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Does time reversal symmetry imply hermicity of s-matrix?

  1. Apr 21, 2014 #1
    If the probability for a state α prepared initially to be in a state β at a later time is given by:

    [tex]S_{\beta \alpha} S_{\beta \alpha}^* [/tex]

    and for a state β prepared intitially to become a state α is: [itex]S_{ \alpha \beta} S_{ \alpha \beta}^* [/itex]

    then in order for the two to be equal (by time-reversal symmetry), then doesn't S have to be Hermitian?

    [tex]S_{ \alpha \beta}=S_{\beta \alpha}^*[/tex]
     
  2. jcsd
  3. Apr 22, 2014 #2

    king vitamin

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    The problem is that you haven't done a complete time-reversal. Not only does T exchange initial and final states (by antiunitarity), it also flips all momenta and spin. Time reversal invariance of your theory actually implies

    [tex]
    S_{\beta,\alpha} = S_{T\alpha,T\beta}.
    [/tex]

    This is what you really mean when picturing the reverse process. In addition, this shows that the processes [itex]\alpha \rightarrow \beta[/itex] and [itex]T\alpha \rightarrow T\beta[/itex] don't proceed at the same rate in general, unlike the similar result for parity invariance.
     
    Last edited: Apr 22, 2014
  4. Apr 22, 2014 #3
    I forgot about the momentum flipping (the helicity flips, but not the spin, right?). What if you combine parity with time reversal, PT, to flip back the momentum and helicity? Does a PT-invariant theory then must have

    [tex]
    S_{\beta,\alpha}=S_{\alpha,\beta}
    [/tex]?

    Also, I recall that the rule for antiparticles is to treat the antiparticle like a particle, but swap initial and final states, and make the 4-momentum negative. That is, an incoming positron of 4-momentum p has [itex]\bar{v}(-p)[/itex] and an outgoing one has [itex]v(-p)[/itex]. Is the antiparticle process a time-reversal transformation for the corresponding particle process? Or is the antiparticle process a PT transformation of the corresponding particle process, since although we insert -p into the spinor, the process is really describing a 4-momentum +p, so we need to flip parity to change -p (from time-reversal) to +p?
     
  5. Apr 22, 2014 #4

    king vitamin

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    You got it backwards - parity flips the spatial momentum, the helicity, but not the spin. In contrast, time reversal flips the spatial momentum, the spin, but not the helicity. As required, the helicity is the product of the spin and angular momentum. So as you see, PT does reverse spin.

    You also need to be careful due to extra phase factors; recall that a symmetry operation only needs to take a state into its symmetric partner modulo a phase factor, and sometimes these phase factors can't be eliminated by a general redefinition. Parity is a really important example of this; a parity operation will generate a factor [itex]\pm1[/itex] for each particle dependent on the species of the particle, and unlike time-reversal cannot be eliminated (I'm a little confused on this point, it seems to be because a parity redefinition can be made using internal symmetry operators). In addition, under time-reversal, a state with total angular momentum j=l+s generates an extra phase [itex](-1)^{j-s}[/itex] (this was implicit in my notation [itex]T\alpha[/itex] above). So although many of these phases cancel out when discussing amplitudes, you can't restrict the values of the S-matrix itself.

    We introduce another operator, C, for transforming particles into antiparticles. If C is a symmetry, we get the same case as for P (an arbitrary [itex]\pm1[/itex] for each species of particle). You asked a question specifically about its action on spinors, I would need to spend more time with references to get into the explicit representation of C. However, the celebrated CPT theorem proves that any interaction which is a Lorentz scalar will commute with CPT, so it's always conserved in relativistic field theories.

    By the way, all of this info is discussed at great length in the opening chapters of Weinberg's QFT Vol.1.
     
  6. Apr 22, 2014 #5
    For a particle with spin-0, one without spin, then it would seem then that PT symmetry would imply [itex]\alpha \rightarrow \beta[/itex] has the same probability as [itex]\beta \rightarrow \alpha[/itex]. Time reversal swaps the two states and makes the momentum negative, but parity restores the momentum? Given a Hamiltonian that doesn't involve spin, H=H(X,P)=P^2+V(X), then if V(X) is invariant under parity, that's sufficient to be invariant under PT. Under time reversal: H(X,-P)=H(X,P) and under parity: H(-X,-P)=H(X,P).


    The phases vanish when taking the probability, so it seems okay.

    What I wanted to argue was that since CPT is a symmetry, CPT=1, where 1 is the identity, and therefore T=PC (ignoring phases). So I wanted to say that time reversal is equivalent to reflection and charge conjugation, or that travelling backwards in time is the same as making all particles antiparticles and mirror-imaging everything. Not sure if this is sound reasoning.
     
  7. Apr 24, 2014 #6

    king vitamin

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    Yes, I agree with all of this. (though I assume you mean your Hamiltonian to be a functional of a scalar field).

    Yes, that all sounds right. In general, CPT invariance implies that a process has the same probability as the inverse process with all spins flipped and particles replaced by antiparticles. The other nice thing you can prove using only CPT invariance (with help of the optical theorem, required by unitarity) is that particles have the same decay rate as their antiparticles.
     
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