# Super Conductors

Exploring Quantum Physics – Week 3, Lecture 7

• Super conducting first observed by Heike Kamerlingh Onnes in 1913
• Bellow 4.2K, resistivity of Hg drops to zero very quickly
• Magnetic flux expulsion (a.k.a. Meissner effect) Magnetic fields “avoid” super conductor
• equivalent to the Anderson-Higgs mechanism
• High temperature super conductor (“High-Tc”), a.k.a. Cooper superconductor
• Theory of High-Tcs not yet discovered

## Particle braiding & spin-statistics theorem

• “Particle braiding” = particles switching positions
• What are the possible values of the quantum statistical phase, $\phi$?
• For bosons (integer spin), $\phi = 0$
• For fermions (half-integer spin), $\phi = \pi$

## Pauli exclusion principle for fermions

• What are the consequences of $\psi(\vec{r}_1, \vec{r}_2) = e^{i\pi} \psi(\vec{r}_2, \vec{r}_1) = -\psi(\vec{r}_2, \vec{r}_1)$?
• An important one for $\vec{r}_1 = \vec{r}_2$ is $\psi(\vec{r}_1, \vec{r}_1) = - \psi(\vec{r}_1, \vec{r}_1) = 0$
• This means the probability of finding two identical fermions in the same point is zero
• More generally: two identical fermions can not occupy the same quantum state.
• No such constraint for bosons
• Given a single particle quantum state $| i \rangle$, the possible occupation numbers for fermions are $n_i^{(f)} = 0, 1$, while for bosons $n_i^{(b)} = 0, 1, 2, \ldots, \infty$

## Ground state of many boson system: Bose-Einstein condensate (BEC)

What happens if we “pour” identical bosons into a prescribed landscape of quantum states (i.e. single-particle states, $| i \rangle$, with energies $E_i$)?

A: At low temperatures, they form the lowest-energy state (ground state), which in the absence of interactions corresponds to putting all bosons into a single lowest-energy state

Superfluidity

• Closely related to BEC
• Condensate of bosons forms a quantum liquid which has zero viscosity and flows without resistance
• Discovered experimentally in Hellium by Pyotr Kapitsa 1937
• Mathematical theory was put together by Lev Landau

## Ground state of a many-fermion system: Fermi gas

Representation of a potential well with the inner energy levels filled. The outer electrons can change energy levels, but the inner cannot due to Pauli

• Fermi Temperature = threshold momentum, below which all the low energy states are occupied (up to the Fermi-level)
• $\frac{p_F^2}{2m}$ = Fermi energy

## Two Particle Schrödinger

• Consider generic case of particles with masses $m_1$ and $m_2$ interacting via potential $V(\vec{r})$
$\displaystyle \left[ - \frac{\hbar^2 \nabla_1^2}{2m_1} - \frac{\hbar^2 \nabla_2^2}{2m_2} + V(\vec{r}_1 - \vec{r}_2) \right] \psi(\vec{r}_1, \vec{r}_2) = E \psi(\vec{r}_1, \vec{r}_2)$
• Change of variables: $(\vec{r}_1, \vec{r}_2) \rightarrow (\vec{R}, \vec{r}) = \left( \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}, \vec{r}_2 - \vec{r}_1 \right)$
• Schrödinger equation:
$\displaystyle \left[ - \frac{\hbar^2}{2} \left( \frac{1}{m_1} \frac{\partial^2}{\partial \vec{r}_1^2} + \frac{1}{m_2} \frac{\partial^2}{\partial \vec{r}_2^2} \right) + V(\vec{r}) \right] \tilde{\psi} (\vec{R}, \vec{r}) = E \tilde{\psi}(\vec{R}, \vec{r})$

