# Commutation of time derrivative

1. Feb 1, 2014

### 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

2. Feb 1, 2014

### WannabeNewton

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.

Last edited: Feb 1, 2014
3. Feb 1, 2014

### 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:
$$[A,f(B)]=i \frac{d}{dB} f(B)$$ ???

Last edited: Feb 1, 2014
4. Feb 1, 2014

### 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 $\varphi$.