# Linear momentum density of light in general case

If I am given a photon wave function of the form:

$\vec{A} = \hat{x} \;u(r,\phi,z)\; e^{-ikz}$

where $u(r,\phi,z)$ is a distribution of the field amplitude (linear polarization).

Then the linear momentum density can be written as

$p = \frac{\epsilon_2}{2}(\vec{E}^* \cdot \vec{B}+\vec{E} \cdot \vec{B}^*) = i\frac{\epsilon_0 \omega}{2}(u^*\nabla u-u\nabla u^*)+ \omega k \epsilon_0 |u|^2\hat{z}$

No matter how long I stare at this, I can't see it. How can I prove the last equality sign?

This is the link to the source article.

My feeling is that essentially it boils down to expressing linear momentum density in terms of vector potential in general case.

• I haven't really worked it out but the last equality seems to come from setting E = -delA/delt = -i omega A and B = curl A. – Raziman T V Jan 22 '17 at 21:03

The equation is not fully correct or at least I cannot reproduce it without further approximation. If you use \begin{align} E& = i\omega A = i\omega u e^{-ikz} \vec{x} \\ B& = \nabla \times A = \partial_z u e^{-ikz} \vec{y} - \partial_y u e^{-ikz}\vec{z} \end{align}

The $E^*\times B$ term in the momentum is given by \begin{align} E^*\times B = i\omega\; u^* (\partial_y\vec{y}+\partial_z\vec{z})u+\omega k |u|^2\vec{z} \end{align} If you insert that into the equation for $p$ you will find the result from the paper except, that the x component of the gradient in $u\nabla u^*$ is missing. The momentum must have a vanishing x-component because E is parallel to A and A points only in the x direction.

• Yeah, I got that one, except in my case $\partial_z$ is missing. He states that he dropped it, I just don't get why he dropped it in grad, but didn't drop it deriving the last term... Anyways, now I have troubles with eqn. (9). Allen doesn't explain the difference between $\phi$ and $\Phi$... I don't get how he obtained this last term there. – MsTais Jan 26 '17 at 15:37
• The idea is typically that $u$ varies very little in propagation (here z) direction and more in transverse direction, i.e. it determines the beam profile. The somewhat puzzeling part is that $\partial_x \vec{x} u$ is added, which is certainly not small. $\phi$ is a coordinate of the polar coordinate system and $\Phi$ is the respective unit vector $(-\sin\phi,\cos\phi,0)$ – Jannick Jan 27 '17 at 8:48