# Why is transverse delta distribution needed for Poisson brackets of classical EM field components?

During reading a German lecture notes (Quanten Optik by Dirk–Gunnar Welsch) about quantization of the EM field, I stumbled over a statement I cannot reproduce in detail:

Generally, it is argued, that for (classical) conjugated fields $\Phi$ and $\Pi$, one can write

$$\Phi(x) = \int d^3x' \Phi(x') \delta (x-x')$$

$$\Pi(x) = \int d^3x' \Pi(x') \delta (x-x')$$

so that \begin{align} \frac{\delta \Phi(x)}{\delta \Phi(x')} &= \delta (x-x')\\ \frac{\delta \Phi(x)}{\delta \Pi(x')} &= 0\\ \frac{\delta \Pi(x)}{\delta \Pi(x')} &= \delta (x-x')\\ \frac{\delta \Pi(x)}{\delta \Phi(x')} &= 0 \end{align}

Hence

$$\{ \Phi(x), \Pi(x') \} := \int d^3x'' \left(\frac{\delta \Phi(x)}{\delta \Phi(x'')} \frac{\delta \Pi(x)}{\delta \Pi(x'') } - \frac{\delta \Phi(x)}{\delta \Pi(x'')} \frac{\delta \Pi(x)}{\delta \Phi(x'') }\right) = \cdots = \delta(x-x')$$

Up to here I can follow.

But now, they apply the formalism to the classic EM field, having $A_k$ and $\Pi_k = \epsilon_o \dot{A}_k$ as conjugated fields under Coulomb gauge $\vec \nabla \cdot \vec A = 0$.

They say (Eqs. 1.81 ff, translated from German and shortened):

Although one might be inclined to assume $$\{ A_k(x), \Pi_{k'}(x') \} = \delta_{k,k'} \cdot \delta(x-x')$$ this is wrong, because of $A_{k, k} = 0$, and, therefore, instead of the regular Delta-distribution we have to apply a so called transverse Delta: $$\{ A_k(x), \Pi_{k'}(x') \} = \delta_{k, k'}^{\bot}(x-x'),$$ with an explicit form of the latter in terms of regular $\delta$ and an extra term: $$\delta_{k, k'}^{\bot}(r) = \delta_{k,k'} \cdot \delta(r) + \frac{1}{4 \pi} \frac{\partial ^2 }{\partial x_k \partial x_{k'} }\frac{1}{|r|}.$$

However, this statement is not motivated in more detail and I cannot follow, why the most general considerations above do not hold in detail below. There must be a tiny detail, giving rise to the extra complication. How can this be mathematically derived based on the physics of EM fields?

I think the answer is related to the fact, that due to choice of Coulomb's gauge the $A_k$ are not indpendent from each other. Indeed, by using some vector calculus, one gets, after doing a little tedious calculation:

$A_i(x) = \int d^3x' \delta_{i, j}^{\bot}(x-x') A_j(x')$

On the other hand I would still expect

$A_i(x) = \int d^3x' \delta(x-x') A_i(x')$

since this second identity is a simple tautology. But this somehow contradicts the first form, because the functional derivates are clearly different...I'm closer to the solution, but this question remainsstill open...

Is the second form wrong, because the constraint is not involved?

• To readers and potential answers: I added the tag constrained-dynamics for organisational purposes. One may presume OP is not familiar with this concept. – AccidentalFourierTransform May 20 '17 at 19:15
• Minor comment to the post (v5): Please consider to mention explicitly author, title, page, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic May 20 '17 at 19:18
• Equations 1.81 ff on page 20, see hint above – michael May 20 '17 at 19:23
• Yes, you are very close to the answer! I suggest you the paragraph 7.7 of this book: Field Quantization by Walter Greiner, Joachim Reinhardt, D.A. Bromley – Jared Aug 14 '17 at 23:26
• Do you have any hints on how to show that $A_i(x) = \int d^3x' \delta_{i, j}^{\bot}(x-x') A_j(x')$? – jak Nov 4 '19 at 9:38

The reason is that Maxwell theory involves gauge invariance which implies the existence of constraints in the system. In the Hamiltonian formulation, the Gauss equation $\phi_1\equiv\nabla\cdot \Pi=0$ is a first class constraint in the system. Now by imposing the Coulomb gauge $\phi_2\equiv\nabla\cdot A=0$, we get two second class constraints with the following Poisson bracket $$[\phi_1,\phi_2]=-\nabla^2\delta^3(x-x'),$$ which can be obtained from the basic Poisson bracket $[A_i(x),\Pi_j(x')]=\delta_{ij}\delta^3(x-x')$.
In this constrained system, the commutators have to be computed using the Dirac brackets instead of Poisson brackets, $$[A,B]_*=[A,B]-[A,\phi_a]\,C^{ab}\,[\phi_b,B],$$ where $C_{ab}\equiv[\phi_a,\phi_b]$ and $C^{ab}$ with upper indices denotes its inverse. In our problem $$\label{Dirac} [A_i,\Pi_j]_*=[A_i,\Pi_j]-[A_i,\phi_1]C^{12}[\phi_2,\Pi_j]$$ To compute $C^{12}$, note that by definition $C^{-1}\cdot C=\mathbf{1}$ or explicitly $$\int d^3y\, C^{ac}(x-y)\, C_{cb}(y-x')=\delta^a_b\delta^3(x-x').$$ Specifically for $a=b=1$ we get \begin{align} \int d^3y \,C^{12}(x-y)\,\nabla^2\delta^3(y-x')=\delta^3(x-x') \end{align} By integration by parts, we put the derivatives on $C^{12}$ and find $\nabla^2 C^{12}(x-y)=\delta^3(x-x')$ from which we infer that $C^{12}$ is nothing but the Green's function \begin{align} C^{12}(x-y)=-\dfrac{1}{4\pi}\dfrac{1}{|x-x'|} \end{align} Going back to the Dirac bracket, we have \begin{align} [A_i,\phi_1]C^{12}[\phi_2,\Pi_j]&=-\int d^3y d^3z \,[A_i(x),\nabla\cdot \Pi(y)]\,\dfrac{1}{4\pi}\dfrac{1}{|y-z|}\,[\nabla\cdot A(y),\Pi_j(x')]\\ &=-\int d^3y d^3z \,\partial_i \delta^3(x-y)\,\dfrac{1}{4\pi}\dfrac{1}{|y-z|}\,\partial_j \delta^3(z-x')\\ &=-\dfrac{1}{4\pi}\partial_i\partial'_j\dfrac{1}{|x-x'|} \end{align} Therefore \begin{align} [A_i(x),\Pi_j(x')]_*&=\delta_{ij}\delta^3(x-x')+\dfrac{1}{4\pi}\partial_i\partial'_j\dfrac{1}{|x-x'|}. \end{align} I hope this helps.