Coherent states

If you are not working on quantum optics, you might be in a similar situation as I was a couple of days ago regarding coherent states: You had encountered them in a homework exercise on the harmonic oscillator where you had to prove that they are eigenstates of the creation operator and have minimal uncertainty. And you know that they are important to quantum optics. At least this was my state of knowledge until very recently. Since then, I have read this review (with which I do not agree in all parts) and have spent some thoughts on the subject I would like to share my current understanding.

Let's suppose we have two hermitean operators A and B and want to find states such that the uncertainty is minimal. To this, let's briefly go through the derivation of the uncertainty relation: You assume any state and from it form new states and similarly for B where is the expectation value of A. For these two new states, you use the Cauchy-Schwarz inequality (which basically says in the form ), expand and find


Finally, we use that the absolute value of the imaginary part of a number is less or equal to that number and realise that as A and be are hermitean the imaginary part of the left expectation value is to arrive at which is the usual uncertainty relation.

If you want to rest for a minute think about the following puzzle: Consider a particle on a circle (or on the interval with periodic boundary conditions). Take the wave function and compute that for this state while . This seems to clash with what we just derived. Where is the flaw? (Hint: see a previous post about quantum mechanics)

Back to the main argument. We want to find a state which saturates the inequality. To have that we have to saturate the inequality in the two places where used inequalities: The Cauchy Schwarz and the abs less Im parts of the argument. Cauchy Schwarz is saturated (the scalar product is maximal for vectors of fixed length) if the two vectors are proportional to each other, that is if there is a complex number such that in our case . We can rearrange that to that is has to be an eigenvector of the operator . Furthermore, for the absolute value of a number to be equal to its imaginary part, the number has to be purely imaginary. In our case, this means has to be purely imaginary which can only be if is purely imaginary.

So we found that the uncertainty of operators A and B in a state is minimal if the state is an eigenvector of an operator with real .

So much for the general theory. Now, we can specialise to the usual case A=x and B=p and conclude that states of minimal uncertainty are eigenstates of for some real . Note that so far we have not talked about the harmonic oscillator at all. We have just picked two operators and asked for states in which they have minimum uncertainty. This was a question at the level of Hilbert space operators and we did not specify any sort of dynamics.

Thus, coherent states are not about the harmonic oscillator at all. It just happens that they are eigenstates of annihilation operators for some harmonic oscillator. Above any real does the job and this translates directly to the frequency of the oscillator: What people call "squeezed states" are just coherent states for a different that can in a similar way be related to the annihilation operators at different frequencies.

This so far is my current understanding. In the above mentioned review there is another generalisation which does involve dynamics which I do not yet fully understand. It somehow splits a Hamiltonian into sums of products of 'elementary' operators and then considers the Lie algebra generated by these elementary operators upon commutators. Then you exponentiate this algebra to a group and consider the orbit of the ground state of that Hamiltonian under the action of this group. The part I do not yet understand is how physical this is and how the different choices on the way (the set of elementary operators for example) influence the result.

Science and arXiv

I was made part of the team that assembles our school's research report involving contributions from all faculty members including references to their publications. My job is mainly merging all contributions into a single LaTeX document and managing the references. We decided to do them in BibTeX so we asked all faculty to provide a list of their papers in BibTeX format. The idea was of course that only a tiny part of this data would have to be typed as most literature databases (such as spires for our field) provide data in this format and other programs like Endnote can export BibTeX as well. Thus with a bit of cut&paste the job would be easy.

I had not expected the amount of computer illiteracy amongst science professors. OK, I knew that most biologists do not use TeX for their papers. But I must admit that I was impressed receiving an MS Word document containing BibTeX entries but for example having all the title="..." fields set in italics. That is not to speak of the many complaints I received about people having to retype their references. Plus the concept of separating content and layout is completely alien to many.

But what I really wanted to talk about is this: I learned during one of these discussions with an experimental surface physicist (who by the way keeps his references typed in a Word document) why he does not submit his preprints to the arXiv: He told me Science does not accept papers which are in electronic archives others than their own! I find this completely ridiculous and could not believe it since at least my only science paper (btw my first paper at all and still the one with most citations although not in high energy)had a preprint on the arXive. But it seems he is right, at least as far as the current policy is concerned.

