# Quantum Mechanics

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Quantum Mechanics

Notes on Quantum Mechanics

1   The Wave Function

1.1   The Schrodinger Equation

Goal of Quantum Mechanics: Determine the wave function $\Psi(x, t)$ of the particle.

Schrodinger Equation:

$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V\Psi$

Given the initial conditions: $\Psi(x, 0)$; the Schrodinger equation determines $\Psi(x, t)$ for all future time.

1.2   Probability

$|\Psi(x, t)|^2\,dx$ represents the probability of finding the particle between $x$ and $x + dx$.

$N(j)$ – number of people with $j$:

$N = \sum_{j = 0}^{\infty} N(j)$

$P(j)$ – probability of getting $j$:

$P(j) = \frac{N(j)}{N}$

$\sum_{j = 0}^{\infty} P(j) = 1$

Expectation value (average):

$\left = \frac{\sum jN(j)}{N} = \sum_{j = 0}^{\infty} jP(j)$

$\left = \frac{\sum jN(j)}{N} = \sum_{j = 0}^{\infty} f(j)P(j)$

Variance:

$\Delta j = j - \left$

$\sigma ^ 2 = \left<(\Delta j)^2\right>$

Standard deviation – $\sigma$

$\sigma ^2 = \left - \left^2$

$\rho(x)\,dx$ – probability that individual is between $x$ and $x + dx$

$\rho(x)$ – probability density

$P_{ab} = \int_{a}^{b} \rho(x)\,dx$

Continuous distributions:

$\int_{-\infty}^{\infty} \rho(x)\,dx = 1$

$\left = \int_{-\infty}^{\infty} x \rho(x)\,dx$

$\left = \int_{-\infty}^{\infty} f(x) \rho(x)\,dx$

$\sigma ^2 = \left - \left^2$

1.3   Normalization

Particle must be somewhere:

$\int_{-\infty}^{\infty} |\Psi(x, t)|^2\,dx = 1$

Non-normalizable solutions cannot represent particles.

Wave function remains normalized as time goes on.

1.4 Momentum

Average of measurements performed on particles all in the state $\Psi$:

$\left = \int_{-\infty}^{\infty} x |\Psi(x, t)|^2\,dx$

Through integration by parts and discarding boundary terms:

$\frac{d\left}{dt} = - \frac{i\hbar}{m} \int \Psi^*\frac{\partial \Psi}{\partial x}\,dx$

Expectation value of velocity:

$\left = \frac{d\left}{dt}$

Expectation value of momentum:

$\left = m \frac{d\left}{dt} = -i\hbar \int \left(\Psi^* \frac{\partial \Psi}{\partial x} \right)\,dx$

Notation with operators:

$\left = \int \Psi^* (x) \Psi\,dx$

$\left = \int \Psi^* \left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right) \Psi\,dx$

Expectation value of dynamical variables:

$\left = \int \Psi^* Q\left(x,\frac{\hbar}{i}\frac{\partial}{\partial x}\right) \Psi\,dx$

1.5   The Uncertainty Principle

de Broglie formula:

$p = \frac{h}{\lambda} = \frac{ 2\pi \hbar}{\lambda}$

Heisenberg uncertainty principle:

$\sigma_x \sigma_p \geq \frac{\hbar}{2}$

2   The Time-Independent Schrodinger Equation

2.1   Stationary States

Separation of Variables:

$\Psi(x, t) = \psi(x)f(t)$

Using the Schrodinger equation:

$i \hbar \frac{1}{f} \frac{df}{dt} = - \frac{\hbar^2}{2m} \frac{1}{\psi} \frac{\partial^2 \psi}{\partial x^2} + V$

Crucial argument: Left side is a function of only $t$, right side is a function of only $x$, therefore both sides are constant denoted by $E$.

