tag:blogger.com,1999:blog-88830342017-04-09T19:50:56.018+02:00atdotdeWhat comes to my mind. Physics, politics, computers, rants. All CC-BY-SA unless otherwise stated.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.comBlogger185125tag:blogger.com,1999:blog-8883034.post-58424659282751179802016-12-09T14:48:00.001+01:002016-12-09T14:48:29.824+01:00Workshop on IoT Liability at 33c3After my recent <a href="https://atdotde.blogspot.de/2016/10/mandatory-liability-for-software-is.html">blog post</a> on the dangers of liability for manufacturers of devices in the times of IoT, I decided I will run a <a href="https://events.ccc.de/congress/2016/wiki/Session:Haftung_f%C3%BCr_Devices_und_Software_gestalten">workshop</a> at 33C3, the annual hacker conference of the Chaos Computer club. I am proud I could convince <a href="https://buermeyer.de/ulf/">Ulf Buermeyer</a> (well known judge, expert in constitutional law, hacker, podcaster) to host this workshop with me.<br /><br />The main motivation for me is that I hope that this will be a big issue in the coming year but it might still be early enough to influence policy before everybody commits herself to their favorite (snake oil) solution.<br /><br />I have started collecting and sorting ideas in a <a href="https://docs.google.com/document/d/1hTYnLYdzDjeZAs_45HFx_HfMaZI2FrYJTV9YaXCHwrw/edit?usp=sharing">Google document</a>. <a href="https://commons.wikimedia.org/wiki/File%3AInternet_of_things_signed_by_the_author.jpg" title="By Wilgengebroed on Flickr [CC BY 2.0 (http://creativecommons.org/licenses/by/2.0)], via Wikimedia Commons"><img alt="Internet of things signed by the author" src="https://upload.wikimedia.org/wikipedia/commons/thumb/0/01/Internet_of_things_signed_by_the_author.jpg/512px-Internet_of_things_signed_by_the_author.jpg" width="512" /></a>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-28507534590825136672016-11-26T17:19:00.002+01:002016-11-26T17:19:58.857+01:00Breaking News: Margarine even more toxic!One of the most popular posts of this blog (as far as resonance goes) was the one on <a href="https://atdotde.blogspot.de/2006/08/scaling-of-price-of-margarine.html">Scaling the Price of Margarine</a>. Today, I did the family weekend shopping and noticed I have to update the calculation:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Hp34RKjWpPk/WDmxh3naCbI/AAAAAAAALUA/WrC0JAlclgM-_f6uc2WJ7ZHL33QAR50agCLcB/s1600/laetta%2B-%2B1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://1.bp.blogspot.com/-Hp34RKjWpPk/WDmxh3naCbI/AAAAAAAALUA/WrC0JAlclgM-_f6uc2WJ7ZHL33QAR50agCLcB/s320/laetta%2B-%2B1.jpg" width="240" /></a></div>At our local Rewe branch, they offer the pound of Lätta for 88 cents while they ask 1.19Euro for half the pound. With the ansatz from the old post, this means the price for the actual margarine is now -9,78Euro/kg. This, by coincidence is approximately also the price you have to pay to get rid of waste oil.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-30445981001602457042016-11-13T14:12:00.001+01:002016-11-13T14:12:31.254+01:00Theoretical diverBesides physics, another hobby of mine is scuba diving. For many reasons, unfortunately, I don't have much time anymore, to get in the water. As partial compensation, I started some time ago to contribute the <a href="https://subsurface-divelog.org/">Subsurface</a>, the open source dive log program. Partly related to that, I also like to theorize about diving. To put that in form, I now started another blog <a href="http://atdotde.de/theoreticaldiver/">The Theoretical Diver</a> to discuss aspects of diving that I have been thinking about.Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-78417764819897241552016-11-13T09:36:00.000+01:002016-11-13T11:52:37.483+01:00OpenAccess: Letter to the editor of Süddeutsche ZeitungIn yesterday's Süddeutsche Zeitung, there is an <a href="http://www.sueddeutsche.de/politik/kolumne-goldener-zugang-1.3244900">opinion piece</a> by historian <a href="https://de.wikipedia.org/wiki/Norbert_Frei">Norbert Frei</a> on the German government's OpenAccess initiative, which prompted me to write a letter to the editor (naturally in German). Here it is:<br /><br /><span style="font-family: "helvetica"; font-size: 12px;">Zum Meinungsbeitrag „Goldener Zugang“ von Norbert Frei in der SZ vom 12./13. November 2016:</span><br /><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Herr Frei sorgt sich in seinem Beitrag, dass der Wissenschaft unter der Überschrift OpenAccess von außen ein Kulturwandel aufgezwungen werden soll. Er fürchtet, dass ihn die Naturwissenschaftler zusammen mit der Politik zwingen, seine Erkenntnisse nicht mehr in papiernen Büchern darlegen zu können, sondern alles nur noch zerstückelt in kleine Artikel-Happen in teure digitale Archive einzustellen, wo sie auf die Bitverrottung waren, da schon in kürzester Zeit das Fortschreiten von Hard- und Software dazu führen wird, dass die Datenformate unlesbar werden.