Calculating $\nabla_1^2/m_1 + \nabla_2^2/m_2$

• Focus on one (x-) component of the Laplacian with
$\displaystyle X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$ and $x = x_2 - x_1$
• Change of variables in the derivatives:
$\displaystyle \frac{\partial}{\partial x_1} = \frac{\partial X}{\partial x_1} \frac{\partial}{\partial X} + \frac{\partial x}{\partial x_1} \frac{\partial}{\partial x} = \frac{m_1}{m_1 + m_2} \frac{\partial}{\partial X} - \frac{\partial}{\partial x}$
$\displaystyle \frac{\partial}{\partial x_2} = \frac{m_2}{m_1 + m_2} \frac{\partial}{\partial X} - \frac{\partial}{\partial x}$
• Change of variables in the kinetic energy (Laplacians):
$\displaystyle \frac{1}{m_1} \frac{\partial^2}{\partial x_1^2} + \frac{1}{m_2} \frac{\partial^2}{\partial x_2^2} = \frac{1}{m_1} \left[ \frac{m_1}{m_1 + m_2} \frac{\partial}{\partial X} - \frac{\partial}{\partial x} \right]^2 + \frac{1}{m_2} \left[\frac{m_2}{m_1 + m_2} \frac{\partial}{\partial X} - \frac{\partial}{\partial x} \right]^2 = \frac{1}{m_1 + m_2} \frac{\partial^2}{\partial X^2} + \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \frac{\partial^2}{\partial x^2}$

Reduced mass, $\mu = \frac{m_1 m_2}{m_1 + m_2}$

Putting it together
$\displaystyle - \frac{\hbar^2}{2m_1} \Delta_1 - \frac{\hbar^2}{2m_2} \Delta_2 = - \frac{\hbar^2}{2(m_1 + m_2)} \Delta_R - \frac{\hbar^2}{2\mu} \Delta_r$
where $Delta_R$ is the Laplacian with respect to the center of mass (R)

$\displaystyle \left[ - \frac{\hbar^2}{2} \left( \frac{\Delta_R}{m_1 + m_2} + \frac{\Delta_r}{\mu} \right) + V(\vec{r}) \right] \tilde{\psi}(\vec{R}, \vec{r}) = E \psi(\vec{R}, \vec{r})$

$\displaystyle \tilde{\psi}(\vec{R}, \vec{r}) = e^{\frac{i \vec{p} \cdot \vec{R}}{\hbar}} \psi(\vec{r})$

Left term looks like $\frac{\vec{p}^2}{2(m_1 + m_2)}$, can move it into the energy (E) on the right:
$\displaystyle \left[ - \frac{\hbar^2}{2\mu} \frac{\partial^2}{\partial \vec{r}^2} + V(\vec{r}) \right] \psi(\vec{r}) = E' \psi(\vec{r})$

## The Cooper problem

Isotope effect

• only mass differs on isotope lattice
• electrons exchange “phonons” (waves running through the lattice) which leads to weak effective phonon mediated attraction between electrons

Spherical-cow model of the phonon-mediated attraction

• Electrons interact by exchanging phonons. Energy and momentum must be conserved!
• The phonon energies available are pretty low compared to the Fermi energy. When converted to temperature, $T_F \sim 10,000K$ and $T_D \sim 400 K$
• The exact interaction is complicated, simplified model:
$\displaystyle V(\vec{p}) = \left\{ \begin{array}{ll} -V_0 & \frac{p^2}{2m} - E_F < \hbar \omega \\ 0 & \mathrm{otherwise} \end{array} \right.$
where the first condition describes electrons on the Fermi shell

Cooper pairing problem

• 2-particle Schrödinger:
$\displaystyle \left[ - \frac{\hbar^2\nabla^2}{m} - V_0 \delta(r) \right] \psi(\vec{r}) = E \psi(\vec{r})$
• Momentum space version (apply Fourier transform):
$\displaystyle \frac{\hbar^2}{m} \vec{p}^2 \psi(\vec{p}) - V_0 \int \frac{d^3p}{(2\pi)^3} \psi(\vec{p}) = E \psi(\vec{p})$
• Result (same method from Lecture 5)
$\displaystyle \nabla \sim - \hbar \omega_D \mathrm{exp} \left[ - \frac{1}{N_0 V_0} \right], \qquad N_0 = \frac{m p_F}{4 \pi^2 \hbar^2}$
Advertisements