Quantum Harmonic Oscillator

Exploring Quantum Physics – Week 4, Lecture 7

Classical Harmonic Oscillator

  • Hook’s force: F = -kx = -\frac{d}{dx} \left( \frac{kx^2}{2} \right) = -\frac{d}{dx} V(x)
  • Newton’s 2nd law: m \frac{d^2x}{dt^2} = F = -kx
    • Rewrite as: \ddot{x} + \omega^2 x = 0, where \omega = \sqrt{k}{m}
  • Solving for x: x = x_0 \sin(\omega t + \phi_0)

Phase portrait of a harmonic oscillator

  • Choose initial condition s.t. \phi_0 = 0, x(t) = x_0 \sin(\omega t)
  • Velocity: v(t) = \frac{dx}{dt} = x_0 \omega \cos(\omega t)
  • Energy: E = \frac{mv^2}{2} + \frac{1}{2} m \omega^2 x^2 = \frac{mv^2}{2} + \frac{1}{2} k x^2
    • Expanding x and v:
      \displaystyle E = \frac{m}{2} x_0^2 \omega^2 \cos^2 \omega t + \frac{m}{2} x_0^2 \omega^2 \sin^2 \omega t = \frac{m}{2} x_0^2 \omega^2 \equiv \mathrm{const}

Going quantum

  • In most situations we can approximate the lowest energy potential as a quadratic, since V'(x_0) = 0
    \displaystyle V(x) \approx V(x_0) + 0 + \frac{V''(x_0)}{2}(x - x_0)^2
  • Classical energy: E = \frac{p^2}{2m} + \frac{m \omega^2 x^2}{2}
  • Usual transformation:
    \displaystyle E \rightarrow \hat{H}, \quad p \rightarrow \hat{p}, \quad x \rightarrow \hat{x} = x
  • Quantum Hamiltonian
    \displaystyle \hat{H} = \frac{\hat{p}^2}{2m} + \frac{m \omega^2 \hat{x}^2}{2}

Creation and annihilation operators

\displaystyle \hat{H} = \frac{\hat{p}^2}{2m} + \frac{m \omega^2 \hat{x}^2}{2}
Making the Hamiltonian dimensionless:
\displaystyle \hat{H}/\hbar \omega = \frac{m \omega \hat{x}^2}{2 \hbar} + \frac{\hat{p}^2}{2m\omega\hbar} = \left( \sqrt{\frac{m\omega}{2\hbar}} \hat{x} \right)^2 + \left( \frac{\hat{p}}{\sqrt{2m\omega\hbar}} \right)^2
Using A^2 - B^2 = (A-B)(A+B), \quad A^2 + B^2 = (A - iB)(A + iB):
\displaystyle \hat{H}/\hbar \omega = \left[ \sqrt{\frac{m\omega}{2\hbar}} \hat{x} - \frac{i\hat{p}}{\sqrt{2m\omega\hbar}} \right] \left[ \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \frac{i\hat{p}}{\sqrt{2m\omega\hbar}} \right] - \frac{i}{2\hbar} [\hat{x}, \hat{p} ]
where [\hat{x}, \hat{p} ] = \hat{x}\hat{p} - \hat{p}\hat{x}

Let \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} - \frac{i\hat{p}}{\sqrt{2m\omega\hbar}} be the creation operator
and \hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \frac{i\hat{p}}{\sqrt{2m\omega\hbar}} be the annihilation operator.

Summary:
\displaystyle \hat{H} =  \frac{\hat{p}^2}{2m} + \frac{m \omega^2 \hat{x}^2}{2} = \hbar \omega \left( \hat{a}^\dagger \hat{a} - \frac{i}{2\hbar} [ \hat{x}, \hat{p}] \right)

Generating the energy spectrum of the quantum harmonic oscillator

Commutation relations

(\hat{x}\hat{p}) \psi(x) = x \cdot \left(-i\hbar \frac{d}{dx} \right) \psi = -i \hbar x \psi'
(\hat{p}\hat{x}) \psi(x) = -i\hbar \frac{d}{dx} (x\psi) = -i \hbar \psi - i \hbar x \psi'
\Rightarrow ([\hat{x}, \hat{p}]) \psi(x) = (\hat{x}\hat{p} - \hat{p}\hat{x}) \psi = + i \hbar \psi