</div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Als Gegenmodell führt er die Gutenberg-Bibel an, von der eine Mehrzahl der Exemplare die Jahrhunderte überdauert haben. Nun weiss ich nicht, wann Herr Frei das letzte Mal in seiner Gutenberg-Bibel geblättert hat, ich habe in meinem Leben nur ein einziges Mal vor einer gestanden: Diese lag in einer Vitrine der Bibliothek von Cambridge und war auf einer Seite aufgeschlagen, keine andere Seite war zugänglich. Dank praktischem OpenAccess ist es aber nicht nur den guten Christenmenschen möglich, eine Kopie zu Hause vorzuhalten. Viel mehr noch, die akademischen Theologen aus meinem Bekanntenkreis arbeiten selbstverständlich mit einer digitalen Version auf ihrem Laptop oder Smartphone, da diese dank Durchsuchbarkeit, Indizierung und Querverweisen in andere Werke für die Forschung viel zugänglicher ist.</div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Geschenkt, dass es bei der OpenAccess-Initiative eine Ausnahme für Monographien geben soll. Niemand will das Bücherschreiben verbieten. Es geht nur darum, dass, wer Drittmittel von der öffentlichen Hand erhalten will, nicht noch einmal die Hand dafür aufhalten soll, wenn sich dann die vor allem wissenschaftliche Öffentlichkeit über die Ergebnisse informieren will. Professorinnen und Professoren an deutschen Universitäten schreiben ihre wissenschaftlichen Veröffentlichungen nicht zu ihrem Privatvergnügen, es ist Teil ihrer Dienstaufgaben. Warum wollen sie die Früchte ihres bereits entlohnten Schaffens dann noch ein weiteres Mal den öffentlichen Bibliotheken verkaufen? </div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Ich kann mich noch gut an meinen Stolz erinnern, als ich das erste Mal meinen Namen gedruckt auf Papier sah, der das Titelblatt meiner ersten Veröffentlichung zierte. Jenseits davon ist es für mich als Wissenschaftler vor allem wichtig, dass das, was ich da herausfinde, von anderen wahrgenommen und weitergetrieben wird. Und das erreiche ich am besten, wenn es so wenig Hürden wie möglich gibt, dieses zu tun.</div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Ich selber bin theoretischer Hochenergiephysiker, selbstredend gibt es sehr unterschiedliche Fächerkulturen. In meinem Fach ist es seit den frühen Neunzigerjahren üblich, alle seine Veröffentlichungen - vom einseitigen Kommentar zu einem anderen Paper bis zu einem Review von vielen hundert Seiten - in arXiv.org, einem nichtkommerziellen Preprintarchiv einzustellen, wo es von allen Fachkolleginnen und -kollegen ab dem nächsten Morgen gefunden und in Gänze gelesen werden kann, selbst viele hervorragend Lehrbücher gibt es inzwischen dort. Diese globale Verbreitung neben einfachem Zugang (ich habe schon seit mehreren Jahren keinen papiernen Fachartikel in unserer Bibliothek mehr in einem Zeitschriftenband mehr nachschlagen müssen, ich finde alles auf meinem Computer) hat so viele Vorteile, das man gerne auf mögliche Tantiemen verzichtet, zumal diese für Zeitschriftenartikel noch nie existiert haben und, von wenigen Ausnahmen abgesehen, verschwinden gering gegenüber einem W3-Gehalt ausfallen und als Stundenlohn berechnet jeden Supermarktregaleinräumer sofort die Arbeit niederlegen ließen. Wir Naturwissenschaftler sind auf einem guten Weg, uns von parasitären Fachverlagen zu emanzipieren, die es traditionell schafften, jährlich den Bibliotheken Milliardenumsätze für unsere Arbeit abzupressen, wobei sie das Schreiben der Artikel, die Begutachtung, den Textsatz und die Auswahl unbezahlt an von der Öffentlichkeit bezahlte Wissenschaftlerinnen und Wissenschaftler delegiert haben und sie sich ausschliesslich ihre Gatekeeper Funktion bezahlen liessen. </div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Und da ich an Leserschaft interessiert bin, werde ich diesen Brief auch in mein Blog einstellen.</div><div><br /></div>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com4tag:blogger.com,1999:blog-8883034.post-73095928251685852732016-10-27T14:56:00.003+02:002016-10-27T15:12:54.729+02:00Daylight saving time about to end (end it shouldn't)Twice a year, around the last Sunday in March and the last Sunday in October, everybody (in particular newspaper journalists) take a few minutes off to rant about daylight savings time. So, for this first time, I want to join this tradition in writing.