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Wigner's theorem, Bargmann's proof

by Fredrik
Tags: bargmann, proof, theorem, wigner
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Fredrik
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Mar8-12, 02:51 PM
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I've been trying to read Bargmann's 1964 proof of Wigner's theorem, but I find it really hard to follow. This is the article. If the link goes dead, I can email a PDF to anyone who PMs me their email address. I really don't understand what he's doing in section 4.5, right after eq. (15b). If anyone understands his proof of the formula he writes as ##\alpha_\rho'=\chi_\rho(\alpha_\rho)##, and can explain it to me, I would really appreciate it. The step where ##\alpha_\rho'## enters the calculation looks like magic to me. Actually, it looks like he's using what he's trying to prove, so I must have misunderstood something.

I have typed up some notes for myself on the earlier parts of the proof, which I believe that I understand, but it's possible that they would just be confusing, since my notation is different. I'll post all of my notes if someone requests it. The following is a very brief summary of the stuff before the section that's causing me problems. (It's probably very hard to follow if you're not already somewhat familiar with Wigner's theorem). Let me know if you want me to clarify some detail in this summary, or in the earlier parts of Bargmann's proof.

We consider equivalence classes of vectors in a Hilbert space H. Members of the same class differ only by a complex factor c such that |c|=1. Bargmann calls these classes rays*. We're given a permutation T of the set of unit rays, which is assumed to preserve probabilities. We begin by extending this map to the set of all rays. Our goal is to define Ux for all x in H in a way that ensures that the map U is either unitary or antiunitary, and also that for all x, Ux is a member of T[x], where [x] is the ray that x belongs to. Bargmann picks an arbitrary unit vector e, and first defines Ux for all x of the form x=e+y, such that y is orthogonal to e. Then he defines Ux for all x that are orthogonal to e. Actually, he calls the "U" defined on that subspace "V" instead of "U" for some reason. Later, he extends the definition to all of H, but before that, he proves a bunch of things about this "V". That's the part that's causing me trouble.

*) I think it's more common to call the 1-dimensional subspaces rays, but I'm using Bargmann's terminology in this post.
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martinbn
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Mar8-12, 05:10 PM
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Quote Quote by Fredrik View Post
I've been trying to read Bargmann's 1964 proof of Wigner's theorem, but I find it really hard to follow. This is the article. If the link goes dead, I can email a PDF to anyone who PMs me their email address. I really don't understand what he's doing in section 4.5, right after eq. (15b). If anyone understands his proof of the formula he writes as ##\alpha_\rho'=\chi_\rho(\alpha_\rho)##, and can explain it to me, I would really appreciate it. The step where ##\alpha_\rho'## enters the calculation looks like magic to me. Actually, it looks like he's using what he's trying to prove, so I must have misunderstood something.
With his notations it is
[tex]1=(f_i\gamma_i,f_i\alpha_i)=(f_i\gamma _i,x)=(Vf_i\gamma _i,Vx)= (f'_i\chi_i(\gamma_i),f'_i\alpha'_i)=\chi_i(\gamma_i)^*\alpha'_i=\chi_i (\alpha_i)^{-1}\alpha'_i=1[/tex]
Fredrik
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Mar8-12, 05:44 PM
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Hey, I actually understood that. Awesome. Thank you. I've been stuck on that detail for...longer than I want to admit. For some reason it didn't occur to me to just apply V to both vectors, even though I knew that I could. Weird.

I'm about to go to bed, so I won't be continue with this article tonight, but I might make another post here tomorrow if I get stuck again on the next detail.

martinbn
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Mar9-12, 06:32 AM
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Wigner's theorem, Bargmann's proof

You mentioned that you were writing notes on the paper. The original paper has notations and fonts that make it very hard to read. It would be nice if you posted the notes, when you are done writing them, of course if you are ok with it.

By the way the link that you gave in the begining doesn't open on my computer.
Fredrik
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Mar9-12, 07:31 AM
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Quote Quote by martinbn View Post
You mentioned that you were writing notes on the paper. The original paper has notations and fonts that make it very hard to read. It would be nice if you posted the notes, when you are done writing them, of course if you are ok with it.
Sure, no problem. I'll post my notes when I've made sure I understand the rest of the paper. Not sure if that will take hours or days, but I'm going to type it up anyway, so I might as well post it here when I'm done.

