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}

Quantum Wells

Exploring Quantum Physics – Week 3, Lecture 5

Types of problems for the Schrödinger equation

Single-particle quantum mechanics, i\hbar \dot{\Psi} = \left[ \frac{\hat{p}^2}{2m} + V(r) \right] \Psi

Different potential functions present different types of problems

Different potential functions present different types of problems

Quantization in a guitar string and quantum well

  • Wavelength “quantization” in a guitar string: String anchored at end nodes, must have multiples of a half-wave between them
    \displaystyle \lambda_n = \frac{2L}{n},\; n=1,2,\ldots
    \displaystyle u_n(x) = A_n \sin \left( \frac{2\pi x}{\lambda_n} \right)
  • Wavelength quantization in an infinite quantum well: potential “walls” anchor the wave function similar to guitar string
    \displaystyle \lambda_n = \frac{2L}{n},\; n=1,2,\ldots
    \displaystyle \psi(x) = A_n \sin \left( \frac{2\pi x}{\lambda_n} \right)

Formulation of the finite potential well problem

\displaystyle U(x) = \left\{ \begin{array}{ll}    U_0, &|x| > a/2 \\ 0, &|x| < a/2 \end{array} \right.

Region II is the well, with finite barriers (I and III) on either side

Region II is the well, with finite barriers (I and III) on either side

  • For |x| > a/2,\quad -\frac{\hbar^2}{2m}\psi''(x) + U_0\psi(x) = E\psi(x)
  • For |x| < a/2,\quad -\frac{\hbar^2}{2m}\psi''(x) = E\psi(x)
  • Find the bound state(s) (with 0 < E < U_0) satisfying continuity constraints (problem definition):
    \displaystyle \psi(\pm a/2 + 0) = \psi(\pm a/2 - 0) \;\mathrm{and}\; \psi'(\pm a/2 + 0) = \psi'(\pm a/2 - 0), \;\mathrm{and}\; \psi(x \rightarrow \pm \infty) \rightarrow 0
    Where +0 means to the right of a point, and -0 means to the left

Using the symmetry

  • If the Hamiltonian commutes with an operator, \hat{A}, then solutions to the Schrödinger equation can be chosen to have definite a and E, \psi_{aE}(x)
    \displaystyle [ \hat{H}, \hat{A} ] = \hat{H} \hat{A} -  \hat{A} \hat{H} = 0
  • Potential is then inversion symmetric, \hat{I}U(x) = U(-x) \equiv U(x). Eigenvalues of \hat{I} are p = \pm 1
  • General solution of \left( \frac{d^2}{dx^2} + k^2 \right) \psi(x) = 0:
    \displaystyle \psi(x) = C_1 e^{ikx} + C_2 e^{-ikx}
  • Can choose solutions with definite parity, i.e.
    \displaystyle \psi_+(x) = C \cos(kx), \quad \psi_-(x) = \tilde(C) \sin(kx)

Using the constraints at infinities

  • General solution of \left( \frac{d^2}{dx^2} - \gamma^2 \right) \psi(x) = 0:
    \displaystyle \psi(x) = A e^{-\gamma x} + B e^{+ \gamma x}
  • E.g. for x > a/2 we must request that \psi \rightarrow 0 as x \rightarrow + \infty. Otherwise probability would not be finite (does not make sense).
  • So drop B term and the solution is
    \displaystyle \psi(x) = A e^{-\gamma x}, \quad x > a/2

Using matching conditions at transition points

Making the self-consistency equation dimensionless

  • The non-linear self-consistency equation is not solvable analytically:
    \displaystyle \tan \left( \frac{ka}{2} \right) = \frac{\gamma}{k} \equiv \sqrt{\frac{U_0}{E} - 1}
    with k = \sqrt{2mE/\hbar^2}
  • Start by introducing dimensionless parameters, x = ka/2 and \xi^2 = \frac{m U_0 a^2}{2\hbar^2}
    \displaystyle \tan x = \sqrt{(\xi / x)^2 - 1}
  • 2 limiting cases:
    • Deep well: \frac{m U_0 a^2}{\hbar^2} >> 1
    • Shallow well: \frac{m U_0 a^2}{\hbar^2} << 1

Solving
\displaystyle \tan x = \sqrt{(\xi / x)^2 - 1}

Solutions to the self-consistency equation.

Solutions to the self-consistency equation.