<br /><br />Until I had kids, I could not care less about the question of changing the time twice a year. But at least for our kids (and then secondary also for myself), I realize biorhythm is quite strong and at takes more than a week to adopt to the 1 hour jet lag (in particular in spring when it means getting out of bed "one hour earlier"). I still don't really care about cows that have to deliver their milk at different times since there is no intrinsic reason that the clock on the wall has to show a particular time when it is done and if it were really a problem, the farmers could do it at fixed UTC.<br /><br />So, obviously, it is a nuisance. So what are the benefit that justify it? Well, obviously, in summer the sun sets at a later hour and we get more sun when being outside in the summer. That sounds reasonable. But why restrict it to the summer?<br /><br />Which brings me to my point: If you ask me, I want to get rid of changing the offset to UTC twice a year and want to permanently adopt daylight saving time.<br /><br />But now I hear people cry that this is "unnatural", we have to have the traditional time at least in the winter when it does not matter as it's too cold to be outside (which only holds for people with defective clothing as we know). So how natural is CET (the time zone we set our clocks to in winter), let's take people living in Munich for an example?<br /><br />First of all: It is not solar time! CET is the "mean solar time" when you live at a longitude of 15 degrees east, which is (assuming the latitude) close to <a href="https://www.google.de/maps/@48.1454127,15,13z?hl=de">Neumarkt an der Ypps</a> somewhere in Austria not too far from Vienna. Munich is about 20 minutes behind. So, this time is artificial as well, and Berlin being closer to 15 degrees, it is probably Prussian.<br /><br />Also a common time zone for Germany was established only in the 1870s when the advent of railways and telegraphs make synchronization between different local times advantageous. So this "natural" time is not that old either.<br /><br />It is so new, that<a href="https://en.wikipedia.org/wiki/Christ_Church,_Oxford"> Christ Church college in Oxford</a> still refuses to fully bow to it: Their clock tower shows Greenwich time. And the cathedral services start according to solar time (about five minutes later) because they don't care about modern shenanigans. ("How many Oxford deans does it take to change a light bulb?" ---- "Change??!??"). Similarly, in Bristol, there is a famous clock with two minute hands.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Exchangeclock.JPG/800px-Exchangeclock.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="238" src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Exchangeclock.JPG/800px-Exchangeclock.JPG" width="320" /></a></div><br />Plus, even if you live in Neumarkt an der Ybbs, your sun dial does not always show the correct noon! Thanks to the tilt of the earth axis and the fact that the orbit of the earth is elliptic, this varies through the year by a number of minutes:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Equation_of_time.svg/512px-Equation_of_time.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="274" src="https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Equation_of_time.svg/512px-Equation_of_time.svg.png" width="320" /></a></div><br />So, "winter time" is in no way more natural than the other time zone. So we should be free to choose a time zone according to what is convenient. At least for me, noon is not the center of my waking hours (it's more 5,5 : 12). So, aligning those more with the sun seems to be a pretty good idea.<br /><br />PS: The title was a typo, but looking at it I prefer it the way it is...<br /><br />Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com2tag:blogger.com,1999:blog-8883034.post-37901975775254653842016-10-24T14:55:00.003+02:002016-10-24T15:02:02.875+02:00Mandatory liability for software is a horrible ideaOver the last few days, a number of prominent web sites including Twitter, Snapchat and Github were effectively unreachable for an extended period of time. As became clear, the problem was that DynDNS, a provider of DNS services for these sites was under a number of very heavy DDoS (distributed denial of service) attack that were mainly coming from compromised internet of things devices, in particular web cams.