Heisenberg's equation of motion

noospace

The equation of motion for an observeable A is given by [latex]\dot{A} = \frac{1}{i \hbar} [A,H][/latex].

If we change representation, via some unitary transformation [latex] \widetilde{A} \mapsto U^\dag A U[/latex] is the corresponding equation of motion now

[latex]\dot{\widetilde{A}} = \frac{1}{i \hbar} [\widetilde{A},U^\dag H U][/latex]
or
[latex]\dot{\widetilde{A}} = \frac{1}{i \hbar} [\widetilde{A},H][/latex]?

I'm asking because I want to write the time derivative of the Dirac representation of the position operator in the Foldy-Wouthusyen representation.

malawi_glenn

If you know how to derive Heisenberg eq of Motion, then you should have no problem to find the answer.

StatusX

They're the same, the first equation of motion for the operator UAU^t gives the second EOM for A.

noospace

Are you saying that the transformed operator satisfies the first equation but not the second?

reilly

If the generator of the unitary transform U depends on t -- like going from Schrodinger picture to the Interaction Picture -- then noospace, you have left out a term. Standard stuff, can be found in most QM or QFT texts.
Regards,
Reilly Atkinson

olgranpappy

see Messiah QM vol 2.