Quote Quote by martinbn View Post
By the way the link that you gave in the begining doesn't open on my computer.
That's odd. It's a link to a pdf file, and it works fine for me. But since you clearly have another way to access the paper, I guess we don't have to solve that mystery.
martinbn
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Mar9-12, 07:42 AM
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Quote Quote by Fredrik View Post
Sure, no problem. I'll post my notes when I've made sure I understand the rest of the paper. Not sure if that will take hours or days, but I'm going to type it up anyway, so I might as well post it here when I'm done.
I didn't mean to put pressure on you. The paper is from 1964, we can all wait for a few more days, weeks or months.
That's odd. It's a link to a pdf file, and it works fine for me. But since you clearly have another way to access the paper, I guess we don't have to solve that mystery.
It works now. I guess the problem was in my browser or intertent connections or something like that.
Fredrik
#7
Mar11-12, 09:43 AM
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I'm not done yet, but I'm attaching my preliminary notes on the stuff from the beginning of the article until just after the detail you helped me with, because there are still things I don't understand, and I think this document will make it easier to discuss them.

Questions:

1. Is my description (on page 1) of the idea behind the proof reasonably accurate?

2. Lemma 1.7 proves that for all ##y,z\in\{e\}^\perp##, ##\langle Uy,Uz\rangle## is either ##=\langle y,z\rangle## or ##=\langle y,z\rangle^*##, but don't we have to prove this for all y,z in the larger set ##\{e+w|w\perp e\}\cup\{e\}^\perp## before we can extend the definition to all of ##\mathcal H##?

3. Is there any reason to think that it's either ##\langle Uy,Uz\rangle=\langle y,z\rangle## for all y,z, or ##\langle Uy,Uz\rangle=\langle y,z\rangle^*## for all y,z? It seems possible that we might have to use the version with complex conjugation for some choices of y,z, and the version without the complex conjugation for other choices of y,z. It seems to me that the entire proof fails if this is the case.

In the notation defined on page 1, question 3 can be stated as follows: We know that for all ##y,z\in\{e\}^\perp##, there's a ##\theta\in\{I,I^*\}## such that ##\langle Uy,Uz\rangle=\langle y,z\rangle^\theta##, but how do we know that it's the same ##\theta## for all y,z? If it's not, then U won't be either unitary or antiunitary.

Keep in mind that this pdf is just a draft. It's likely that it contains mistakes. I may want to change a lot once I've understood the whole proof.

Edit: I suspect that the answer to question 3 will have something to do with the fact that definition 1.4 says "Let ##e'\in T[e]## and ##f'\in T[e]## be arbitrary". The U we're defining depends on the choice of the representatives e' and f' from the unit rays T[e] and T[f], so I think it's going to be a matter of choosing those representatives wisely. But I don't think Bargmann did that in his article. I'm going to have to look again.
Fredrik
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Mar12-12, 03:37 PM
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I have continued to think about this, but I haven't made much progress. The idea I had about clever choices of e' and f' seems to be wrong. I can prove that the functions ##\chi_y## defined by ##U(ay)=\chi_y(a)Uy## are independent of y. So I will denote them by ##\chi##, and write ##U(ay)=\chi(a)Uy##. I can prove that ##\chi(a)\in\{a,a^*\}##, but it still seems to me that an essential part of the proof is just missing from Bargmann's article. (If I'm right, this is pretty weird, since this is the number one reference for a proof of one of the most important theorems in QM). Bargmann doesn't seem to prove that ##\chi## is either the identity or complex conjugation. There could be a set ##A\subset\mathbb C## such that ##\chi(a)=a## for all ##a\in A## and ##\chi(a)=a^*## for all ##a\in A^c##.

I suspect that what I should be trying to do is to prove that ##\chi## is a continuous field automorphism of ℂ, because I think I read somewhere that the only continuous automorphisms of ℂ are the identity and complex conjugation. I will try that after I've rewritten a few things I'd like to change in my notes.

If anyone who understands these things well would like to comment, I would appreciate it.
Fredrik
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Mar13-12, 03:24 AM
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I think I figured it out. I'll prove that ##\chi## is continuous. (No need to prove that it's an automorphism at this point). Then I'll use that to prove that the sets A and Ac are both closed. This result will imply that either ##A\in\{\emptyset,\mathbb C\}## or ℂ is not connected. Since ℂ is connected, A is either ∅ or ℂ. So ##\chi## is either the identity or the complex conjugation operation.

Edit: Hm, maybe not. It's not obvious that U is continuous, and I was thinking I could use that to prove that ##\chi## is continuous. I need to think this through. ...OK, I've thought about it. I think I can prove that the restriction of U to the subspace ##\{e\}^\perp## is continuous. That should be enough to prove that ##\chi## is continuous.
Fredrik
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Mar13-12, 03:34 PM
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My idea about how to prove that U is continuous on ##\{e\}^\perp## failed. So I still have no idea how to complete this proof. Also, what I said earlier about how ##\chi_y## is independent of y was probably wrong. I can only prove that ##\chi_y=\chi_z## when y,z are orthogonal to both e and each other.

