# Unitary time operator

1. Jul 25, 2013

### michael879

This has been bugging me for a while, so I'm really hoping someone can give me a good answer. Please get as technical as necessary, I'm a 4th year HE physics grad student so I do know my stuff!

Why is the time-reversal operator made anti-unitary in quantum mechanics?

It is very straight forward to show that an object with positive rest mass, when viewed from a reference frame going the opposite direction through time (still time-like though), will have negative energy. Negative energy particles don't quite behave as you might naively expect though, and their interactions with positive energy particles are very foreign (two particles with opposite energy and opposite velocities will have a space-like 4-momentum). QFT is a relativistic theory, so I'm completely baffled why negative energy states have , almost artificially, been ruled out by making the time-reversal operator anti-unitary!

I was taking a look at what the time-reversal operator does in certain situations, and the complex conjugation component really just seems to effectively reverse the sign of the rest mass. So basically the time-reversal of a state with positive rest mass and positive energy will have negative rest mass and positive energy. This all seems fine, except for the interpretation that antiparticles are equivalent to particles under time reversal, since the two particles would have opposite signed rest mass!

2. Jul 25, 2013

### The_Duck

I don't think you should try to think about "reference frames going backward through time." I'm not sure what that would mean. A simpler way to think about time reversal is that it is a fancy name for momentum reversal. The time reversal operation simply reverses all momenta, leaving positions unaffected. This is a straightforward symmetry operation: it is easy to see what it does to a given system and you don't have to get caught up in metaphysical ideas like "reversing time."

In the language of QM, then, time reversal is implemented by an operator $T$ has the following relations with the generators of the Poincare group:

$T^{-1}\vec{P}T = -\vec{P}$ (momenta are reversed)

$T^{-1}HT = H$ (energy is unaffected by reversing all momenta, if T is a good symmetry)

$T^{-1}\vec{J}T = -\vec{J}$ (angular momenta are reversed, as expected from the above two)

$T^{-1}\vec{K}T = -\vec{K}$ (boosts are reversed)

For T to be a good symmetry, we want $e^{-i H t} T | \psi \rangle = T e^{i H t} |\psi \rangle$. That is, reversing all momenta and then evolving for a time $t$ is the same as evolving *backward in time* for a time $t$ and then reversing all momenta.

Contrast this to the the case of, say, spatial translation symmetry, where we want the translation operator $S$ to satisfy $e^{-i H t} S | \psi \rangle = S e^{-i H t} | \psi \rangle$. That is, translating and then evolving for a time $t$ is the same as evolving for a time $t$ and then translating. Note the extra minus sign here.

In both cases we want the unitary operators $S$ and $T$ that implement the symmetries to commute the the Hamiltonian, since neither spatial translations nor reversal of momenta should affect the energy of a system. Once we have decided that $[S, H] = [T, H] = 0$, the signs in exponentials above require $S$ to be unitary but $T$ to be antiunitary.

3. Jul 29, 2013

### michael879

Yes, that is how T is typically defined in QM. However, in relativity time reversal means something very different! Backwards time travel is perfectly defined relativistically, even though it may be a strange concept, and the theory remains self-consistent even if you allow for it. And its not like this concept of time travel hasn't made its way into QM, as anti-particles are actually defined as particles going backwards through time (although T is used to define time-reversal).

Reversing the proper time of a particle (i.e. the direction through time in which it is traveling) is very easy to define, and it's actually not that hard to imagine conceptually. Its space-like trajectories that really confuse things, time-like time travel is pretty tame. However, if you reverse the proper time of a particle its not just the 3-momentum that reverses sign: the entire 4-momentum vector changes sign, including energy!

4. Jul 29, 2013

### michael879

O yea, and some more relevant things about time-reversal:

In non-relativistic mechanics (both classical and quantum) time-reversal reverses the sign of all velocities, and since p = mv and m is invariant, momenta too

In relativistic mechanics (classical, but not QFT which is the reason for this post) time-reversal reverses the sign of all momenta, but not the 3-velocities! Of course there is a slight ambiguity as to what time-reversal means in relativity, but here I'm taking it to mean the reversal of dτ for all objects but not the observer. Identically, you could take it to mean the reversal of the observers time only.

