The explanations of quantizing the Electric/Magnetic (E/M) fields that I've read have all basically worked by using the Coulomb gauge in free space to define the vector potential in some volume as

$$ \mathbf{A}(\mathbf{x},t)=\sum_{\mathbf{k},r}\sqrt{\frac{\hbar c^2}{2V\omega_{\mathbf{k}}}}\mathbf{e}_r(\mathbf{k})[a_{r,\mathbf{k}}e^{i(\mathbf{k}\cdot\mathbf{x}-\omega_{\mathbf{k}}t)}+a^*_{r,\mathbf{k}}e^{-i(\mathbf{k}\cdot\mathbf{x}-\omega_{\mathbf{k}}t)}] $$

where the classical amplitudes $a$, $a^*$ are essentially engineered to be directly replaced with the creation/annihilation operators $\hat{a}$, $\hat{a}^\dagger$ to give the quantized form of the fields, the Hamiltonian etc. This is neat and easy to follow but I don't see any physical motivation for replacing those particular objects with the creation/annihilation operators.

By contrast, in most pedagogical treatments, the quantum harmonic oscillator is first given in terms of the physical quantities position and momentum, these are then replaced with their quantum versions and the creation/annihilation operators are derived from their anticommutation relations. I find this to be a more satisfying procedure because the $\hat{a}$, $\hat{a}^\dagger$ formalism is shown to naturally emerge from switching to the quantum version of these physical observables. In particular, you have the relations

$$ \hat x = \sqrt{\frac{\hbar}{2m\omega}}(\hat a+\hat a^\dagger),\quad \hat p = i\sqrt{\frac{\hbar m\omega}{2}}(\hat a+\hat a^\dagger), $$


$$ \hat a = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x + \frac{i}{m\omega}\hat p\right), \quad \hat a ^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x - \frac{i}{m\omega}\hat p\right). $$

Naturally I was wondering if there are analogous physical quantities for the EM field where similar relations to the above hold. Hoping to gain some insight, I considered a single mode $\mathbf{k}$ and wrote $$ \hat a = \hat Q + i\hat P $$ which gets us $$ \hat a e^{i(\mathbf{k}\cdot\mathbf{x}-\omega_{\mathbf{k}}t)}+\hat a ^\dagger e^{-i(\mathbf{k}\cdot\mathbf{x}-\omega_{\mathbf{k}}t)} = 2\hat Q \cos(\mathbf{k}\cdot\mathbf{x}-\omega_{\mathbf{k}}t) - 2\hat P \sin(\mathbf{k}\cdot\mathbf{x}-\omega_{\mathbf{k}}t). $$ I now have an expression in terms of hermitian operators but I'm not sure what they are supposed to be nor do I know why we should expect them to have commutation relations similar to position and momentum. Can anyone clear this up for me or point me to a source which treats the quantized EM field in this way?

  • $\begingroup$ Check out Quantum and Atom Optics by Dan Steck and also Robert Littlejohns Quantum Mechanics Lecture Notes. $\endgroup$
    – Jagerber48
    Commented Feb 15, 2022 at 21:15

2 Answers 2


You can actually find that your $\hat{Q},\hat{P}$s are very similar to the electric and magnetic field parts, if you're using the linear polarization basis where ${\bf k}\times{\bf e}={\bf k}\times{\bf \bar{e}}$.

Check the Wikipedia page.

$$\begin{aligned} &\mathbf{A}(\mathbf{r})=\sum_{\mathbf{k}, \mu} \sqrt{\frac{\hbar}{2 \omega V \epsilon_{0}}}\left\{\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}}+\overline{\mathbf{e}}^{(\mu)} a^{\dagger^{(\mu)}}(\mathbf{k}) e^{-i \mathbf{k} \cdot \mathbf{r}}\right\} \\ &\mathbf{E}(\mathbf{r})=i \sum_{\mathbf{k}, \mu} \sqrt{\frac{\hbar \omega}{2 V \epsilon_{0}}}\left\{\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}}-\overline{\mathbf{e}}^{(\mu)} a^{\dagger^{(\mu)}}(\mathbf{k}) e^{-i \mathbf{k} \cdot \mathbf{r}}\right\} \\ &\mathbf{B}(\mathbf{r})=i \sum_{\mathbf{k}, \mu} \sqrt{\frac{\hbar}{2 \omega V \epsilon_{0}}}\left\{\left(\mathbf{k} \times \mathbf{e}^{(\mu)}\right) a^{(\mu)}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}}-\left(\mathbf{k} \times \overline{\mathbf{e}}^{(\mu)}\right) a^{\dagger^{(\mu)}}(\mathbf{k}) e^{-i \mathbf{k} \cdot \mathbf{r}}\right\} \end{aligned}$$

