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.

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