Please, please, somebody tell me this interpretation is not true and the greed of the AAAS is not in the way of good scientific practices.

Paper cranes

Yesterday, I served as jury member for the exciting physics competition which is part of the WellenWelten ('wave worlds') physics exhibition in Bremen.

Children (possibly in teams) from 10 to 19 could choose from six construction tasks (announced two months earlier) and present their result in the Congress Centre.

I had to judge paper cranes. The task was to use only paper, glue, sand and twine to build a crane. It should only touch the table in an area of A4 size and be able to hold a 400g weight 40cm above the table and 25cm in front of the base. Furthermore, the crane had to be stable both with and without the weight. With these constraints the task was to build the crane as light as possible.

We had to judge the cranes not only on their stability and weight but also on design, presentation and construction. The level of the 15 submissions was very high and it was not easy to determine the winners. In the end we settled for this crane



which was constructed by two girls from 9th grade (13 years). The three runners up are



.

You find all the pictures here (two pages). Red t-shirts indicate participants and their teachers and dark blue is the jury.

Seminars while you drive

If I drive longer distances and get bored of listening to the radio I love audio books. Too bad they are usually quite expensive. But i discovered an alternative: Listen to seminars.

A good place to start is the KITP, for example their Blackboard Lunches.

The audio formats they offer are realaudio and ipod. As I do not own one of these white gadgets, I have to use the other option. My car radio can play mp3's (from its USB port or CDs). So I have to convert the .rm files to mp3. Here is how you can do that (so you don't have to spend as much time on it as I did):

In case it is not done already, install mplayer.

When you call it like

mplayer -quiet -vo null -vc dummy -af volume=0,resample=44100:0:1   -ao pcm:waveheader http://online.itp.ucsb.edu/download/bblunch/dine.rm

it will create a file audiodump.wav which in turn you can convert to an mp3 with bladeenc.

After you've done that for a couple of talks, use

mp3burn *mp3

to write the seminars to a CD and you can hit the road, Jack!

Update: Of course you don't use mp3burn as that would produce an ordinary audio CD with maximally 72 Minutes playing time. You want a data disk with the mp3's on it so you really use a program like xroastcd.

Two Sudoku Problems

Here are two Sudoku meta-problems I have been thinging about for a while.

The first is about the normal form of a sudoku. The rules of sudoku have a huge symmetry group of order (3!)^8 x 2 x 9! = 1.22E12. It is generated by permutations of rows in a group of three rows, permutations of these groups, same thing for columns, transposition of the grid and permutations of the numbers 1,...,9 which are just labels for nine different things. So for each grid, there are 1.22E12 grids which are basically the same. Is there an easy way to determine if two puzzles are related by the symmetry and even better, is there a normal form (a distinct element for each orbit)?

There is a trivial solution to this problem: You just write out all 10^12 puzzles you obtain by acting with the symmetry group and then sort them according to some lexicographic ordering. This gives a normal form.

What I am asking for is a more direct construction. One which sounds more like: Use the 9! permutations of the symbols to make the first row read 123456789. Then use row permutations to make the square below the 1 as small as possible. Then use row permutations to make the next squares under that square as small as possible... Unfortunately, starting to permute colums screws up what was achieved with this ordering prescription. So how would a better algorithm read?

The other problem is how to rate the difficulty of a puzzle? Question one really applied after the puzzle is solved, this question is about puzzles to be done. Newepaper which publish these puzzles often give ratings such as "simple", "intermediate", "hard". But I found these ratings differ significantly between papers (what Zeit online considers hard is much much easier compared to what The Guardian calls a hard sudoku) but are also not consistent amongst themselves.

Earlier, I have talked about the perl program I wrote to solve sudokus. It recursively figures out for all squares which numbers are still allowed and then takes the square with the least number and tries to put these numbers there. If there is an empty square with no allowed numbers remaining it backtracks.

