# A quick question

1. Nov 27, 2006

### 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

Last edited: Nov 27, 2006
2. Nov 27, 2006

### dextercioby

What does it mean that an operator is conserved ?

Daniel.

3. Nov 27, 2006

### 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

4. Nov 27, 2006

### dextercioby

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

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

Daniel.

5. Nov 27, 2006

### 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......

6. Nov 27, 2006

### 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.

7. Nov 27, 2006

### 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

Last edited: Nov 27, 2006
8. Nov 27, 2006

### 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.