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## Main Question or Discussion Point

In the Heisenberg picture, we move the time dependence away from the states and incorporate them in the operators. That is, if we write the time dependent state in the Schrodinger picture as [tex]|\Psi(t)\rangle=e^{-iHt}|\Psi\rangle[/tex], then an expectation value for an operator Q at time t, which we would write in the Schrodinger picture as

[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

[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?