Bound status in quantum potential wells

Fourier transforms

  • Can solve for shallow potential well, \sim U(x) = - \alpha \delta(x) using Fourier transform:
    \displaystyle F[g(x)] = \tilde{g}(k) = \int_{-\infty}^{\infty} g(x) e^{ikx} dx, \qquad     g(x) = \int_{-\infty}^{\infty} \tilde{g}(k) e^{-ikx} \frac{dk}{2\pi}
  • Specifically for Dirac \delta-function:
    \displaystyle F[\delta(x)] = \int_{-\infty}^\infty \delta(x) e^{ikx} dx = 1, \qquad     \delta(x) = \int_{-\infty}^\infty e^{-ikx} \frac{dk}{2\pi}
  • Use these identities to solve stationary Schrödinger equation for \psi(x) (with \Psi(x,t) = \psi(x) e ^{-\frac{i}{\hbar} Et})

Shallow level in the delta-well

\displaystyle \left[ - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} - \alpha \delta(x) \right] \psi(x) = E \psi(x)

\displaystyle \psi(x) = \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{\psi}_k e^{-ikx}
\displaystyle \psi''(x) = \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{\psi}_k (-ik)^2 e^{-ikx}
\displaystyle \delta(x)\psi(x) = \psi(0) \delta(x) = \psi(0) \int_{-\infty}^\infty \frac{dk}{2\pi} e^{-ikx}
\displaystyle \Rightarrow \left[ - \frac{\hbar^2}{2m} \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{\psi}_k (-ik)^2 e^{-ikx} - \alpha \psi(0) \int_{-\infty}^\infty \frac{dk}{2\pi} e^{-ikx} \right] = E \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{\psi}_k e^{-ikx}

\displaystyle \left[ \frac{\hbar^2 k^2}{2m} \tilde{\psi}_k - \alpha \psi(0) \right] = E \tilde{\psi}_k
\displaystyle \left( \frac{\hbar^2 k^2}{2m} - E \right) \psi_k = \alpha \psi(0)
\displaystyle \Rightarrow \tilde{\psi}_k = \frac{\alpha \psi(0)}{\frac{\hbar^2 k^2}{2m} - E}

What is E?
\displaystyle \psi(0) = \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{\psi}_k

Take the integral of \tilde{\psi}_k:
\displaystyle \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{\psi}_k = \psi(0) = \alpha \psi(0) \int_{-\infty}^\infty \frac{dk}{2\pi} \frac{1}{\frac{\hbar^2 k^2}{2m} - E}


\displaystyle 1 = \frac{\alpha}{\pi} \int_0^\infty \frac{dk}{\frac{\hbar^2 k^2}{2m} + |E|}

Final result:
\displaystyle E = - \frac{m \alpha^2}{2 \hbar^2}

Increasing to D dimensions

\displaystyle \left[ - \frac{\hbar^2}{2m} \nabla^2 - \alpha \delta(\vec{r}) \right] \psi(\vec{r}) = E \psi(\vec{r})
\displaystyle \Rightarrow \frac{\hbar^2 \vec{k}^2}{2m} \tilde{\psi}_{\vec{k}} - \alpha \psi(0) = E \tilde{\psi}_{\vec{k}}
\displaystyle 1 = \alpha \int \frac{d^D k}{(2\pi)^D} \frac{1}{\frac{\hbar^2 k^2}{2m} + |E|}

In 2 dimensions: d^2 k = dk\; k\;  d\phi_{\vec{k}}, where \phi_k is the angle of the k-vector
\displaystyle 1 = \alpha \int \frac{d^2 k}{(2 \pi)^2} \ldots = \frac{\alpha}{(2 \pi)^2} \int_0^\infty dk\; k \ldots \int_0^{2\pi} d\phi_k

In D dimensions:
\displaystyle 1 = \frac{\alpha S_{D-1}}{(2\pi)^D} \int \frac{dk\; k^{D-1}}{\frac{\hbar^2 k^2}{2m} + |E|}
where S_0 = 2,\; S_1 = 2\pi,\; S_2 = 4\pi from the integral over the angle \phi

Critical dimension (=2)
\displaystyle \frac{1}{\alpha} = \frac{S_{D-1}}{(2\pi)^D} \int_0^{1/a} dk \frac{k^{D-1}}{\frac{\hbar^2 k^2}{2m} + |E|}

  • Main question: is there a bound state in a weak potential (\alpha \rightarrow 0)?
  • Can we make \int_0^{1/a}  \frac{k^{D-1}}{\frac{\hbar^2 k^2}{2m} + |E|} arbitrarily large by choosing E ?
    • Equivalent to asking whether \int_0^{1/a} dk\; k^{D-3} diverges
    • For D=2, \int \frac{dk}{k} = \ln \frac{1}{0} = \infty
  • If D \leq 2, it diverges (there is a bound state), otherwise it is finite (no bound state).