The other possible meaning of relativistic time reversal would be to reverse the time of the entire universe, including the observer. In this case velocities and momenta would be reversed, but energies would not. This is most likely where the standard T-reversal operator comes from in quantum. However, the T-reversal operator only takes this form if it is applied to the ENTIRE universe! Applying it to a single state assumes the definition above, where energy AND momentum are reversed and velocity is not

5. Jul 29, 2013

### DrDu

Sternberg's book on group theory gives an argument that relativistically there is a possibility to represent time reversal unitary but it would have the consequence that the hamiltonian is unbounded from below which is unphysical. Something similar happens when you compare Dirac's theory of the electron with QED.
In Dirac's theory, the Hamiltonian is not bounded from below while in QED it is. Hence Charge conjugation is represented unitary in one theory but anti-unitary in the other.

6. Jul 29, 2013

### Bill_K

What about particles that are their own antiparticles. Which way do they travel?

7. Jul 30, 2013

### michael879

The quote you took of mine was referring to the QM definition of time-reversal, which I'm arguing is completely different from any classical concept of time-reversal. Applying the time reversal operator to one of these particles will leave you with the exact same particle.

8. Aug 1, 2013

### michael879

So I've been thinking about this further, and I'm still very confused... It seems to me that there are 3 kinds of time "operators" classically:

1) Newtonian time reversal: reverse the time coordinate of the observer and watch everything in reverse. Basically it reverses velocities and momenta, but leaves energies, masses, charges, and accelerations intact.

2) Proper time reversal: applied to a single worldline it reverses the proper time of that worldline. This reverses the sign of the Lorentz factor for that world line, but leaves the mass, charge, and velocity intact (3-velocity is dx/dt and is unaffected by proper time reversal). The sign change of the Lorentz factor reverses the sign of energy and momenta though, and effectively reverses the sign of mass and charge. i.e. it would have the same energy/momentum/electromagnetic field as a particle with opposite mass/charge traveling in the other direction through time.

3) Relativistic time reversal: reverses the time coordinate of the observer, like the Newtonian operation. Unlike the Newtonian operation though, this also reverses the sign of the Lorentz factor for each worldline in the universe (unless of course you apply #2 as well). So what you find is that 3-velocities reverse, but 4-velocities don't. Momenta and 4-velocities remain the same while energies and electromagnetic fields flip sign.

Now in QM the time reversal operator is chosen so that it has the properties of #1. However, #1 is a purely non-relativistic operation and really doesn't seem to fit into a relativistic quantum theory... The only justification I see for it is if you apply #2 to every particle in the universe and apply #3. The result would basically be the same as #1 and only 3-velocities and momenta would change.

The problem I see with this is that if you just look at the quantum description of an electron (for example), the CPT operator evaluates to 1. So applying just the T operator is equivalent to applying C and P, which is exactly where Feynman's view of positrons being electrons traveling backwards in time comes from. However, this application of T resembles #3 where charge and parity flip (and Feynman's description of the operator resembles #2). Certainly this operator can't be #2 and #3 combined, because how would you explain the change in sign of charge!

9. Aug 12, 2013

### michael879

^ bump, can anyone shed some light on this? The classical time-reversal operator is a combination of the 2 relativistic time-reversal operators which each independently reverse effective charge. So if the QM time-reversal operator is derived using the classical definition I don't understand how it can be equivalent to a charge-parity reversal...

10. Aug 13, 2013

### vanhees71

11. Aug 13, 2013

### michael879

Do you have a translation? I don't speak german...

I haven't read your notes but I think its safe to assume they are very similar to what I learned in QFT, which really doesn't shed much light on this issue. Energies are assumed to remain positive under time reversal, which is why the time-reversal operator is made antiunitary. I really don't see a good justification for this assumption though, except for the problems that can occur when negative energies are introduced (which IMO isn't a good enough justification).

