Prove T Not Conserved When [H,T] = 0

[benfrombelow]

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

 

[dextercioby]

What does it mean that an operator is conserved ?

Daniel.

 

[benfrombelow]

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

 

[dextercioby]

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

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

Daniel.

 

[benfrombelow]

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...

 

[dextercioby]

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

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

to

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

Daniel.

 

[benfrombelow]

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

 

[dextercioby]

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

\langle \psi (t), \psi(t) \rangle \neq \langle \psi (0), \psi (0) \rangle

Daniel.