Very shallow level in 2D quantum well
\displaystyle \frac{1}{\alpha} = \frac{1}{2\pi} \int_0^{1/a} dk \frac{k}{\frac{\hbar^2 k^2}{2m} + |E|} = \frac{m}{\pi \hbar^2} \int_0^{1/a} \frac{dk\; k}{k^2 + \left( \frac{\sqrt{2m |E|}}{\hbar} \right)^2}
\approx \frac{m}{\pi \hbar^2} \int_{\frac{\sqrt{2m|E|}}{\hbar}}^{1/a} \frac{dk}{k}    = \frac{m}{\pi \hbar^2} \ln \left( \frac{1}{a} \frac{\hbar}{\sqrt{2m|E|}} \right)

Result:
\displaystyle |E| \sim \frac{\hbar^2}{ma^2} \exp \left[ - \frac{2\pi\hbar^2}{\alpha} \right]
As \alpha \rightarrow 0, this result has no Taylor expansion.

Using the Feynman path integral

Exploring Quantum Physics – Week 2, Lecture 4

Laplace’s Method (saddle-point approximation)

  • Gaussian integral: \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}
  • There is no closed analytical expression for \int e^{-f(x)}dx. But if there is a small parameter in the exponential, \int e^{\frac{1}{\epsilon}f(x)}dx,\; \epsilon \rightarrow 0, one can simplify things.
  • The max of e^{-\frac{1}{\epsilon} f(x)} occurs at the min of f(x)
  • As \epsilon \rightarrow 0, \; e^{\frac{1}{\epsilon} f(x)} approaches Gaussian
  • E.x: \displaystyle  I = \int_{-\infty}^\infty e^{\frac{1}{\epsilon}[x^2 + 1/x^2]}dx \approx e^{\frac{-1}{\epsilon} f_{min}} \sqrt{\frac{2\pi\epsilon}{|f''(x_0)|}} = \sqrt{-\frac{2}{\epsilon}}\frac{\sqrt{\pi\epsilon}}{2}
    where f_min is the minimum value of the function (stationary point), so there is no first derivative, and the 2nd order term of the Taylor expansion becomes the sqrt in the above

The principle of least action

  • Quantum effects hinge on interference phenomena. Classical limit implies suppressing them, which happens if
    \displaystyle \lambda = \frac{2\pi\hbar}{p} \rightarrow 0 or, formally, \hbar \rightarrow 0
    Can think of this as increasing the scale, or shrinking the Planck scale.
  • Feynman path integral in the classical limit, \int \mathscr{D}  \vec{r}(t)e^{iS/\hbar \rightarrow 0}, is “collected” from the trajectories for which the action is minimal.
    • Consequence of destructive interference when \delta S \neq 0, since S_t \;\&\; S_{t+\epsilon} will be out of phase.
  • I.e. When \hbar = 0 the Feynman path integral reduces to the principle of stationary action.

How to find the “special” trajectory

  • To determine a point x_0 where a function f(x) is at a minimum:
    \displaystyle f(x_0 + \delta x) = f(x_0) + \left[ f'(x_0)\delta x\right] + \frac{1}{2} f''(x_0) \delta x^2 + \ldots
    where f'(x_0) = 0, so drop first-order term
  • To determine a trajectory x_{cl}(t) where functional-action S(x) is minimal:
    \displaystyle S[x_{cl}(t) + \delta x(t)] = S[x_{cl}(t)] + \delta S + \ldots
    and \delta S = 0
  • So set the first variation of the action to 0: \delta S = 0

Recovering Newton’s 2nd law

  • Action: S[\vec{r}(t)] = \int_0^t \left[ \frac{m\vec{v}^2}{2} - V(\vec{r}) \right] dt
  • Calculate first variation:
    \displaystyle \begin{array}{ll}  S[\vec{r}_cl(t) + \delta \vec{r}(t)]   &= \int_0^t \left\{ \frac{m}{2} \left[ \frac{d}{dt}(\vec{r}_cl + \delta \vec{r}) \right]^2 - V(\vec{r}_cl + \delta \vec{r}) \right\} dt \\  &= \int_0^t \left\{ \frac{m}{2} \vec{r}_cl^2 - m \dot{\vec{r}}_cl \delta \dot{\vec{r}} - V(\vec{r}_cl) - \frac{\partial V}{\partial \vec{r}} \cdot \delta \vec{r} \right\} dt \\  &= S_cl - \int_0^t \left\{ m \ddot{\vec{r}}_cl + \frac{\partial V}{\partial \vec{r}} \right\} \partial \vec{r}\; dt  \end{array}
    The integrand in the last equation is \delta S, which is 0 (by the principle of least action), therefore m \ddot{\vec{r}}_cl = - \frac{\partial V}{\partial \vec{r}}, m\vec{a} = \vec{F}

Conductivity of a metal – 1900 theory

  • Simplistic Drude model:
    \displaystyle m\vec{a} = q\vec{E} + \vec{F}_{fr}
  • With “friction” force
    \displaystyle \vec{F}_{fr} = - \gamma \vec{v} = - \frac{\vec{p}}{\tau}
  • In equilibrium, \vec{a} = 0
    \displaystyle \frac{d\vec{p}}{dt} = q\vec{E} - \frac{\vec{p}}{\tau} = 0
    \displaystyle \vec{J} = qn\vec{v} = \frac{nq^2\tau}{m} \vec{E}
    where \tau is the expected time between collisions w/ imperfections

