# Crystal magnetic response only skin deep?

The Hamiltonian for a single electron in a magnetic field reads $$H=\left(\frac{{\bf p}^{2}}{2m_{e}}+q_{e}\phi\right)+\mu_{B}\left({\bf \hat{L}}+g{\bf \hat{S}}\right)\cdot{\bf B}+\frac{e^{2}}{8m_{e}}\left({\bf B}\times{\bf r}\right)^{2}$$ where symmetric gauge ${\bf A}=\frac{1}{2}{\bf B \times r}$ is used and where the last two term are responsible for paramagnetism and diamagnetism respectively.

The many-body Hamiltonian, weak electron-electron interaction considered, reads $${\cal H} = \sum_{\sigma}\int d{\bf r}\,\left[-\left(\psi_{\sigma}^{\dagger}({\bf r})\frac{\hbar^{2}}{2m^{*}}{\nabla}^{2}\psi_{\sigma}({\bf r})\right)+\boldsymbol{\mu}({\bf r})\cdot({\nabla}\times{\bf A})+\hat{n}_{e}({\bf r})\frac{e^{2}}{2m^*}{\bf A}({\bf r})^{2}\right]$$ where $\boldsymbol\mu({\bf r})=\sum_\sigma \mu_B \psi^\dagger({\bf L}+g{\bf S})\psi$ is local magnetic moment density of electron of total electron angular momentum and define $\mathbf M_\text{para} \equiv \langle \boldsymbol \mu\rangle$

Now that we vary $\bf A$, we should obtain the response current $${\bf j}={\nabla}\times{\mathbf M_\text{para}}+\frac{n_{e}e^{2}}{m^*}{\bf A}$$

It is rather strange to me that (i) the last term is essentially the same as the one in London's equation, which gives a penetration depth $\lambda$, and that (ii) the first term is non-zero only on the crystal surface if $M$ is uniform deep inside, which I think is just the magnetization current we saw in textbooks.

If (i) is true, one may conclude that magnetic response for all crystals (neglect interactions) happens in a thin layer near to the surface (a few nanometer) since magnetic field can only penetrate that far. The length scale that electron density drops to zero is a few angstroms, so we are able to neglect the first term in an appropriate region.

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Are you sure that $\bf{A}$ is penetrating here only skin-depth? In the London Equation this was true, but there the first term in the current density was negligible. Possibly is the analogy flawed. –  Lupercus Sep 16 '12 at 11:52
I am not saying so, but the equation tells. $\nabla\times\nabla\times A = \nabla \times B = j = A/\lambda^2$ and notice $\nabla \cdot A = 0$ holds for my choice of gauge. –  ChenChao Sep 16 '12 at 13:35
That is correct, you gave the derivation of the Meissner effect. But my point was that your $j$ contains the also the cross product terms, and what their effect could be has still to be worked out. After all each current density will contain a London-type term, yet superconductivity is seldom. I am afraid we deal in this case with a similar situation (i.e. no skin effect). –  Lupercus Sep 16 '12 at 14:22
I see. You have a good point. A superconductor demonstrates "complete" diamagnetism, no paramagnetism terms at all. I'm not sure whether my formalism about para- part is correct. I will work it out. –  ChenChao Sep 16 '12 at 15:15
I modified my expressions a little bit. And I think the paramagnetic surface current live in a much thinner layer and hence we can neglect that in deeper region. It is analogous to a Type I superconductor. –  ChenChao Sep 17 '12 at 8:19

The substitution you made, replacing $B \times r$ with $A$, is not valid for doing variations with respect to A. The electron operators "r" in $B \times r$ are not the same type of thing as the classical function $A(r)$--- this is a formal error. You can't replace $B \times r$ (the classical function $\times$ an electron position operator) with A (the classical function), because then you are ignoring the fact that r is an operator that acts on electron wavefunctions. This is what comes back to bite you when you take the variation.
You should sort it out physically--- in a superconductor, making an A field produces a coherent bosonic current which cancels the A. In a normal metal, you rearrange the Fermi sea into Landau levels, which are then filled up in opposite senses of current, up to the Fermi energy, so that the net current gives a diamagnetic response. It is easiest to say the error in field theory--- the Fermi field operator which has an expected value is a neutral bilinear $\bar{\psi}\psi$ which counts the number of electrons, not the charged bilinear $\psi\psi$ as in a superconductor, which gives a classical charged field response.
The correct variation to take is using the A in the kinetic term for the many-electron Hamiltonian, replacing $p$ with $p-eA$, and then varying with respect to A. In this case, you get the current from the electron motion. The amount of this current is not proportional to A, unless the electrons are coherently superposed bosons. But the reason you got the boson answer in the Fermi case is that you substituted for the psi field as if they were a classical field, and then you do get the same answer as for a classical charged field, and you get the exponential decay.
Actually the $(\frac{1}{2} B\times r)^2$ comes from $A^2$. –  ChenChao Sep 17 '12 at 15:23