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A quick question

  1. Nov 27, 2006 #1
    If [H,T] = 0 (where H is the Hamiltonian and T is the (antiunitary) time-reversal operator) prove that T is NOT conserved

    This is for a homework assignment due is less than 12 hours....

    I'm guessing it has to with the property of antiunitary operators: T|a> = |a'> then <a|b> = <a'|b'>* but it's late and I'm lazy
     
    Last edited: Nov 27, 2006
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  3. Nov 27, 2006 #2

    dextercioby

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    What does it mean that an operator is conserved ?

    Daniel.
     
  4. Nov 27, 2006 #3
    Usually, it means that the operator is constant, in the Heisenberg picture, which means that [H,A]=0 but that isn't true when A is anti-linear
     
  5. Nov 27, 2006 #4

    dextercioby

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    And what does "conserved" mean for an antilinear operator ?

    If you don't know that, you can't solve it.

    Daniel.
     
  6. Nov 27, 2006 #5
    The best I can come up with is to look at the expectation value of T in an arbitrary state: <Psi(t)|T|Psi(t)> = <Psi(0)|(exp(iHt/h))T(exp(-iHt/h))|Psi(0)> = <Psi(0)|(exp(2*iHt/h))T|Psi(0)> = <Psi(-2t)|T|Psi(0)> = <Psi(0)|T|Psi(2t)>

    which clearly isn't constant, but it also isn't very satisfying......
     
  7. Nov 27, 2006 #6

    dextercioby

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    I don't seem to get the line of thought in your calculations. Could you care to explain how did you get from

    [tex] \langle \Psi(0)|\mbox{(exp(iHt/h))}T\mbox{(exp(-iHt/h))}|\Psi(0)\rangle [/tex]

    to

    [tex] \langle \Psi(0)|\mbox{(exp(2*iHt/h))}T|\Psi(0) \rangle [/tex]

    Daniel.
     
  8. Nov 27, 2006 #7
    Well, basically, since T is antilinear, TiH = -iTH. But T commutes with H, so TiH = -iHT.

    edit: To elaborate, because TiH = -iHT, it follows that T(exp(-iHt/h)) = (exp(iHt/h))T since the exponential is a series in powers of H. When moving T to the right of H, the coefficient of each term in the series must be replaced with its complex conjugate
     
    Last edited: Nov 27, 2006
  9. Nov 27, 2006 #8

    dextercioby

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    Yes, i can see now. One thing to notice is the improper use of Dirac's bra-ket formalism for T, since this operator is not self adjoint.
    Thus the last equality does not follow. You managed to find (neglecting the last equality) that

    [tex] \langle \psi (t), \psi(t) \rangle \neq \langle \psi (0), \psi (0) \rangle [/tex]

    Daniel.
     
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