  • $\begingroup$ Sure, but what I'm really wondering is why I should expect the quantized versions of these "parts" to not commute trivially. $\endgroup$ Commented Feb 15, 2022 at 19:31
  • 1
    $\begingroup$ I think you're asking why we impose the commutation relation to a pair of conjugate variables in classical Hamiltonian (aka canonical quantization)? This is how people tried to generalize the Poisson brackets into quantum versions. Basically, the idea is every normal mode (harmonic oscillator-like things) corresponds to a boson and therefore people imply the bosonic commutation rule. For here, the EM wave is obviously a very good example which quantized as photon modes. $\endgroup$ Commented Feb 15, 2022 at 19:50

If one starts from $Q$ and $P$ in order to quantize the electromagnetic field, one can use an analogy to the quantization of classical mechanics.

The quantization of classical mechanics or differently said the transition of classical mechanics to quantum mechanics is done by requiring the famous commutation relation between $Q$ and $P$ (we assume here $\hbar=1$):

$$[Q,P]= i$$

which is the basis for Heisenberg's uncertainty relation. Then the definition

$$a = Q + iP \quad \text{and} \quad a^\dagger = Q - iP $$

can be used to find the commutation relation between the creation and annihilation operators.

Actually it is better to use the following definitions:

$$c = \frac{(\omega Q + iP)}{\sqrt{2\omega}}\quad \text{and} \quad c^\dagger = \frac{(\omega Q - iP)}{\sqrt{2\omega}}$$

one can find the commutation relation for the creation and annihilation operators:

$$[c,c^\dagger]= 1$$

The formalism also allows to find the Hamiltionian expressed in creation and annihilation operators. In order to get there in case of the electromagnetic field one starts off from the classical expression of the energy of the electromagnetic field:

$$ H \equiv E =\frac{1}{8\pi}\int (\mathbf{E}^2 + \mathbf{H}^2) d^3x $$

where the electrical and magnetic field are given by the vector potential:

$$ \mathbf{E} = -\dot{\mathbf{A}}\quad\text{and}\quad \mathbf{H} = rot A$$

Using the development of the vector potential in creation and annihilation operators as it is given in the first formula of the post plugged in the definition of $\mathbf{E}$ and $\mathbf{H}$, furthermore these two plugged in the definition of the field energy yields the Hamiltonian expressed in creation and annihilation operators after a longer calculation:

$$H = \frac{1}{2}\sum (P^2 +\omega^2 Q^2) = \frac{1}{2}\sum \omega (c c^\dagger + c^\dagger c)$$

So far the fact that the EM-field actually not only has one set of canonical variables $Q$ and $P$, but infinitely many was neglected. The summation sign in the above expression is put in order to fix this.

We can observe that the quantities $P$ and $Q$ fulfill the role for momentum and position in a harmonic oscillator. So they deserve to be considered as (generalized) momentum $P$ and position $Q$.

This means one is led rather naturally to the operator property of the creation and annihilation operators with the right commutation relation by using the analogy of the quantization of classical mechanics.

This answer only schematically shows the way to go. In order to see the calculation in all details and with the right normalization I recommend to read volume 4 (Relativistic Quantum Theory) of the series of tomes edited by Landau & Lifshitz. More modern books in particular those of pure Quantum Field Theory use often a Lorentz covariant formalism where these details are not shown. But nevertheless there are books around where this particular approach can be found.


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