$\frac{df}{dt} = - \frac{iE}{\hbar}f$

$f(t) = e^{-iEt/\hbar}$

Time-independent Schrodinger equation:

$- \frac{\hbar^2}{2m} \frac{d^2 \psi}{d x^2} + V \psi = E \psi$

$\Psi(x,t) = \psi(x)e^{-iEt/\hbar}$

Separable Solutions

1. They are stationary states – the probability density does not depend on time.

1. They are states of definite total energy.

Hamiltonian:

$H(x, p) = \frac{p^2}{2m} + V(x)$

$\left = E$

$\sigma^2_H = 0$

3. The general solution is a linear combination of separable solutions. There is a different wave function for each allowed energy.

$\Psi_1(x, t) = \psi_1(x)e^{-iE_1t/\hbar}$

$\Psi_2(x, t) = \psi_2(x)e^{-iE_2t/\hbar}$

Time-dependent Schrodinger equation has the property that any linear combination of solutions is itself a solution.

General solution:

$\Psi(x, t) = \sum_{n = 1}^{\infty} c_n \psi_n(x) e^{-iE_nt/\hbar}$

2.2 The Infinite Square Well

$V = 0$ if $0 \leq x \leq a$, and $V = \infty$ otherwise.

Outside the well: $\psi(x) = 0$

Inside the well: $V = 0$

Simple harmonic oscillator:

$\frac{d^2\psi}{dx^2} = -k^2\psi$

$k = \frac{\sqrt{2mE}}{\hbar}$

General solution:

$\psi(x) = A\sin(kx) + B\cos(kx)$

Boundary Conditions:

$\psi(0) = \psi(a) = 0$

Hence:

$\psi(x) = A\sin(kx)$

$ka = 0, \pm \pi,, \pm 2\pi, \pm 3\pi, ....$

Distinct solutions:

$k_n = \frac{n\pi}{a}$

$E_n = \frac{\hbar^2 k^2_n}{2m} = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$

Normalize $\psi$:

$A = \sqrt{\frac{2}{a}}$

Solutions:

$\psi_n(x) = \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi}{a} x \right)$

$\psi_1$ – ground state, other waves are excited states

Important properties of $\psi_n(x)$

1. They are alternately even and odd (true when potential is an even function).

2. Each successive state has one more node (universal).

3. They are mutually orthogonal (quite general).

$\int \psi_m(x)^* \psi_n(x)\,dx = 0, m \neq n$

$\int \psi_m(x)^* \psi_n(x)\,dx = \delta_{mn}$

$\delta_{mn}$ – Kronecker delta

1. They are complete, in the sense that any other function, $f(x)$, can be expressed as a linear combination of them

$f(x) = \sum_{n = 1}^{\infty} c_n \psi_n(x) = \sqrt{\frac{2}{a}}\sum_{n = 1}^{\infty} c_n \sin\left(\frac{n \pi}{a}x\right)$

Expansion coefficients can be evaluated by Fourier’s trick:

$\int \psi_m(x)^* f(x) d = \sum_{n = 1}^{\infty} c_n \int \psi_m(x)^* \psi_n(x)\,dx = \sum_{n = 1}^{\infty} c_n \delta_{mn} = c_m$

$c_m = \int \psi_m(x)^*f(x)\,dx$

Stationary States for Infinite Square Well:

$\Psi_n(x, t) = \sqrt{\frac{2}{a}}\sin\left(\frac{n \pi}{a}x\right)e^{-i(n^2 \pi^2 \hbar/2ma^2)t}$

General Solution:

$\Psi(x, t) = \sum_{n = 1}^{\infty} c_n \sqrt{\frac{2}{a}}\sin\left(\frac{n \pi}{a}x\right)e^{-i(n^2 \pi^2 \hbar/2ma^2)t}$

$c_n = \sqrt{\frac{2}{a}} \int_{0}^{a} \sin\left(\frac{n \pi}{a} x\right) \Psi(x, 0)\,dx$

2.3 The Harmonic Oscillator

Practically any potential is approximately parabolic. Virtually any oscillatory motion is approximately simple harmonic.