12. Aug 13, 2013

### Avodyne

A unitary time-reversal operator T would have to obey $THT^{-1}=-H$. The problem with this is that it is incompatible with the energy being bounded below. Consider an energy eigenstate, $$H|\psi\rangle = E|\psi\rangle$$ Now act on this with T, and, on the left-hand side, insert I=T-1T between H and the state:
$$TH|\psi\rangle = ET|\psi\rangle$$
$$THT^{-1}T|\psi\rangle = ET|\psi\rangle$$
$$-HT|\psi\rangle = ET|\psi\rangle$$
$$H(T|\psi\rangle) = -E(T|\psi\rangle)$$
That is, if $|\psi\rangle$ is an eigenstate of H with eigenvalue E, then $T|\psi\rangle$ is an eigenstate of H with eigenvalue -E. Hence if T exists (and does not annihilate every energy eigenstate), then H cannot be bounded below (if it is not bounded above).

The only known consistent formalism for relativistic quantum mechanics (that is not in conflict with experiment) is quantum field theory. In quantum field theory, the energy is bounded below and not above. Therefore, a unitary T operator does not exist.

13. Aug 13, 2013

### vanhees71

Argh, I posted the wrong link. It's

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

That's in English. Sorry for the confusion.

14. Aug 14, 2013

### michael879

Yes I understand that a unitary time reversal operator will make energies negative, I just don't understand why everyone assumes that's impossible. Negative energies are allowed in, and even predicted by, relativity under time reversal operations. Just because negative energies are allowed though, doesn't mean they are readily accessible for particles to "fall" into...

15. Aug 14, 2013

### Ilmrak

I liked the view presented in these notes (p.96), where antiunitarity is derived from the definition of the effect of time reversal on asymptotic states.

You are perfectly right, infact there are conservation laws that prevent particles to decay in their own antipartcles

Ilm

16. Aug 14, 2013

### Avodyne

Because, as I just explained, there is no known theory that is consistent with experiment that allows negative energies.

17. Aug 14, 2013

### michael879

Special Relativity...?

18. Aug 14, 2013

### The_Duck

In QFT (and QM in general) we want a spectrum of states whose energy is strictly bounded below so that there is a stable ground state, namely the vacuum. We do not want the spectrum of the theory to contain negative energy states. Then the vacuum would be unstable, and a theory without a stable vacuum would look very different from the world we see around us.

Symmetry operators, such as $T$, have to act on the Hilbert space of the theory. If we have decided, for the above reasons, that the Hilbert space should not contain negative energy states, then acting with $T$ on a positive energy state had better not produce a negative energy state.

The fact that the theory is relativistic does not mean that it should predict negative energy states. I like Weinberg's perspective on what it means to have a relativistic QFT. A relativistic QFT is a regular quantum mechanical system with the requirement that there must exist operators $H, \vec{P}, \vec{J}, \vec{K}$ (the generators of time translations, space translations, rotations, and boosts) which act on the Hilbert space and have the right commutation relations to be a representation of the Lorentz group. If you can define these operators, you have a relativistic quantum theory. Nothing about this forces $H$, the Hamiltonian, to have negative eigenvalues. Relativity does not require the existence of negative energy states.

What do you mean by this? That there are negative energy states but it is impossible or unlikely to transition to those states?

Not in general. The $K^0$ meson rapidly and spontaneously turns into its antiparticle, the $\bar{K^0}$ (which then turns back, etc.). This is possible because the $K^0$ carries no conserved quantum numbers.

In any case, antiparticles have positive energy, just like regular particles.

19. Aug 14, 2013

### Avodyne

*Sigh* Your original question was "Why is the time-reversal operator made anti-unitary in quantum mechanics?"

And the answer is, the only known way to combine quantum mechanics and special relativity (in a way that's consistent with experiment) is quantum field theory, with a hamiltonian that is bounded below and unbounded above, and that requires time reversal to be anti-unitary.

Now, if you drop "consistent with experiment", you could consider the single-particle Dirac equation, with ψ(x) interpreted as a wave function (and not as a field operator). Then the hamiltonian in unbounded both above and below, and it may be possible to define a unitary time-reversal operator. I say may because in this case there might be some other issue; I don't know (though I'm sure someone does). IMO it's not a very interesting question, because the single-particle Dirac equation is not consistent with experiment (except as a sometimes useful approximation to quantum electrodynamics in certain restricted situations).

20. Aug 15, 2013

### andrien

There are other problems associated with single particle dirac eqn.,the most famous one is Klein paradox which shows that particle encountering a potential step larger than the kinetic energy will transmit,where it is expected that it will reflect.The problem is resolved properly only in qft where ψ is interpreted as field.