# Magnetic Vector Potential Operator [closed]

How do magnetic vector potential $\vec{A}$ act on a ket? What is the basis representation of $\vec{A}$ when it acts on a ket. Can we say that it acts directly (multiplied) on the ket just like a normal scalar potential?

• – Kyle Kanos Sep 19 '17 at 10:05

A vector potential acts on a ket by multiplying it by a vector. For example, say your vector potential is $\vec{A}=(x,2x,z)$. Then $\vec{A}|\psi\rangle=(\hat{X}|\psi\rangle, 2\hat{X}|\psi\rangle,\hat{Z}|\psi\rangle)$. You might be bothered by the fact that you've acted on a single ket and gotten a vector's worth of kets, but that's the same thing that happens with the momentum operator, $\vec{P}$, and the spin operator, $\vec{S}$. Fortunately, the Hamiltonian only involves terms like $\vec{A}\cdot\vec{P}$, $\vec{P}\cdot\vec{A}$, $\vec{P}\cdot\vec{P}$, and $\vec{A}\cdot\vec{A}$, all of which act on a single ket and give back a single ket, not a vector of kets.
In case you mean the $\hat{\vec{A}}$ from free QFT, then you first need the right expansion in terms of creation and annihilation operators, then any Fock space ket can be acted upon.