupon only with probability $p$. Let's see which $p$ optimizes your expectation assuming the other nine player follow the same strategy: You only get the money if you send the letter (probability $p$) and all nine other don't (probablity $(1-p)^9$) so the expectation is $E=p(1-p)^9$. Setting to zero the $p$ derivative of $E$ gives $0=(1-p)^9-9p(1-p)^8=(1-p)^8(1-p-9p)$ thus $p=1/10$. So you could prepare ten envelopes but only one with the coupon and mail a random one of these to optimize your expectation.<br /><br />But with this idea of taking into account also mixed strategies we can go back to the prisoner's dilemma and see what happens when both players defect with probability $p$ (this is the new part of the story I came up with this morning under the shower. Of course, I do not claim any originality here). Then the expected number of years I spend in prison is $p^2M+Lp(1-p)+(1-p)^2$. Quick check for $p$ being 0 or 1 I get back the two deterministic values. So can I do better? Obviously, this is a quadratic function of $p$ going through $(0,1)$ and $(1,M)$. So it has is minimum in the interior of the range $p\in[0,1]$ if the slope at $p=0$ is negative (remember $M>S=1$). But the slope is $2(M-L+1)p+L-2$ which is positive as long as $L>2$. But this is really the interesting parameter range for the game since for $L<2$ it is better for both players to always switch between cooperate-defect and defect-cooperate since the average sentence in the asymmetric case is shorter than the one year sentence of both cooperating. So, unless that is the case, always cooperating is still the better symmetric strategy of superrational players than the probabilistic ones.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-58862771576989714682011-03-15T16:34:00.001+01:002011-03-15T16:35:32.530+01:00Formulas in BloggerTo include formulas in blogger.com I have so far used <a href=http://www.forkosh.com/mimetex.html>mimetex</a> which uses an external server running a cgi-script to convert TeX-style formulas to picutres.<br /><br />This did its job most of the time except that mathphys, the old machine in Bremen that hosted my mimetex service died a couple of months ago and that the formulas have that stupid box around them which is particularly annoying for single symbols (this could probably be fixed by investing some time staring at the stylesheet for this blog). This is very much 8bit pixel style and does not scale nicely but I never touched it since it allowed you to read what I wrote.<br /><br />Now, some reader suggested <a href=http://mnnttl.blogspot.com/2011/02/latex-on-blogger.html>MathJax</a> which I try out here:<br /><br />Let's start witha wave function $\psi$, we define the velocity field $\vec v= \frac1{2m}\Im(\frac{\nabla \psi}{\psi})$. This leads to a conserved current:<br />$$\frac{\partial\rho}{\partial t}= -\vec\nabla\cdot (\bar\psi\psi\vec v).$$<br />At first, I thought it does not work but it just takes some time to reload.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com5tag:blogger.com,1999:blog-8883034.post-16651841694947501352011-03-02T17:46:00.001+01:002011-04-01T16:28:31.471+02:00Bohmian mechanics threatend by Occam's razorLast semester, I have been running a seminar on "Foundation of Quantum Mechanics" (<a href=https://wiki.physik.uni-muenchen.de/TMP/index.php/Foundations_of_Quantum_Mechanics_Seminar_WS_10/11>wiki page</a>) for TMP students that had been disappointed that "Mathematical Quantum Mechanics" was not on foundations. <br /><br />Overall, I am quite satisfied with the outcome. We had covered several approaches to foundational issues, in particular the relation of quantum to classical physics and here specifically the "measurement problem" (which I am convinced is not a problem but is explained withing quantum theory by decoherence). We will produce a reader with all the contributions and I myself will write some introduction (which I will post here as well once it is finished). <br /><br />But today, want to discuss Bohmian mechanics which was one of the topics and which has strong support by some <a href=http://www.mathematik.uni-muenchen.de/~duerr/>local experts</a>. I never really cared about this approach (being one of the Gallic villages where a small group of people <em>know</em> they are doing it better than the rest of the ignorant world, much like algebraic QFT or loop quantum gravity) being satisfied with quantum physics without any extras. <br /><br />But now was the time to find out what Bohmian mechanics is really about and in this post I would like to share my findings. The big question everybody asks really is "do they make any predictions that differ from usual quantum mechanics i.e. can it be distinguished by some sort of experiment or is it just an alternative interpretation?" but unfortunately I do not have a final answer. But more below. <br /><br />Before I start, let me put it a bit in perspective: Inequalities of Bell type (an I would include the Kochen-Specker theorem and GHZ type experiments) show in effect that the world cannot be both "realistic" and "local". Realistic means here that all properties have values at any instant of time irrespective of whether they are measured or not while local means that any decision I take here and now (for example whether I measure the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" alt="x"> or <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?y" alt="y"> component of the spin of my half of an EPR singlet state) cannot influence measurements that are so far away that they cannot be reached even at the speed of light. <br /><br />Thus one has to give up either realism or locality. The common interpretation of quantum mechanics gives up realism, the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" alt="x"> component of the spin does not have a value when I measure the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?y" alt="y"> component but is local. In some of the popular literature you will find statements to the contrary but they are mistaken: It is true, there can be non-local correlations. But this is no different from classical physics: Most of the time the color of the sock on my right foot is correlated with the color of the sock on the left foot, even at the same instant of time (when they are space-like to each other). But the question of locality is not about states (which are always global) it is about operators or measurements. And measuring the color (as compared for example to the size) of one of the socks does not influence the other sock, the local operators do commute. <br /><br />Bohmian mechanics insists on realism and the price it has to pay is to give up is locality. It does not violate causality in an obviously measurable way but doing the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" alt="x">- or <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?y" alt="y">-measurement here influences what happens far far away. But enough of these philosophical remarks, let's look at some formulas. <br /><br />In its pure form, Bohmian mechanics is about non-relativistic systems of <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?N" alt="N"> particles with Hamiltonian of the form <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?H=\sum_i p_i^2 + V(x_1,\ldots,x_N)" alt="H=\sum_i p_i^2 + V(x_1,\ldots,x_N)">. Everybody knows that the norm-squared wave function in position representation <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\rho(x_1,\ldots,x_N)=|\psi(x_1,\ldots,x_N)|^2" alt="\rho(x_1,\ldots,x_N)=|\psi(x_1,\ldots,x_N)|^2"> gives the probability distribution of finding particle 1 at <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x_1" alt="x_1">, particle 2 at <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x_2" alt="x_2"> etc. and there is a conserved current <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?j(x_1,\ldots,x_N)=Im(\bar\psi\nabla\psi)" alt="j(x_1,\ldots,x_N)=Im(\bar\psi\nabla\psi)"> for this density. That is if you start with some distribution <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\rho" alt="\rho"> at an initial time then wait a bit while you flow according to the current you end up with the new <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\rho" alt="\rho"> at a later time. <br /><br />The new thing for the Bohmians is to interpret this current as an actual current of particles with velocities <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\dot Q(x_1,\ldots,x_N) = v =j/\rho= Im(\nabla\psi/\psi)" alt="\dot Q(x_1,\ldots,x_N) = v =j/\rho= Im(\nabla\psi/\psi)">. According to the Bohmians, these particles with joint coordinates <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> are dots that for example show up on the screen of a double slit experiment. Obviously, if you start with a probability distribution of particle positions given by <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?|\psi|^2" alt="|\psi|^2"> at an initial time and follow the deterministic flow equation for <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> above, then at any later time the particles will be distributed according to <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?|\psi|^2" alt="|\psi|^2">. The Bohmians claim, that there are really particles and at any instant of time their position is <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> and the velocity is <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\dot Q" alt="\dot Q"> no matter whether they are measured or not. That's it. <br /><br />A few trivial remarks: This theory is non-local as the velocity of the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?i" alt="i">-th particle does depend via the wave function on the positions of all the other particles. Bohmians say that this is not to worry about since their theory is non-relativistic and this is like for example the Coulomb interaction in non-relativistic quantum mechanics where the force on one electron depends on the instantaneous positions of the other charged particles. <br /><br />The next remark is that in Bohm's theory there is also the wave function that follows the same Schroedinger equation as in usual quantum mechanics. Thus any question involving only the wave function trivially gives the same answer as in quantum mechanics. The equation of motion for the particle positions <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> which are the new ingredient in the Bohm theory depend on the wave function but not the other way around. There is no feed-back and the wave function does not know about the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q">. Any quantum mechanical measurement that in the end measures position (like for example Stern-Gerlach) gives the same result as the <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?q" alt="q"> follow the wave function that determines the outcome in the usual interpretation. <br /><br />All observables that are functions of the coordinates <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x_i" alt="x_i"> at one instant of time do commute with each other and one can thus give them all sharp values at that instant of time. Thus there is no problem with claiming those positions are <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> even in the usual interpretation. <br /><br />Position measurements at different times in general do not commute and thus they have no common meaning. Thus the only hope to find disagreement is in experiments that in the Bohmian interpretation require sharp positions at different instants of time. <br /><br />When it comes to spin I have the impression that the Bohmians cheat a bit: They declare that "spin is no a property of a point-like particle" meaning that realism does not apply to the different components and like in the usual interpretation, the components do not have a meaning unless measured. One can read this as a manifestation of the preferred role the Bohmians give to observables that are a function of the position operators over all other operators. In effect they claim only those position observables deserve realism. <br /><br />Of course, one can reformulate the Bell type experiments mentioned above in terms of positions (e.g. by translating spins into positions via Stern-Gerlach set-ups) but then the non-local flow equation seems to prevent any obvious contradictions with quantum mechanics. <br /><br />There are more formal problems: For time-reversal invariant Hamiltonians, one can always choose the eigenfunctions of the Hamiltonian to be real. Thus for the wave-function to be such an eigenfunction <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\dot Q=0" alt="\dot Q=0">, the particles don't move, even in i.e. the Coulomb field of a hydrogen atom. You may say that this is not the classical world but the quantum world and there are other equations of motion but I must say I find particles standing still even in the presence of forces a bit strange. <br /><br />That that brings us to my main criticism: It is not clear to me how to observe the particle at <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q">. Do experiments measure the wave function (via <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?