$V(x) \cong \frac{1}{2}V''(x_0)(x-x_0)^2$

$V(x) = \frac{1}{2}m\omega^2x^2$

Solve:

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2x^2\psi = E\psi$

2.3.1   Algebraic Method

$\frac{1}{2m}\left[\left(\frac{\hbar}{i}\frac{d}{dx}\right)^2 + (m\omega x)^2\right] \psi = E \psi$

$a_{\pm} = \frac{1}{\sqrt{2m}}\left(\frac{\hbar}{i}\frac{d}{dx} \pm im\omega x \right)$

$a_-a_+ = \frac{1}{2m} \left[\left(\frac{\hbar}{i}\frac{d}{dx}\right)^2 + (m\omega x)^2\right] + \frac{1}{2}\hbar \omega$

$a_+a_- = \frac{1}{2m} \left[\left(\frac{\hbar}{i}\frac{d}{dx}\right)^2 + (m\omega x)^2\right] - \frac{1}{2}\hbar \omega$

$a_-a_+ - a_+a_- = \hbar \omega$

Schrodinger equation:

$(a_-a_+ - \frac{1}{2}\hbar \omega)\psi = E\psi$

$(a_+a_- + \frac{1}{2}\hbar \omega)\psi = E\psi$

Important: If $\psi$ satisfies the Schrodinger equation with energy $E$, then $a_+\psi$ satisfies the Schrodinger equation with energy $(E+\hbar \omega)$. $a_-\psi$ is a solution with energy $(E - \hbar\omega)$.

There must exist a “lowest rung”:

$a_-\psi_0 = 0$

$\frac{1}{\sqrt{2m}}\left(\frac{\hbar}{i}\frac{d\psi_0}{dx} - im\omega x \psi_0\right) = 0$

$\psi_0 = A_0e^{-\frac{m\omega}{2\hbar}x^2}$

$E_0 = \frac{1}{2}\hbar\omega$

Excited states:

$\psi_n(x) = A_n(a_+)^ne^{-\frac{m \omega}{2\hbar}x^2}$

$E_n = (n + \frac{1}{2})\hbar \omega$

2.3.2   Analytic Method

Substitution:

$\xi = \sqrt{\frac{m\omega}{\hbar}}x$

Schrodinger equation:

$\frac{d^2\psi}{d\xi^2} = (\xi^2 - K)\psi$

$K = \frac{2E}{\hbar \omega}$

Large $\xi$:

$\frac{d^2\psi}{d\xi^2} \approx \xi^2 \psi$

$\psi(\xi) \approx Ae^{-\xi^2/2} + Be^{+\xi^2/2}$

Asymptotic form:

$\psi(\xi) \rightarrow ( )e^{-\xi^2/2}$

$\psi(\xi) = h(\xi) e^{-\xi^2/2}$

$\frac{d^2h}{d\xi^2} - 2\xi\frac{dh}{d\xi} + (K=1)h = 0$

Look for a solution in the form of a power series:

$h(\xi) = a_0 + a_1\xi + a_2\xi^2 + ... = \sum_{j = 0}^{\infty} a_j\xi^j$

Recursion formula:

$a_{j+2} = \frac{(2j + 1 - K)}{(j+1)(j+2)}a_j$

The power series must terminate for the solution to be normalizable. One series must truncate, the other must be zero from the start.

$K = 2n + 1$

$E_n = (n + \frac{1}{2})\hbar\omega$

$a_{j+2} = \frac{-2(n - j)}{(j+1)(j+2)}a_j$

Stationary states:

$\psi_n(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^nn!}}H_n(\xi)e^{-\xi^2/2}$

2.4   The Free Particle

$V(x) = 0$ everywhere.

