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Is there a physical intuition for diamagnetic inequality?

+ 4 like - 0 dislike
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Diamagnetic inequality implies, quantum mechanically, for a charged particle without intrinsic magnetic moment(or to say ignoring spin-magnetic field interaction) in some potential \(V(x)\), when subjected to some arbitrary external magnetic field \(A(x)\) (vector potential), the ground state energy(more generally the infimum of all energy expectation values) is always bigger than the one without magnetic field$^1$. 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 \(V(x)\), 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 $\frac{1}{2}\hbar\omega$.

Crossposted from my own post at stackexchange: Is there a physical intuition for diamagnetic inequality?


1.Stablity of Matter in Quantum Mechanics, E.Lieb and R. Seiringer, Cambridge University Press. Lemma 4.1, and parts of Theorem 3.3, discussing when ground state wavefunction can be chosen to be real and positive.

asked Apr 4, 2014 in Theoretical Physics by Jia Yiyang (2,465 points) [ no revision ]

2 Answers

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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
answered Mar 12, 2014 by Qmechanic (2,790 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-04-04 05:13 (UCT), posted by SE-user Jia Yiyang
+ 3 like - 0 dislike

Consider the path integral in imaginary time. Here, the wavefunction at time t is the sum over all paths of a weight. The weight coming from the kinetic energy term is the (real-valued) probability of a random walk taking this path, it's a product of Gaussians when you discretize. The potential energy at position x is an additional positive or negative decay rate in time for each path whenever it is going through x.

If you consider the ground state wavefunction, you weight the starting points with the weight equal to the ground state, and you reproduce the ground state at time t, rescaled by exp(-Et) where E is the ground state energy. So the rate of shrinkage is the ground state energy. The ground state is real and positive without a magnetic field.

Now when you turn on a magnetic field, you add a term which is a phase (it is complex in imaginary time too). It adds a complex phase on each path equal to the line-integral of the vector potential along the path. The only possible effect is to reduce the value of the integral at time T, since adding together paths with phase is always less than adding together paths without phases. That means all states decay faster in the presence of magnetic field, and that means the ground state energy went up.

answered Apr 5, 2014 by Ron Maimon (7,535 points) [ no revision ]

Thanks for the reply, it is good to have you answering my question again. I need a few clarifications: If I undestand you correctly, you are talking about the path integral representation of \(\langle x|e^{-tH}|0\rangle= \langle x|0\rangle e^{-tE}\), where \(|0\rangle\) is the ground state. (1)You claim "adding together paths with phase is always less than adding together paths without phases.", I don't understand why this is true, as a naive counterexample, if we have two positive real numbers \(a\) and \(b\), \(a+b\) is not necessarily bigger than \(|ae^{i\alpha}+be^{i\beta}|\)(Edit: This is silly, it is necessarily bigger).(2)When magnetic field is turned on, the wavefunction of ground state would change, so we must consider a path integral with a different choice of initial weight, does it jeopardize some of your arguments?

Ah I see what you did now, you basically argued \(\langle\psi|e^{-tH_B}|\psi\rangle\leq \langle\psi|e^{-tH}|\psi\rangle \) for all  \(\psi\) with positive wavefunction, where the subscript B means the presence of a magnetic field. Because it holds for all t, in first order we must have\(\langle\psi|-tH_B|\psi\rangle\leq \langle\psi|-tH|\psi\rangle \), i.e. \(\langle\psi|H_B|\psi\rangle\geq \langle\psi|H|\psi\rangle \), which is just diagmagnetic inequality, which of course indicates a increase in ground state energy! In essence you proved diagmagnetic inequality using path integral. This is cool and definitely adds something to my intuitition. +1.

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