Thus the search can be represented by a tree where each node represents a square to be filled out and there are as many branches from that node as there are numbers which are not yet rules out. What I am looking for is a numerical rating for a puzzle which is a predictor of how hard I find to do the puzzle and for example correlates with the time it takes me to solve it. Even if I use a different strategy when doing these puzzles by hand I would expect the information could be obtained form the tree. Do you have any good idea for such a function from trees to the reals, say? Obviously the trees all have a depth given by the number of empty squares in the puzzle and each node can have at most nine branches but typically has much less (even for "hard" puzzles most of the nodes have only one branch).

An easy guess is of course the number of nodes or the number of leaves but I found those at least not be proportional to my manual solution time. To give you an idea: Today's hard puzzle from Die Zeit

..573.864
.4...8...
.83.....1
71....2..
3..6921.7
4...7..9.
....4.97.
..6..5..8
.......1.

has 52 nodes (four times the program encounters situtations with two possibilities, all others are unique or dead ends, manually it took me exactly 6:30) while

.98......
....7....
....15...
1........
...2....9
...9.6.82
.......3.
5.1......
...4...2.

has 2313 nodes and took me well over an hour some months ago.



Of course, if early on you have several possibilities and learn only much later which ones do not work this is much worse than having many possibilities which are ruled out immediately.

UPDATE: In case anybody is interested, I put up the decision tree for the difficult puzzle.

Cheap quantum cryptography

Quantum information theory is a fascinating subject. By applying the simple computational rules of quantum mechanics it is often possible to process information much better than with just classical devices. A famous example being Shor's algorithm for the factorisation of integers relevant for breaking many popular public key encryption schemes such as RSA. The speed up is not exponential but "only" by a square root but this can already be substantial. Formal computer scientists amuse themselves by investigating how the many complexity classes change if you have access to some quantum computations, the Complexity Zoo gives a nice overview.

A beatiful introduction into the subject are the lecture notes by John Preskill. A more formal aspect (investigating which quantum machines can be constructed, how does the impossible quantum copier differ from possible devices and that in the language of operator algebras) is treated in the notes by Reinhard Werner.

However, all this, much like string theory, is only theory and does not have real world applications. As far as real experiments go, IBM has been able to factor 15 with a quantum computer in 2001.

Another potentially applicabel area of application is cryptography: It is possible to construct quantum channels that are immune to eavesdropping: If somebody in the middle listens in the information does not reach the intended recipient anymore. This has been demonstrated in a real experiment by Anton Zeilinger: He managed to transmit the keys of a classical encryption scheme via entangled photons (with a bitrate of 400-800bit/s and 3% error).

This still involves a considerable experimental set-up and remember: You only transmit the keys, as the quantum channel is my far too slow to transmit real data (you probably read much faster than 400bit/s). But there are cheap alternatives (not in the theoretical but in the practical sense) which are as well impossible to crack: One-time pads. Assume I have 1GB of data which I want to transmit to you securely. All I need are 1GB of random numbers which I share with you beforehand and then I xor the data with the random numbers transmit the result (which is just noise for anybody in between) and xor the encrypted message to recover the original data.

This is not elegant as we have to share the random numbers beforehand, we have to share as many bits as we want to transmit. But for many practical applications this is easily possible (headquater of some company who want to transmit construction plans to factory, a spy who wants to phone home, you name it): Cheap small harddrives have more room than all the information you would like to transmit secretly in all you life. When you construct the factory you just bring there the harddrive in a sealed box and practially all future communication is secure, the spy carries a sealed usb-stick and has more shared randomness than all the secret pictures he is goint to take and all the reports he has to send. This is not elegant but dead cheap and efficient.

And if you like you can even produce the randomness using quantum physics to make it physically safe: For example you could sample the timing of the ticks of a Geiger counter and have pure quantum randomness.

Here is a small homework: Take t1, t2,...,tN to be the times of N clicks. By themselves, they are not pure random numbers but the time difference follows a Poisson distribution. Assume that the ti are specified with B bits each, so the possible time resolution is dt and the decay constant of the probe is lambda. How many bits of pure randomness can you extract and how do you do it?

