Solution Expansion of ##A^{\mu}## when ##L \rightarrow \infty##

[JD_PM]

TL;DR Summary

I wanted to discuss the generalization of free-Maxwell's solution. More insight is very welcomed.

Please note that

I'll write $A^{\mu} := A^{\mu}(x)$ for simplicity.

I'll work in natural units.

The Lagrangian density $\mathscr{L} = -\frac 1 2 (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$ has equations of motion

$$\Box A^{\mu}=0 \tag{1}$$

We expand $A^{\mu}$ in a complete set of solutions of the harmonic equation $(1)$

$$A^{\mu}=A^{\mu+}+A^{\mu-} \tag{2.1}$$

Where

$$A^{\mu+}=\sum_{r=0}^3 \sum_{\vec k}\Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} \epsilon_r^{\mu}(\vec k) a_r(\vec k)e^{-ik \cdot x} \tag{2.2}$$

$$A^{\mu-} = \sum_{r=0}^3 \sum_{\vec k}\Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} \epsilon_r^{\mu}(\vec k) a_r^{\dagger}(\vec k)e^{ik \cdot x} \tag{2.3}$$

And

$$V=L^3, \ \ \ \ \omega_{\vec k}:= |\vec k|=k^0$$

Alright. I started wondering why there was a volume term involved in the solution expansion. After reading a bit I understood why. It has to do with the concept of a field (which is defined at every spacetime point and describes mechanical systems with infinite degrees of freedom) and making the problem simpler.

We are dealing with the vector potential $A(\vec x , t)$ which, at a given instant of time $t$, must take a value at every spatial point $\vec x$.

To simplify the problem then, the mechanical system is enclosed in a cube of volume $V=L^3$ and some periodic boundary conditions are imposed

$$x \approx x+L, \ y \approx y+L, \ z \approx z+L, \tag{3}$$

I understand that by doing that, we go from having an arbitrary 3-vector $\vec k \in \Bbb R^3$ to $\vec k \in 2\pi/L$ (where the $2\pi$ factor is chosen so that the solution-expansion term $\exp(\mp ik \cdot x)=\exp(\mp ik^0 \cdot x^0) \exp(\pm i \vec k \cdot \vec x)$ is single-valued).

Thus I understand that $(2.2)$ and $(2.3)$ are solutions of the mechanical system enclosed in a 3D lattice.

Once here: what if we wanted to generalize $(2.2)$ and $(2.3)$ for an infinitely large volume $V_{\infty}$? Well, then we should take $L \rightarrow \infty$ in our periodic boundary conditions $(3)$. In such conditions the summation becomes an integral; i.e. (Mandl & Shaw page 12, footnote $9$)

$$\frac{1}{V} \sum_{\vec k} \rightarrow \frac{1}{(2 \pi)^3} \int d \vec k \tag{*}$$

So I'd say that $(2.2)$ and $(2.3)$ generalize to

$$A^{\mu+}= \frac{1}{\sqrt{16 \pi^{3}}} \sum_{r=0}^3 \int d \vec k \Big(\frac{1}{\omega_{\vec k}} \Big)^{1/2} \epsilon_r^{\mu}(\vec k) \tilde a_r(\vec k)e^{-ik \cdot x} \tag{4.1}$$

$$A^{\mu-} = \frac{1}{\sqrt{16 \pi^{3}}} \sum_{r=0}^3 \int d \vec k \Big(\frac{1}{\omega_{\vec k}} \Big)^{1/2} \epsilon_r^{\mu}(\vec k) \tilde a_r^{\dagger}(\vec k)e^{ik \cdot x} \tag{4.2}$$

Comparing to $(2.2), (2.3)$ we notice that the ladder operators get rescaled

$$a_r (\vec k) \rightarrow \tilde a_r (\vec k) = \sqrt{\frac{V}{(2\pi)^3}} a_r(\vec k) \tag{5.1}$$

And

$$\tilde a_r^{\dagger} (\vec k) =\Big( \tilde a_r(\vec k) \Big)^{\dagger} \tag{5.2}$$

Do you agree with the generalization $(4.1), (4.2)$?

Extra insight is very welcomed!

Thanks

 

[Demystifier]

I agree with your (4.1) and (4.2). In fact, that's pretty standard in QFT literature.

 

[vanhees71]

But there should be only 2 polarization degrees of freedom, at least when you use the simpler solution of the quantization problem starting from a completely fixed Coulomb gauge, i.e., $\vec{\nabla} \cdot \vec{A}^{+}|\text{phys} \rangle=0$. In this gauge it automatically follows for the the free field that $A^0=0$ and you have only the two transverse polarization degrees of freedom. You can choose them as helicity eigenfunctions $\pm 1$ (corresponding to left- and right-circularly polarized plane waves).

 

[Fred Wright]

As the volume of your box goes infinity your ladder operators blow up. Infinity isn't a number, it's a place. You can't impose boundary conditions with a non-number. That's why people usually require the field to vanish at infinity.