# 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.

# Using the Feynman path integral

Exploring Quantum Physics – Week 2, Lecture 4

• 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.

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

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.

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

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

• Photo electric effect: beam of light hitting charged plates with a potential across them creates current.
• Expected: amount of current is independent of frequency of the light, dependent on the intensity.
• Actual: current depends on frequency of light (no current below frequency threshold), independent of intensity.
• Einstein proposed photons (nobel prize)
• Electron diffraction: (Davisson-Germer experiment)
• Crystalline structure acts as a diffraction grating, electron beam demonstrates wave interference like light.
• Experiment supports DeBroglie hypothesis:
$\lambda = h / p$, $E = \frac{hc}{\lambda} = \hbar \omega$
• For a derivation of the DeBroglie wave length, see
http://www.chip-architect.com/physics/deBroglie.pdf

## 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.

## 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:
$\displaystyle \delta x \cdot \delta p \geq \hbar$

# Infinite series

http://www.physics.miami.edu/~nearing/mathmethods/series.pdf

## Series worth knowing

Geometric Series
$\sum\limits_0^\infty x^n = 1 + x + x^2 + x^3 + \ldots = \frac{1}{1-x}$
Consider: $(1 + x + x^2 + \ldots + x^N)(1-x) = 1 - x^{N+1}$
so $\sum\limits_0^N x^N = \frac{1-x^{N+1}}{1-x}$

Power Series

• $e^x = 1 + x + \frac{x^2}{2!} + \ldots = \sum\limits_0^\infty \frac{x^k}{k!}$
• $\sin x = x - \frac{x^3}{3!} + \ldots = \sum\limits_0^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$
• $\cos x = 1 - \frac{x^2}{2!} + \ldots = \sum\limits_0^\infty (-1)^k \frac{x^{2k}}{(2k)!}$
• $\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots = \sum\limits_1^\infty (-1)^{k+1} \frac{x^k}{k}$ for $|x| < 1$
• $(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)x^2}{2!} + \ldots = \sum\limits_0^\infty \frac{\alpha(\alpha-1) \ldots (\alpha - k + 1)}{k!} x^k$ for $|x| < 1$ (Binomial series)

Taylor Series
$\displaystyle \begin{array}{rl} f(x) &= f(0) + xf'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \ldots \\ &= f(t_0) + (t - t_0)f'(t_0) + \frac{(t - t_0)^2}{2!} f''(t_0) + \ldots \end{array}$

## Convergence Tests

Comparison test
Let $u_k, v_k$ be sequences of positive reals and $\exists k_0 < \infty$ s.t. $\forall_{k < k_0} u_k < v_k$ and $v_k$ converges, then $u_k$ converges

Ratio test
If for large $k$, $\frac{u_{k+1}}{u_k} \leq x \le 1$, then $\sum u_k$ converges. Derived from comparison test with geometric series.

Integral test
If $f(x)$ is decreasing positive and $\int_0^\infty f(x) dx$ converges, $\sum\limits_0^\infty f(x)$ converges

## Stirling’s approximation

$\displaystyle n! = \Gamma (n + 1) = \int_0^\infty t^n e^{-t} dt = \int_0^\infty e^{-t + n\ln t}$

Using newton’s method to approximate this to the 2nd order:

$\displaystyle \Gamma (n + 1) \sim n^n e^{-n} \int_{-\infty}^\infty e^{-(t-n)^2 / 2n} dt = n^n e^{-n} \sqrt{2\pi n}$ (for moderately large n)

The last step in the above makes use of the Gaussian integral:

$\displaystyle \int_{-\infty}^\infty e^{- \alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$

using parametric differentiation:

$\displaystyle \int_{-\infty}^\infty x^2 e^{-\alpha x^2} dx = -\frac{d}{d\alpha} \sqrt{\frac{\pi}{\alpha}} = \frac{1}{2} \left( \frac{\sqrt{\pi}}{\alpha^{3/2}} \right)$