Final answer:
The expression for the probability current j under an EM field is derived using the continuity equation and the Schrödinger equation, considering that the momentum operator P and the vector potential A do not commute. This results in the term (P - q/c A)² which needs to be expanded in the Hamiltonian. The final expression for j reflects contributions from the probability amplitude of the wave function and its interaction with the EM field.
Step-by-step explanation:
To derive the expression for the probability current when a particle is under the influence of an electromagnetic (EM) field, one must begin with the continuity equation and the time-independent Schrödinger equation. The continuity equation in quantum mechanics is given by the partial derivative of the probability density ρ with respect to time (∂ρ/∂t) plus the divergence of the probability current j being equal to 0:
∂ρ/∂t + ∇ ⋅ j = 0
The probability density is given by ρ = |ψ|², where ψ is the wave function. Taking the time derivative of ρ involves considering the time-dependent Schrödinger equation. When a particle is under an EM field, the Hamiltonian operator includes the term (P - q/c A)², where P is the momentum operator, q is the charge, c is the speed of light, and A is the vector potential. Since P and A do not commute, expanding the term yields P² - (q/c) (P ⋅ A + A ⋅ P) + (q/c)² A².
The non-commutation of P and A is due to the fact that the momentum operator involves a derivative with respect to position, and since A is a function of position, the order of operations matters, which is characteristic of non-commuting operators.
Using the expanded Hamiltonian in the Schrödinger equation, and working through the algebra, one finds that the probability current j is indeed given by the expression ψ*(Ĩ/m) ∂ψ - (q/mc) A|ψ|², reflecting the contribution from both the wave function and the EM field interaction.