Path-integral view of electron model

  • Probability to propagate \vec{r}_i \rightarrow \vec{r}_f
    \displaystyle \omega_{i\rightarrow f} = \left| \sum_l e^{ \frac{i}{\hbar} S_l } \right|^2 = e^{ \frac{i}{\hbar} S_1 } e^{ \frac{i}{\hbar} S_2 } \ldots + \mathrm{c.c.} \equiv 2\cos\left( \frac{S_1 + S_2 + \ldots}{\hbar} \right)
  • Typical action, S_l \sim \int \frac{m \vec{v}^2}{2}dt \sim pL
  • Quantum (interference) terms
    \displaystyle \omega_{i\rightarrow f}^{quant.} \propto \sum_{l_1 \neq l_2} e^{ \frac{i}{\hbar} p_F(L_1 - L_2)} + \mathrm{c.c.} = 2 \sum_{l_1 \neq l_2} \cos \left( \frac{p_F \Delta L}{ \hbar} \right)
  • A “loop” in the path (a.k.a. “weak localization”) tries to bring electron back to where it came from
    • Loop can be traversed in either direction, giving 2 paths of equal length, so the quantum interference is 0

Diffusion Equation

  • If many particles experience a random walk, their (average) density satisfies the diffusion equation:
    \displaystyle \frac{\partial \rho}{\partial t} = D\nabla^2 \rho
    where \rho is a function \rho( \vec{r}, t)
  • Density spread from a point (e.g. the origin)
    \displaystyle \rho( \vec{r}, t) = \frac{1}{(2\pi D t)^{d/2}} \exp \left[ - \frac{ \vec{r}^2}{4Dt} \right]
    where d is the dimensionality of the space

Probability of self-crossing trajectory

  • Probability (density) to return to a close vicinity of the origin in a time t
    \displaystyle \rho( \vec{r}, t) = \frac{1}{(2\pi D t)^{d/2}} \exp \left[ - \frac{ \vec{r}^2}{4Dt} \right] \;\overrightarrow{r \rightarrow 0}\; \rho( \vec{r}, t) = \frac{1}{(2\pi D t)^{d/2}}
  • Remember probability of electron to diffuse from \vec{R}_i \rightarrow \vec{R}f:
    \displaystyle \omega_{i\rightarrow f}^{quant.} = 2 \sum_{l_1 \neq l_2} \cos \left( \frac{p_F \Delta L}{ \hbar} \right)

\displaystyle \Rightarrow \mathscr{P}_{total} \propto \int_{t_{min} \sim \tau}^{t_{max} \rightarrow \infty} \frac{dt}{t^{d/2}} = \int_\tau^\infty \frac{dt}{t^{d/2}} = \left\{  \begin{array}{ll} \mathrm{finite} & d=3 \\ \infty & d=1,2 \end{array} \right.

Conclusion:
At “high” temperatures, the phase is disrupted and the particle behaves as described by the Drude model, but at low temperatures the quantum effects matter.

Feynman Path Integral

Exploring Quantum Physics – Week 2, Lecture 3

Developed in Feynman’s 1948 paper: Space-Time Approach to Non-Relativistic Quantum Mechanics

Summary

For a particle traveling from a coordinate \vec{R}_i \longrightarrow \vec{R}_f
The probability of going to \vec{R}_f is the sum of the probability (or weight) of taking a particular path, over all possible paths:
\displaystyle \vec{R}_f = \omega_{i\shortrightarrow f} = \left| \sum_l e^{i \frac{S_l}{\hbar}} \right|^2
\displaystyle \mathrm{where}\; S_l = \mathrm{action\; of\; trajectory\;} l = \int_0^t [K - V] \;dt

Trajectories

  • Consider a particle localized at \vec{R}_i at time t =0, that is | \Psi(0) \rangle = | \vec{R}_i \rangle
  • Evolve according to Schrödinger:
    \displaystyle i \hbar \partial_t | \Psi \rangle = \hat{H} | \Psi \rangle = \left[ \frac{\hat{\vec{p}}^2}{2m} + V(\hat{\vec{R}}) \right] |\Psi\rangle
  • What “part” will propagate to \vec{R}_f?
    i.e. what is the overlap with \Psi, \langle \vec{R}_f | \Psi{t} \rangle = ??
  • How will it get there? (In Q.M. it follows all possible trajectories)

Evolution Operator

\displaystyle  i \hbar | \Psi(t) \rangle = \hat{H} | \Psi(t) \rangle  \\  | \Psi(0) \rangle = | \Psi_0 \rangle = | \vec{R}_i \rangle  \\  | \Psi(t) \rangle = \hat{U}(t) | \Psi_0 \rangle
Where \hat{U}(t) is the evolution operator

Evolution operator must preserve the vector norm (according to the Born interpretation), so it looks like a rotation.

