# I Little issue regarding physical states

1. Nov 27, 2017

### Staff: Mentor

I wonder how Pauli missed this and spent all those years unsuccessfully trying to rigorously derive it. Feynman even got into the act - and failed. That theorem is not won easily at all:
http://www.worldscientific.com/worldscibooks/10.1142/3457

Its one of the most important, but well known difficult to prove theorems in all of QFT. As far as I know the first actually valid proof was given by Luders in 1958:
http://www.sjsu.edu/faculty/watkins/spinstats10.htm

Thanks
Bill

Last edited: Nov 27, 2017
2. Nov 27, 2017

### mikeyork

In the Heisenberg picture, the variation of the observable is only in the time dependence. I am saying something much more general -- as you'll see if you go back an look at #6.

3. Nov 27, 2017

### mikeyork

There have been a zillion "proofs" of the theorem, nearly all of them relying on an assumption (that differs from case to case, sometimes explicit, sometimes hidden) that effectively gives the desired result.

The apparent difficulty is due to Dirac's original formulation in which he supposed that the "exchange" operator had two distinct eigenvalues. Then people became obsessed with this and the symmetry versus anti-symmetry description of the theorem. But, as Feynman correctly noted, the sign change for fermions is due to a $2\pi$ rotation on one particle. (If I recall correctly he actually attempted a proof on the basis of this rotation originating in T-invariance.). Many authors have published proofs that you obtain anti-symmetry for fermions from such a rotation when you do the "exchange". (The list includes Steven Weinberg, E C G Sudarshan and Michael Berry.). But they nearly all make an assumption about single-valued state vectors or wave functions that is not obviously valid. The resolution of this problem comes from recognizing the importance of distinguishing frames related by a $2\pi$ rotation in specifying a basis (a specific case of my more general suggestion in this thread). And this is the part I never see in any textbook. If it were there, the author would easily have seen the spin-statistics proof.

4. Nov 28, 2017

### Staff: Mentor

I know Wigner's Theroem

It has noting to do with if states are frame invariant.

Thsnks
Bill

5. Nov 28, 2017

### Staff: Mentor

OK - the link I gave to what was considered the first valid proof - did it make that assumption - it lists them. Exactly which assumption would anyone care to doubt and why?

Like I said many have tried and failed with that one, but in recent times proofs have appeared that are generally considered to contain no loopholes - and I don't think it was the one you said was 'In fact, since a frame rotation about the spin projection axis is a simple phase change, this observation enables us to remove arbitrary phase factors from state vectors and reduce them to a single global phase factor for the entire Hilbert space. And this alone enables us to deduce the spin-statistics theorem'

That pure states are invariant to phase changes follows immediately from the definition of a pure state ie its of the form |u><u|. It strangely leads to EM by making the symmetry local - not just global - that to me is something that is almost a rabbit pulled out of a hat and may be trying to tell us something important - don't know what though. But proving the spin-statistics theorem from it - that looks a stretch. Simple, but not 100% rigerous proofs of it aren't that hard - loophole free ones - much more difficult eg for a simple one:
http://merlin.fic.uni.lodz.pl/concepts/2008_2/2008_2_281.pdf

If its a reasonably simple, but not quite rigorous one - like the above - then maybe - but I cant see it - relativity seems a key ingredient. Overall however its a very difficult thing to prove correctly.

As an aside, yes beginning texts often simply assume it - but more advanced texts are much more careful about the issue.

Thanks
Bill

Last edited: Nov 28, 2017
6. Nov 28, 2017

### mikeyork

There are two ways to describe a change of frame: (1) simply transform the frame itself, regardless of any observable or its value, or (2) change the observable value to what it would be if you had transformed the frame. If (as in every QM textbook I have seen) you do not explicitly include the frame in the choice of basis then, as per Wigner, you can only choose (2) and get a unitary operator. Without the explicit reference to the frame in the chosen basis, it is as if the observable had magically changed its value by itself. But if you make the frame explicit then you can choose (1) instead and get the inverse operator. So, by including the frame explicitly, you can apply both operators and maintain a frame invariant state vector -- as I described in #6.

