# Is there a physical intuition for diamagnetic inequality?

Diamagnetic inequality implies, quantum mechanically, for a charged particle without intrinsic magnetic moment(or to say ignoring spin-magnetic field interaction) in some potential $V(\vec{x})$, when subjected to some arbitrary external magnetic field $\vec{A}(\vec{x})$(vector potential), the ground state energy(more generally the infimum of all energy expectation values) is always bigger than the one without magnetic field. The statement seems simple enough to make it tempting to find a physical intuition. A classical picture just does not help, since the situation becomes trivial: the ground state is always such that the particle is sitting still at the minimum of $V(\vec{x})$, when applying a magnetic field, this is still the the ground state. Does anyone has an idea about what the intuition should be, if any?

What I want is something more physically pictorial, like "magnetic field curves the electron's trajectory and hence...". But I admit this might possibly be too much to ask.

Update: The Landau level seems to be a good exactly-solvable example, without the magnetic field, we are dealing with a free particle of which the ground state energy (again the infimum of all energy expectation values, if we wish to stay in Hilbert space)is 0, while with a uniform magnetic field turned on the ground state energy becomes $\frac{1}{2}\hbar\omega$.

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:

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

• Thanks for the reply, and +1 for putting a proof. I am aware of this kind of proof. What I really want is something more physically pictorial, like "magnetic field curves the electron's trajectory and hence...". But I admit this might possibly be too much to ask. – Jia Yiyang Mar 13 '14 at 1:19

I am not an ace in these topics but I have thought a bit about the Bohr-van Leuuwen theorem that is sort of connected to it.

I think that what is going on is that in absence of the magnetic field there is a particular length scale which is that of the confining potential (the walls of the box in Landau diamagnetism). Now, when applying the magnetic field, a new length scale enters the picture which is the Larmor radius that essentially confines the particle on a circular trajectory if looked from above say. If the Larmor radius is smaller than the length scale of the confining potential and if the particle wavelength is of the order of the Larmor radius then Heisenberg inequality will increase momentum fluctuations when compared to the original situation and the ground state kinetic energy with it.

This is super ultra hand waving so do not hesitate to shoot at sight.