## time independent operators and Heisenberg eq - paradox?

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?
 Recognitions: Science Advisor First, your Heisenberg equation is not quite correct. It should be $$\frac{dA}{dt}=\frac{i}{\hbar}[H,A]+ \frac{\partial A}{\partial t}$$ 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 $da_0/dt=0$.
 I think I see that. Thanks.

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## time independent operators and Heisenberg eq - paradox?

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

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