Question

Does E commute with B (in second quantization)?

Answer 1

 



 



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:

Ex(z,t)=∑jAjqj(t)sin(kjz)Ex(z,t)=∑jAjqj(t)sin⁡(kjz)

By(z,t)=ϵ0μ0kj∑jAjq˙j(t)cos(kjz)By(z,t)=ϵ0μ0kj∑jAjq˙j(t)cos⁡(kjz)

where Aj=(jπcL)(2mjVϵ0)2Aj=(jπcL)(2mjVϵ0)2

Now for the quantum mechanical part, we postulate that qjqj and q˙jq˙j are operators that don't commute:

[qi,mjq˙j]=iℏδij[qi,mjq˙j]=iℏδij

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:

[Ex,By]≠0⟹[qi,mjq˙j]=iℏδij[Ex,By]≠0⟹[qi,mjq˙j]=iℏδij

And in fact, it seems like E and B mustcommute in order for q and q˙q˙ to commute.

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

E(r,t)=∑kϵkEkake−iνkt+ik⋅r+H.c.E(r,t)=∑kϵkEkake−iνkt+ik⋅r+H.c.

B(r,t)=∑k(k×ϵ^k)νkEkake−iνkt+ik⋅r+H.c.B(r,t)=∑k(k×ϵ^k)νkEkake−iνkt+ik⋅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 [qi,mjq˙j]=iℏδij[qi,mjq˙j]=iℏδij as an axiom

Answer 2

the answer given above is correct