3
$\begingroup$

"The CPT theorem says that CPT symmetry holds for all physical phenomena, or more precisely, that any Lorentz invariant local quantum field theory with a Hermitian Hamiltonian must have CPT symmetry." says CPT symmetry Wikipedia article, suggesting that all processes have CPT-symmetric analogues.

However, there are popular doubts regarding some, especially measurement:

  1. Example of wavefunction collapse is atom deexcitation, releasing energy - it is reversible, but it requires providing energy e.g. in form of photon to excite back an atom. Can measurement be seen this way - that there is always some accompanying process like energy release, which would need to be also reversed? For example in Stern-Gerlach: spin tilting to parallel or anti-parallel alignment to avoid precession in strong magnetic field - does it have some accompanied process like energy release? Can it be observed?

  2. Another somehow problematic example is stimulated emission used in laser - causing photon emission, which finally e.g. excites a target, later by light path. Does it have time/CPT-symmetric analogue: some stimulated absorption - causing photon absorption, which e.g. deexcites a target, earlier by light path?

  3. Quantum algorithms usually start with state preparation: all 0/1 qubits are initially fixed to let say <0|. Could there be time/CPT analogue of state preparation: fixing values but at the end (as |0>)?

  4. One of cosmological examples is Big Bang: which hypothesis of the point of start of time seems in disagreement with CPT theorem - instead suggesting some symmetric twin of Big Bang before it, like in cyclic model of universe. Is hypothesis of the point of start of time in agreement with CPT theorem? Could these two possibilities be distinguished experimentally?

What other processes are seen as problematic from time/CPT symmetry perspective?

Which can be defended, and which essentially require some fundamental asymmetry?

Violating CPT theorem would require pointing its incorrect assumption - which one is the most likely?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Sure, but the most controversial is time, and most of theories have this symmetry, e.g. classical mechanics, EM, GR, unitary QM evolution. CPT is only required for QFT level. The main question here is if measurement is a time or CPT-reversible process? $\endgroup$ – Jarek Duda Aug 30 '19 at 12:37
  • $\begingroup$ Some earlier comment to which you were apparently replying has been deleted, but, by "QFT level", I'm assuming you mean "at the deepest level of physics", as the only equivalent I could find in a Google search was "at the Heisenberg level", and the derivation of his Uncertainty Principle is notoriously complex. Is my assumption correct? $\endgroup$ – Edouard Sep 6 '19 at 16:43
  • 1
    $\begingroup$ @Edouard, Lagrangian mechanics (e.g. QFT, EM,GR) is deterministic, time/CPT reversible. Practical QM has e.g. wavefunction collapse which is not - but it models only e.g. a single electron in atom, neglecting everything else - needing to "close eyes" during interaction with this environment e.g. deexcitation. However, increasing the considered system up to the universe, its wavefunction should no longer have neglected environment - should have unitary, time reversible evolution. Anyway, theoretical physics suggests there is some deterministic level of nature - fully governed by physics. $\endgroup$ – Jarek Duda Sep 7 '19 at 6:42
1
$\begingroup$

I think it would be useful to make a distinction between two different, but related concepts. On one side, you have what is called reversibility, information conservation or determinism (unitarity in the context of quantum mechanics). On the other side, you have time-reversal symmetry.

Let's start with reversibility. It is an essential feature of our microscopic theories, classical and quantum. It is embodied by the fact that the future and past states of a system are uniquely determined by its present state. In quantum mechanics, it means that there is a well-defined operator $U(t)^{-1} = U(-t)$ associated with the evolution operator. If we take conservation of probability as a fundamental assumption, reversibility follows directly, the operator being $U(t)^\dagger = U(-t)$. It also follows that the evolution operator can be represented as $U(t) = e^{-iHt}$.

Time reversal symmetry goes a step further. It is basically the statement that we are not able to distinguish between a state evolving in the future and the same state with momenta reversed evolving in the past. In quantum mechanics, it translates to $\mathcal{T}^{-1}U(t)\mathcal{T} = U(-t)$, where $\mathcal{T}$ is the anti-unitary time-reversal operator. It implies $[\mathcal{T},H]$ like for other symmetry. As you know the laws of physics are not T symmetric. However, the lighter statement that they are CPT symmetrical seems told. That is, a state evolving in future is indistinguishable from the same state with momenta, positions and charges reversed evolving in the past.

