I Time-reversal symmetry in QM

462
9
Summary
What is it and why should it apply to qm?
I've bumped into a few interesting papers talking about time-reversal symmetry in QM (eg: https://arxiv.org/abs/1507.07745) but I can't seem to wrap my head around the concept.

1) What does it mean for one to say that standard QM isn't time-reversal symmetric? Does this have to do with the "collapse" post-measurement?

2) Why should we look for quantum theory that is time-reversal symmetric? Are the reasons empirical or theoretical?

Regarding 2), I found a passage on wikipedia that says: "In other words, time is said to be non-symmetric, or asymmetric, except for special equilibrium states when the second law of thermodynamics predicts the time symmetry to hold. However, quantum noninvasive measurements are predicted to violate time symmetry even in equilibrium, contrary to their classical counterparts,". Does this mean that these theorists are looking to modify QM such that time symmetry holds for these "equilibrium states"? The paper I cited, on the other hand, doesn't say anything about such equilibrium states.

Thanks in advance!

Edit: What role does an anti-unitary operator have to play in all this?
 
Last edited:

jambaugh

Science Advisor
Insights Author
Gold Member
2,174
230
Last first, do you mean non-unitary vs anti-unitary? Non-unitary operators can be used to represent dissipative and hence irreversible evolution.

To your main questions, there are two components to standard QM. The first is the dynamics which is expressed via unitary operators acting on the system description (wave function, "state" vector, or density operator). This is time reversal symmetric.

The second is the interpretation of the system description in terms of acts of observation. Observables are expressed as Hermitian operators, their available values being the eigenvalues. An act of observation is an irreversible process. One is amplifying the "signal" contained in one of the system's observables and that is a intrinsically thermodynamic process. The time asymmetry manifests in the "collapse" i.e. the projection onto the eigenspace of the observable contingent on the actual value observed. The magnitude of the projection relative to the prior description tells us the probability of observing that specific value via Born's formula.

The measurement process is not time symmetric due to the 2nd law of thermodynamics.

(Grandiose speculation warning!)
Now I believe it is possible to reformulate the 2nd law in a time symmetric fashion and I think that might be what the paper is about implicitly.

If you imagine that you sharply measure some system with a large numbers of components, say a box full of many hydrogen atoms, then you would write down a sharp description of the system at that time. The system would have entropy 0. If you then assert the system was coupled to a larger high entropy environment and unitarily evolve backward in time you would find the entropy of the description would increase as time devolves.
Keep in mind that entropy is not a substance or energy. It is a measure of how imprecisely we
We lose knowledge about both past and future systems relative to the present when we allow them to evolve/have evolved while coupled, however weakly, to the environment.






 
462
9
Thanks for your response

The second is the interpretation of the system description in terms of acts of observation. Observables are expressed as Hermitian operators, their available values being the eigenvalues. An act of observation is an irreversible process. One is amplifying the "signal" contained in one of the system's observables and that is a intrinsically thermodynamic process. The time asymmetry manifests in the "collapse" i.e. the projection onto the eigenspace of the observable contingent on the actual value observed. The magnitude of the projection relative to the prior description tells us the probability of observing that specific value via Born's formula.
This is the part I don't quite get in the paper. What would a series of operations in T-symmetric quantum theory, say, state preparation and then measurement in the ##+t## direction look like in the ##-t## direction? And if indeed this lack of reversibility is due to thermodynamics, why would the authors feel the need to formulate a T-symmetric quantum theory?

Cheers.
 

DrChinese

Science Advisor
Gold Member
7,181
999
(Grandiose speculation warning!)
Now I believe it is possible to reformulate the 2nd law in a time symmetric fashion and I think that might be what the paper is about implicitly.

If you imagine that you sharply measure some system with a large numbers of components, say a box full of many hydrogen atoms, then you would write down a sharp description of the system at that time. The system would have entropy 0. If you then assert the system was coupled to a larger high entropy environment and unitarily evolve backward in time you would find the entropy of the description would increase as time devolves.
Keep in mind that entropy is not a substance or energy. It is a measure of how imprecisely we
We lose knowledge about both past and future systems relative to the present when we allow them to evolve/have evolved while coupled, however weakly, to the environment.
I think it is fair to say that [Boltzmann] entropy increases in both time directions. A measurement would represent a local maximum of knowledge. That is not something normally considered (entropy to the past), but I have seen a few references in the literature over the years.
 

vanhees71

Science Advisor
Insights Author
Gold Member
12,971
4,997
Time reversal symmetry is somewhat special. I've not followed the above quoted paper, which seems to claim something different to the well-established standard treatment of this symmetry, according to which standard QT is time-reversal invariant. So here I refer to the standard treatment, known since the very beginning of QT.

First of all one has to take into account that the Hamiltonian is a very special observable in QT since it defines the time evolution of the system. One conclusion is that a system can only be stable if the Hamiltonian is bounded from below, i.e., there should be a state (or maybe several degenerate states) of minimum energy.

The first important consequence of this is that you cannot ##\hat{H}## interpret as the canonical conjugate momentum of time and thus time is a "c-number"-parameter in quantum theory, labelling the causality order of events (as in classical physics too).

Now, as proven by Wigner, a single symmetry transformation can always be lifted to either a unitary or antiunitary transformation. For a review, see


If you have an entire continuous (e.g., as usual in physics even a Lie group) the symmetry transformations are necessarily unitary, while discrete symmetries may be antiunitary.

Now the Hamiltonian for a stable system is bounded from below, and since it generates time translations the time-reversal operator must act as
$$\hat{T}(\mathrm{i} \hat{H})\hat{T}^{\dagger}=-\mathrm{i} \hat{H}.$$
Thus, in order for ##\hat{H}## not flip sign, we must necessarily assume that ##\hat{T}## is anti-unitary rather than unitary.
 

Want to reply to this thread?

"Time-reversal symmetry in QM" You must log in or register to reply here.

Related Threads for: Time-reversal symmetry in QM

  • Posted
Replies
2
Views
911
  • Posted
Replies
8
Views
12K
Replies
5
Views
3K
Replies
5
Views
2K
  • Posted
Replies
12
Views
3K
  • Posted
Replies
3
Views
1K
  • Posted
Replies
8
Views
2K
  • Posted
Replies
4
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top