$\frac{d^2\psi}{dx^2} = -k^2\psi$

$k = \frac{\sqrt{2mE}}{\hbar}$

$\Psi(x, t) = Ae^{ik(x - \frac{\hbar k}{2m}t)} + Be^{-ik(x+\frac{\hbar k}{2m}t)}$

Special combination: $(x \pm vt)$ represents a wave of fixed profile, traveling in the $\mp x$ direction, at speed $v$.

$\Psi_k(x, t) = Ae^{i(kx - \frac{\hbar k^2}{2m}t)}$

$k > 0$: Wave traveling to the right

$k < 0$: Wave traveling to the left

$v_{quantum} = \frac{\hbar|k|}{2m} = \sqrt{\frac{E}{2m}}$

A free particle cannot exist in a stationary state; there is no such thing as a free particle with a definite energy.

Wave packet:

$\Psi(x, t) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx - \omega t)}\,dk$

$\omega = \frac{\hbar k^2}{2m}$

Plancherel’s theorem:

$\phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Psi(x, 0)e^{-ikx}\,dx$

$v_{group} = \frac{d\omega}{dk}$

$v_{phase} = \frac{\omega}{k}$

2.5   The Delta-Function Potential

Bound State: If $V(x)$ rises higher than the particle’s total energy $E$ on either side, then the particle is stuck in the potential well.

Scattering State: If $E$ exceeds $V(x)$ on one side or both, the the particle comes in from infinity and returns to infinity.

Tunneling allows the particle to leak through any finite potential barrier.

$E < V(-\infty)$ and $V(+\infty)$ – Bound State

$E > V(-\infty)$ or $V(+\infty)$ – Scattering State

Real life: Most potentials go to zero at infinity.

$E < 0$: Bound State

$E > 0$: Scattering State

Infinite square well and harmonic oscillator admit bound states only. Free particle only allows scattering states.

Dirac delta function

$\delta(x) = 0, x \neq 0$

$\delta(x) = \infty, x = 0$

$\int_{-\infty}^{\infty} \delta(x)\,dx = 1$

$f(x)\delta(x-a) = f(a)\delta(x-a)$

$\int_{-\infty}^{\infty} f(x)\delta(x-a)\,dx = f(a) \int_{-\infty}^{\infty} \delta(x-a)\,dx = f(a)$

Potential:

$V(x) = -\alpha\delta(x)$

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} - \alpha\delta(x)\psi = E\psi$

2.5.1   Bound States: $E<0$

$x<0, V(x) = 0$

$\frac{d^2\psi}{dx^2} = \kappa^2\psi$

$\kappa = \frac{\sqrt{-2mE}}{\hbar}$

General Solution:

$\psi(x) = Be^{\kappa x}, x<0$

$\psi(x) = Fe^{-\kappa x}, x>0$

Boundary Conditions:

1. $\psi$ is always continuous.
2. $\frac{d\psi}{dx}$ is continuous except at points where the potential is infinite.

$F = B$

Integrate the Schrodinger equation from $-\epsilon$ to $\epsilon$ and take the limit as $\epsilon \rightarrow \infty$:

$\Delta \left(\frac{d\psi}{dx}\right) = \frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0}\int_{-\epsilon}^{\epsilon} V(x)\psi(x)\,dx = -\frac{2m\alpha}{\hbar^2}\psi(0) = -2B\kappa$

$\kappa = \frac{m\alpha}{\hbar^2}$

$B = \frac{\sqrt{m\alpha}}{\hbar}$

One bound state:

$\psi(x) = \frac{\sqrt{m\alpha}}{\hbar}e^{-m\alpha|x|\hbar^2}$

$E = -\frac{m\alpha^2}{2\hbar^2}$

2.5.2   Scattering States: $E>0$

$\frac{d^2\psi}{dx^2} = -k^2\psi$

$k = \frac{\sqrt{2mE}}{\hbar}$

$\psi(x) = Ae^{ikx} + Be^{-ikx}, x<0$

$\psi(x) = Fe^{ikx} + Ge^{-ikx}, x>0$

In a typical scattering experiment particles are fired in from one direction. In that case the amplitude of the wave coming in from the right will be zero.

