# Given expectation values for E and B, can you find an associated state?

When we quantize the electromagnetic field, we develop the concept of the field operator $A(\vec{r},t)$ and the simultaneous eigenstates of momentum and the free field Hamiltonian (i.e., each eigenstate is given by specifying the number of photons with momentum $k$ and polarization $\mu$). We can then construct the operators for the electric and magnetic fields, and we can calculate their expectation values for an arbitrary state.

Now, suppose the expectation value of the electric field is $E(\vec{r},t)$ and the magnetic field is $B(\vec{r},t)$. Assuming $E$ and $B$ obey Maxwell's Equations, can we construct a state that has these expectation values? Is it unique, or could there be multiple states with the same expectation value for $E(\vec{r},t)$ and $B(\vec{r},t)$?

What if the expectation values are time independent (i.e., static fields $E(\vec{r},t)=E(\vec{r},0)$ and $B(\vec{r},t)=B(\vec{r},0)$ for all $t$)?

I can prove the non-uniquess of such states just by giving an example: all states with a definite number of photons have zero expectation values $\langle \vec{E} \rangle = \langle \vec{B} \rangle = 0$. This follows because $\vec{E}$ and $\vec{B}$ are operators which change the photon number.
• Interesting! Upvote! I was looking for a closed form for the state that produces a specific $\vec{E}$ and $\vec{B}$, but if it's not unique... How do you physically interpret the statement that having a definite number of photons results in $\langle \vec{E} \rangle = \langle \vec{B} \rangle =0$? And what quantum state corresponds to what we would call a "uniform electric field"? – ChickenGod Apr 10 '13 at 3:54
• @ChickenGod The fields average to zero because there is a reflection symmetry. Think of a single harmonic oscillator. The equivalent statement is that an eigenstate $|n\rangle$ has zero average position and momentum. The Hamiltonian is reflection invariant under $q\to-q,p\to-p$. But the position and momentum operators have odd parity under inversion $q\to-q,p\to-p$. Similary in QED you can take $A_\mu\to-A_\mu$ so for any state with a definite number of photons (i.e., an eigenstate under this inversion) must have zero expectation value of any operator made of an odd number of $A_\mu$ fields. – Michael Brown Apr 10 '13 at 5:32
• @ChickenGod Continuing... the fields average to zero, but their correlation functions $\langle \vec{E}(x)\vec{E}(y)\rangle$ etc. are in general nonzero, and the energy momentum tensor is nonzero as well. It is an instructive exercise to work these quantities out. :) A state $|psi\rangle$ with a definite value of the field $\vec{\hat{E}}(x) |\psi\rangle = \vec{E}(x) |\psi\rangle$ is called a coherent state. Again you should construct these for a harmonic oscillator first before going to field theory. They are the states that correspond most closely to a "large" classical motion. – Michael Brown Apr 10 '13 at 5:37