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.

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  • $\begingroup$ But $\mathbf{x}$ is not an operator. Here the operator is $\mathbf{V}(\mathbf{x})$. Am I missing something? $\endgroup$ – NarcosisGF Nov 3 '18 at 6:21
  • $\begingroup$ @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 $)...$. $\endgroup$ – J C Nov 3 '18 at 6:24
  • $\begingroup$ I think I understand what you're trying to say. I had not seen it from that point of view. Thanks for your comment. $\endgroup$ – NarcosisGF Nov 3 '18 at 6:40
  • $\begingroup$ @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 $\endgroup$ – J C Nov 3 '18 at 6:42

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