# Commutation relation in quantized electromagnetic field theory

I have a question regarding a proposed problem (Problem 4.8) in Rodney Loudon's book "The Quantum Theory of Light". Let $$U(t)$$ be an operator defined by $$U(t)=\exp\left\lbrace\frac{i}{\hbar}\int\text{d}^3x\,\mathbf{V}(\mathbf{x})\cdot\mathbf{A}(t,\mathbf{x})\right\rbrace,$$ where $$\mathbf{A}(t,\mathbf{x})$$ is the quantized vector potential of the electromagnetic field and $$\mathbf{V}(\mathbf{x})$$ is any vector field operator that commutes with the electric field and vector potential operators. One is asked to prove that $$U^{-1}(t)\mathbf{E}(t,\mathbf{x})U(t)=\mathbf{E}(t,\mathbf{x})-\frac{1}{\varepsilon_0}\mathbf{V}_\text{T}(\mathbf{x}),$$ where $$\mathbf{E}(t,\mathbf{x})$$ is the quantized electric field operator and $$\mathbf{V}_\text{T}(\mathbf{x})$$ is the transversal part of the vector field, defined by $$\mathbf{V}_\text{T}(\mathbf{x})=\frac{1}{(2\pi)^{3/2}}\int\frac{\text{d}^3k}{k^2}\,[\mathbf{k}\times\hat{\mathbf{V}}(\mathbf{k})]\times\mathbf{k}\,e^{i\mathbf{k}\cdot\mathbf{x}},$$ where $$\hat{\mathbf{V}}(\mathbf{k})=\frac{1}{(2\pi)^{3/2}}\int\text{d}^3x\,\mathbf{V}(\mathbf{x})e^{-i\mathbf{k}\cdot\mathbf{x}}.$$ I tried to solve this problem by applying the following consequence of the Baker-Campbell-Hausdorff formula: $$U^{-1}(t)\mathbf{E}(t,\mathbf{x})U(t)=\mathbf{E}(t,\mathbf{x})+[X(t),\mathbf{E}(t,\mathbf{x})]+\frac{1}{2}[X(t),[X(t),\mathbf{E}(t,\mathbf{x})]]+\cdots,$$ where $$X(t)=-\frac{i}{\hbar}\int\text{d}^3x\,\mathbf{V}(\mathbf{x})\cdot\mathbf{A}(t,\mathbf{x}).$$ Taking into account the canonical commutation relation $$[A_i(t,\mathbf{x}^\prime),-\varepsilon_0{E}_j(t,\mathbf{x})]=i\hbar{\delta_\text{T}}_{ij}(\mathbf{x}-\mathbf{x}^\prime),$$ where $${\delta_\text{T}}_{ij}(\mathbf{x})$$ is the transverse delta-function (defined on page 145 of the book), with the property that $$\sum_j\int\text{d}^3x^\prime{\delta_\text{T}}_{ij}(\mathbf{x}-\mathbf{x}^\prime)V_j(\mathbf{x}^\prime)={V_\text{T}}_i(\mathbf{x}),$$ I've been able to get to the expression $$[X(t),\mathbf{E}(t,\mathbf{x})]=-\frac{1}{\varepsilon_0}\mathbf{V}_\text{T}(\mathbf{x}),$$ so the problem is reduced to showing that $$\mathbf{V}_\text{T}(\mathbf{x})$$ commutes with $$X (t)$$, but I don't know how to do it. Could someone tell me how to get to the requested result?

Your $$V_T(x)$$ and $$X(t)$$ are commute, because $$V_T(x)$$ onlys has a $$x$$ in expotentional, use series expansion that the component $$\frac{(i k\cdot x)^n}{n!}$$ and $$X(t)$$ is commute, that $$\frac{(i k\cdot x)^n}{n!}X(t)=X(t)\frac{(i k\cdot x)^n}{n!}$$, so you can pull $$X(t)$$ to the right of $$V_T(x)$$ or the left of $$V_T(x)$$ without any misunderstanding.
• But $\mathbf{x}$ is not an operator. Here the operator is $\mathbf{V}(\mathbf{x})$. Am I missing something? – NarcosisGF Nov 3 '18 at 6:21
• @NarcosisGF I know, I think I it saw somewhere but bascially you are in position space and $X(t)$ being an operator doesn't changes $x$ being left or right. $X(t)|x>...=xX(t)...=(|x>$ in position space $)...$. – J C Nov 3 '18 at 6:24
• @NarcosisGF here, en.wikipedia.org/wiki/Position_operator , $X(t)$ is hermitian and the question made it very specific that $A,E$ and $V$ are commute, and check the wiki page note 3 – J C Nov 3 '18 at 6:42