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.

## Monday, November 27, 2006

## Wednesday, November 15, 2006

### 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.

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.

## Friday, November 10, 2006

### 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.

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.

## Wednesday, November 08, 2006

### 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

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

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.

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 *mp3to 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.

## Tuesday, November 07, 2006

### 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

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

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.

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.

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