# Electomagnetic Field Quantization

From Quantum Field Theory by Franz Mandl and Graham Shaw page 4.

When we are expanding the vector potential as a Fourier series;

$\renewcommand{\vec}[1]{\mathbf{#1}}\vec{A}(\vec{x},t) = \sum\limits_{k}\sum\limits_{r}(\frac{\hbar c^2}{2V\omega_k})^{1/2}\vec{\varepsilon_r}(\vec{k})[a_r(\vec{k},t)e^{i\vec{k}\cdot\vec{x}}+a_r^*(\vec{k},t)e^{-i\vec{k}\cdot\vec{x}}]$

I did not understand how we can determine the constant term $(\frac{\hbar c^2}{2V\omega_k})^{1/2}$. Thanks in advance.

• The point is that $\langle \mathbf{k}_1| \mathbf{k}_2 \rangle$ should be Lorentz invariant. This is ensured if: $$\langle \mathbf{k}_1| \mathbf{k}_2 \rangle \propto E_{\mathbf{p}}$$ I don't own the book you are referring to, but you should be able to verify the above equation by finding the commutation relations $[a(\mathbf{k}_1),a(\mathbf{k}_2)]$. Commented Mar 1, 2014 at 22:13

That factor is there because they quantize the field in a box and not in the continuum. When you take the continuum limit one gets $$\int d^3k \rightarrow \sum_k \sqrt{\frac{\hbar c^2}{V}}$$ by dimensional analysis.
The $1/\sqrt{2\omega_k}$ is related to Lorentz invariance. A factor of $1/2\omega_k$ makes the measure $d^3k/2\omega_k$ Lorentz invariant (see this post) and the remaining $\sqrt{2\omega_k}$ goes with the creation operators so that the definite momentum states $$|k\rangle = \sqrt{2\omega_k} a^\dagger(k)|0\rangle$$ have a Lorentz invariant measure too, as @Hunter explains.