I) The diamagnetic inequality
|→∇|ψ(→r)|| ≤ |(→∇+i→A(→r))ψ(→r)|
is proven rigorously in Lieb & Loss, Analysis, sections 7.19 - 7.22, as user Willie Wong explains in his mathoverflow answer.
II) However, it seems that OP is not asking for rigor, but rather intuition. Here is a heuristic proof. As good physicists let us assume that all involved functions are smooth/differentiable.
It turns out that the absolute value |ψ(→r)| of the wave function ψ(→r) cannot be differentiable and have a zero |ψ(→r)|=0 unless the zero is of at least second order, i.e, a stationary point →∇|ψ(→r)|=→0. In that case the inequality (1) is trivially satisfied.
III) Let us therefore assume from now on that the wave function ψ(→r)≠0 does not have a zero. Then we may locally polar decompose the wave function
ψ(→r) = R(→r)eiθ(→r),R(→r) > 0,θ(→r) ∈ R.
The square of the inequality (1) becomes a triviality:
|→∇|ψ(→r)||2 = |→∇R(→r)|2 ≤ |→∇R(→r)|2+R(→r)2|→∇θ(→r)+→A(→r)|2
= |eiθ(→r){→∇R(→r)+iR(→r)(→∇θ(→r)+→A(→r))}|2 =
= |(→∇+i→A(→r))R(→r)eiθ(→r)|2 = |(→∇+i→A(→r))ψ(→r)|2.
This post imported from StackExchange Physics at 2014-04-04 05:13 (UCT), posted by SE-user Qmechanic