Consider the usual commutation relations of two scalar fields
[ϕm(t,x),πn(t,y)]=iδmnδ(x−y),
[ϕm(t,x),ϕn(t,y)]=[πm(t,x),πn(t,y)]=0.
What's the commutator of [∂iϕm(t,x),ϕn(t,y)], where ∂i≡∂/∂xi is one of the three spatial derivatives?
What about [∂iϕm(t,x),πn(t,y)] ?
Attempt 1:
[∂iϕ(t,x),ϕ(t,y)]=∂i[ϕ(t,x),ϕ(t,y)]+[∂i,ϕ(t,y)]ϕ(t,x)=[∂i,ϕ(t,y)]ϕ(t,x)=(∂iϕ(t,y))ϕ(t,x)−ϕ(t,y)∂iϕ(t,x)=?
Answer
Since we're not taking time derivatives, this is actually a pretty simple thing, but something that, for some reason, doesn't really pop out on a first viewing of a problem like this.
The confusion perhaps arises from the fact that you have two types of operators acting on different spaces. You have the derivative operator ∂i acting on the space of functions from Rn to some general algebra of fields. You also have the field operators themselves, acting on your Hilbert space H. Since these two operators act on different spaces, then we have
[∂∂xiϕ(x,t),ϕ(y,t)]=∂∂xi[ϕ(x,t),ϕ(y,t)]=0.
That is to say, you can pull out the derivative since only the first term in the commutator depends on x. Similarly, we have
[∂∂xiϕ(x,t),π(y,t)]=∂∂xi[ϕ(x,t),π(y,t)]=i∂∂xiδ(x−y).
I don't know what it is about this question that trips people up (including myself the first time I was faced with something like this), but it's a lot simpler than it's made out to be.
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