Evolution operator must preserve the vector norm (according to the Born interpretation), so it looks like a rotation.

Plug into Schrödinger:
\displaystyle i\hbar \partial_t \hat{U}(t) = \hat{H}\hat{U}(t), \qquad \hat{U}(0) = \hat{1}
i.e. \hat{U} does not depend on the initial condition of | \Psi_0 \rangle
\displaystyle \hat{U}(t) = e^{-\frac{i}{\hbar}\hat{H}t}
Verify:
\displaystyle i \hbar \partial_t \left(e^{-\frac{i}{\hbar}\hat{H}t}\right) = \hat{H} \left(e^{-\frac{i}{\hbar}\hat{H}t} \right) = \hat{H}\hat{U}(t)

\Rightarrow | \Psi(t) \rangle = e^{-\frac{i}{\hbar}\hat{H}t} | \vec{R}_i \rangle

Asside: define the exponential of an operator, \hat{X} with it’s Taylor series:
e^{\hat{X}} = 1 + \hat{X} + \frac{\hat{X}^2}{2!} + \frac{\hat{X}^3}{3!} + \ldots

Propagator

a.k.a. the transition amplitude

\displaystyle \langle \vec{R}_f | \Psi(t) \rangle = \langle \vec{R}_f | e^{-\frac{i}{\hbar}\hat{H}t} | \vec{R}_i \rangle = G(\vec{R}_i, \vec{R}_f; t) = \mathrm{"Green function"}

Exponential of an operator is messy (infinite sum, see above) so simplify using:

  • \int | x \rangle \langle x | \; dx = \hat{1}

  • \begin{array}{ll}     | \Psi(t) \rangle &= \hat{U}(t) | \Psi(0) \rangle \\                       &= \hat{U}(t - t_1) | \Psi(t_1) \rangle \\                       &= \hat{U}(t - t_1) \hat{U}(t_1) | \Psi(0) \rangle     \end{array}

Combining the above:
\displaystyle \langle x_f | \hat{U}(t) | x_i \rangle = \langle x_f | \hat{U}(t/2) \hat{U}(t/2) | x_i \rangle = \int \langle x_f | \hat{U}(t/2) | x \rangle \langle x | \hat{U}(t/2) | x_i \rangle \; dx

Consider all possible intermediate points on the path from x_i to x_f

Consider all possible intermediate points on the path from x_i to x_f

Reapplying this N times gives:
\displaystyle \hat{U}(t) = \hat{U}(t/N)^N = \hat{U}(t/N) \times \hat{U}(t/N) \times \ldots \times \hat{U}(t/N)

Let N \rightarrow \infty, \quad \Delta t = t/N becomes small, and using a first order approximation:
\displaystyle e^{-\frac{i}{\hbar N} \hat{H}t} \approx 1 - \frac{i}{\hbar} \hat{H} \Delta t
\displaystyle \mathrm{i.e.}\; e^{-\frac{i}{\hbar}\hat{H}t} = \left( 1 - \frac{i}{\hbar} \hat{H} \frac{t}{N} \right)^N, \quad N \rightarrow \infty

Propagator becomes a product:
\displaystyle \langle x_f | \hat{U}(t) | x_i \rangle = \int dx_1 \ldots x_{N-1} \prod_{k=0}^N \langle x_{k+1} | 1 - \frac{i}{\hbar N} \hat{H} t | x_k \rangle

Asside: difference between | \Psi \rangle, \; \Psi(x)

| \Psi \rangle – Represents an “abstract” state of the system, coordinate-independent
\Psi(x) – Represents | \Psi \rangle in a coordinate system. \Psi(x) = \langle x | \Psi \rangle

Asside: the Dirac delta-function

  • \delta_{ij} = \vec{e}_i \cdot \vec{e}_j = \left\{  \begin{array}{ll}0 & i \neq j \\ 1 & i = j \end{array} \right.
  • Extension to a “continuum” basis \{ | x \rangle \}
    \displaystyle \langle x | x' \rangle = \delta(x - x') = \left\{  \begin{array}{ll} 0 & x \neq x' \\ C & x=x' \end{array} \right.
    where C normalizes: \int_{-\infty}^{\infty} dx \; \delta(x - x') = 1
    \Rightarrow C = \infty
  • Fourier transform:
    \displaystyle \delta(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} e^{ikx}

Breaking down the propagator equation

  • \displaystyle \langle x | P \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{i}{\hbar}px}
  • \displaystyle \langle x_{k+1} | x_k \rangle = \delta(x_{k+1} - x_k)
  • \displaystyle \langle x_{k+1} | V(\hat{x}) | x_k \rangle = V(x_k) \delta (x_{k+1} - x_k)
    because | x_k \rangle is an eigen state of the coordinate system
  • \displaystyle \langle x_{k+1} | \frac{\hat{p}^2}{2m} | x_k \rangle = \int dp_k \langle x_{k+1} | \frac{\hat{p}^2}{2m} | p_k \rangle \langle p_k | x_k \rangle = \int \frac{1}{2\pi\hbar} \frac{p_k^2}{2m} e^{\frac{ip_k}{\hbar}(x_{k+1} - x_k)}

Putting it all together

Illustration of 2 possible trajectories.