[\hat{a},\hat{a}^\dagger ] = \left[ \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \frac{i\hat{p}}{\sqrt{2m\omega\hbar}}, \sqrt{\frac{m\omega}{2\hbar}} \hat{x} - \frac{i\hat{p}}{\sqrt{2m\omega\hbar}} \right]  = 2 \sqrt{\frac{m \omega}{2 \hbar}} \left( - \frac{i}{\sqrt{2 m \omega \hbar}} \right) [\hat{x}, \hat{p}] = 1
\Rightarrow \hat{a}\hat{a}^\dagger = 1 + \hat{a}^\dagger \hat{a}

\displaystyle \Rightarrow \hat{H} = \hbar \omega ( \hat{a}^\dagger \hat{a} + 1/2)

Generating the spectrum

\hat{a}^\dagger \hat{a} | n \rangle = n | n \rangle, so \hat{H} | n \rangle = \hbar \omega (n + 1/2) | n \rangle

Creation: latex \hat{a}^\dagger \hat{a} ( \hat{a}^\dagger | n \rangle) = \hat{a}^\dagger (1 + \hat{a}^\dagger \hat{a}) |n\rangle = \hat{a}^\dagger ( n + 1) | n \rangle = (n + 1)(hat{a}^\dagger | n \rangle)$
Annihilation: \hat{a}^\dagger \hat{a} ( \hat{a} | n \rangle) = (n - 1) ( \hat{a} | n \rangle )

Ground state: \hat{a} | 0 \rangle = 0, \quad \hat{a}^\dagger \hat{a} | 0 \rangle = 0
\displaystyle \Rightarrow E_0  = \frac{\hbar \omega}{2}

Harmonic oscillator wave-functions

Deriving ground-state wave-function
For the ground state, \hat{a} | 0 \rangle = 0
\displaystyle \hat{a} \psi_0(x) \equiv \left( \sqrt{\frac{m \omega}{2\hbar}} \hat{x} + \frac{i \hat{p}}{\sqrt{2 m \omega \hbar}} \right) \psi_0(x) = 0

Expanding operators:
\displaystyle \left( \sqrt{\frac{m \omega}{2\hbar}} x + \frac{\sqrt{\hbar}}{\sqrt{2 m \omega}} \frac{d}{dx} \right) \psi_0(x) = 0

Let x_0 = \sqrt{\frac{\hbar}{m \omega}}
\displaystyle \left( \frac{x}{x_0} + x_0 \frac{d}{dx} \right) \psi_0(x) = 0
\displaystyle \Rightarrow \psi_0' = - \frac{x}{x_0^2} \psi_0

Result
\displaystyle \psi_0 (x) = C e^{- \frac{x^2}{2 x_0^2}}

Where C is the normalizing coefficient:
\displaystyle \int_{-\infty}^{+\infty} \psi_0^2(x) \;dx = 1 \Rightarrow C = \left( \frac{m \omega}{\pi \hbar} \right)^{1/4}

Excited states

“Hermite polynomials”:
\displaystyle \hat{a}^\dagger | 0 \rangle = | 1 \rangle
\displaystyle \frac{\hat{a}^\dagger}{\sqrt{n + 1}} | n \rangle = | n + 1 \rangle

Hermite polynomials

Hermite polynomials

Summary

  • Hamiltonian
    \displaystyle \hat{H} =  \frac{\hat{p}^2}{2m} + \frac{m \omega^2 \hat{x}^2}{2} = \hbar \omega ( \hat{a}^\dagger \hat{a} + 1/2)
  • Energy spectrum
    \displaystyle E_n = \hbar \omega \left( n + \frac{1}{2} \right), \quad n = 0, 1, 2, \ldots
  • Relation between (normalized) eigenfunctions
    \displaystyle \hat{a}^\dagger | n \rangle = \sqrt{n + 1} | n + 1 \rangle, \qquad  \hat{a} | n \rangle = \sqrt{n - 1} | n - 1 \rangle
  • Ground-state wave function:
    \displaystyle \psi_0 (x) = \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{- \frac{x^2}{2 x_0^2}}
  • Excited-state wave functions:
    \displaystyle \psi_1 (x) = \hat{a}^\dagger \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{- \frac{x^2}{2 x_0^2}}, etc.
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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

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}