# Dirac's Quantum Mechanics - the definition of the time evolution operator

1. Aug 7, 2012

### Loro

Dirac's "Quantum Mechanics" - the definition of the time evolution operator

I have a question about the way he defines the time evolution operator in the book. Either there's a mistake or I'm missing something.

In chapter 27 he says (eqn 1) that $\hat{T}$ is defined such that:

$|P(t)> = \hat{T} |P(0)>$

Where |P(0)> is a ket at time t=0 , and |P(t)> - at time t
Or equivalently |P(0)> is a ket in the Heisenberg picture, and |P(t)> - in the Schrodinger picture.

So this implies that:

$<P(t)| = <P(0)| \hat{T}^{\dagger}$

And then in chapter 32, eqn 45 implies that:

$<P(t)| = <P(0)| \hat{T}$

And I understand, that we can define it both ways, since it's a unitary operator. But we should stick to one way of defining it, and I'm sure Dirac does. So what it is here, that I'm not understanding properly?

Last edited: Aug 7, 2012
2. Aug 8, 2012

### strangerep

Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

3. Aug 8, 2012

### Loro

Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

Thanks again,

But here $\hat{T}$ isn't self-adjoint. In fact it's unitary.

4. Aug 8, 2012

### dextercioby

Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

Yes, the T's are not identical, one is the adjoint of the other.

5. Aug 8, 2012

### Loro

Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

Thanks,

Yes. I'm sorry, but I still don't understand how this connects to my question.