# 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

## 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

• 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.

## 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.