Sunday, January 20, 2019

quantum field theory - What is the commutator of an operator and its derivative?


Is it possible to calculate in a general way the commutator of an operator O which depends on some variable x and the derivative of this O with respect to x? O=O(x)[xO(x),O(x)]=?

To be honest I don't quite understand, what the derivative of an operator is. When one plugs in the standard definition of the derivative of a function, it looks as though the operator commutes with its derivative, but I don't quite know how to feel about that "proof".


In practice I need this to calculate the commutator of the field operator of a free scalar field and any of its four derivatives: [μϕ(x,t),ϕ(x,t)]

(I'm looking at scalar field theory described by a Lagrangian density L=μϕμϕm2ϕϕ)


One can of course simply compute this by plugging in the field operator, but I was wondering about the general situation.


And what about the more general case of two operators, whose commutator is known? [O(x),U(x)]=A(x,x)[xO(x),U(x)]=?



Answer



The derivative of an operator: Let X(t),RX, where X is some normed linear space, say a Banach or Hilbert space. Then we can define the derivative in the usual way: tX(t)=limδ0X(t+δ)X(t)δ.

However, on X different topologies exist, strong, weak, uniform, etc. Hille and Phillips, "Functional Analysis and Semi-Groups" contains a quite readable discussion of these matters.



In general the commutator does not vanish. Consider


X(ξ)=Ucosξ+VsinξdXdξ=Usinξ+Vcosξ

then one gets, after some algebras


[X(ξ),dXdξ]=[U,V]


Then it is clear that one recovers the usual result for commuting U and V operators.


Edit: the result is correct ! Thanks to the comments.


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