\langle O \rangle= \langle\psi|O|\psi\rangle" alt="\langle O \rangle= \langle\psi|O|\psi\rangle">) or do they measure <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q">? And if so, can I prepare (and later measure) <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> without significantly disturbing the wave function? If that is the case I can of course check whether I put an electron in a hydrogen atom in an energy eigenstate at some <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> and later check whether I find it at some other place (which quantum mechanics would predict). <br /><br />There are if course ways to wiggle out: You could argue that this experiment is impossible since I would always disturb the wave function significantly by placing a particle at <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> and thus everything get screwed up. <br /><br />But this excuse is pretty much equivalent to "you cannot observe <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> (directly)". But then we are adding something (the particles at positions <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q">) to our theory which is not observable. And that sounds to me to be directly threatened by Occam's razor. <br /><br />Anyway. Unless somebody explains to me how to measure <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q">, I maintain that adding <img src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?Q" alt="Q"> to the theory is as good as adding invisible angels.<br /><br /><b>Update:</b> The promised write-up is <a href=https://wiki.physik.uni-muenchen.de/TMP/images/5/5a/Foundations_introduction.pdf>here</a>.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com7tag:blogger.com,1999:blog-8883034.post-28034073875052613372010-11-16T16:14:00.001+01:002010-11-16T16:18:37.674+01:00Picking the bigger number of two even if one is unknown<div style="text-align: left;">Here is a nice problem from <a href="http://blog.xkcd.com/2010/02/09/math-puzzle/">the xkcd blog</a>: Two real numbers, A and B, are drawn using some unknown, possibly probabilistic process and written on papers that go into two envelopes. You randomly pick one and open it to find some number on it. You now have to decide whether you want to receive that number as an amount in dollars or rather the number that is in the other envelope (which is still sealed). Can you come up with a process that with probability >50% picks the larger amount?</div><div style="text-align: left;"><br /></div><div style="text-align: left;">Think about it.</div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;">SPOILER ALERT</div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;">You can. You will need some function f that maps the real numbers to the open interval (0,1) in a strictly monotonic way. You could for example take f(x) = (1+tanh(x))/2. Assume the number that you found in the first envelope was X. Then throw an unfair coin such that with probability f(X) you keep X and otherwise take the other envelope. Obviously (?), if you started with the envelope with the smaller number you are more likely to switch than if you had started with the larger number.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">This sounds a bit counter intuitive. How can you increase your expected payoff if you know nothing about the number in the second envelope?</div><div style="text-align: left;"><br /></div><div style="text-align: left;">You might have the idea that something fishy is going on. What comes to mind is for example that you can produce paradoxes when assuming that there is a uniform probability distribution on the reals (or integers). But I believe, that this is not what is going on here since I did not say how the numbers were picked. They could have been picked with any perfectly fine probability measure on the reals, nobody said all numbers were equally likely. Below I will compute the expected outcome for any probability distribution that might have been used and it always works, not just in average.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">More precisely, I think this is unrelated to a similar puzzle: In that second puzzle, there are also two envelopes that contain numbers representing payout but none is opened but instead it is known that the number in one envelope is twice the number in the other. You just don't know if you have the half oder double. There, assuming your envelope contains X then you could be tempted to argue that with probability 50% the other contains 2X and with 50% it contains X/2 and thus the expectation value is 50% x 2 X + 50% x X/2= 5/4 X and thus you could increase your expectation by 25% by switching. But then you could increase it by another 25% by using the same argument again and switching back. </div><div style="text-align: left;"><br /></div><div style="text-align: left;">In the second puzzle, it is really the implied uniform distribution of X's that is the origin of the paradox: You can see this by giving the additional information that both numbers are definitely smaller than 100 trillion dollars. That sounds like a trivial information but note that the calculation of the expectation value changes: If X is greater than 50 trillion dollars, you know with certainty that the other number cannot be 2X and thus the expectation of taking the other envelope is nor 125%X but X/2. If you now carefully go through the expectation value calculation you will find that averaged over all values of X the expectation for switching is the same as for keeping the first envelope.<br /><br />Some of my readers will notice that the second puzzle is related to recent arguments that were made in the Landscape scenario about the imminent end of the world.<br /><br />Back to the first game. Let's do some calculation to compute the expectation of the outcome. We will assume that the numbers were picked according to some probability measure rho(x)dx and that has a finite expectation value, i.e. the integral E=E(X)=int x rho(x)dx converges.<br /><br />Then the expected outcome of the strategy above is X with probability f(X) and E with probability (1-f(X)) (as in that case we take the number in the second envelope).