Commutation of time derrivative

[Abigale]

Hi I regard,

$$[\partial_t \Psi, \Psi]=0$$
but \Psi is a field-operator.

I don't understand why the commutation of the derrivative of the operator \Psi by itself should be zero?
THX

 

[WannabeNewton]

I don't understand why the commutation of the derrivative of the operator \Psi by itself should be zero?

Could you provide a little more context? The commutator shouldn't vanish in general. For example if $\Psi$ is a (free) real Klein Gordon field then from the canonical commutation relations we have $[\Psi(x,t), \partial_t \Psi (x',t)] = i\delta^{3}(x - x')$ so clearly they don't commute in this case.

 

[Abigale]

I regard a Klein-Gordon-field. But for the Klein-Gordon-Field, the commutation-relation is $$ [\Psi^\dagger (x,t), \partial_t \Psi (x',t)] = -i\hbar\delta(r-r')$$
I even know that: $$ [\Psi^\dagger(x,t), \Psi (x',t)] = 0$$
and $$ [\partial_t\Psi^\dagger(x,t), \partial_t \Psi (x',t)] = 0$$

and for sure $$ [\partial_t\Psi^\dagger(x,t), \Psi (x',t)] =-i\hbar\delta(r-r') $$
Maybe with the formula:
$$[A,f(B)]=i \frac{d}{dB} f(B)$$ ?

 

[WannabeNewton]

I'm not sure what you're asking. If you're talking about complex Klein Gordon fields as opposed to real Klein Gordon fields then $[\varphi, \pi^{\dagger}] = [\varphi, \partial_t \varphi] = 0$ is a trivial statement of second quantization; $\varphi$ and $\varphi^{\dagger}$ are regarded as independent fields and $\pi^{\dagger} = \partial_t \varphi$ is the conjugate momentum of $\varphi^{\dagger}$ not that of $\varphi$.