Quick question:
I want to find an expression for the (electric) current density of an electron, in quantum mechanics. Either a single electron or a general charge distribution ρ.
Classically j=ρ v.
What should I use here?
Maybe the electric charge multiplied by the probability current?
Thanks.
Answer
Yes, →ȷ(x,y,z) should be defined as e times the Schrodinger probability current. →ȷ=eℏ2mi(Ψ∗∂Ψ∂x−(∂Ψ∗∂x)Ψ),e<0.
One may be puzzled because the expression for →ȷ above isn't an operator – it is quadratic in the wave function. But in quantum field theory, it is an operator – an observable – because it is a function of the field operators Ψ.
If we consider non-relativistic quantum mechanics with fixed coordinates of particles and we still want to define →ȷ(x,y,z) as a linear operator, an observable, we must appreciate that this operator is only nonzero is the particle is located in the infinitesimal vicinity of the point (x,y,z). So we have ρ(x0,y0,z0)=eδ(3)(ˆ→r−→r0)
If there are N charged particles, the operators ˆ→r and ˆ→p acquire an extra index from 1 to N and ρ(x0,y0,z0) and →ȷ(x0,y0,z0) must be written as a sum of the expressions over this index.
One may verify that e.g. for wave packets, the integrals over →r0 (some regions) give us what we would expect.
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