# Confusion (4) from Weinberg's QFT.(Time reversal))

1. Jul 31, 2011

### kof9595995

This question came across my mind when I read section 2.6, but it does not have close relation to the content of the book.
If we apply a time reversal to 4-momentum, the 0-component(i.e. the energy) will become negative, and this is why time-reversal operator on quantum states must be antilinear according to Weinberg. I wonder how we interpret the negative energy if there were no quantum mechanics, i.e. if we only have classical description, it seems energy will inevitably become negative after time reversal, how do we resolve this?

2. Jul 31, 2011

### Fredrik

Staff Emeritus
In classical mechanics, time reversal is just to replace t with -t. This changes the direction of the momentum, but energy depends only on its magnitude.

3. Jul 31, 2011

### Bill_K

In Minkowski geometry there are two independent reflections, space reflection and time reflection. After you've specified the transformation properties of a quantity under restricted Lorentz transformations you still have to say how it transforms under the two reflections.

For example the space reflection is (x, ct) → (-x, ct), and we distinguish proper vectors for which (v, v0) → (-v, v0) and pseudovectors for which (v, v0) → (v, v0).

When it comes to time reflection, again two behaviors are possible, and the point is that the position vector and the energy-momentum vector transform differently under time reflection. For the position vector, (x, ct) → (x, -ct), while for the energy-momentum vector, (p, E/c) → (-p, E/c).

4. Jul 31, 2011

### kof9595995

Emm, if it's like what you said, how would you comment on the way Weinberg argue time-reversal operator on Hilbert space must be antilinear(section 2.6, page75~76)? In case you don't have the book in hand, I'll briefly describe it:
From the group property and unitarity of the symmetry transformation we can derive $TiP^{\rho}T^{-1}=i{\cal {T}}^{\;\;\rho}_{\mu}P^{\mu}$......(2.6.6)
where all symbols on the left are operators on quantum states and on the right are 4-vectors and corresponding Lorentz transformations(time reversal in this case). Now inspect the 0-component of $\rho$, we have
$TiHT^{-1}=-iH$..................(*)
So to avoid negative energy state we must require T to be antillinear so that
$THT^{-1}=H$........(2.6.13)
blablabla.......
So you see in deriving (*) Weinberg obviously assumed time-reversal acts the same way on 4-momentum as 4-vector of space and time.

Last edited: Jul 31, 2011
5. Aug 1, 2011

### Bill_K

Yes, I have the book. I don't see there's a difference, except that Weinberg is doing it all in second-quantized framework, and I was thinking in terms of Schrodinger. We both wind up in the same place, namely TpT-1 = -p, THT-1 = +H.

From my point of view, since xμ and kμ transform differently, k'·x' = - k·x, and to maintain the form ψ ~ exp(ik·x) you need to complex conjugate the wavefunction.

6. Aug 3, 2011

### kof9595995

Can you elaborate？ I completely don't understand what you said. Like I don't see why they are the same. and what does it have anything to do with second-quantization or schrodinger

7. Aug 3, 2011

### vanhees71

The reason why the time-reversal transformation has to be antiunitary has been already given in posting 4. By assumption the Hamiltonian is an operator that is bounded from below (i.e., you have a state of minimal energy, i.e., a stable ground state). Then the calculation in posting 4 shows that necessarily the time-reversal operator must be antiunitary since, if you assume it to be unitary, the transformed Hamiltonian would be -H. If you have time-reversal symmetry, the eigenvalues time-reveresed Hamiltonian would have to be energy eigenvalues, but the eigenvalues of -H are usually not bounded from below since the free-particle Hamiltonian is not bounded from above and so -H cannot be bounded from below. By assumption, however, there must be a stable ground state, and thus the time-reversal operator must be antiunitary.

8. Aug 3, 2011