# Heisenberg's equation of motion

1. Mar 1, 2008

### 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.

Last edited: Mar 1, 2008
2. Mar 1, 2008

### malawi_glenn

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

3. Mar 1, 2008

### StatusX

They're the same, the first equation of motion for the operator UAUt gives the second EOM for A.

4. Mar 1, 2008

### noospace

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

5. Mar 2, 2008

### 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

6. Mar 2, 2008

### olgranpappy

see Messiah QM vol 2.