<br /><br />Even though I do not see a lot of benefit from being able to change the color of my bedroom light via internet, I love the idea to have lots of cheap devices (I continue to have a lot of fun with C.H.I.P.s, full scale Linux computers with a number of ports for just 5USD, also for Subsurface, in particular those open opportunities for the mobile version), there are of course concerns how one can economically have a stable update cycle for those, in particular once they are build into black-box customer devices.<br /><br />Now, after some dust settled comes of course the question "Who is to blame?" and should be do anything about this. Of course, the manufacturer of the web cam made this possible through far from perfect firmware. Also, you could blame DynDNS for not being able to withstand the storms that from time to time sweep the internet (a pretty rough place after all) or the services like Twitter to have a single point of failure in DynDNS (but that might be hard to prevent given the nature of the DNS system).<br /><br />More than once I have now heard a <a href="http://www.taz.de/Kommentar-Hackerangriff/!5347856/">call for new laws</a> that would introduce a liability for the manufacturer of the web cam as they did not provide firmware updates in time that prevent these devices from being owned and then DDoSing around on the internet.<br /><br />This, I am convinced, would be a terrible idea: It would make many IT businesses totally uneconomic. Let's stick for example with the case at hand. What is the order of magnitude of damages that occurred to the big companies like Twitter? They probably lost ad revenue of about a weekend. Twitter recently made $6\cdot 10^8\$ $ per quarter, which averages to 6.5 million per day. Should the web cam manufacturer (or OEM or distributor) now owe Twitter 13 million dollars? I am sure that would cause immediate bankruptcy. Or just the risk that this could happen would prevent anybody from producing web cams or similar things in the future. As nobody can produce non-trivial software that is free of bugs. You should strive to weed out all known bugs and provide updates, of course, but should you be made responsible if you couldn't? Responsible in a financial sense?<br /><br />What was the damage cause by the <a href="https://en.wikipedia.org/wiki/Heartbleed">heart bleed bug</a>? I am sure this was much more expensive. Who should pay for this? OpenSSL? Everybody that links against OpenSSL? The person that committed the wrong patch? The person that missed it code review?<br /><br />Even if you don't call up these astronomic sums and have fixed fine (e.g. an unfixed vulnerability that gives root access to an attacker from the net costs 10000$) that would immediately stop all open source development. If you give away your software for free, do you really want to pay fines if not everything is perfect? I surely wouldn't.<br /><br />For that reason, the GPL has the clauses (and other open source licenses have similar ones) stating<br /><br /><blockquote class="tr_bq"> 11. BECAUSE THE PROGRAM IS LICENSED FREE OF CHARGE, THERE IS NO WARRANTY<br />FOR THE PROGRAM, TO THE EXTENT PERMITTED BY APPLICABLE LAW. EXCEPT WHEN<br />OTHERWISE STATED IN WRITING THE COPYRIGHT HOLDERS AND/OR OTHER PARTIES<br />PROVIDE THE PROGRAM "AS IS" WITHOUT WARRANTY OF ANY KIND, EITHER EXPRESSED<br />OR IMPLIED, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF<br />MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. THE ENTIRE RISK AS<br />TO THE QUALITY AND PERFORMANCE OF THE PROGRAM IS WITH YOU. SHOULD THE<br />PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF ALL NECESSARY SERVICING,<br />REPAIR OR CORRECTION.<br /> 12. IN NO EVENT UNLESS REQUIRED BY APPLICABLE LAW OR AGREED TO IN WRITING<br />WILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MAY MODIFY AND/OR<br />REDISTRIBUTE THE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES,<br />INCLUDING ANY GENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING<br />OUT OF THE USE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED<br />TO LOSS OF DATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY<br />YOU OR THIRD PARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER<br />PROGRAMS), EVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE<br />POSSIBILITY OF SUCH DAMAGES.