Heisenberg's equation of motion

  • Thread starter noospace
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  • #1
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The equation of motion for an observeable A is given by [itex]\dot{A} = \frac{1}{i \hbar} [A,H][/itex].

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

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

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

  • #2
malawi_glenn
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If you know how to derive Heisenberg eq of Motion, then you should have no problem to find the answer.
 
  • #3
StatusX
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They're the same, the first equation of motion for the operator UAUt gives the second EOM for A.
 
  • #4
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Are you saying that the transformed operator satisfies the first equation but not the second?
 
  • #5
reilly
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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
olgranpappy
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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.
 

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