<br /><br /><div style="text-align: left;">We can now compute the expectation <f(x)x> E(f(X) X + (1-f(X))E)= E(f(X)X) + E - E E(f(X))<f(x)>. For simplicity assume that E=0. Otherwise we could pay out E immediately and then subtract E from all number in the envelopes. Thus the expected payout of our strategy is E(f(X)X) <f(x)x> but it is easy to see that this is positive (and thus we make more than the average E=0): In computing</f(x)x></f(x)></f(x)x><br /><br /><f(x)x> E(f(X)X) = int f(x) x rho(x) dx</f(x)x><br /><br />we can for x<0 overestimate f(x) by f(0) and for x>0 underestimate f(x) by f(0) and then conclude (unless rho(x) = delta(x) and we always spit out 0s)<br /><br /><f(x) x="">E(f(X)X) > f(0)E(X) <x> = 0</x></f(x)><br /><br />Thus we do better on the average than by deciding for one envelope or the other not taking into account the contents of the first one.<br /><br />Not that the difference to the first puzzle is that this works for any rho(x), we did not have to assume some (non existent) uniform rho(x) and the effect does not go away as soon as a cut-off is introduced contrary to the other puzzle. </div></div>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-35004282408533841882010-10-15T13:51:00.000+02:002010-10-15T13:51:40.175+02:00Is there a prisoner's dilemma in vaccination?Recently, I have been thinking about vaccination strategies as I was confronted with opinions which I consider at least very risk inviting. Not to spill oil in the fire I will thus anonymize illnesses and use made up probabilities. But let me assure you that for the illness I have in mind the probabilities are such that the story is similar.<br /><br />Let's consider illness X. For simplicity assume that if you meet somebody with that illness you'll have it yourself a bit later with 100% probability. If you have X then in 1 in 2000 cases you will develop complication C which is lethal. But C itself is not contagious. <br /><br />Luckily, there exists a vaccination against X that is 100% effective, i.e. if vaccinated you are immune to X. But unfortunately, the veccination itself causes the deadly C in 1 in a million cases.<br /><br />So, the question is: Should you get vaccinated?<br /><br />Unfortunately, the answer is not clear: It depends on the probability that if not vaccinated you will run into somebody spreading X. If X is essentially eradicated there is no point in taking the vaccination risk but if X is common it is much safer to vaccinate.<br /><br />The break even is obviously when that probability is 1 in 500. If it's less likely to meet somebody then it would be to your advantage not to vaccinate.<br /><br />Unfortunately, the probability of meeting an X infected person depends on how well people are vaxinated: As X is so contagious, if the vaccination rate drops the probability of meeting somebody with X dramatically increases. That is, not vaccinating yourself might be profitable for you but if everybody follows the same strategy the vaccination rate might drop and the society as a whole will see many more cases of C.<br /><br />If you assume in addition that your information is not perfect and you might be wrong in estimating the probabilities involved it is not clear to me to which fixed point this system evolves. <br /><br />But it seems likely to me that there are situations where for the society as a whole it is much better if you get vaccinated even if this increases you personal risk of encountering C.<br /><br />Opinions?Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-58740915178839552322010-09-14T17:24:00.000+02:002010-09-14T17:24:54.922+02:00Tensor factor subalgebras questionAfter some serious arm-twisting performed by some TMP students I accepted to run a seminar on <a href=https://wiki.physik.uni-muenchen.de/TMP/index.php/Foundations_of_Quantum_Mechanics_Seminar_WS_10/11>Foundations of Quantum Mechanics</a> under the condition that it would be a no-nonsense class. This is in addition to the String Theory Lectures I have to teach as well. <br /><br />After coming back from our vacation and workshop I found myself in the situation that I had much more fun doing some reading for the seminar (reviews on decoherence etc) than preparing for the string class (given that I have already twice taught Intro to Strings and that there are David Tong's wonderful <a href=http://arxiv.org/abs/arXiv:0908.0333>lecture notes</a> which give you the impression that you could take a few pages of those and be well prepared for class).<br /><br />In the preparation, I came across a wonderful video of a lecture by a well known physicist (I will link this later as I might take part of this as a quiz for the first session. Let me just mention that it contains the clearest version of Bell's inequality that I a am aware of).<br /><br />I am convinced that a lot of possible confusion about quantum physics together with locality (let me only mention the three letters E, P and R) comes from the fact that people confuse the roles of observables and states: Observables can be local and causality is built in by asking operators localised at space like distances to commute while states are always global objects. There is nothing like "the wave function of electron 1" or only in the approximation where you ignore all the other particles. You cannot use it when talking about correlations etc. But this is not bad, even in classical (statistical) physics, there are non-local correlations, like the colors of the socks on my two feet. The fact that in addition to correlations, there can be entanglement in the quantum theory does not change that.<br /><br />Furthermore, I find it helpful to think (of course I did not come up with this approach) of the Hilbert space (and its wave functions) as a secondary object and take the observables as a starting point (and not derived as the operators acting on the wave functions). Those then are the elements of a (C*)-algebra and the Hilbert space only arises as a representation of that algebra. Stone and von Neumann for example then tell you that there is essentially a unique representation if the algebra is that of canonical commutation relations.<br /><br />States are then functionals w that map each observable A to a complex number w(A) (interpreted as the expectation value). This linear function has to be normalised, w(1)=1 and positive meaning that for all A one has w(A^* A)>=0 (did I tell you that formuals are broken?). Then the GNS construction is similar to a highest weight representation: Using w and the algebra, one can construct a Hilbert space: As a vector space you can take the algebra. It is a representation after defining the action to be simply left multiplication. The scalar product of the elements A and B can be given by w(A^* B). Positivity of w tells you this is at least positive semi-definite. One can quotient out the zero-space to obtain something potitive definite and then employ some C*-magic to show that the action by left multiplication can be lifted to the quotient. I have suppressed some topological fine-print here like taking completions etc.