Time derivative of a Schrodinger-picture operator?

[pellman]

Given an operator Q, how do we derive the relationship

[tex]\frac{dQ}{dt}=i\left[H,Q \right]+\frac{\partial Q}{\partial t}[/tex]

?

I had thought that this was only true in the Heisenberg picture. But Greiner has it here (eq 8.19) for an operator in the Schrodinger picture.

No need to show the derivation. Just tell me about it.

 

[arkajad]

Greiner is not "deriving" this formula. He is "defining" dL/dt by the formula (8.6). The derivation serves as way of justifying such a definition.

 

[pellman]

In that section Greiner is starting with the dL/dt expression, not deriving it. But I don't see that he is justifying it either. the expression for dL/dt is his starting assumption for that section. I can't find where he justifies the expression earlier in the book.

If this expression is true in the Schrodinger picture, then we would have

[tex]\frac{d\hat{p}}{dt}=-\frac{\partial\hat{V}}{\partial\hat{x}}\neq 0[/tex]

assuming the partial derivative of p with respect to time is zero. But that can't be right. The basis states in the Schrodinger picture are time-independent. So if

[tex]\hat{p}|p'\rangle = p'|p'\rangle[/tex]

then the LHS would be time-dependent and the RHS constant.

So I am missing something. Probably I don't understand the proper meaning of the expression in OP.

Help me out here, somebody.

 

[xepma]

He's working in the Heisenberg picture. He's just not mentioning it. It makes no sense to talk about the time evolution of the operator in the Schroedinger picture.

If you look closely, the motivation that he uses for this time evolution of the operator comes from section 8.1. More precisely, he states that the time evolution of the expectation value of the operator L is given by 8.5. This expression is true, regardless of the picture you are working in.

In the next line, however, he chooses to interpret the time evolution of the expectation value as a time evolution of the operator itself. And this is precisely where he makes a switch in the picture he is working with: from Schroedinger to Heisenberg.

 

[pellman]

This explanation makes sense. I was afraid I was missing something though. Thank you, xepma.

It makes no sense to talk about the time evolution of the operator in the Schroedinger picture.

I don't think this is correct, though. We can have time-dependent operators in the Schroedinger picture.

 

[xepma]

This explanation makes sense. I was afraid I was missing something though. Thank you, xepma.



I don't think this is correct, though. We can have time-dependent operators in the Schroedinger picture.

Sure, but those are driven by some external mechanism. I was referring to the time evolution due to commutation with the Hamiltonian.

 

[pellman]

Sure, but those are driven by some external mechanism. I was referring to the time evolution due to commutation with the Hamiltonian.

Right. We're on the same page then. Thanks for the replies.