- #1

- 677

- 4

[tex]\langle\Psi(t)|Q|\Psi(t)\rangle=(\langle\Psi|e^{+iHt})Q(e^{-iHt}|\Psi\rangle)[/tex]

we think of it in the Heisenberg picture as

[tex]\langle\Psi|Q(t)|\Psi\rangle=\langle\Psi|(e^{+iHt}Qe^{-iHt})|\Psi\rangle[/tex]

Same answer of course. But instead of having a state which satisfies the Schrodinger equation

[tex]H|\Psi(t)\rangle=i\hbar\frac{\partial}{\partial t}|\Psi(t)\rangle[/tex]

we have an operator obeying the Heisenberg equation of motion

[tex][H,Q]=-i\hbar \frac{\partial Q}{\partial t}[/tex]

at least this is the case for bosons. (I write all this preface material just in case my understanding thus far is mistaken. Now the question.)

But what about fermions? Do operators related to fermions obey a Heisenberg equation with

*anti-commutators*instead of commutators?