Little issue regarding physical states

[Tio Barnabe]

Consider the QM postulate which states that physical states are represented by rays in a Hilbert space. Consider a ray $R$. An observer from other frame will have a correspoding $R'$ which can be either

- equal to $R$ or,
- not equal to $R$

Suppose the two frames are inertial frames. Consider the Relativity principle that "the laws of nature are the same in all inertial frames". This is translated to the statement that the rays are the same, i.e. $R' = R$, correct?

Then only the first scenario above would satisfy Relativity. What if it turns out that the second case is meet?

 

[bhobba]

Well first consider physically what states are. They are used, with observable's, to predict probabilities. Do you expect probabilities to change with frame? This is used by Ballentine in his text to derive Schrödinger's equation etc, and yes of course it easily's implied states are frame independent.

Thanks
Bill

 

[Tio Barnabe]

it easily's implied states are frame independent

Then we should ask ourselves in what sense are they frame independent. To me they do change from one frame to the other, but they would go along the same ray in the Hilbert space. So they are not equal, but they are equivalent, because same ray means same physical state.

 

[bhobba]

Then we should ask ourselves in what sense are they frame independent. To me they do change from one frame to the other, but they would go along the same ray in the Hilbert space. So they are not equal, but they are equivalent, because same ray means same physical state.

I explained the exact sense; remember they encode the probabilities of outcomes of observations and even imply it via Gleason's theorem. If they were frame dependent then the probabilities of observations would vary between frames. Different observers would register different probabilities of outcomes of observations. Imagine the outcome on a digital readout. All observers would observe the same readout and hence the same probabilities. That they would be different is physical non-sense. Strictly speaking you are invoking the POR, but aside from that it would be downright nonsensical and inconsistent even.

Thanks
Bill

 

[PeterDonis]

Consider the QM postulate which states that physical states are represented by rays in a Hilbert space.

You aren't just assuming that. You are assuming that the Hilbert space is a space of rays that correspond to "the state of some system at some time". In other words, something that gets transformed from frame to frame. (At least you are allowing for that possibility.)

But if we are talking about QM in the context of relativity, we are talking about quantum field theory, and in QFT, the Hilbert space is a space of rays that correspond to "quantum field operators at some spacetime event". In other words, there is nothing to transform because everything is at one event. If you are talking about observations made by different observers in different states of motion at that event, you are talking about different operators; that is how different "frames" are realized. So you don't "transform" anything when you go from frame to frame; you just switch which operators you are talking about.

 

[mikeyork]

Consider the QM postulate which states that physical states are represented by rays in a Hilbert space. Consider a ray $R$. An observer from other frame will have a correspoding $R'$ which can be either

- equal to $R$ or,
- not equal to $R$

Suppose the two frames are inertial frames. Consider the Relativity principle that "the laws of nature are the same in all inertial frames". This is translated to the statement that the rays are the same, i.e. $R' = R$, correct?

Not in the conventional construction of a Hilbert space. The reason is, as Wigner showed, that a frame transformation is represented by a unitary operator.

Sometimes people will argue that the transformed ray is in a distinct Hilbert space -- as if each Hilbert space had an associated frame/observer. But that is not consistent with a frame transformation being a unitary operator in a single space.

So, even though the physical state is observer-independent, the ray (state vector) representing it differs between frames/observers if we stick with a single Hilbert space picture.

There is a way around this and that is to recognize that the change in an observable that corresponds to a frame transformation is represented by the inverse of the unitary operator representing the frame transformation. Consider a spatial translation in which the origin is shifted by $+a$. Then the change in x is given by $x\rightarrow x' = x-a$ because a positive shift in the origin results in a negative shift in the coordinate.

Now a coordinate base vector representing coordinate $x$ is usually written $|x\rangle$. But notice the frame of reference is not explicitly specified; it is just assumed as implicit. Now suppose we make a frame transformation such that $x\rightarrow x'$. Clearly $|x'\rangle$ is a different base vector. Now suppose we make the frame of reference $F$ explicit by specifying the base vector in $F$ as $|x,F\rangle$. Clearly then $|x,F\rangle \ne |x',F\rangle$ and $|x,F\rangle \ne |x,F'\rangle$. But now we are free to equate $|x,F\rangle = |x',F'\rangle$ so that we get an observer-independent state vector. In effect, the unitary operator representing $x\rightarrow x'$ is canceled by the unitary operator representing $F\rightarrow F'$

 

[A. Neumaier]

The situation is complicated because ordinary quantum mechanics is nonrelativistic, and relativistic elements can only be introduced in a heuristic way. See papers by Peres and Terno (e.g., https://arxiv.org/abs/quant-ph/0212023). In particular, the Hilbert spaces of observers related by a Lorentz boost are not directly comparable, hence one cannot talk about the same ray!