Now we see that CPT symmetry implies reversibility but that the contrary is not true. We could very well imagine reversible systems in which this symmetry doesn’t hold. I insist on this difference because the examples 1, 2, 3 you give (I’m not sure about the fourth so I won’t say anything on this one) don’t just challenge CPT symmetry but more generally reversibility.

Now, the question is whether they are real examples of a failure of reversibility :

1.This one is still a controversial topic of research. Different interpretations of QM give different explications for what happens in a measurement. According to objective collapse interpretations, an irreversible process is really taking place, but to my knowledge, all the other interpretations salvage reversibility in a way or another.

2 and 3. Both of these phenomena have the same explanation (it should not be too surprising because you can implement state preparation with spontaneous emission). They look irreversible in the fact that many different states seem to evolve to the same state $\left|{0}\right>$. But in reality, reversibility is restored when you consider the full picture of the system plus its environment. The apparent loss of reversibility is an artifact of discarding correlations with the environment.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Especially 2 is problematic: standard laser causes outgoing photon beam - exciting a target later, so shouldn't its CPT analogue cause incoming photon beam - deexcitng a target earlier? It seems realizable by free electron laser ( physics.stackexchange.com/questions/308106/… ) - could activating such lasar increase deexcitation rate in a target (e.g. sodium lamp)? $\endgroup$ – Jarek Duda Sep 7 '19 at 6:52
  • $\begingroup$ I don't think there is anything special about the laser problem. It is a standard case of thermodynamic entropy increase. Incoming laser radiation comes in an "ordered" way and is dissipated on a target (environment) in a "disordered" way. This process is, in reality, reversible, however, only a particular configuration of the microscopic degrees of freedom in the target will allow it to emit laser radiation. It is thus extremely unlikely to observe it spontaneously, thus the apparence of irreversibility. $\endgroup$ – Undead Sep 7 '19 at 12:00
  • $\begingroup$ This is mostly energy-related effect: standard laser releases energy in photons, causing excitation of target later. Shouldn't its CPT analogue cause deexcitation of target earlier? Sure such target would need to be excited in the first place, hence deexcite spontaneously, but the question is if this deexcitation rate shouldn't be increased if hitting it with such "CPT analogue of laser"? $\endgroup$ – Jarek Duda Sep 7 '19 at 17:11
  • $\begingroup$ No it shouldn't. Like I said, it is not only the laser part of the system that is important when considering this phenomenon time-reversed. The target or environment is just as important. The problem is that you cannot produce conditions in which a target will spontaneously desexcite to produce laser radiation. $\endgroup$ – Undead Sep 9 '19 at 16:17
  • $\begingroup$ Think about a device that propels an hammer then used to break an egg. Seing the time reversed process of an egg reforming is extremely unlikely right? Does having your hammer device run backwards will make it any more likely for the egg to reform? No. $\endgroup$ – Undead Sep 9 '19 at 16:20
0
+50
$\begingroup$

In reality there is no physical way to show that all processes are $CPT$ symmetric, though you could in principle show that it's not by showing $CPT$-violation in an experiment. You can prove mathematically that a given physical theory is $CPT$-symmetric. The $CPT$-theorem does this for (Lorentz-invariant) quantum field theories.

Since the standard model of particle physics satisfies the conditions of the theorem, all processes that can ultimately (at least in principle) be described by the standard model, are $CPT$-symmetric.

Since there is no good theory of quantum gravity, it is impossible to say if such a theory would be $CPT$-invariant. For the main candidate, string theory, there is no $CPT$-theorem. Since the standard model should appear as a low-energy limit of string theory, it could only be violated at very high energies.

Finally, note that $CPT$-symmetry does not imply time symmetry, and indeed the standard model is not $T$-invariant, namely, $CP$-violation was seen to occur in nature, and found a place in the standard model. As $CP$ is violated, $CPT$ can be a symmetry only if $T$-invariance is violated as well.

