Minimal coupling Hamiltonian

Question

A charged particle in an em field can be described by the following Hamiltonian (in CGS units):
$$H = frac {(vec{p}  + frac {q}{c}vec{A})^2}{2m} + U(r)$$
But... what does it mean to square the sum of two vectorial operators? Isn't it supposed to be the modulus squared?

Vadim · Answer

It should be understood as this:
$$
(mathbf{p}+frac{e}{c}mathbf{A})^2 = (mathbf{p}+frac{e}{c}mathbf{A})cdot (mathbf{p}+frac{e}{c}mathbf{A})
= mathbf{p}^2 + frac{e^2}{c^2}mathbf{A}^2 + frac{e}{c}(mathbf{p}mathbf{A}+mathbf{A}mathbf{p})
$$
Remark: Note that we are working here in the context of the canonical quantization - the quantities are replaced by their operators. Thus, one could expand the square before quantizing and should obtain the same result (although one would need explicitly order the operators, to make the result Hermitian).

Minimal coupling Hamiltonian

One Answer

Add your own answers!

Ask a Question