<br /><br />The states correspond in general to density matrices (or reducible representations) and as always can be convex combined as x w1 + (1-x) w2, the extremal states corresponding to irreducible representations. <br /><br />In quantum information applications (as well as EPR and decoherence), one often starts with a Hilbert space that is a tensor product H = H1 x H2. Restricting attention to the first factor only corresponds to taking the partial trace over H2 and in general turns pure states on H into mixed states on H2. This has the taste of "averaging over all possible states of H2" but in the algebraic formulation if becomes clear that one is only restricting a state w to the subalgebra of operators of the form A1 x id where id is the identitiy on H2.<br /><br />What I do not understand yet and where I am asking your help is the following: How does the splitting into tensor factors really work on the algbraic side? In particular, assume I have a C*-algebra C and a pure state w. Now I take some subalgebra C1 of C and obtain a new state w1 on C1 by restricting w to this subalgebra. What is the relation of the two Hilbert spaces H and H1 I obtain from the GNS construction on w and and w1 respectively? What is a sufficient condition on C1 that I can regard H1 as a tensor factor of H as above? <br /><br />A necessary condition are obviously dimensions in the finite dimensional case: Here, the C*-algebras are just the complex matrix algebras of size n x n and the irreducible representation is on C^n. This is only a non-trivial tensor product if n is not prime. But nothing stops me for example to start with the big algebra being the 17x17 matrices and the subalgebra being those matrices that have the last row and column filled with zeros. But C^16 is definitely not a tensor-factor of C^17.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-90323716560071740582010-09-14T16:22:00.000+02:002010-09-14T16:22:44.165+02:00Back from silenceIt has been very silent here recently (or not so recently) but there is no particular reason for this except that I have been busy with other things (including an update on my facebook relationship status) and small things have been posted to <a href=http://twitter.com/atdotde>Twitter</a> rather than the blog.<br /><br />And if one is not constantly taking care of things they tend to degrade. So is this blog. What happened is that mathphys.jacobs-university.de, the computer that I have been using to host (background and other) images and the mimetex installation that was serving formulas for this blog as well as a number of other CGI scripts has died or at least is being turned off. Anyway, I have to relocate these things and I am still looking for a good solution. It should be a computer with a static, routed IP address on which I can install programs and in particular cgi-scripts of my liking. Here at LMU, this is probably not going to happen for reasons of security paranoia on the sysadmin side. In addition, mathphys was handling my email traffic, meaning that currently spam reaches my inbox and messages are threatened to be deleted by well meaning service providers. But this just means that the suffering is strong enough that I will be looking for a solution in the very near future. The solution will most likely be renting some virtual linux server. Suggestions in this direction would be more than welcome.<br /><br />Not long ago, I have been attending the 40th incarnation of the <a href=http://people.physik.hu-berlin.de/~ahoop/>Ahrenshoop Symposium</a> once more organised by the Humboldt Uni crowd. This get together had a particularly interesting selection of talks many of which I really enjoyed. In particular I learned a lot and updated my options on F-Theory GUTs and AdS-Condensed Matter. Many thanks to the organisers! As you would expect, PDFs are online except for Sean's who gave a flip-chart talk (on four flip charts). <br /><br />At that meeting I was asked what had happened to this blog and this post is supposed to be the answer to this question. I hope of course that more content will be here, soon. I was also asked to mention that it was Martin Rocek who got all the soap.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-66057201864336317502010-02-10T14:37:00.003+01:002011-03-02T14:41:03.757+01:00How to obtain a polymer Hilbert spaceOn Monday, I will be at HU Berlin to give a seminar on <a href="http://arxiv.org/abs/0912.3011">my loop cosmology paper</a> (at 2pm in case you are interested and around). Preparing for that I came up with an even more elementary derivation of the polymer Hilbert space (without need to mention C*-algebras, the GNS-construction etc). Here it goes: <br /><br />Let us do quantum mechanics on the line. That is, the operators we care about are <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> and <img alt="p" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?p" />. But as you probably know, those (more precisely, operators with the commutation relation <img alt="[x,p]=i" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?[x,p]=i" />) cannot be both bounded. Thus there problems of domains of definition and limits. One of the (well accepted) ways to get around this is to instead work with Weyl operators <img alt="U(a)=\exp(iax)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U%28a%29=%5Cexp%28iax%29" /> and <img alt="V(b)=\exp(ibp)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?V%28b%29=%5Cexp%28ibp%29" />. As those will be unitary, they have norm 1 and the canonical commutation relations read (with the help of B, C and H) <img alt="U(a)V(b)=V(b)U(a)e^{iab}" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U%28a%29V%28b%29=V%28b%29U%28a%29e%5E%7Biab%7D" />. If you later want, you can go back to <img alt="x=dU(a)/da|_{a=0}" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x=dU%28a%29/da%7C_%7Ba=0%7D" /> and similar for <img alt="p" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?p" />. <br /><br />Our goal is to come up with a Hilbert space where these operators act. In addition, we want to define a scalar product on that space such that <img alt="U" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U" /> and <img alt="V" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?V" /> act as unitary operators preserving this scalar product. We will deal with the position representation, that is wave