I have reorganized my notes somewhat. I guess I'll upload a new version tomorrow, for anyone who might want to help me figure this out.
Fredrik
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Mar15-12, 02:25 PM
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I realize that I'm probably just talking to myself, but I'll say this anyway, just in case someone (maybe years from now) is interested in seeing a proof of Wigner's theorem and finds this thread.

Unless something very unexpected occurs to me in the next few days, I'm abandoning this attempt to learn how to prove Wigner's theorem by studying Bargmann's article. My conclusion is that there are two big holes in his attempted proof. One problem is that he just proves that ##\chi_y(a)\in\{a,a^*\}## for all ##a\in\mathbb C##, when the statement we need is that ##\chi_y## is either the identity or the complex conjugation operation. This is the problem I was the most concerned about at first, but I think I know how to fix it: Let f be an arbitrary unit vector, orthogonal to e. Suppose that A and B are two subsets of ℂ such that ##\chi_f(a)=a## for all ##a\in A##, and ##\chi_f(a)=a^*## for all ##a\in B##. Clearly, ##A\cap B=\mathbb R##. Now let ##a\in A## and ##b\in B## be arbitrary. We have
\begin{align}
&U(af)=\chi_f(a)Uf=aUf\\
&U(bf)=\chi_f(b)Uf=b^*Uf
\end{align}
and
\begin{align}
\newcommand{\re}{\operatorname{Re}}
&\re\langle U(af),U(bf)\rangle=\re\langle af,bf\rangle=\re(a^*b)\\
&\re\langle U(af),U(bf)\rangle=\re(\chi_f(a)^*\chi_f(b))=\re(a^*b^*)
\end{align}
It's not hard to show that this implies that at least one of a,b must be real. So either ##A\subset B## or ##B\subset A##, and this means that either A=ℂ or B=ℂ. This doesn't completely solve the problem, but I think it solves the hard part of it, and that what's left are only minor details.

The other problem seemed less significant at first, but now I think it completely kills this proof. It's that to prove that the definition of Ux for arbitrary x will give us a unitary or antiunitary U, we need to use that the partially defined U (i.e. the U we defined on a proper subset of the Hilbert space) is unitary or antiunitary on its domain. Bargmann only proves that ##\langle Ux,Uy\rangle## is equal to either ##\langle x,y\rangle## or ##\langle x,y\rangle^*## for all x,y in a subspace ##\{e\}^\perp## that's a proper subset of that domain, and his proof method can't be extended to the whole domain, because the domain isn't a subspace. Maybe this part of the proof can be fixed. I don't know. But since it can't be fixed by using essentially the same method that Bargmann used, I have to consider his attempted proof a failure.

It's of course possible that I'm completely wrong, but I've spent a week on this now, and I think it's time to move on. I guess I'll check out Weinberg's proof instead.

I won't be posting my updated version of my notes, since I now think that Bargmann's proof strategy doesn't work.
Fredrik
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Mar23-12, 10:18 PM
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I certainly didn't expect this, but I was finally able to make sense of Bargmann's proof. I returned to it after reading about half of Weinberg's proof and thinking that his proof is probably going to work, but if I can fix Bargmann's proof, it will be prettier, and not require the assumption that the Hilbert space is separable.

So I'm uploading a new version of my notes. This should be a complete proof, that I hope will be much easier to follow than any of the proofs that appear in books and articles. Comments are welcome. Feel free to bump the thread if you find a mistake, even if the thread is old when you do. I would also like to know if an argument is hard to follow, or if you hate my notation or something.

Anyone who's interested in this should probably study the projection theorem first, or at least make sure that you understand what it says. It says that if M is a Hilbert subspace of a Hilbert space H, then for all x in H, there's a unique pair (y,z) of vectors in H such that x is in M, y is orthogonal to every member of M, and x=y+z.
Fredrik
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Apr6-12, 11:37 PM
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The first definition in the pdf uploaded with the previous post contains a ridiculous mistake. Obviously it should say that U is said to be θ-linear if ##U(ax+by)=a^\theta Ux+b^\theta Uy## for all ##x,y\in\mathcal H## and all ##a,b\in\mathbb C##. I'm uploading a new version with that mistake fixed. (The new version is otherwise identical to the previous one).

I deleted the two versions that I uploaded before.
Attached Files
File Type: pdf wigner.pdf (80.6 KB, 20 views)


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