time independent operators and Heisenberg eq - paradox?

pellman

Suppose we have time-dependent operator [tex]a(t)[/tex] with the equal-time commutator

[tex][a(t),a^{\dag}(t)]=1[/tex]

and in particular

[tex][a(0),a^{\dag}(0)]=1[/tex]

with Hamiltonian

[tex]H=\hbar \omega(a^\dag a+1/2)[/tex]

The Heisenberg equation of motion

[tex]\frac{da}{dt}=\frac{i}{\hbar}[H,a]=-i\omega a[/tex]

implies that [tex]a(t)=a_0e^{-i\omega t}[/tex] where [tex]a_0[/tex] is a constant operator. Thus [tex]a^\dag a=e^{+i\omega t}a^\dag_0 a_0 e^{-i\omega t}=a^\dag_0 a_0[/tex] and so

[tex]H(t)=\hbar \omega(a^\dag_0 a_0+1/2)[/tex]

for all times. Since [tex]a_0=a(0)[/tex],

[tex][a_0,a^{\dag}_0]=1[/tex]

means that

[tex]\frac{i}{\hbar}[H(t),a_0]=-i\omega a_0[/tex]

for all times. But this it so say that

[tex]\frac{da_0}{dt}=-i\omega a_0\neq 0[/tex]

contradicting that a_0 is a constant. Where did I go wrong?

Avodyne

First, your Heisenberg equation is not quite correct. It should be

[tex]
\frac{dA}{dt}=\frac{i}{\hbar}[H,A]+ \frac{\partial A}{\partial t}
[/tex]

where the last term accounts for any explicit time dependence in the definition of the operator [itex]A=A(q,p,t)[/itex], where [itex]q[/itex] and [itex]p[/itex] are the canonical coordinates and momenta.

Then, we have [itex]a=(m\omega/2\hbar)^{1/2}(q+ip/m\omega)[/itex]. This has no explicit time dependence, so [itex]{\partial a}/{\partial t}=0[/itex].

But now [itex]a_0=e^{+i\omega t}a[/itex]. And this does have explicit time dependence. So if you plug it into the corrected Heisenberg equation, you will find [itex]da_0/dt=0[/itex].

pellman

I think I see that. Thanks.

Demystifier

Isn't it just a trivial mistake in elementary calculus? I mean
[tex] \frac{df(0)}{dt}=0[/tex]
but
[tex] \frac{df(t)}{dt}|_{t=0} \neq 0[/tex]

pellman

Thanks, Demystifier. Always good to run into you here.

Well, yes, that is just what I thought. But the way the time dependent factors cancel in the Hamiltonian, and since the time-dependent factors are c-numbers so the commutator is all in the constant operators a_0, it as actually the constant operators which have the non-zero commutator with the Hamiltonian, not just that it is true at t=0. Hence my question.