Heisenberg's equation of motion

[noospace]

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

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

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

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 Schrödinger 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]

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

see Messiah QM vol 2.