# Does E commute with B (in second quantization)?

In second quantization, there's a standard procedure where we first find solutions to Maxwell's Equations. After doing so we apply quantum mechanical properties to these solutions.

So for some boundary conditions we can solve Maxwell's Equations and get:

$$E_x (z,t) = \sum_j A_j q_j(t) \sin(k_j z)$$

$$B_y(z,t) = \frac{\epsilon_0}{\mu_0 k_j} \sum_j A_j \dot q_j(t) \cos(k_j z)$$

where $A_j = \left(\frac{j\pi c}{L} \right) \left( \frac{2 m_j}{V \epsilon_0} \right) ^2$

Now for the quantum mechanical part, we postulate that $q_j$ and $\dot q_j$ are operators that don't commute:

$$[q_i, m_j \dot q_j ] = i \hbar \delta_{ij}$$

My origional thought when looking at this was that we are imagining that E and B are simultaneous observables, and simply extracting the essential parts of the E & B fields that don't commute (the other parts of E & B aren't operators). That is:

$$[E_x, B_y] \neq 0 \implies [q_i, m_j \dot q_j ] = i \hbar \delta_{ij}$$

And in fact, it seems like E and B must commute in order for q and $\dot q$ to commute.

But strangely after working out the second-quantized plane wave expansion of the electric and magnetic fields, we obtain:

$$\color{Mahogany} E(r, t) \color{black} = \sum_\textbf{k} \epsilon_\textbf{k} \mathscr{E}_\textbf{k} a_\textbf{k} e^{-i \nu_\textbf{k} t + i \textbf{k} \cdot \textbf{r}} + H.c.$$

$$B(r, t) \color{black} = \sum_\textbf{k} \frac{(\textbf{k}\times \hat \epsilon_\textbf{k})}{\nu_k} \mathscr{E}_\textbf{k} a_\textbf{k} e^{-i \nu_\textbf{k} t + i \textbf{k} \cdot \textbf{r}} + H.c.$$

It appears as though E and B only differ in their polarization, and appear to commute.

If E and B really do commute in the general case, is there any intuition for choosing $[q_i, m_j \dot q_j ] = i \hbar \delta_{ij}$ as an axiom?

• The hermitian conjugate part will contain $a^\dagger$ that will not commute with $a$. – ZeroTheHero Jan 23 '17 at 3:20
• There will be cross terms that are commutators of creation and annihilation operators that are nonzero. You cannot do the commutator without the Hermitian conjugate part. – Paul Malinowski Jan 23 '17 at 3:24
• Working out the commutator, it does indeed appear that they do not necessarily commute (and that both of you are correct)! I find the result to be a little surprising (that when you want to quantize an observable, an observable that's different by only a constant might not commute with that observable). – Steven Sagona Jan 23 '17 at 4:13
• In your last equation you have vectors! Collinear components commute. Ex and By don't – borilla Jan 23 '17 at 6:34
• @Steven I realized I had answered a little too fast without reading the comments. Your original question and your last comment are interesting. So doing the computations for the 2nd-quantized expansion you saw that $[E(r,t),B(r',t)]$ is null for r=r' but not necessarily null otherwise if I am right. Because in this case there's a phase shift that applies differently to $a_{k}$ and $a^{+}_{k}$. I never saw the computations done so I am not 100% sure. But all this doesn't answer your original question. You can see from your equations with the boundary conditions you have $[E(z,t),B(z,t)]\not=0$ – borilla Jan 28 '17 at 15:41