7. Nov 28, 2017

### mikeyork

Discussion of the spin-statistics theorem in this thread is getting off-topic. I mentioned it only to show the power of explicitly including the frame of reference when choosing a basis that enables frame-invariant state vectors. I'll address your disagreement about this one last time.

Any "proof" that relies on demonstrating that the state vector, in x or p representation, must be anti-symmetric (and relates the spin projection frame to the x/p frame) will be incomplete because it is trivially easy to show that there is an equivalent symmetric state vector for which "exchange" does not involve a $2\pi$ rotation on the spin frame of one particle relative to the other*. The usual $2\pi$ rotation originates in a geometric asymmetry. And this asymmetry can be eliminated by specifying states in a symmetric way. This is the basis of all the $2\pi$ and single-valued state vector proofs. If, however, the proof applies to the SU(2) (l,m) representation instead of x or p, for which the $2\pi$ ambiguity is already removed by the usual convention, then yes you will always get anti-symmetric states for identical half-integer spin. Since the proof you mention is an x proof, it necessarily contains the geometric asymmetry but it will be hidden. Not being a field theory expert I could not say how that applies to the explicit assumptions in the proof you mention. Usually, in field theory proofs, it is hidden in the choice of spinors (e.g. Weinberg).

Of course. It is fundamental that the state itself does not care about an observer's frame of reference. It is the selection of a unique invariant (i.e. single-valued) state vector that is not possible if the frame is not explicitly included in the basis specification, because the state vector $|x,s_z\rangle$, where $s_z$ is spin, will always be multi-valued (up to an arbitrary phase factor) until you include some way to distinguish the situations that differ by rotations of the spin projection frame (not necessarily the same as the co-ordinate frame) about the z-axis or if you specify all axes of the complete spin projection frame, then, when $s_z$ is half-integer spin, it will still be double-valued because of a possible $2\pi$ rotation. To remove this multi-valuedness you have to specify a unique rotation which takes the co-ordinate frame into the spin projection frame.

*Symmetrizing The Symmetrization Postulate'', AIP Proceedings 545 (2000) Spin Statistics Connection And Commutation Relations'', ed. Hilborn and Tino, pp 104-110. (http://arxiv.org/pdf/quant-ph/0006101.pdf)

8. Nov 28, 2017

### Staff: Mentor

Wigner says any symmetry transformation can be represented by a unitary or anti-unitary operator. It says absolutely nothing about if the state changes or not - only restrictions on if it does. If the state does not change - then guess what the transformation is that wonderful transformation the identity transformation.

Basis and that other stuff you mention - why you even mention it is beyond me.

If one takes the Heisenberg picture then in normal non-relativistic QM which implies the Galilean transformations, by the very definition of what the Heisenberg Picture is, states do not change - of course observable's do when you frame jump.

Relativistic QM can evidently be done in the Heisenberg picture - but I don't think anyone ever does it that way - they generally use the Dirac picture. Even in usual QM I don't see the Heisenberg used much - but we are speaking of matters of principle here.

Thanks
Bill

9. Nov 28, 2017

### mikeyork

Once more, as you seem to have missed it yet again, the issue is NOT whether the state changes, but whether the state vector (in the Dirac picture) changes. If you do not make the frame explicit in the choice of basis then it does. To get a state vector that does not change you must make the frame explicit in the basis specification.

Unfortunately much discussion of QM goes off the rails because of the loose language that confuses a state (obervables in a frame of reference) with its state vector (the Hilbert space vector that represents the state).

10. Nov 28, 2017

### Staff: Mentor

The state is NOT a vector. Its a positive operator of unit trace. Now this whole thread is if states change. When a pure state |u><u| is represented by the vector |u> a phase change is irrelevant so best to stick with what states actually are rather than cloud the issue by thinking of them as vectors.