$G = 0$

$A$ – amplitude of the incident wave

$B$ – amplitude of the transmitted wave

$\beta = \frac{m\alpha}{\hbar^2k}$

$B = \frac{i\beta}{1 - i\beta}A$

$F = \frac{1}{1 - i\beta}A$

Reflection Coefficient

$R = \frac{|B|^2}{|A|^2} = \frac{\beta^2}{1 + \beta^2} = \frac{1}{1 + (2\hbar^2E/m\alpha^2)}$

Transmission Coefficient

$T = \frac{|F|^2}{|A|^2} = \frac{1}{1 + \beta^2} = \frac{1}{1 + (m\alpha^2/\hbar^2E)}$

$R + T = 1$

2.6   Finite Square Well

$V(x) = -V_0$ for $-a < x < a$
$V(x) = 0$ for $|x| > a$

2.6.1   Bound States: $E<0$

$x<-a$, $V(x) = 0$:

$\frac{d^2\psi}{dx^2} = \kappa^2\psi$

$\kappa = \frac{\sqrt{-2mE}}{\hbar}$

$\psi(x) = Be^{\kappa x}$

$-a, $V(x) = V_0$:

$\frac{d^2\psi}{dx^2} = l^2\psi$

$l = \frac{\sqrt{2m(E+V_0}}{\hbar}$

$\psi(x) = C\sin(lx) + D\cos(lx)$

$x > a$, $V(x) = 0$:

$\psi(x) = Fe^{-\kappa x}$

1. Wide, deep well

$E_n + V_0 \cong \frac{n^2\pi^2\hbar^2}{2m(2a)^2}$

2. Shallow, narrow well: There is always one bound state, no matter how weak the well becomes.

2.6.2   Scattering States: $E>0$

$x<-a$:

$\psi(x) = Ae^{ikx} + Be^{-ikx}$

$k = \frac{\sqrt{2mE}}{\hbar}$

$-a:

$\psi(x) = C\sin(lx) + D\cos(lx)$

$l = \frac{\sqrt{2m(E+V_0}}{\hbar}$

$x>a$ (assuming there is no incoming wave in this region):

$\psi(x) = Fe^{ikx}$

$B = i\frac{\sin(2la)}{2kl}(l^2 - k^2)F$

$F = \frac{e^{-2ika}A}{\cos(2la)-i\frac{\sin(2la)}{2kl}(k^2+l^2)}$

$T^{-1} = 1 + \frac{v^2_0}{4E(E+V_0}\sin^2\left(\frac{2a}{\hbar}\sqrt{2m(E+V_0}\right)$

Energies for perfect transmission:

$E_n + V_0 = \frac{n^2\pi^2\hbar^2}{2m(2a)^2}$

2.7   The Scattering Matrix

Arbitrary Localized Potentials

Region 1:

$\psi(x) = Fe^{ikx} + Ge^{-ikx}$

Region 2:

$\psi(x) = Cf(x) + Dg(x)$

Region 3:

$\psi(x) = Fe^{ikx} + Ge^{-ikx}$

$B = S_{11}A + S_{12}G$

$F = S_{21}A + S_{22}G$

Scattering Matrix

$S = \left(\begin{tabular}{cc} S11 & S12 \\ S21 & S22 \end{tabular}\right)$

$\left(\begin{tabular}{c} B \\ F \end{tabular}\right) = S \left(\begin{tabular}{c} A \\ G \end{tabular}\right)$

Scattering from the left:

$R_l = |S_{11}|^2$

$T_l = |S_{21}|^2$

Scattering from the right:

$R_r = |S_{22}|^2$

$T_r = |S_{12}|^2$

If you want to locate the bound states, put in $k \rightarrow i\kappa$.