Deriving Jeans' Result from Magnetic Vector Potential

[noospace]

I'm trying to get from the magnetic vector potential

$\vec{A}(\vec{x},t) = \frac{1}{\sqrt{\mathcal{V}}}\sum_{\vec{k},\alpha=1,2}(c_{\vec{k}\alpha}(t) \vec{u}_{\vec{k}\alpha}(\vec{x}) + c.c.)$

where
$c_{\vec{k}\alpha}(t) = c_{\vec{k}\alpha}(0) e^{-i\omega_{\vec{k}\alpha}t}$
$\vec{u}_{\vec{k}\alpha}(t) = \vec{u}_{\vec{k}\alpha}(0) e^{i \vec{k} \cdot \vec{x}}$
c.c. = complex conjugate.

to Jeans' result

$H := \frac{1}{2}\int d^3\vec{x} (E^2 + B^2) = \sum_{k,\alpha}\left(\frac{\omega_{k\alpha}}{c}\right)^2 2c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha}$.

The first question I have is why we add the complex conjugate term at all. In classical electrodynamics, I'm used to dealing with the complex fields?

Tod derive the result, my approach was to define the complex vector potential

$\widetilde{\vec{A}}(\vec{x},t) = \frac{1}{\sqrt{\mathcal{V}}}\sum_{\vec{k},\alpha=1,2}c_{\vec{k}\alpha}(t) \vec{u}_{\vec{k}\alpha}(\vec{x})$

and use the relation

$H = \frac{1}{2}\int d^3\vec{x} \frac{1}{2} (\Re[\widetilde{\vec{E}}\cdot \widetilde{\vec{E}}^\ast]+\Re[\widetilde{\vec{B}}\cdot \widetilde{\vec{B}}^\ast])$

where $\widetilde{\vec{E}} = - \frac{1}{c}\frac{\partial \widetilde{\vec{A}}}{\partial t}, \;\widetilde{\vec{B}} = \nabla \times \widetilde{\vec{A}}$.

The first term is easy to deal with and gives (after using the normalization $\frac{1}{\mathcal{V}}\int d^3\vec{x} \vec{u}_{\vec{k}\alpha}\cdot\vec{u}_{\vec{k'}\alpha'}^\ast = \delta_{\vec{k}\vec{k}'}\delta_{\alpha\alpha'}$):

$\frac{1}{4}\int d^3 \vec{x}\Re[\widetilde{\vec{E}}\cdot \widetilde{\vec{E}}^\ast] = \frac{1}{4}\sum_{k,\alpha}\left(\frac{\omega_{k\alpha}}{c}\right)^2 c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha}$

The second term is a bit more problematic. Using the identity

$(\vec{k} \times \vec{u})\cdot(\vec{k}' \times \vec{u}') = \vec{k} \cdot\vec{k}' \vec{u} \cdot\vec{u}' - \vec{k} \cdot{u}' \vec{k}' \cdot\vec{u}$

I get

$\widetilde{\vec{B}} = \nabla \times \widetilde{\vec{A}} = \frac{1}{\sqrt{V}}\sum_{\vec{k},\alpha} i c_{\vec{k}\alpha} \vec{k} \times \vec{u}_{\vec{k}\alpha}$

so

$\frac{1}{2}\int d^3\vec{x} \frac{1}{2}\Re[\widetilde{\vec{B}}\cdot \widetilde{\vec{B}}^\ast]$

$= \frac{1}{4}\sum_{\vec{k},\alpha=1,2}\left(\frac{\omega_{\vec{k}\alpha}}{c}\right)^2 c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha} - \frac{1}{4}\frac{1}{V}\sum_{\vec{k},\vec{k}',\alpha,\alpha'}\int d^3 \vec{x} c^\ast_{\vec{k}'\alpha'}c_{\vec{k}\alpha} (\vec{k}\cdot \vec{u}^\ast_{\vec{k}'\alpha'})(\vec{k'}\cdot \vec{u}_{\vec{k}\alpha})$

where I have used the normalization condition to evaluate the first term. I'm stuck on how to evaluate this last integral. I know that I have to use the Gauge condition that $\vec{k} \cdot \vec{u}_{\vec{k}\alpha} = 0$ but I'm not sure how to incorporate it.

Any help would be greatly appreciated.

 

[noospace]

I suppose what I want to show is that the term

$\sum_{\vec{k},\vec{k}',\alpha,\alpha'}\int d^3 \vec{x} c^\ast_{\vec{k}'\alpha'}c_{\vec{k}\alpha} (\vec{k}\cdot \vec{u}^\ast_{\vec{k}'\alpha'})(\vec{k'}\cdot \vec{u}_{\vec{k}\alpha})$

vanihses. For then,

$\frac{1}{2}\int d^3 \vec{x}\frac{1}{2}\Re[\widetilde{\vec{E}}\cdot\widetilde{\vec{E}}^\ast + \widetilde{\vec{B}}\cdot\widetilde{\vec{B}}^\ast] = \frac{1}{2}\sum_{\vec{k},\alpha}\left(\frac{\omega_{\vec{k}\alpha}}{c}\right)^2 c^\ast_{\vec{k}\alpha}c_{\vec{k}\alpha}$

which is uncannily similar to what I set out to show. I know $\vec{k} \cdot\vec{u}_{\vec{k}\alpha}$ so I can reduce the sum to a sum over $\sum_{\vec{k} \neq \vec{k}',\alpha,\alpha}$

 

[denian]

I would like to provide some clarification and guidance on your approach to deriving Jeans' result from the given magnetic vector potential.

Firstly, the addition of the complex conjugate term is necessary because we are dealing with complex fields in quantum electrodynamics, which is the theory that this result is derived from. In classical electrodynamics, we typically deal with real fields, but in quantum electrodynamics, we must use complex fields to accurately describe the behavior of particles.

Secondly, your approach to defining the complex vector potential and using the relation for H is correct. However, the second term in the integral for H can be simplified by using the gauge condition \vec{k} \cdot \vec{u}_{\vec{k}\alpha} = 0. This condition is necessary to ensure that the vector potential is properly normalized and avoids any mathematical issues with diverging integrals.

Using this condition, the second term in the integral becomes:

\frac{1}{4}\frac{1}{V}\sum_{\vec{k},\vec{k}',\alpha,\alpha'}\int d^3 \vec{x} c^\ast_{\vec{k}'\alpha'}c_{\vec{k}\alpha} (\vec{k}\cdot \vec{u}^\ast_{\vec{k}'\alpha'})(\vec{k'}\cdot \vec{u}_{\vec{k}\alpha}) = 0

since \vec{k} \cdot \vec{u}_{\vec{k}\alpha} = 0. This simplification allows us to arrive at the desired result:

H := \frac{1}{2}\int d^3\vec{x} (E^2 + B^2) = \sum_{k,\alpha}\left(\frac{\omega_{k\alpha}}{c}\right)^2 2c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha}.

I hope this helps clarify the derivation process and the role of the complex conjugate term in the result. Keep up the good work in your studies!