# 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

## 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.

# 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)
$\displaystyle \nabla \sim - \hbar \omega_D \mathrm{exp} \left[ - \frac{1}{N_0 V_0} \right], \qquad N_0 = \frac{m p_F}{4 \pi^2 \hbar^2}$