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<j\right> = \frac{\sum jN(j)}{N} = \sum_{j = 0}^{\infty} jP(j) 

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

Variance:

\Delta j = j - \left<j\right>

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

Standard deviation – \sigma 

\sigma ^2 = \left<j^2\right> - \left<j\right>^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<x\right> = \int_{-\infty}^{\infty} x \rho(x)\,dx 

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

\sigma ^2 = \left<x^2\right> - \left<x\right>^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<x\right> = \int_{-\infty}^{\infty} x |\Psi(x, t)|^2\,dx

Through integration by parts and discarding boundary terms:

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

Expectation value of velocity:

\left<v\right> = \frac{d\left<x\right>}{dt}

Expectation value of momentum:

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

Notation with operators:

\left<x\right> = \int \Psi^* (x) \Psi\,dx

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

Expectation value of dynamical variables:

\left<Q(x,p)\right> = \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<H\right> = 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

1

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 

Ladder Operators:

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

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<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<x<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 .

For more information see: Introduction to Quantum Mechanics (David J. Griffiths)