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


  • 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

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}

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s