Admitting my ignorance

I don't know what is you attitude towards rigorous functional analysis but mine could be summarised as "I know it exists. There are subtleties but they don't bite as long as you are not asking for it. So, for everyday quantum mechanics, it's enough to remember that operators are not just matrices and there might be convergence issues (otherwise taking the trace would show that [x,p]=i cannot work)".

I knew that most of the time we are dealing with unbounded operators which are thus not continuous and mathematicians might be worried about their domains of definition (which can only be a proper subset of the Hilbert space) but if you do the natural things (and implicitly work on the proper dense subset of the Hilbert space), you will be ok. Furthermore, the spectrum of an operator can be a bit tricky as everybody knows that the 'eigenfunctions' for example of the momentum operator are plane waves which are not square integrable and similarly, eigenfunctions of x are 0 as elements in L^2. But every child knows that the proper definition of the spectrum of A are those z for which (A-z) is not invertible and any physics argument involving eigenfunctions can be made precise using wave packets which are not exactly eigenfunctions but if one wanted to one could control the error and after a long and messy argument you could prove what the physicist had known right from the beginning.

But, as I have learned, sometimes the subtleties are also physically relevant: The first time I realised this was in my oral diploma exam: I was asked to discuss the particle in a piecewise constant potential (and compute reflection and transmission coefficients etc). I was asked why I picked particular boundary conditions of my wave function at the jumps of the potential. Luckily, instead of parroting what I had read in some textbook ('the probability current has to be continuous so no probability gets lost') I had one of my very few bright moments and realised (I promise I came up with this myself, I had not heard or read it before) that this comes from requiring the Hamiltonian (esp. the kinetic term) to be self-adjoined: If you check this property, you have to integrate by parts and the boundary terms vanish exactly if you assume the appropriate continuity conditions of the wave function.

More recently I learned when the distinction between continuous and point spectrum is physically important: Long ago, in some advanced quantum mechanics class, we were shown some strange, seemingly unmotivated calculation with a random potential which after some time showed that that the eigenfunctions of the Hamiltonian have exponential decay. And "thus, even with arbitrarily small randomness the conductor turns into an isolator." I had never quite understood how this calculation was supposed to imply this conclusion. Only a few months ago, I understood in a seminar by Hajo Leschke that what was really meant was "with probability 1 the spectrum is a pure point spectrum and thus there are no scattering states". For more information check out a PhysRept by Fröhlich and Spence or, if you are particularly brave, the discussion of the RAGE-theorem in Reed Simon vol. III.

But this is not what I came to tell you about. I came to tell you that yesterday over lunch I was reading quant-ph/0609163 by H. Nicolic about myths and facts about quantum mechanics. I could comment on many sections but one particular argument stroke me. It goes back to Pauli and shows that if your Hamiltonian is bounded from below there is no time operator.

Here I will give you a slightly modified version: Consider quantum mechanics on the half line . In the zeroth approximation you would take as your Hilbert space. Obviously, in this space x is a positive operator. From the above reasoning it follows that you probably want to ask your wave functions to vanish at 0 as otherwise p is not symmetric:.


Now take some wave function
such that the expectation value of x is finite, say . Now, you can convice yourself that by applying a translation operator
you produce a new state for which x has the negative expectation !

How did that happen, wasn't x supposed to be a positive operator? The solution can be found in chapter 2.5 of Thirring's text book vol. 3 (no link from Amazon) as pointed out to me by Wolfgang Spitzer. The solution comes really from the functional analysis fine print: p is only symmetric but not self-adjoined. The domain of definition of is strictly larger as it does not require the vanishing condition at 0! There is no self-adjoined extension of p which is still hermitean. And therefore you cannot form the translation operator: It is not defined.

Another way to see this is to realise that for
you need all powers of p. And those are only symmetric (vanishing boundary terms) if actually all derivatives of the wave function vanish at 0. And as the translation operator works nicely only on analytic functions (after all it's just the Taylor series) that requirement does not leave us with too many functions.

Therefore you really have to worry about the finer points of functional analysis not to translate wave packets to where they should not be!