# What is the meaning of the diamagnetic current in linear response theory?

When we consider the response of a quantum lattice model with Hamiltonian $$H=H_{kin}+H_{int}$$ to an applied vector potential $$\mathbf{A}(\mathbf{r},t)$$ we obtain the current operator $$\mathbf{j}(\mathbf{r})$$ by expanding the kinetic energy $$H_{kin}(\mathbf{A})$$ to second order in $$\mathbf{A}$$ and taking the derivative, $$j_x=-\frac{\delta H_{kin}(\mathbf{A})}{\delta A_x}$$, see for example PRL 68 2830, 1992.

Since we have expanded $$H_{kin}$$ to the second order, this gives us two terms for the current operator, $$j_x=j_x^p+j_x^d A_x$$, the paramagnetic and the diamagnetic current, respectively.

What is the physical meaning of the diamagnetic current $$\mathbf{j}^d$$? What does it describe and why/when is it important? What does the fact that it arises from the second order in $$\mathbf{A}$$ tell us?

Two extra questions: In quantum lattice systems this procedure to find the current operator seems to be somewhat questionable (PRL 66 365, 1991), why? Why does the diamagnetic current play an important role in superconductors (Phys. Rev. 117 648, 1960), does it have something to do with the fact that a superconductor is a perfect diamagnet?

• Wow, zero answers despite a sizable bounty, I didn't expect that the diamagnetic current is so mysterious. Meanwhile, the only useful information I found was in Chapter 10.2 of "Introduction to many-body physics" by P. Coleman. I recommend it as a starter for an answer to my question. – Fitzgerald Creen Jun 25 '19 at 6:42
• Is doesn't seem meaningful (at least naively to me) to separately think about the current operator like that unless you are planning to work in a specific gauge. Do you have a gauge in mind? – KF Gauss Jul 4 '19 at 7:49
• For example, for describing a photon plane wave, I've seen the paramagnetic term be dropped by the choice of $\nabla\cdot \mathbf{A}=0$ – KF Gauss Jul 4 '19 at 7:54
• "It doesn't seem meaningful to separately think about the current operator like that ...". This is precisely the reason for my question. Since the two contributions arise at separate orders in $\mathbf{A}$, it isn't me who makes a distinction between them. Why do they arise separately? – Fitzgerald Creen Jul 4 '19 at 15:48

For the kinds of systems that you're considering, the current needs to satisfy: 1) gauge invariance, and 2) a continuity equation (in this case, consider the Heisenberg equation of motion for $$\psi(x)^{\dagger}\psi(x)$$; specifically, I recommend that you calculate $$i[H,\psi(x)^{\dagger}\psi(x)]$$ with gauge-invariant Hamiltonian $$H$$ to see how both terms in the current come about).
To understand the physical consequence of the calculation (which you just did, right?), consider Ampere's law in terms of $$A$$ and $$j$$ (fix $$\nabla \cdot A=0$$). Since the "bare" gauge-invariant current operator has a term that is linear in $$A$$, you expect that $$A$$ will exponentially decrease in the system. In other words, one might expect that external field cannot penetrate deeply into the system (see https://en.wikipedia.org/wiki/London_penetration_depth).
However, for the above phenomenon to occur, the charge carriers should respond to the field in a way that cancels the field except for a layer with a small thickness. This response depends on the material, the temperature, etc. When you write down the expectation of the gauge-invariant current in the presence of the external field (this requires a little many-body theory, as usual), you will find that there are two terms: one coming from $$j^{d}$$, and one coming from the transverse part of the current-current response function of $$j^{p}$$. The magnitude of that last part is the "normal fluid density" $$\rho_{n}$$. If the equilibrium state of the charge carriers is superfluid (such as occurs in a superconductor), this is very small. For static, low-energy external fields, one finds that the average gauge-invariant current is proportional to $$A$$ with a prefactor that depends on $$\rho_{n} = \rho - \rho_{s}$$. Then, going back to Ampere's law, one finds that a large (small) $$\rho_{s}$$ corresponds to a small (large) penetration depth. This is the Meissner effect.