functions <img alt="\psi(x)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Cpsi%28x%29" />. <img alt="U" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U" /> and <img alt="V" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?V" /> then act in the usual way, <img alt="V(b)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?V%28b%29" /> by translation <img alt="(V(b)\psi)(x)=\psi(x-b)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%28V%28b%29%5Cpsi%29%28x%29=%5Cpsi%28x-b%29" /> and <img alt="U(a)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U%28a%29" /> by multiplication <img alt="(U(a)\psi)(x)=e^{iax}\psi(x)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%28U%28a%29%5Cpsi%29%28x%29=e%5E%7Biax%7D%5Cpsi%28x%29" />. Obviously, these fulfil the commutation relation. You can think of <img alt="U" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U" /> and <img alt="V" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?V" /> as the group elements of the Heisenberg group while <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> and <img alt="p" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?p" /> are in the Lie algebra. <br /><br />Here now comes the only deviation from the usual path (all the rest then follows): We argue (motivated by similar arguments in the loopy context) that since motion on the real line is invariant under translation (at least until we specify a Hamiltonian) is invariant under translations, we should have a state in the Hilbert space which has this symmetry. Thus we declare the constant wave function <img alt="|1\rangle=\psi(x)=1" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%7C1%5Crangle=%5Cpsi%28x%29=1" /> to be an element of the Hilbert space and we can assume that it is normalised, i.e. <img alt="\langle 1|1\rangle=1" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Clangle%201%7C1%5Crangle=1" />. <br /><br />Acting now with <img alt="U(a)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U%28a%29" />, we find that linear combinations of plane waves <img alt="e^{ikx}" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?e%5E%7Bikx%7D" /> are then as well in the Hilbert space. By unitarity of <img alt="U(a)" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?U%28a%29" />, it follows that <img alt="\langle e^{ikx}| e^{ikx}\rangle =1" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Clangle%20e%5E%7Bikx%7D%7C%20e%5E%7Bikx%7D%5Crangle%20=1" />, too. It remains to determine the scalar product of two different plane waves <img alt="\langle e^{ikx}|e^{ilx}\rangle" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Clangle%20e%5E%7Bikx%7D%7Ce%5E%7Bilx%7D%5Crangle" />. This is found using the unitarity of <img alt="V" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?V" /> and sesquilinearity of the scalar product: <img alt="\langle e^{ikx}|e^{ilx}\rangle = \langle V(b) e^{ikx}|V(b)e^{ilx}\rangle = e^{ib(l-k)}\langle e^{ikx}|e^{ilx}\rangle" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Clangle%20e%5E%7Bikx%7D%7Ce%5E%7Bilx%7D%5Crangle%20=%20%5Clangle%20V%28b%29%20e%5E%7Bikx%7D%7CV%28b%29e%5E%7Bilx%7D%5Crangle%20=%20e%5E%7Bib%28l-k%29%7D%5Clangle%20e%5E%7Bikx%7D%7Ce%5E%7Bilx%7D%5Crangle" />. This has to hold for all <img alt="b" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?b" /> and thus if <img alt="k\ne l" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?k%5Cne%20l" /> it follows that the scalar product vanishes. <br /><br />Thus we have found our (polymer) Hilbert space: It is the space of (square summable) linear combinatios of plane waves with a scalar product such that the <img alt="e^{ikx}" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?e%5E%7Bikx%7D" /> are an orthonormal basis. <br /><br />Now, what about <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> and <img alt="p" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?p" />? It is easy to see that <img alt="p" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?p" /> when defined by a derivative as above acts in the usual way, that is on a basis element <img alt="pe^{ikx}=ke^{ikx}" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?pe%5E%7Bikx%7D=ke%5E%7Bikx%7D" /> which is unbounded as <img alt="k" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?k" /> can be arbitrarily large. The price for having plane waves as normalisable wave functions is, however, that <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> is not defined: It would be <img alt="xe^{ikx} = \lim_{\epsilon\to 0}\frac{e^{i(k+\epsilon}x}-e^{ikx}}{\epsilon}" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?xe%5E%7Bikx%7D%20=%20%5Clim_%7B%5Cepsilon%5Cto%200%7D%5Cfrac%7Be%5E%7Bi%28k+%5Cepsilon%29x%7D-e%5E%7Bikx%7D%7D%7B%5Cepsilon%7D" />. But for <img alt="\epsilon\ne 0" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Cepsilon%5Cne%200" /> the two exponentials in the denominator are always orthogonal and thus not "close" as measured by the norm. The denominator always has norm 2 and thus the limit is divergent. Another way to see this is to notice that <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> would of course act as multiplication by the coordinate <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" />, but <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> times a plane wave is no longer a linear combination of plane waves. <br /><br />To make contact with loop cosmology one just has to rename the variables: What I called <img alt="p" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?p" /> for a simplicity of presentaion is the volume element <img alt="v" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?v" /> in loop cosmology while the role of <img alt="x" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?x" /> is played be the conjugate momentum <img alt="\beta" src="http://euve10195.vserver.de/rch-cgi-bin/mimetex?