Transverse and Longitudinal Fields

In non-relativistic Quantum Mechanics, the static Electric field is represented by a scalar potential, magnetic fields by the vector potential, and the radiation field also through the vector potential. It will be convenient to keep this separation between the large static atomic Electric field and the radiation fields, however, the equations we have contain the four-vector \bgroup\color{black}$A_\mu$\egroup with all the fields mixed. When we quantize the field, all E and B fields as well as electromagnetic waves will be made up of photons. It is useful to be able to separate the E fields due to fixed charges from the EM radiation from moving charges. This separation is not Lorentz invariant, but it is still useful.

Enrico Fermi showed, in 1930, that \bgroup\color{black}$A_\parallel$\egroup together with \bgroup\color{black}$A_0$\egroup give rise to Coulomb interactions between particles, whereas \bgroup\color{black}$A_\perp$\egroup gives rise to the EM radiation from moving charges. With this separation, we can maintain the form of our non-relativistic Hamiltonian,

\bgroup\color{black}$\displaystyle H=\sum\limits_j{1\over 2m_j}\left(\vec{p}-{e\...
...sum\limits_{i>j}{e_ie_j\over 4\pi\vert\vec{x}_i-\vec{x}_j\vert}+H_{rad} $\egroup
where \bgroup\color{black}$H_{rad}$\egroup is purely the Hamiltonian of the radiation (containing only \bgroup\color{black}$\vec{A}_\perp$\egroup), and \bgroup\color{black}$\vec{A}_\perp$\egroup is the part of the vector potential which satisfies \bgroup\color{black}$\vec{\nabla}\cdot\vec{A}_\perp=0$\egroup. Note that \bgroup\color{black}$\vec{A}_\parallel$\egroup and \bgroup\color{black}$A_4$\egroup appear nowhere in the Hamiltonian. Instead, we have the Coulomb potential. This separation allows us to continue with our standard Hydrogen solution and just add radiation. We will not derive this result.

In a region in which there are no source terms,

\begin{displaymath}\bgroup\color{black} j_\mu=0 \egroup\end{displaymath}

we can make a gauge transformation which eliminates $A_0$ by choosing \bgroup\color{black}$\Lambda$\egroup such that

\begin{displaymath}\bgroup\color{black} {1\over c}{\partial\Lambda\over\partial t}=A_0. \egroup\end{displaymath}

Since the fourth component of \bgroup\color{black}$A_\mu$\egroup is now eliminated, the Lorentz condition now implies that

\begin{displaymath}\bgroup\color{black} \vec{\nabla}\cdot\vec{A}=0. \egroup\end{displaymath}

Again, making one component of a 4-vector zero is not a Lorentz invariant way of working. We have to redo the gauge transformation if we move to another frame.

If \bgroup\color{black}$j_\mu\neq 0$\egroup, then we cannot eliminate \bgroup\color{black}$A_0$\egroup, since \bgroup\color{black}$\Box A_0={j_0\over c}$\egroup and we are only allowed to make gauge transformations for which \bgroup\color{black}$\Box\Lambda=0$\egroup. In this case we must separate the vector potential into the transverse and longitudinal parts, with

\vec{A}&=&\vec{A}_\perp+\vec{A}_\parallel \\
\vec{\nabla}\cdot\vec{A}_\perp&=&0 \\
\vec{\nabla}\times\vec{A}_\parallel&=&0 \\

We will now study the radiation field in a region with no sources so that \bgroup\color{black}$\vec{\nabla}\cdot\vec{A}=0$\egroup. We will use the equations

\vec{B}&=&\vec{\nabla}\times\vec{A} \\
\vec{E}&=&-{1\over c}{...
...^2\vec{A}-{1\over c^2}{\partial\vec{A}\over\partial t^2}&=&0 \\

Jim Branson 2013-04-22