# I Little issue regarding physical states

1. Nov 21, 2017

### 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?

2. Nov 21, 2017

### Staff: Mentor

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 Schrodinger's equation etc, and yes of course it easily's implied states are frame independent.

Thanks
Bill

3. Nov 21, 2017

### Tio Barnabe

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.

4. Nov 21, 2017

### Staff: Mentor

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

5. Nov 21, 2017

### Staff: Mentor

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.

6. Nov 22, 2017

### mikeyork

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 cancelled by the unitary operator representing $F\rightarrow F'$

7. Nov 22, 2017

### 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).

8. Nov 22, 2017

### mikeyork

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 $x$.

9. Nov 22, 2017

### A. Neumaier

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

10. Nov 23, 2017

### 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.)

Last edited: Nov 23, 2017
11. Nov 23, 2017

### Staff: Mentor

The other thing that needs to be mentioned is the Heisenberg and Schrodinger 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

Last edited: Nov 23, 2017
12. Nov 23, 2017

### Tio Barnabe

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

13. Nov 26, 2017

### vanhees71

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).

14. Nov 26, 2017

### vanhees71

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.

15. Nov 26, 2017

### A. Neumaier

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.

16. Nov 26, 2017

### mikeyork

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.

17. Nov 27, 2017

### 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 ;-).

18. Nov 27, 2017

### vanhees71

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).

19. Nov 27, 2017

### mikeyork

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.

20. Nov 27, 2017

### Staff: Mentor

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 Schrodinger, 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

Last edited: Nov 27, 2017