Illustration of 2 possible trajectories.

\displaystyle \begin{array}{lll}  \langle x_{k+1} | 1 - \frac{i}{\hbar} \hat{H} \Delta t | x_k \rangle &\propto  \int dp_k\; e^{\frac{i}{\hbar}p_k \Delta x_k} \left[ 1 - \frac{i}{\hbar} \left( \frac{p_k^2}{2m} + V(x_k) \right) \Delta t \right] & \\   &= \int dp_k\; e^{\frac{i}{\hbar}p_k \Delta x_k} \exp \left[ -\frac{i}{\hbar} \left( \frac{p_k^2}{2m} + V(x_k) \right) \Delta t \right] & (\mathrm{using}\; e^\epsilon = 1 + \epsilon) \\  &= \exp \left[ \frac{i}{\hbar} \left( \frac{m}{2} \frac{\Delta x_k^2}{\Delta t^2} - V(x_k) \right) \Delta t \right] & (\mathrm{recognizing\; Gaussian\; integral}) \\  &= e^{\frac{i}{\hbar} \mathscr{L}_k \Delta t} \end{array}

where \mathscr{L} is the Lagrangian

\displaystyle \begin{array}{ll}  \Rightarrow \langle x_f | \hat{U}(t) | x_i \rangle  &= \int dx_1\;dx_2 \ldots\; e^{\frac{i}{\hbar} \mathscr{L}_1 \Delta t} e^{\frac{i}{\hbar} \mathscr{L}_2 \Delta t} \ldots \\  &= \int e^{\frac{i}{\hbar} \sum_k \mathscr{L}_k \Delta t} = \int e^{\frac{i}{\hbar} \int_0^t \mathscr{L} dt} = e^{\frac{i}{\hbar}S} \\  &\equiv \int \mathscr{D}x(t) e^{\frac{i}{\hbar} \int_0^t \mathscr{L} dt}  \end{array}

A physical interpretation of Quantum theory

Exploring Quantum Physics – Week 1 Lecture 2

Born Interpretation of the Schrödinger equation

Born Rule: |\Psi(x, y, z; t)|^2 dx dy dz is the probability of finding the quantum particle (described by \Psi(\vec{r}, t)) in the volume dV = dx dy dz at time t.

Continuity Equation

\displaystyle \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{\jmath} = 0    \qquad    \rho(\vec{r}, t) = \left| \Psi(\vec{r}, t) \right|^2
Integrate over volume:
\displaystyle \frac{\partial \rho}{\partial t} + \int_V d^3r \nabla \cdot \vec{jmath} = 0
Recognizing Gauss’ theorem in the integral, we get:
\displaystyle \frac{\partial \rho}{\partial t} + \oint_{\partial V} \vec{\jmath} \cdot d \vec{s}= 0

From Born’s Rule, calculate the probability flux:
\displaystyle \dot{P}_v = \frac{\partial}{\partial t} \int_V \Psi^*\Psi d^3r = \int_V \left[ \dot{\Psi}^*\Psi + \Psi^*\dot{\Psi} \right] d^3r
Remember |\Psi|^2 = \Psi^*\Psi

Apply to Schrödinger:
\displaystyle \begin{array}{ll}  i \hbar \dot{\Psi} = \hat{H}\Psi \Rightarrow & \dot{\Psi} = -\frac{i}{\hbar} \hat{H}\Psi \\                                               & \dot{\Psi}^* = \frac{i}{\hbar} \hat{H}\Psi^*  \end{array}
\displaystyle \Rightarrow \dot{P}_V = \int_V \left[ \frac{i}{\hbar} \left( \hat{H}\Psi^* \right) \Psi - \frac{i}{\hbar} \Psi^* \left( \hat{H} \Psi \right) \right] d^3r
Expanding the Hamiltonian and canceling the potential gives:
Remember K.E. is an operator: \hat{k} = \frac{\hat{\vec{p}}^2}{2m} = -\frac{\hbar^2}{2m} \nabla^2

\displaystyle \begin{array}{ll}  \dot{P}_V &= -\frac{i\hbar}{2m} \int_V \left[ \left( \nabla^2 \Psi^* \right) \Psi - \Psi^* \left( \nabla^2 \Psi \right) \right] d^3r \\ \\            &= -\frac{i\hbar}{2m} \int_V \nabla \cdot \left[ \left( \nabla \Psi^* \right) \Psi - \left( \nabla \Psi \right) \Psi^* \right] d^3r \\ \\            &= - \oint \vec{\jmath} \cdot d\vec{s}  \end{array}

\vec{\jmath} is the probability current \rho \vec{v}
\displaystyle \Rightarrow \vec{\jmath} = \frac{1}{2} \left( \Psi^* \frac{\hat{\vec{p}}}{m}\Psi - \Psi \frac{\hat{\vec{p}}}{m}\Psi^* \right)

This connects the change in the probability of finding a particle in V with the flux of the probability current flowing through the surface.

