Saturday, September 24, 2016

quantum field theory - 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 $t$)?



Answer



Of course one can construct states with any desired expectation values. This is no different from constructing a state of a simple harmonic oscillator with the desired expectation value of position and momentum, repeated for each field mode. Just make a wavepacket centered on the desired position and with the right phases. Note however that you cannot make a simultaneous eigenstate of the electric and magnetic fields since they don't commute with each other, but you can fix the expectation values.


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.


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