</blockquote><div>(capitalization in the original). Of course, there is "required by applicable law" but I cannot see people giving you software for free if you later make them pay fines.</div><div><br /></div><div>And for course, it is also almost impossible to make exceptions in the law for this. For example, a "non-commercial" exception does not help as even though you do not charge for open source software a lot of it is actually provided with some sort of commercial interest.</div><div><br /></div><div>Yes, I can understand the tendency to make creators of defective products that don't give a damn about an update path responsible for the stuff they ship out. And I have the greatest sympathy for consumer protection laws. But here, there collateral damage would be huge (we might well lose the whole open source universe every small software company except the few big one that can afford the herds of lawyers to defend against these fines).<br /><br />Note that I only argue for mandatory liability. It should of course always be a possibility that a provider of software/hardware give some sort of "fit for purpose" guarantee to its customers or a servicing contract where they promise to fix bugs (maybe so that the customer can fulfill their liabilities to their customers herself). But in most of the cases, the provider will charge for that. And the price might be higher than currently that for a light bulb with an IP address.</div><div><br /></div><div>The internet is a rough place. If you expose your service to it better make sure you can handle every combination of 0s and 1s that comes in from there or live with it. Don't blame the source of the bits (no matter how brain dead the people at the other end might be).</div><div><br /></div><div><br /></div>Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-86039624289600381512016-10-07T10:15:00.003+02:002016-10-07T12:07:57.756+02:00My two cents on this year's physics Nobel prizeThis year's <a href="http://www.nobelprize.org/nobel_prizes/physics/laureates/2016/press.html">Nobel prize</a> is given for quite abstract concepts. So the popular science outlets struggle in giving good explanations for what it is awarded for. I cannot add anything to this, but over at math overflow, mathematicians <a href="http://mathoverflow.net/questions/251470/topology-and-the-2016-nobel-prize-in-physics/251590#251590">asked for a mathematical explanation</a>. So here is my go of an outline for people familiar with topology but not so much physics:<br /><br />Let me try to give a brief explanation: All this is in the context of Fermi liquid theory, the idea that you can describe the low energy physics of these kinds of systems by pretending they are generated by free fermions in an external potential. So, all you need to do is to solve the single particle problem for the external potential and then fill up the energy levels from the bottom until you reach the total particle number (or actually the density). It is tempting (and conventional) to call these particles electrons, and I will do so here, but of course actual electrons are not free but interacting. This "Fermi Liquid" explanation is just and effective description for long wavelength (the IR end of the renormalization group flow) where it turns out, that at those scales the interactions play no role (they are "irrelevant operators" in the language of the renormalization group).<br /><br />The upshot is, we are dealing with free "electrons" and the previous paragraph was only essential if you want to connect to the physical world (but this is MATH overflow anyway).<br /><br />Since the external potential comes from a lattice (crystal) it is invariant under lattice translations. So Bloch theory tells you, you can restrict your attention as far as solving the Schrödinger equation to wave functions living in the unit cell of the lattice. But you need to allow for quasi-periodic boundary conditions, i.e. when you go once around the unit cell you are allowed to pick up a phase. In fact, there is one phase for each generator of the first homotopy group of the unit cell. Each choice of these phases corresponds to one choice of boundary conditions for the wave function and you can compute the eigenvalues of the Hamiltonian for these given boundary conditions (the unit cell is compact so we expect discrete eigenvalues, bounded from below).