%5Cbeta" />. <br /><br />If you want you can find my notes for the blackboard talk at HU here (<a href="http://homepages.physik.uni-muenchen.de/%7Ehelling/LQCtalk.pdf">pdf</a> or <a href="http://homepages.physik.uni-muenchen.de/%7Ehelling/LQCtalk.djvu">djvu</a>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com7tag:blogger.com,1999:blog-8883034.post-79544415402385014702010-01-27T13:38:00.002+01:002010-01-27T13:55:35.594+01:00Entropic EverythingThe <a href=http://arxiv.org/abs/1001.0785>latest paper</a> by Eric Verlinde on gravity as an entropic force makes me wonder whether I am getting old: Let me admit it: I just don't get it. Is this because I am conservative or lack imagination or too narrow minded? If it were not for the author, I would have rated it as pure crackpottery. But maybe I am missing something. Today, there were three follow-up <a href=http://arxiv.org/abs/1001.4585>p</a>a<a href=http://arxiv.org/abs/1001.4677>p</a>e<a href=http://arxiv.org/abs/1001.4786>r</a>s dealing with cosmological consequences (the idea being roughly that Verlinde uses the equipartition of energy between degrees of freedom each getting a share of 1/2 kT which is not true quantum mechanically at low temperatures as there the system is in the ground state with the ground state energy. As in this business temperature equals acceleration a la Unruh this means the argument is modified for small accelerations which is a modification of MOND type). <br /><br />Maybe later I try once more to get into the details and might have some more sensible comments then but right now the way different equations from all kinds of different settings (Unruh temperature was already mentioned, E=mc^2, one bit per Planck area, etc) are assembled reminds me of this:<br /><br /><img src=http://www.scripting.com/images/mathematicalProofWomen.gif>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com13tag:blogger.com,1999:blog-8883034.post-69744442890599314802010-01-19T15:16:00.000+01:002010-01-19T15:17:30.625+01:00Instability of the QED vacuum at large fine structure constantToday, in the "Mathematical Quantum Mechnics" lecture, I learned that the QED vacuum (or at least the quantum mechanical sector of it) is unstable when the fine structure constant gets too big. <br /><br />To explain this, let's go back to a much simpler problem: Why is the hydrogen-like atom stable? Well, a simple answer is that you just solve it and find the spectrum to be bounded above <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?-13.6Z^2{\rm eV}" alt="-13.6Z^2{\rm eV}">. But this answer does not extend to other problems that cannot be diagonalised analytically. <br /><br />First of all, what is the problem we are considering? It's the potential energy of the electron which in natural (for atomic physics) units is <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?V(r)=-\alpha Z/r" alt="V(r)=-\alpha Z/r">. And this goes to negative infinity when <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?r" alt="r"> goes to 0. But quantum mechanics saves you. Roughly speaking (this argument can be made mathematically sound in terms of Hardy's inequality), if you essentially localise the electron in a ball of radius <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?R" alt="R"> and thus have the potential energy <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?V\le-\alpha Z/R" alt="V\le-\alpha Z/R">, Heisenberg's uncertainty implies the momentun is at least of the order <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?1/R" alt="1/R"> and thus the kinetic energy is at least of the order <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?+1/R^2" alt="+1/R^2">. Thus, when <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?R" alt="R"> becomes small and you seem to approach the throat of the potential the positive kinetic energy wins and thus the Hamiltonian of the hydrogen atom is bounded from below. This is the non-relativistic story. <br /><br />Close to the nucleus however, the momentum can be so big that you have to think relativistically. But then trouble starts as at large momenta the energy grows only linearly with momentum and thus the kinetic energy only scales like <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?+1/R" alt="+1/R"> which is the same as the potential energy. Thus a more careful calculation is needed. The result of it is that it depends on <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\alpha Z" alt="\alpha Z"> which term eventually wins. Above a critical value (which happens to be of order one) the atom is unstable and one can gain an infinite amount of energy by lowering the electron into the nucleus and quantum mechanics is not going to help. <br /><br />Luckily, nuclei with large enough <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?Z" alt="Z"> do not exist in nature. Well, with the exception of neutron stars which are effectively large nuclei. And there it happens. All the electrons are sucked into the nuceus and fuse with the protons to neutrons. In fact, the finite size of the nucleon is what regulates this process as the <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?1/r" alt="1/r"> nature of the Coulomb potential is smeared out in the nucleus. But such a highly charged atom would be only of the size of the nucleus (about some femto meters) rather than the size of typical atoms. <br /><br />But now comes QED with the possibility of forming electron-positron pairs out of the vacuum. The danger I am talking about is the fact that they can form a relativistic, hydrogen like bound state. And both are (as far as we know) point like and thus there is no smearing out of the charge. It is only that <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\alpha Z\approx 1/137" alt="\alpha Z\approx 1/137"> in this case which luckily is less than one. If it would be bigger you could create this infinte amount of energy from the vacuum by pair creation and bringing them on-shell in their relative Coulomb throat. What a scary thought. Especially, since <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\alpha" alt="\alpha"> is probably only the vev of some scalar field which can take other values in other parts of the multiverse which would then disappear with a loud bang. <br /><br />Some things come to my mind which in principle could help but which turn out to make things worse: <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\alpha" alt="\alpha"> is not a constant but it's running and QED has asymptotic slavery which means at short distances (which we are talking about) it gets bigger and makes things worse. Further, we are treating the electromagnetic field classically which of course is not correct. But my mathematical friends tell me that quantizing it also worsens things. <br /><br />We know, QED has other problems like the Landau pole (a finite scale where <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\alpha" alt="\alpha"> goes to infinity due to quantum effects). But it seems to me that this is a different problem since it already appears at <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\alpha\approx 1" alt="\alpha\approx 1">. <br /><br />Any ideas or comments?Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com5tag:blogger.com,1999:blog-8883034.post-24323029904559971602009-10-09T14:41:00.002+02:002009-10-09T14:44:27.919+02:00Download compete twitter timelineUpon popular request, I wrote a small script to download all tweets of a given twitter id. Have fun!<br /><br /><pre><br />#!/usr/bin/perl<br /><br />use Net::Twitter;<br />$|=1;<br /><br />unless(@ARGV){<br /> print "Usage: $0 twitter_id [sleep_seconds]\n";<br /> exit 0;<br />}<br /><br />my ($follow,$sleeper) = @ARGV;<br /><br /># No account needed for this.<br />my $twit = Net::Twitter->new(username => 'MYNAME', password => 'XXX');<br /><br />$p=1;<br />while(1){<br /> my $result = $twit->user_timeline({id => $follow, page => $p});<br /> <br /> foreach my $tweet (@{$result}){<br /> print "At ", $tweet->{'created_at'},"\n";<br /> print $tweet->{'text'},"\n\n";<br /> }<br /> ++$p;<br /> sleep $sleeper if $sleeper;<br />}<br /></pre><br /><br />You might have to install the Net::Twitter module. This is most easily done as<br /><br /><pre><br />sudo perl -MCPAN -e shell<br /></pre><br /><br />and then (possibly after answering a few questions)<br /><br /><pre><br />install Net::Twitter<br /></pre>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com2tag:blogger.com,1999:blog-8883034.post-83296284403219724672009-10-05T22:52:00.003+02:002009-10-06T14:55:29.069+02:00Not so canonical momentumTwo weeks ago, I was on Corfu where I attended the <a href=http://www.physics.ntua.gr/corfu2009/>conference/school/workshop on particles, astroparticles, strings and cosmology</a>. This was a three week event, the first being on more conventional particle physics, the second on strings and the last on loops and non-commutative geometry and the like. I was mainly there for the second week but stayed a few days longer into the loopy week. <br /><br />I think it was a clever move by the organisers of the last week to give five hours to the morning lecturers rather than one or two as in the string week. So they had the time to really develop their subjects rather than just mention a few highlights. <a href=http://math.ucr.edu/home/baez/week280.html>John Baez</a> has already reported on some of the lectures. <br /><br />I would like to mention something I learned about elementary classical mechanics and quantum mechanics which was just a footnote in Ashtekar's first lecture but which was new to me: One canonical variable can have several canonical conjugates! In the loopy context, this appears as both the old and the new connection variables have the same canonical momentum although they differ by the Imirzi parameter times the second fundamental form (don't worry if you don't know what this is in detail, what's important that the 'positions' are different in the two sets of variables although they have the same canonical momentum). <br /><br />How can this be? I always thought that if <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?x" alt="x"> is a canonical variable the conjugate variabel is determined by <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?p=\partial L/\partial\dot x" alt="p=\partial L/\partial\dot x">. What I had not realized is that you could for example take <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?p'=p+x" alt="p'=p+x"> and obtain the same fundamental Poisson brackets (and consequently commuation relations after quantization). Similarly, you could add any function <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?f(x)" alt="f(x)"> to the momentum without changing the commutation relations. <br /><br />The origin of this abiguity can be found in the fact that also the Lagrangian is not unique: You can always add a total derivative without changing the action (at least locally, see below). For example, to obtain <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?p'=p+f(x)" alt="p'=p+f(x)"> by the derivative formula, add <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?d/dt (\int f)" alt="d/dt (\int f)"> to the action. The most general change would be to add <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?d/dt \Lambda(x)" alt="d/dt \Lambda(x)">. <br /><br />What about the quantum theory? This is most easily seen by realising that upon a gauge transformatio <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\delta A=d\Lambda" alt="\delta A=d\Lambda">, the action of a charge <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?Q" alt="Q"> particle changes by <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?Q\int d\Lambda" alt="Q\int d\Lambda">. Thus our change in Lagrangian (with a corresponding change in the canonical momentum) can be viewed as a gauge transformation (even if no gauge field is around one could add a trivial one). Correspondingly, the wave function would have to be changed to <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\psi'=\psi e^{i\Lambda}" alt="\psi'=\psi e^{i\Lambda}"> as acting on <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\psi'" alt="\psi'"> by a canocially quantized <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?p'" alt="p'"> is the same as <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\psi" alt="\psi"> acted on by <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\psi" alt="\psi">. <br /><br />So, it seems as if you would get exactly the same physics in the primed variables as in the unprimed ones. But we know that not all total derivatives have no influence on the qunatum theory the <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\theta" alt="\theta">-angle <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?\int F\wedge F" alt="\int F\wedge F"> being the most prominent example. How would that appear in our much simpler quantum mechanics example? Here, it is important to remember that one should only use gauge transformations that are trivial at infinity. Here, if you change the phase of the wave function too wildly at <img src="http://mathphys.jacobs-university.de/rob-cgi-bin/mimetex?x=\pm\infty" alt="x=\pm\infty"> you might leave the good part of the Hilbert space: For example the kinetic energy being an unbounded operator is not defined on all of Hilbert space but only on a dense subspace (most often taken to be some Sobolev space). And that you might leave by adding a wild phase and end up in a different self adjoint extension of the kinetic energy. <br /><br />I have no idea if all this is relevant in the loopy case and the old and new variables or the variables are related by a (generalized) gauge transformation but at least I found in amusing to learn that the canonical conjugate is not canonical.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com5