ADDED IN EDIT

I'll try to say something about your examples.

  1. De-excitation of an atom. This is an ordinary time-evolution of a quantum system (although before measurement the system would be in a superposition of being excited and not). This is indeed reversible, but the process is not and example of what is usually referred to as wavefunction collapse. Wavefunction collapse is a linear projection (due to a measurement), hence it can not be invertible, and it can not be described by quantum dynamics.

  2. Stimulated emission and stimulated absorption. This should indeed be reversible, though that may be problematic for statistical (rather than fundamental) reasons. Better ask an expert on lasers if you want to be sure.

  3. For quantum algorithms it depends: some algorithms involve measurement in intermediate steps, which result in a (non-invertible) collapse. A pure quantum circuit (consisting of quantum gates) can indeed be run in reverse.

  4. The physics to describe the big bang is unknown and experimentally inaccessible to us. By extrapolation the hypothesized validity of known physics can be extended into scales that are beyond our present experimental range (at $10^{-11}$ s after the big bang), most optimistically to some $10^{-43}$ s after the big bang, see the wikipedia articles on cosmological epochs. In any case, going back in time there is a point where all our known physics breaks down, and there is no physical reason to assume reversibility or $CPT$-invariance anymore.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @JarekDuda, if the measurement is a physical process at an energy scale explored with the Standard Model, then CPT invariance is a very strong assumption, being tested in particle physics: pdg.lbl.gov/2019/reviews/rpp2018-rev-cpt-invariance-tests.pdf $\endgroup$ – Daniel Aug 31 '19 at 7:41
  • $\begingroup$ 1. Practical QM neglects environment, making excited states stable - deexcitation requires interaction with this environment: formalism "closes eyes" here, but it generally seems time-reversible. 2. The problem is that while stimulated emission produces photon finally e.g. exciting the target later, stimulated absorption should analogously cause dexcitation of target earlier: physics.stackexchange.com/questions/308106/… 3. Time-reversed state preparation would allow to enforce measurement outcome. $\endgroup$ – Jarek Duda Sep 2 '19 at 12:55
  • $\begingroup$ @JarekDuda 1. Indeed, this kind of apparent collapse, usually referred to as decoherence, is reversible. I looks like a collapse in the small system, but it is a unitary evolution of the system along with the environment. 2. I think the problem is only the extreme unlikelihood, just like a house of cards reassembling if after collapse we invert all velocities etc. 3. The output of a pure quantum circuit is not a measurement outcome, but the result of a computation (on which at some moment measurement will be performed). An input state can indeed be prepared to produce any outcome. $\endgroup$ – doetoe Sep 2 '19 at 13:37
  • $\begingroup$ 2. Stimulated emission in free electron laser (FEL) is just synchrotron radiation in presence of photos - getting its CPT analogue seems simple. However, while stimulated emission excites target, stimulated absorption should deexite it - what requires it was excited in the first place, what is rarely true - but could be tested. 3. State preparation enforces initial state - its CPT analogue should enforce final state instead e.g. to |0> - instead of taking random value in measurement, it would kind of fix in the future, but I don't have idea how to realize it in practice. $\endgroup$ – Jarek Duda Sep 2 '19 at 14:52
0
$\begingroup$

Example of wavefunction collapse is atom deexcitation, releasing energy

You're mixing up two things here. The collapse of a wavefunction is a feature of the Copenhagen interpretation of quantum mechanics, and in that interpretation, it occurs when there is a measurement. Emission of a photon by an atom need not involve any measurement, and measurement need not involve collapse, except in the Copenhagen interpretation. The CPT theorem is proved using the standard postulates of quantum mechanics. Wavefunction collapse lies outside that theory.

Your other quantum-mechanical examples are similar.

One of cosmological examples is Big Bang

The CPT theorem is a theorem proved within a certain theory: quantum mechanics. Our models of the big bang use a completely different theory, which is general relativity. These theories are not compatible. General relativity does not have global discrete symmetries like C, P, and T.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.