## Macroscopic quantum model

Many of the properties of superconductors can be derived from the relatively simple Macroscopic Quantum Model that was introduced by Orlando and Delin.1 The model considers a solution to the Schrödinger equation for a charged particle in electric and magnetic fields. We calculate the probability current density of this wave function and assume that all superconducting electrons are in the same quantum state. Then the supercurrent density is the probability current density times the density of charged particles in the superconducting state times the charge of the particle. From the form of the supercurrent density we can derive the London equations. The first London equation implies that a current can flow in a superconductor without dissipation. The London equations can be used to show that superconductors must exhibit the Meissner effect. Flux quantization can also be derived from the expression for the supercurrent density.

### Probability current

$$$\ -i\hbar\frac{\partial\psi}{\partial t}=\frac{1}{2m}(-i\hbar\nabla-q\vec{A})^2\psi+V\psi.$$$

Here $\vec{A}$ is the vector potential and $V$ is the electrostatic potential. Writing out the $(-i\hbar\nabla-q\vec{A})^2\psi$ term yields,

$$$\ -i\hbar\frac{\partial\psi}{\partial t}=\frac{1}{2m}(-\hbar^2\nabla^2+i\hbar q \vec{A}\cdot\nabla+i\hbar q\nabla \cdot\vec{A}+q^2A^2)\psi+V\psi.$$$

If the wave function is written in polar form,

$$$\psi = |\psi |e^{i\theta },$$$ $$$\nabla \psi = \nabla |\psi |e^{i\theta }+i\nabla \theta |\psi |e^{i\theta },$$$ $$$\nabla ^2\psi = \nabla ^2|\psi |e^{i\theta }+2i\nabla \theta \nabla |\psi |e^{i\theta }+i\nabla ^2\theta |\psi |e^{i\theta }-(\nabla \theta )^2|\psi |e^{i\theta},$$$

the Schrödinger equation becomes,

$$$i\hbar\frac{\partial |\psi |}{\partial t}-\hbar|\psi |\frac{\partial \theta}{\partial t}=\frac{1}{2m}\left[-\hbar^2\left( \nabla ^2|\psi |+2i\nabla \theta \nabla |\psi |+i\nabla ^2\theta |\psi |-(\nabla \theta )^2|\psi | \right) +i\hbar q \vec{A}\cdot\nabla \left(\nabla |\psi |+i\nabla \theta |\psi | \right)+i\hbar q\nabla\cdot\vec{A}|\psi |+q^2A^2|\psi |\right]+V|\psi |.$$$

The real terms have to be equal to each other and the imaginary terms have to be equal to each other so this can be separated into two equations. The real terms yield,

$$$-\hbar |\psi |\frac{\partial \theta}{\partial t}=-\frac{\hbar ^2}{2m}\left(\nabla ^2-\left(\nabla \theta-\frac{q}{\hbar }\vec A \right)^2 \right) |\psi | +V|\psi |,$$$

and the imaginary terms yield,

$$$\hbar\frac{\partial |\psi |}{\partial t} = \frac{1}{2m}\left[-\hbar ^2\left(2\nabla \theta \nabla |\psi |+i\nabla ^2\theta |\psi |-\left( \nabla \theta \right) ^2 |\psi |\right) +2\hbar q\vec{A}\cdot\nabla |\psi |+\hbar q|\psi |\nabla \cdot\vec{A}\right].$$$

By multiplying the imaginary part by $|\psi |$ and rearranging, the equation can be put in the form,

$$$\frac{\partial }{\partial t}|\psi |^2+\nabla \cdot \left[ \frac{\hbar}{m}|\psi |^2\left(\nabla \theta -\frac{q}{\hbar}\vec{A} \right) \right] = 0,$$$

This is a continuity equation for the probability density $P=|\psi|^2$,

$$$\frac{\partial P}{\partial t}+\nabla \cdot \vec{S} = 0,$$$

where the probability current is given by,

$$$\vec{S} = \frac{\hbar}{m}|\psi |^2\left(\nabla \theta -\frac{q}{\hbar }\vec{A}\right).$$$

This result holds for all charged particles in a magnetic field. In superconductivity, the particles are Cooper pairs $q=-2e$, $m=2m_e$, $|\psi |^2=n_{cp}$. Here $n_{cp}$ is the density of Cooper pairs, $-e$ is the charge of an electron and $m_e$ is the mass of an electron. All superconducting electrons are in the same state so the supercurrent density $\vec{j}_s$ is proportional to the probability current density.

$$$\vec{j}_s = -2e\vec{S}.$$$

Substituting $q=-2e$ and $m=2m_e$ into the expression for $\vec{S}$ yield,

$$$\boxed{\vec{j}_s = -\frac{en_{cp}\hbar }{m_e}\left(\nabla \theta +\frac{2e}{\hbar }\vec A\right)}.$$$

