Time independent operators and Heisenberg eq - paradox?

[pellman]

Suppose we have time-dependent operator a(t) with the equal-time commutator

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

and in particular

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

with Hamiltonian

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

The Heisenberg equation of motion

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

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

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

for all times. Since a_0=a(0),

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

means that

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

for all times. But this it so say that

\frac{da_0}{dt}=-i\omega a_0\neq 0

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

 

[Avodyne]

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

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

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

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

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

 

[pellman]

I think I see that. Thanks.

 

[Demystifier]

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

 

[pellman]

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

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.