And the Dirac picture? I am talking of the Heisenberg picture.

I don't think this is going anywhere so I will leave it at that - others may get what you are trying to say - I don't.

Thanks
Bill

11. Nov 28, 2017

### mikeyork

And, of course, no one said it was. I said the state vector is a vector. I was even at pains to point out in my last post how loose language confusing a state with a state vector can destroy rational dialogue. But that seems to have been lost on you.
No it is NOT! Go back and look at #1. It was about whether the ray (effectively the state vector) changes.
You cannot define $|u\rangle\langle u|$ without $|u\rangle>$ -- which is what the op was asking about and how it is affected by a frame change. And certain frame changes do indeed affect the phase.
Well, since you have taken issue with what I wrote, shouldn't you follow my context? Your introduction of the Heisenberg picture was irrelevant.

12. Nov 28, 2017

### Staff: Mentor

The point of the Heisenberg picture is in it the state doesn't change, thus directly answering the OP's question. As far as I am concerned that's the end of the issue, but it seems you have a different context. This, I conjecture, is what has led to us 'talking past each other'. In my experience when that happens its best not to really continue since, yes eventually context etc will all be clear - but with much 'teeth gnashing' on each side. I like to avoid that unless the issue is important, which I don't think this one is.

So I will simply be an observer if anyone else wishes to continue the thread.

Thanks
Bill

13. Nov 29, 2017

### mikeyork

No. The op's question was about an invariant ray (state vector). I really don't understand how you don't get this distinction.

14. Nov 29, 2017

### Fra

One point of the debate is what "physical states" is supposed to mean? Either its some space of distinguishable states as per some observer or measurement device, if you take the instrumentalist perspective, or it is an equivalence class of spaces from all possble observers (or typically from a SUBSET of all possilbe observers). Which is the real thing and what is just "gauge"? You can make reasonable arguments for both views, and the differences also roots in differences between GR and QM. Its like a chicken and egg situation. What is the more fundamental starting point? the equivalence class with constraints or its members that negotiates?

Both ways has its own problems.

Just changing the frame of references, clearly can not exhaustively generate all possible observers because an observer is more than moving origos. In particular what is missing is complexity and different histories that distinguishes observers and constitutes part of the observer "background".

Renormalisation flow is not a general solution to this either. While it is true that decreasing observer complexity, necessarily is a lossy transformation, the situation is necessarily not just a simplistic block-renormalisation. Also the other way of the transformation - the invserse of the lossy transformation must be seen as an evoutionary learning making "theory space" inseparable from the tangent space of the differentially evolving state space.

For example, how are two observers, litteraly living at two different complexity levels supposed to be able to compare a truth statement? Here i argue that - after the natural UV and IR cuttof implied by the observers complexity, we might without contradiction arrive at a situation that these observers disagree about logical statements. If we defined this in termas of a truncated computation.

Just wanting to support the OT as a good question if meant in the most general sense, but i think without conceptually clear answer. I get the impression that this is a kind of beginners question but my own experiensce is that it happens that such questions are handled with a deceptively trivialising attitude by teachers. I rememer this myself, but after a while you figure out that the reason for this is that there IS no good answer, and then its better to acknowledge this instead of denying the question.

QM and QFT as they stand can not handle these questions properly though, but that makes the question no less important.

/Fredrik

15. Nov 29, 2017

### vanhees71

I've never seen that notation either, and I don't know what it should be good for, and I've also no clue how you come to conclude from what I wrote before, I'd suggest to use it.

16. Nov 29, 2017

### vanhees71

of course, whether the ray is time dependent or not depends on the chosen picture of time evolution. In the Heisenberg picture it's time-independent by definition.

17. Nov 29, 2017

### mikeyork

#6 shows what it is good for.

18. Nov 29, 2017

### mikeyork

The OP was about a much more general frame transformation -- not just time evolution.