<br /><br />But these eigenvalues depend on the boundary conditions and you can think of the as a function of the phases. Each of the phases takes values in U(1) so the space of possible phases is a torus and you can think of the eigenvalues as functions on the torus. Actually, when going once around an irreducible cycle of the torus not all eigenvalues have to come back to themselves, you can end up with a permutation it this is not really a function but a section of a bundle but let's not worry too much about this as generally this "level crossing" does not happen in two dimensions and only at discrete points in 3D (this is Witten's argument with the 2x2 Hamiltonian above).<br /><br />The torus of possible phases is called the "Brioullin zone" (sp?) by physicists and its elements "inverse lattice vectors" (as you can think of the Brioullin zone as obtained from modding out the dual lattice of the lattice we started with).<br /><br />Now if your electron density is N electrons per unit cell of the lattice Fermi Liquid theory asks you to think of the lowest N energy levels as occupied. This is the "Fermi level" or more precisely the graph of the N-th eigenvalue over the Bioullin zone. This graph (views as a hyper-surface) can have non-trivial topology and the idea is that by doing small perturbations to the system (like changing the doping of the physical probe or changing the pressure or external magnetic field or whatever) stuff behaves continuously and thus the homotopy class cannot change and is thus robust (or "topological" as the physicist would say).<br /><br />If we want to inquire about the quantum Hall effect, this picture is also useful: The Hall conductivity can be computed to leading order by linear response theory. This allows us to employ the Kubo formula to compute it as a certain two-point function or retarded Green's function. The relevant operators turn out to be related to the N-th level wave function and how it changes when we move around in the Brioullin zone: If we denote by u the coordinates of the Brioullin zone and by $\psi_u(x)$ the N-th eigenfunction for the boundary conditions implied by u, we can define a 1-form<br />$$ A = \sum_i \langle \psi_u|\partial_{u_i}|\psi_u\rangle\, du^i = \langle\psi_u|d_u|\psi\rangle.$$<br />This 1-form is actually the connection of a U(1) bundle and the expression the Kubo-formula asks us to compute turns out to be the first Chern number of that bundle (over the Brioullin zone).<br /><br />Again that, as in integer, cannot change upon small perturbations of the physical system and this is the explanation of the levels in the QHE.<br /><br />In modern applications, an important role is played by the (N-dimensional and thus finite dimensional) projector the subspace of Hilbert space spanned by the eigenfunctions corresponding to he N lowest eigenvalues, again fibered over the Brioullin zone. Then one can use K-theory (and KO-theory in fact) related to this projector to classify the possible classes of Fermi surfaces (these are the "topological phases of matter", as eventually, when the perturbation becomes too strong even the discrete invariants can jump which then physically corresponds to a phase transition).Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-88880790578901583592016-10-07T10:15:00.002+02:002016-10-07T10:42:17.891+02:00My two cents on this years physics Nobel prizeThis year's <a href="http://www.nobelprize.org/nobel_prizes/physics/laureates/2016/press.html">Nobel prize</a> is given for quite abstract concepts. So the popular science outlets struggle in giving good explanations for what it is awarded for. I cannot add anything to this, but over at math overflow, mathematicians <a href="http://mathoverflow.net/questions/251470/topology-and-the-2016-nobel-prize-in-physics/251590#251590">asked for a mathematical explanation</a>. So here is my go of an outline for people familiar with topology but not so much physics:<br /><br />Let me try to give a brief explanation: All this is in the context of Fermi liquid theory, the idea that you can describe the low energy physics of these kinds of systems by pretending they are generated by free fermions in an external potential. So, all you need to do is to solve the single particle problem for the external potential and then fill up the energy levels from the bottom until you reach the total particle number (or actually the density). It is tempting (and conventional) to call these particles electrons, and I will do so here, but of course actual electrons are not free but interacting. This "Fermi Liquid" explanation is just and effective description for long wavelength (the IR end of the renormalization group flow) where it turns out, that at those scales the interactions play no role (they are "irrelevant operators" in the language of the renormalization group).