Fourier expansion electromagnetic field

The quantization of the EM fields leads to photons, light quanta of well-defined energy and momentum. In field quantization a key role is played by the Fourier expansion of the different vector fields. Hence, we will now discuss the Fourier expansions of the fields E, B, and A. It will be seen that the expansion of the vector potential A yields immediately the expansions of the fields E and B.

Fourier expansion of a vector field

For an arbitrary real vector field F its Fourier expansion is the following:

$\mathbf {F} (\mathbf {r} ,t)=\sum _{\mathbf {k} }\left(\mathbf {f} _{k}(t)e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {f} }}_{k}(t)e^{-i\mathbf {k} \cdot \mathbf {r} }\right)$

where the bar indicates complex conjugation. Such an expansion, labeled by a discrete (countable) set of vectors k, is always possible when F satisfies periodic boundary conditions, i.e., F(r + p,t) = F(r,t) for some finite vector p. To impose such boundary conditions, it is common to consider EM waves as if they are in a virtual box of finite volume V. Waves on opposite walls of the box are enforced to have the same value (usually zero). Note that the waves are not restricted to the box: the box is replicated an infinite number of times in x, y, and z direction.

Vector potential and its expansion

The magnetic field B is a transverse field and hence can be written as

$\mathbf {B} (\mathbf {r} ,t)={\boldsymbol {\nabla }}\times \mathbf {A} (\mathbf {r} ,t),$

in which the vector potential A is introduced. Also the electric field E is transverse, because earlier we assumed absence of charge distributions. The electric field E also follows from A,

$\mathbf {E} (\mathbf {r} ,t)=-{\frac {\partial \mathbf {A} (\mathbf {r} ,t)}{\partial t}}.$

The fact that E can be written this way is due to the choice of Coulomb gauge for A:

${\boldsymbol {\nabla }}\cdot \mathbf {A} (\mathbf {r} ,t)=0.$

By definition, a choice of gauge does not affect any measurable properties (the best known example of a choice of gauge is the fixing of the zero of an electric potential, for instance at infinity). The Coulomb gauge makes A transverse as well, and clearly A is in the same plane as E. (The time differentiation does not affect direction.) So, the vector fields A, B, and E are all in a plane perpendicular to the propagation direction and can be written in terms of e_x and e_y (in the definition of figure 1). It is more convenient to choose complex unit vectors:

$\mathbf {e} ^{(1)}\equiv {\frac {-1}{\sqrt {2}}}(\mathbf {e} _{x}+i\mathbf {e} _{y})\quad {\hbox{and}}\quad \mathbf {e} ^{(-1)}\equiv {\frac {1}{\sqrt {2}}}(\mathbf {e} _{x}-i\mathbf {e} _{y})$

which are orthonormal,

$\mathbf {e} ^{(\mu )}\cdot {\bar {\mathbf {e} }}^{(\mu ')}=\delta _{\mu ,\mu '}\quad {\hbox{with}}\quad \mu ,\mu '=1,\,-1.$

The Fourier expansion of the vector potential reads

$\mathbf {A} (\mathbf {r} ,t)=\sum _{\mathbf {k} }\sum _{\mu =-1,1}\left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)\,e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\,e^{-i\mathbf {k} \cdot \mathbf {r} }\right).$

The vector potential obeys the wave equation. The substitution of the Fourier series of A into the wave equation yields for the individual terms,

$-k^{2}a_{\mathbf {k} }^{(\mu )}(t)={\frac {1}{c^{2}}}{\frac {\partial ^{2}a_{\mathbf {k} }^{(\mu )}(t)}{\partial t^{2}}}\quad \Longrightarrow \quad a_{\mathbf {k} }^{(\mu )}(t)\propto e^{-i\omega t}\quad {\hbox{with}}\quad \omega =kc.$

It is now an easy matter to construct the corresponding Fourier expansions for E and B from the expansion of the vector potential A. For instance, the expansion for E follows from differentiation with respect to time,

$\mathbf {E} (\mathbf {r} ,t)=i\sum _{\mathbf {k} }\sum _{\mu =-1,1}\omega \left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)\,e^{i\mathbf {k} \cdot \mathbf {r} }-{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\,e^{-i\mathbf {k} \cdot \mathbf {r} }\right)$

Fourier-expanded energy

The electromagnetic energy density $\scriptstyle {\mathcal {E}}_{\mathrm {Field} }$, defined earlier in this article, can be expressed in terms of the Fourier coefficients. We define the total energy (classical Hamiltonian) of a finite volume V by

$H=\iiint _{V}{\mathcal {E}}_{\mathrm {Field} }(\mathbf {r} ,t)\mathrm {d} ^{3}\mathbf {r} .$

The classical Hamiltonian in terms of Fourier coefficients takes the form

$H=V\epsilon _{0}\sum _{\mathbf {k} }\sum _{\mu =1,-1}\omega ^{2}{\big (}{\bar {a}}_{\mathbf {k} }^{(\mu )}(t)a_{\mathbf {k} }^{(\mu )}(t)+a_{\mathbf {k} }^{(\mu )}(t){\bar {a}}_{\mathbf {k} }^{(\mu )}(t){\big )}$

with

$\omega \equiv |\mathbf {k} |c=2\pi \nu =2\pi {\frac {c}{\lambda }}$

and ε₀ is the electric constant. The two terms in the summand of H are identical (factors commute) and may be summed. However, after quantization (interpretation of the expansion coefficients as operators) the factors do no longer commute and according to quantum mechanical rules one must depart from the symmetrized classical Hamiltonian.

Fourier-expanded momentum

The electromagnetic momentum, P_EM, of EM radiation enclosed by a volume V is proportional to an integral of the Poynting vector (see above). In SI units:

$\mathbf {P} _{\textrm {EM}}\equiv {\frac {1}{c^{2}}}\iiint _{V}\mathbf {S} \,{\textrm {d}}^{3}\mathbf {r} =\epsilon _{0}\iiint _{V}\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t)\,{\textrm {d}}^{3}\mathbf {r} .$