# Quantization of the Electromagnetic Field

1. Mar 28, 2014

### MisterX

We have recently begun learning about quantization of the electromagnetic field and I would to understand more. It is tempting to want to connect the number states of a definite momentum and polarization to the concept of a classical plane wave. However it seems less straightforward than I might have thought.
$$\vec{E}_{classical} = C\cos\left(\mathbf{k}\cdot \mathbf{r}-\omega t + \phi\right)$$
For the quantized fields we have been using
$$\mathbf{A}\left(\mathbf{r}, t\right) \propto \int \frac{d^3k}{\sqrt{\omega}} \sum_\alpha \boldsymbol{\epsilon}_\alpha a_\alpha\left(\mathbf{k}\right)e^{i\mathbf{k}\cdot \mathbf{r}}e^{-i\omega t} + \boldsymbol{\epsilon}^*_\alpha a^\dagger_\alpha\left(\mathbf{k}\right)e^{-i\mathbf{k}\cdot \mathbf{r}}e^{i\omega t}$$
$$\mathbf{E}\left(\mathbf{r}, t\right) = -\frac{\partial}{\partial t} \mathbf{A} \;\;\;\;\;\;\;\; \mathbf{B}\left(\mathbf{r}, t\right) = \boldsymbol{\nabla}\times \mathbf{A}$$
I notice that these operators all change the occupation numbers by $\pm 1$. What is the significance of this? I would like to somehow see an oscillating, measurable quantity with frequency $\omega$ , perhaps akin to the classical plane wave. I suppose we might be interested in the above operators, and perhaps also the average number of photons with some particular $\mathbf{k}$ .
I have read of the "coherent states," which would be eigenvectors of $a_\alpha\left(\mathbf{k}\right)$ .
$$\mid \alpha \rangle = e^{\left|\alpha\right|^2/2}\sum_n \frac{\alpha^n}{\sqrt{n!}}\mid n \rangle \;\;\;\;\;\;\;\;a\mid \alpha \rangle = \alpha\mid \alpha \rangle$$
Which evolve in time as
$U\left(t, 0\right)\mid \alpha \rangle = \mid \alpha e^{-i\omega t} \rangle$. However, $a$ is not an observable. I am not sure how to interpret the coherent states, or the action $\mathbf{A}$ upon them. I worked out that $$a^\dagger\mid \alpha \rangle = e^{\left|\alpha\right|^2/2}\sum_n \frac{n\alpha^{n-1}}{\sqrt{n!}}$$
I guess my questions are relating to how to interpret the operators $\mathbf{A}, \mathbf{E},\, \text{and}\, \mathbf{B}$ . How do we obtain something resembling the classical oscillating quantities? What states correspond most closely with a classical plane wave? What are the eigenvectors of $\mathbf{A}$?

2. Mar 28, 2014

### The_Duck

The coherent states are the states that are most similar to classical plane waves.

Recall the 1D harmonic oscillator. The coherent states of the 1D harmonic oscillator have exactly the same form as the ones you wrote out for the electromagnetic field. They are the states in which both the position and and velocity of the particle have small uncertainties, so that the particle "looks classical," at least if the parameter $\alpha$ is $\gg 1$ (which is the condition for the amplitude of the motion to be much larger than the uncertainty in the position). The coherent states of the simple harmonic oscillator are the states most similar to classical simple harmonic motion.

Similarly, the coherent states of the EM field are the states for which both the fields and their time derivatives have small uncertainties, so that the field "looks classical" if $\alpha \gg 1$ (which is the condition for the uncertainty in the fields to be much smaller than the magnitude of the fields).

Try evaluating the expectation value

$\langle \alpha | \vec{E}(\vec{r}, t) | \alpha \rangle$

in some coherent state $|\alpha\rangle$. This will give you the expectation value of the electric field in the coherent state; it should look like a plane wave with an amplitude proportional to $\alpha$ and a wave number, frequency, and polarization determined by which creation operator $a^{\dagger}_{\lambda}$ you used to construct the coherent state $|\alpha\rangle$.

They are analogous to the eigenvectors of $x$ in the case of the 1D harmonic oscillator. Eigenvectors of $x$ have definite position; eigenvectors of $\vec{A}$ have definite values of $\vec{A}$. But in each case, these states are rather unphysical. For instance, they have infinite energy; a quantum particle with 0 position uncertainty has infinite kinetic energy because there is an uncertainty relation between the position and momentum. Similarly a quantum field has an uncertainty relation between its value and its time derivative, so that if there is 0 uncertainty in the field value the time derivative has infinite uncertainty, which gives an infinite energy.

More physical (and more interesting) states include the eigenstates of the Hamiltonian and the coherent states.

Last edited: Mar 28, 2014