I) The diamagnetic inequality

$$\tag{1} \left|\vec{\nabla}|\psi(\vec{r})|\right|
~\leq~ \left|(\vec{\nabla}+i\vec{A}(\vec{r}))\psi(\vec{r})\right| $$

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 $|\psi(\vec{r})|$ of the wave function $\psi(\vec{r})$ cannot be differentiable and have a zero $|\psi(\vec{r})|=0$ unless the zero is of at least second order, i.e, a stationary point $\vec{\nabla}|\psi(\vec{r})|=\vec{0}$. In that case the inequality (1) is trivially satisfied.

III) Let us therefore assume from now on that the wave function $\psi(\vec{r})\neq 0$ does not have a zero. Then we may locally polar decompose the wave function

$$\tag{2}\psi(\vec{r})~=~R(\vec{r})e^{i\theta(\vec{r})}, \qquad R(\vec{r})~>~0 , \qquad \theta(\vec{r})~\in~\mathbb{R}.$$

The square of the inequality (1) becomes a triviality:

$$\left|\vec{\nabla}|\psi(\vec{r})|\right|^2
~=~\left|\vec{\nabla}R(\vec{r})\right|^2
~\leq~ \left|\vec{\nabla}R(\vec{r})\right|^2+ R(\vec{r})^2\left|\vec{\nabla}\theta(\vec{r})+\vec{A}(\vec{r})\right|^2 $$
$$~=~ \left|e^{i\theta(\vec{r})}\left\{\vec{\nabla}R(\vec{r})+iR(\vec{r})(\vec{\nabla}\theta(\vec{r})+\vec{A}(\vec{r}))\right\}\right|^2~=~ $$
$$\tag{3} ~=~ \left|(\vec{\nabla}+i\vec{A}(\vec{r}))R(\vec{r})e^{i\theta(\vec{r})}\right|^2~=~\left|(\vec{\nabla}+i\vec{A}(\vec{r}))\psi(\vec{r})\right|^2. $$

This post imported from StackExchange Physics at 2014-04-04 05:13 (UCT), posted by SE-user Qmechanic