Quantum operators

Classical properties (momentum, location, etc.) become operators acting on the wave function

  • momentum: \hat{\vec{p}} = -i\hbar \nabla
  • location: \hat{\vec{r}} = \vec{r} \times (multiplication operator)
  • kinetic energy: \hat{K} = -\frac{\hbar^2 \nabla^2}{2m}

E.g. Angular momentum \vec{L} = \vec{r} \times \vec{p} becomes \hat{\vec{L}} = \vec{r} \times \hat{\vec{p}} = \vec{r} \times - i \hbar \nabla

How do you measure an operator?

Consider measuring particle position…
x becomes a probability function F(x)

Notation: \langle x \rangle = \mathbb{E}(x) = \int dx\; xF(x)

\displaystyle \begin{array}{ll}  \Rightarrow \langle \vec{r} \rangle &= \int dV \; \vec{r} | \Psi(\vec{r}) |^2 \\                                      &= \int dV \; \Psi^*(\vec{r})\vec{r} \Psi(\vec{r})  \end{array}

This generalizes to:
For a property X with operator \hat{X}:
\displaystyle \langle X \rangle = \int dV \; \Psi^*(\vec{r}) \hat{X} \Psi(\vec{r})

Time-independent schrödinger

Recall the time-dependent Schrödinger equation:
\displaystyle i \hbar \partial_t \Psi(\vec{r}, t) = \hat{H}\Psi(\vec{r}, t) = \left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\vec{r}) \right] \Psi(\vec{r}, t)

Often, \hat{H} is time independent, so we can separate the variables:
\displaystyle \Psi(\vec{r}, t) = \Psi(\vec{r})e^{-\frac{i}{\hbar}Et}
\displaystyle \Rightarrow i \hbar \partial_t \left( \Psi(\vec{r})e^{\frac{-iEt}{\hbar}} \right) = E \Psi(\vec{r})e^{\frac{-iEt}{\hbar}} = (\hat{H} \Psi(\vec{r})) e^{\frac{-iEt}{\hbar}}
Canceling the exponential gives us the time-independent Schrödinger equation:
\displaystyle E \Psi(\vec{r}) = (\hat{H} \Psi(\vec{r}))
Notice that this looks like an eigenvalue equation, where E is the eigen value and \Psi is the eigen vector.

Hermitian operators

For any operator, \hat{A}, acting in a space of functions \Psi(\vec{r}), one can define its Hermitian-adjoint operator \hat{A}^\dagger with:
\displaystyle \int \Psi^*(\vec{r}) \left[ \hat{A} \Psi(\vec{r}) \right] d^3r = \int \left[ \hat{A}^\dagger \Psi(\vec{r}) \right]^* \Psi(\vec{r})  d^3r
And \hat{A} is “Hermitian” if \hat{A}^\dagger = \hat{A}

Eigenvalue properties: \hat{A} \Psi_a(\vec{r}) = a \Psi_a(\vec{r})

  • eigenvalues (a) are real if \hat{A} is Hermitian
  • eigenvectors, \Psi_a(\vec{r}), form a basis

Physical Significance

  • Physical observables in quantum mechanics are described by Hermitian (a.k.a. self-adjoint, though there are subtle differences we will ignore) operators, \hat{A}^\dagger = \hat{A}
  • Eigenvalues of a physical operator determine possible values of the observable that actually can be measured in an experiment.
    \displaystyle \hat{A}\Psi_a(\vec{r}) = a \Psi_a(\vec{r})
  • Eigenvectors form a basis in the sense that a wave-function can be expressed as their linear combination.
    \displaystyle \Psi(\vec{r}) = \sum_a c_a \Psi_a(\vec{r})

Super position principle in Q.M.

