# The Orbital and Spin Momenta of Light

1. Feb 15, 2014

### kderakhshani

Hi everybody,

In most classical or quantum optics texts an angular momentum is considered for the EM radiation as the following:

$J = ε_0 ∫_V r × [E(r, t) × B(r, t)] d^3 r$

Then it is claimed that:

"Using the usual formula for a double vector product and integrating by parts, bearing in
mind the assumption that the fields are zero at the surface of volume V introduced for the
mode expansion, one finds that JR can be written as a sum of two terms:

$J = L + S ,$
given by
$L = ε_0 ∑_{j=(x,y,z)} ∫d^3r Ej (r, t)(r × ∇)Aj (r, t) ,$
$S = ε_0 ∫d^3r E(r, t) × A(r, t)$
"

Thank you

2. Feb 27, 2015

### optophotophys

Hi kderakhshani,

The expressions of L and S include the vector potential. So, first, let's use the vector potential,
$$J=\epsilon_0 \int_Vr\times[E\times [\nabla \times A]]d^3 r.$$
We can proceed by using the formula $a\times(b\times c)=b(a\cdot c)-c(a\cdot b)$. There are two possibilities, $a=r$ or $a=E$. After some consideration, it seems that the former is the right way to go,
\begin{align} J&=\epsilon_0 \int_Vr\times[ E_j \nabla A_j -E_j \nabla_j A ]d^3 r,\\ &=L+\epsilon_0 \int_Vr\times[ -E_j \nabla_j A ]d^3 r,\\ &=L+\epsilon_0 \int_V [-E_j \nabla_j (r\times A)+E\times A ]d^3 r,\\ &=L+S+\epsilon_0 \int_V [-E_j \nabla_j (r\times A) ]d^3 r, \end{align}
where I omit the summation symbol.
After the integration by parts, we have
\begin{align} J&=L+S+\epsilon_0 \int_V [(\nabla_j E_j )(r\times A) ]d^3 r,\\ &=L+S+0, \end{align}
where we assume there is no charge.