A fully correct treatment should involve quantum field theory. But observers break Lorentz invariance, due to their own preferred frame, and in quantum field theory, the observer cannot be represented, only the change of frame, which is a Lorentz transformation applied to the field according to its transformation properties (which depend on its spin).

 

[mikeyork]

The situation is complicated because ordinary quantum mechanics is nonrelativistic, and relativistic elements can only be introduced in a heuristic way. See papers by Peres and Terno (e.g., https://arxiv.org/abs/quant-ph/0212023). In particular, the Hilbert spaces of observers related by a Lorentz boost are not directly comparable, hence one cannot talk about the same ray!

Wigner treats a Lorentz boost as a translation of a momentum frame. It can therefore be treated in the same way as a spatial translation in my post #6 simply by employing a momentum basis instead of a coordinate basis and substituting $p$ for $x$.

 

[A. Neumaier]

Wigner treats a Lorentz boost as a translation of a momentum frame. It can therefore be treated in the same way as a spatial translation in my post #6 simply by employing a momentum basis instead of a coordinate basis and substituting $p$ for $x$.

A boost is _not_ a translation in the momentum frame, but a more complicated transformation!

 

[mikeyork]

Technically, a Lorentz boost is a translation in velocity space not momentum space. But in the case of a momentum basis it is a distinction without a difference because for a mass that is a single-valued function of velocity, the application of such a boost to a state of definite momentum is exactly equivalent to a translation in momentum space. (And yes, of course, in the case of spin it will also involve a Wigner rotation. But the result in terms of state vectors in a fixed basis is still a unitary operator that is the inverse of the operator performed on the frame.)

 

[bhobba]

The other thing that needs to be mentioned is the Heisenberg and Schrödinger pictures.

I will let the OP look them up and see the obvious relation to his question and how it 'mucks up' even unambiguously making sense of it.

And to make matters worse there is even the Dirac picture which is sort of a combination of the two - its maddening these foundational things.

Guess which picture my answer referred to?

Thanks
Bill

 

[Tio Barnabe]

Thank you all.
I need to carefully read the answers and I will do so as soon as possible.

 

[vanhees71]

The situation is complicated because ordinary quantum mechanics is nonrelativistic, and relativistic elements can only be introduced in a heuristic way. See papers by Peres and Terno (e.g., https://arxiv.org/abs/quant-ph/0212023). In particular, the Hilbert spaces of observers related by a Lorentz boost are not directly comparable, hence one cannot talk about the same ray!

A fully correct treatment should involve quantum field theory. But observers break Lorentz invariance, due to their own preferred frame, and in quantum field theory, the observer cannot be represented, only the change of frame, which is a Lorentz transformation applied to the field according to its transformation properties (which depend on its spin).

Well, only because you have an observer you don't break Lorentz (or Poincare) invariance. You can't also break Lorentz invariance simply because there is an equilibrated medium. Unfortunately it's sometimes claimed that this is the case in some textbooks on thermal field theory, but you can very easily formulate everything in a manifestly covariant way by e.g., writing the standard grand-canonical statistical operator in a covariant way,
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta u \cdot \hat{P}-\beta \sum_i \mu_i \hat{Q}_i), \quad Z=\mathrm{Tr} \exp(\dots).$$
Here $u=(u^{\mu})$ is the four-velocity of the rest frame of the heat bath, $\beta$ the inverse temperature, and $\mu_i$ the chemical potentials with respect to a set of conserved charges $\hat{Q}_i$.

So far there's no hint at violation of Poincare symmetry (as long as gravity is neglected).

 

[vanhees71]

Technically, a Lorentz boost is a translation in velocity space not momentum space. But in the case of a momentum basis it is a distinction without a difference because for a mass that is a single-valued function of velocity, the application of such a boost to a state of definite momentum is exactly equivalent to a translation in momentum space. (And yes, of course, in the case of spin it will also involve a Wigner rotation. But the result in terms of state vectors in a fixed basis is still a unitary operator that is the inverse of the operator performed on the frame.)

Lorentz boosts are NOT translations in velocity space. They do not even commute with each other, nor do they build a group! Only together with the rotations they build the proper orthochronous Lorentz group. Restricting to one dimension it's a translation in rapidity space.