  • If \Psi_1(\vec{r}, t) and \Psi_2(\vec{r}, t) are solutions to the Schrödinger equation,
    \displaystyle i\hbar \frac{\partial \Psi(\vec{r}, t)}{\partial t} = \left[ - \frac{\hbar^2 \nabla^2}{2m} + V(\vec{r}) \right] \Psi(\vec{r}, t),
    Then \Psi(\vec{r}, t) = c_1 \Psi_1(\vec{r}, t) + c_2 \Psi_2(\vec{r}, t) is also a solution.
  • This motivates the notion of a Hilbert space – a linear vector space where quantum states live.
  • The wave function, \Psi(\vec{r}, t), is a specific representation of a quantum state (much like coordinates of a vector).
  • Dirac notation for “vectors” of quantum states: \langle \Psi | and | \Psi \rangle
  • For a basis {|q\rangle}:
    \displaystyle \sum_q |q\rangle \langle q| = 1 \quad \mathrm{or} \quad \int_q |q\rangle \langle q| = 1

How to choose a basis?

  • Physical observables in quantum mechanics are associated with linear Hermitian operators.
  • For a generic operator, \hat{A}, the eigenvalue problem \hat{A} |a\rangle = a|a\rangle defines eigenvectors that form a basis in the Hilbert space.
  • \Psi(a) = \langle a | \Psi \rangle is the wave-function in the a-representation.
  • Standard choices are:
    • coordinate representation: \Psi(x) = \langle x | \Psi \rangle
    • momentum representation: \Psi(p) = \langle p | \Psi \rangle

Wave functions and Schrödinger equation

Exploring Quantum Physics – Week 1 Lecture 1

Early Experiments

Wave Equation

Wave equation:
\displaystyle \begin{array}{ll}  u(x, t) &= A_0 \sin(kx - \omega t + \phi_0) \\          &= A_0 \Im e^{i(kx - \omega t + \phi_0)}  \end{array}
Where k is the wave number

\displaystyle \lambda = \frac{2 \pi}{k} \qquad \omega = c k
Where c is the wave velocity

We can represent particles as a wave (but not the other way around) via the Fourier transform:

\displaystyle f(x) = \int dk A_k e^{ikx}
Where A_k is the “fourier harmonics” (amplitudes)

Represent a particle as a Gaussian wave with a sharp peak:

Particle as a Gaussian wave.

Particle as a Gaussian wave.

The Schrödinger wave equation

\displaystyle \Psi(x,t) = C e^{i(kx - \omega t)}
Where \Psi is the wave function, and C is the constant of proportionality, ignore for now.

Since p = \hbar k and E = \hbar \omega, can write as:
\displaystyle \Psi(x,t) = C e^{\frac{i}{\hbar}(px - Et)}

Taking the derivative:
\displaystyle \frac{\partial}{\partial t} \Psi = \frac{-i}{\hbar} E \Psi \qquad  \frac{\partial}{\partial x} \Psi = \frac{-i}{\hbar} p \Psi
(or for a momentum vector: \nabla \Psi = \frac{i}{\hbar} \vec{p} \Psi)

For a free particle (not in a potential field), E = \mathrm{k.e.} = \frac{mv^2}{2} = \frac{p^2}{2m}, so:
\displaystyle E \Psi = i \hbar \frac{\partial}{\partial t} \Psi = \frac{\vec{p}^2}{2m} \Psi = \frac{(-i \hbar \nabla)^2}{2m} \Psi = \frac{\hbar^2 \nabla^2}{2m} \Psi

Giving the free schrödinger equation:

\displaystyle \left[ i \hbar \frac{\partial}{\partial t} + \frac{\hbar^2 \nabla^2}{2m} \right] \Psi(\vec{r}, t) = 0

\displaystyle \hat{H} = \mathrm{Hamiltonian} = \mathrm{K.E.} + \mathrm{Potential} = \frac{\vec{p}^2}{2m} + V(\vec{r})

Using the Hamiltonian for energy gives the fundamental equation of quantum physics:

\displaystyle i \hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi

Particle delocalization

Take a Gaussian wave-packet \Psi(x, 0) = A e^{-x^2 / 2d^2}
Solve schrödinger with this gives
\displaystyle \left| \Psi(x,t) \right|^2 \propto \exp \left[ -\frac{x^2}{d^2 (1+t^2/\tau^2)} \right]
where \tau = md^2 / \hbar = delocalization time

\displaystyle \Psi(x, 0) = Ae^{-\frac{x^2}{2d^2}} = \int \frac{dp}{h} \phi_p e^{\frac{i}{\hbar} px}

Recognize the Gaussian integral: \int_{-\infty}^\infty e^{-\alpha x^2 + \beta x} dx = \sqrt{\frac{\pi}{\alpha}} e^{\frac{\beta^2}{4\alpha}}

Gives wave-function in momentum space:
\phi_p = \int dx \Psi(x, 0)e^{-\frac{i}{\hbar}px} \propto e^{\frac{p^2d^2}{2\hbar^2}}

Notice the uncertainty \delta x \sim d, but \delta p \sim \hbar / d. This is a manifestation of the Heisenberg uncertainty principle:
\displaystyle \delta x \cdot \delta p \geq \hbar