It is possible to perform a gauge transformation, $\vec{A}' = \vec{A} +\nabla \gamma$, $V' = V -\frac{\partial \gamma}{\partial t}$, where $\gamma$ is any scalar function. Since $\theta$ is a scalar function, we can add $-\hbar\nabla\theta /2e$ to the vector potential so that the supercurrent density is proportional to the vector potential. This is known as London gauge,

$$$\vec{j}_s = -\frac{2e^2n_{cp} }{m_e}\vec{A}' = -\frac{e^2n_{s}}{m_e}\vec{A}'$$$

Here $n_s=2n_{cp}$ is the density of superconducting electrons. Recall that the electric field is the minus gradient of the electrostatic potential minus the time derivative of the vector potential, $\vec{E}=-\nabla V'-\frac{\partial \vec{A}'}{\partial t}$. If the density of superconducting electrons $n_s$ is constant and $\nabla V' =0$, the time derivative of the supercurrent density is,

$$$\frac{d\vec j_s}{dt}=-\frac{e^2n_{s}}{m_e}\frac{d\vec A}{dt}=\frac{e^2n_{s}}{m_e}\vec E$$$

This is known as the first London equation.

$$$\boxed{\frac{d\vec j_s}{dt}=\frac{e^2n_{s}}{m_e}\vec E}$$$

The first London equation implies that the superconducting electrons move with no dc electrical resistivity. Consider how electrons would move in an electric field if there were no scattering. Newton's law for this case is,

$$$\vec{F} = -e\vec E= m\vec{a} = m\frac{d\vec v}{dt}=-\frac{m}{n_se}\frac{d\vec j_s}{dt}$$$ $$$\frac{d\vec j_s}{dt}=\frac{e^2n_{s}}{m_e}\vec E$$$

For the derivation of the second London equation we take $\nabla \times \vec{j}$

$$$\nabla \times \vec j_s=-\frac{e^2n_{s}}{m_e}\nabla \times \vec A$$$

Using $\nabla \times \vec A=\vec B$ we obtain the second London equation:

$$$\boxed{\nabla \times \vec j_s=-\frac{e^2n_{s}}{m_e}\vec B}$$$

Combine second London equation with Ampere's law

$$$\nabla \times \vec B = \mu _0\vec j_s$$$ $$$\nabla \times \nabla \times \vec B = -\frac{e^2n_{s}\mu _0}{m_e}\vec B$$$ $$$\nabla \times \nabla \times \vec B = \nabla \left( \nabla \cdot \vec B\right) -\nabla ^2\vec B$$$

Helmholtz equation:

$$$\lambda ^2\nabla ^2 \vec B = \vec B$$$

With the London penetration depth:

$$$\lambda = \sqrt{\frac{e^2n_s\mu _0}{m_e}}$$$

The solution to the Helmholtz equation reads:

$$$\vec B = \vec B_0\textrm{exp}\left(-\frac{x}{\lambda }\right)\hat z$$$ $$$\nabla \times \vec B = \mu _0\vec j_s \Rightarrow \vec j_s = \frac{\vec B_0}{\mu _0\lambda }\textrm{exp}\left(-\frac{x}{\lambda }\right)\hat y$$$

Typical values for $\lambda$ are:

Al: $\lambda = 50$ nm
In: $\lambda = 65$ nm
Sn: $\lambda = 50$ nm
Pb: $\lambda = 40$ nm
Nb: $\lambda = 85$ nm

### Flux quantization

The current density of superconducting electrons is:

$$$\vec j_s = -\frac{en_{cp}\hbar }{m_e}\left(\nabla \theta -\frac{q}{\hbar }\vec A\right)$$$

For a ring much thicker than the penetration depth, $\vec j_s$ is $0$ along a circular path inside the ring.

$$$0 = \left(\nabla \theta -\frac{q}{\hbar }\vec A\right)$$$

For an integration along the path we receive (using Stokes theorem):

$$$\oint \nabla \theta \cdot d\vec l = -\frac{2e}{\hbar }\oint \vec A\cdot d\vec l = -\frac{2e}{\hbar }\int_S \nabla \times \vec A \cdot d\vec s = -\frac{2e}{\hbar }\int_S\vec B \cdot d\vec s = -\frac{2e}{\hbar }\Phi$$$

Where $\Phi$ is the magentic flux.

The flux quantisation can be achieved when quantizing the number of turns $n$:

$$$\oint \nabla \theta \cdot d\vec l = 2\pi n = -\frac{2e}{\hbar }\Phi$$$ $$$n = ...-2,-1,0,1,2,...$$$ $$$2\pi n = \frac{2e}{\hbar }\Phi = \frac{\Phi }{\Phi _0}$$$ $$$\boxed{\Phi = n\Phi _0}$$$

With the superconducting flux quantum

$$$\Phi _0 = \frac{h}{2e}=2.0679\cdot10^{-15}\textrm{ [W=Tm^2]}$$$
1. T. P. Orlando and Ka. A. Delin, Foundations of Applied Superconductivity, Addison-Wesley 1991.