 

[A. Neumaier]

Well, only because you have an observer you don't break Lorentz (or Poincare) invariance. You can't also break Lorentz invariance simply because there is an equilibrated medium. Unfortunately it's sometimes claimed that this is the case in some textbooks on thermal field theory, but you can very easily formulate everything in a manifestly covariant way by e.g., writing the standard grand-canonical statistical operator in a covariant way,

Of course the whole theory is covariant, but how do you define an observer in QFT? It is not just a frame, because a frame cannot observe anything. Thus the observer must be a part of the interacting system, just like in QM when one models the observation process as for the nonrelativistic case in the work by Allahverdyan et al. reviewed by me on PO and discussed here on PF.

 

[mikeyork]

Lorentz boosts are NOT translations in velocity space. They do not even commute with each other, nor do they build a group! Only together with the rotations they build the proper orthochronous Lorentz group. Restricting to one dimension it's a translation in rapidity space.

And the physical significance of your quibble is?

Regardless, my point stands: the frame transformation is a unitary transformation which can be viewed either as a change in observables such as momentum (as per Wigner) or the inverse transformation of the frame itself and, if you include both the observable and the frame in specifying the basis, then you have state vectors that are unchanged -- which I believe is what the OP was asking about.

And, BTW, this true for any frame transformation; the unitary operator and its inverse will always cancel each other. Only the state description (observable, frame) changes.

 

[vanhees71]

Well, I simply wanted to correct a misconception. If you are not interested, just ignore it. Of course, what you wrote, is correct. Invariant objects are, big surprise, invariant ;-).

 

[vanhees71]

Of course the whole theory is covariant, but how do you define an observer in QFT? It is not just a frame, because a frame cannot observe anything. Thus the observer must be a part of the interacting system, just like in QM when one models the observation process as for the nonrelativistic case in the work by Allahverdyan et al. reviewed by me on PO and discussed here on PF.

Of course, "an observer" is some device interacting with the system to provide finally information about the measured system to us. It still doesn't break the fundamental symmetries of nature, because it's part of nature (assumed the symmetries are really realized in nature, of course).

 

[mikeyork]

Invariant objects are, big surprise, invariant ;-).

How often do you see it explicitly stated in QM textbooks that if you explicitly include the frame in the basis specification (so that $|x,F\rangle and |x,F'\rangle$ are each base vectors in differing bases in the same Hilbert space) then every state can be ascribed a unique frame-independent state vector? I have never seen a textbook that says this. 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 whereas just about every textbook treats it as some arbitrary rule imposed by nature. So its hardly as trivial as you suggest.

 

[bhobba]

How often do you see it explicitly stated in QM textbooks that if you explicitly include the frame in the basis specification (so that $|x,F\rangle and |x,F'\rangle$ are each base vectors in differing bases in the same Hilbert space) then every state can be ascribed a unique frame-independent state vector? I have never seen a textbook that says this

Errr - ever heard of the Heisenberg picture where the state remains the same and the observable varies? Its in just about every single textbook.

In standard QM the Galilean transformations are assumed (see chapter 3 - Ballentine) so this implies very simply, the state vector stays the same between frames - in that picture.

As I said in a previous post this is not a simple yes/no answer because of the Schrödinger, Dirac and Heisenberg pictures. In my original answer I implicitly assumed the Heisenberg picture - I really should have made that explicit - hence my second post. I didn't precisely spell it out because I wanted the OP to think about it a bit - maybe that's my bad - I don't know - but what you figure out for yourself you understand better and in my answers I try to get people thinking a bit.

This however is very very basic QM.

Thanks
Bill

 

[bhobba]

And this alone enables us to deduce the spin-statistics theorem whereas just about every textbook treats it as some arbitrary rule imposed by nature. So its hardly as trivial as you suggest.

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

 

[mikeyork]

Errr - ever heard of the Heisenberg picture where the state remains the same and the observable varies? Its in just about every single textbook.

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.

 

[mikeyork]

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

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.

 

[bhobba]

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.

I know Wigner's Theroem

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

Thsnks
Bill

 

[bhobba]

But they nearly all make an assumption about single-valued state vectors or wave functions that is not obviously valid..

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 can't 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

 

[mikeyork]

I know Wigner's Theroem

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

Thsnks
Bill

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.

 

[mikeyork]

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?

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).

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|.

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)

 

[bhobba]

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.

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

 

[mikeyork]

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.

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).

 

[bhobba]

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

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