<br /><br />The upshot is, we are dealing with free "electrons" and the previous paragraph was only essential if you want to connect to the physical world (but this is MATH overflow anyway).<br /><br />Since the external potential comes from a lattice (crystal) it is invariant under lattice translations. So Bloch theory tells you, you can restrict your attention as far as solving the Schrödinger equation to wave functions living in the unit cell of the lattice. But you need to allow for quasi-periodic boundary conditions, i.e. when you go once around the unit cell you are allowed to pick up a phase. In fact, there is one phase for each generator of the first homotopy group of the unit cell. Each choice of these phases corresponds to one choice of boundary conditions for the wave function and you can compute the eigenvalues of the Hamiltonian for these given boundary conditions (the unit cell is compact so we expect discrete eigenvalues, bounded from below).<br /><br />But these eigenvalues depend on the boundary conditions and you can think of the as a function of the phases. Each of the phases takes values in U(1) so the space of possible phases is a torus and you can think of the eigenvalues as functions on the torus. Actually, when going once around an irreducible cycle of the torus not all eigenvalues have to come back to themselves, you can end up with a permutation it this is not really a function but a section of a bundle but let's not worry too much about this as generally this "level crossing" does not happen in two dimensions and only at discrete points in 3D (this is Witten's argument with the 2x2 Hamiltonian above).<br /><br />The torus of possible phases is called the "Brioullin zone" (sp?) by physicists and its elements "inverse lattice vectors" (as you can think of the Brioullin zone as obtained from modding out the dual lattice of the lattice we started with).<br /><br />Now if your electron density is N electrons per unit cell of the lattice Fermi Liquid theory asks you to think of the lowest N energy levels as occupied. This is the "Fermi level" or more precisely the graph of the N-th eigenvalue over the Bioullin zone. This graph (views as a hyper-surface) can have non-trivial topology and the idea is that by doing small perturbations to the system (like changing the doping of the physical probe or changing the pressure or external magnetic field or whatever) stuff behaves continuously and thus the homotopy class cannot change and is thus robust (or "topological" as the physicist would say).<br /><br />If we want to inquire about the quantum Hall effect, this picture is also useful: The Hall conductivity can be computed to leading order by linear response theory. This allows us to employ the Kubo formula to compute it as a certain two-point function or retarded Green's function. The relevant operators turn out to be related to the N-th level wave function and how it changes when we move around in the Brioullin zone: If we denote by u the coordinates of the Brioullin zone and by $\psi_u(x)$ the N-th eigenfunction for the boundary conditions implied by u, we can define a 1-form<br />$$ A = \sum_i \langle \psi_u|\partial_{u_i}|\psi_u\rangle\, du^i = \langle\psi_u|d_u|\psi\rangle.$$<br />This 1-form is actually the connection of a U(1) bundle and the expression the Kubo-formula asks us to compute turns out to be the first Chern number of that bundle (over the Brioullin zone).<br /><br />Again that, as in integer, cannot change upon small perturbations of the physical system and this is the explanation of the levels in the QHE.<br /><br />In modern applications, an important role is played by the (N-dimensional and thus finite dimensional) projector the subspace of Hilbert space spanned by the eigenfunctions corresponding to he N lowest eigenvalues, again fibered over the Brioullin zone. Then one can use K-theory (and KO-theory in fact) related to this projector to classify the possible classes of Fermi surfaces (these are the "topological phases of matter", as eventually, when the perturbation becomes too strong even the discrete invariants can jump which then physically corresponds to a phase transition).Robert Hellinghttps://plus.google.com/118220336522940810893noreply@blogger.com0tag: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.com1tag: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.com1