# A Quantized E field, Coulomb Gauge with Interactions

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1. Oct 25, 2016

### MisterX

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

2. Oct 26, 2016

### A. Neumaier

The true Hamiltonian H is always the generator of translations, no matter how it is split into free and interacting pieces. Hence in the interacting case, (2) and (3) must feature H, not H_0.

3. Oct 26, 2016

### vanhees71

Note that for perturbation theory you usually work in the interaction picture, where the time evolution of the field operators (and the observables built with them) is due to $\hat{H}_0$ and that of the states due to $\hat{H}_{\text{int}}$.

4. Oct 26, 2016

### A. Neumaier

But then the field operators are also those of the free field!

5. Oct 26, 2016

### vanhees71

6. Oct 26, 2016

### MisterX

I understood how it is typical to work in the interaction picture. By doing this are we are saying, what we mean by E is the free field E and not the true conjugate field of A? If we did not mean this, would we have to expand E in perturbation as well so the commutator remained intact up to the given order? Perhaps I am wrong about this.

Maybe the free conjugate field is is the interesting object even when not using perturbation theory. Maybe it is akin to saying what I mean by momentum states are eigenstates of kinetic energy and not the generalized momentum states. For example $p_{\text{gen.}} = p_{\text{kinetic}}+qA.$. We have ascribed a particular meaning to $p_{\text{kinetic}}$ that exists regardless of what the Lagrangian is. So I wondering if $\mathbf{E}$ and $\mathbf{B}$ could be the same way.

Certainly I should read further on QED if I am to understand if we even can answer this question

Last edited: Oct 26, 2016
7. Oct 27, 2016

### vanhees71

Sure, the electromagnetic-field operators (!) obey the free Maxwell equations. In the canonical formalism you start in the Heisenberg picture with the Coulomb-gauge condition
$$\vec{\nabla} \cdot \hat{\vec{A}}=0.$$
Then you realize that the canonical field momentum of $A^0$ vanishes identically, and thus that $A^0$ has to be expressed due to the equation of motion,
$$\Delta A^0=-\rho=-q \bar{\psi} \gamma^0 \psi=-q \psi^{\dagger} \psi,$$
as
$$A^0(t,\vec{x})=\int \mathrm{d}^3 \vec{x} \frac{\rho(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
This you put into the full action. Then $A^0$ is eliminated, and you can go to the Hamilton formalism in the usual way. You have the fields $\vec{A}$ and it's canonical momenta $\vec{\Pi}$.

Using this Hamiltonian you switch to the interaction picture and define the perturbative S-matrix elements in the usual way, leading to the Feynman rules in (old-fashioned) Coulomb gauge. Using current conservation, you can easily transform it into the modern form of the Coulomb-gauge Feynman rules. For the free field, the above solution of the field equation leads automatically to $A^0=0$, i.e., the radiation gauge, as it must be.

For a detailed treatment, see Landau&Lifshitz vol. 4, although I recommend to rather learn the modern path-integral formalism to quantize the em. field in any gauge, including the covariant Lorenz gauge.

8. Oct 27, 2016

### A. Neumaier

Don't they satisfy the Maxwell equations with a source term? The latter needs to be renormalized, hence is not so easy to define.

9. Oct 27, 2016

### vanhees71

I was referring to the interaction picture, where the field operators obey the free Maxwell equations. Of course, you are right, that the interaction picture doesn't really exist (Haag's theorem) and that one has to renormalize perturbation theory. That's the next step after having derived the "naive" Feynman rules ;-). If you want to do better, you need to use the more complicated Epstein-Glaser approach an "smear the distributions". See, e.g., the book by Scharff, Finite QED.