The common presentation for free field quantization proceeds with the Lorentz and Coulomb (##\phi = 0, \,\nabla \cdot \mathbf{A} = 0 ##) constraints. Then ##A## can be defined(adsbygoogle = window.adsbygoogle || []).push({});

$$\mathbf{A} \propto \iint \frac{d^3 p}{\sqrt{2\omega_p}}\sum_{\lambda} \Big(e^{i\mathbf{p}\cdot \mathbf{x}}\boldsymbol{\epsilon}_\lambda a(\mathbf{p}, \lambda) +e^{-i\mathbf{p}\cdot \mathbf{x}}\boldsymbol{\epsilon}^*_\lambda a^\dagger (\mathbf{p}, \lambda) \Big) $$

##\mathbf{E} ## could be defined

\begin{align}

\mathbf{E} & = -\nabla\phi -\frac{\partial}{\partial t}\mathbf{A}\\

\frac{\partial}{\partial t}\mathbf{A} &\Leftrightarrow \frac{1}{i\hbar}[\mathbf{A}, H_0] \\

\mathbf{E} &\propto -\frac{1}{i\hbar}[\mathbf{A}, H_0],\, \nabla\phi = 0 \\

H_0 &\propto \int \sum_\lambda d^3 p\, E_{\mathbf{p}}a^\dagger (\mathbf{p}, \lambda) a(\mathbf{p}, \lambda) \\

H_0 &\propto \int d^3 x\, \epsilon_0 \mathbf{E}^2 + \frac{1}{\mu} \mathbf{B}^2

\end{align}

My question is about what happens when ##H = H_0 + H_1##. If we are to define the ##\mathbf{E}## operator as the rate of change of ##\mathbf{A}##, if we are to keep using ##\mathbf{E} = -\frac{\partial}{\partial t}\mathbf{A}## Then I suppose we might argue

$$ \mathbf{E} \propto -\frac{1}{i\hbar}[\mathbf{A}, H_0 + H_1]$$

On the other hand we can argue ##\mathbf{E}## to be an operator independent of ## H## and always has the same form as with the free fields. I am not sure which is correct or what the consequences will be.

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# A Quantized E field, Coulomb Gauge with Interactions

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