Day: March 16, 2022

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202203160924 Brief on Quantum Mechanics

(This brief outline is based on the manuscript of 2016-2017 PHYS4351 Advanced Quantum Mechanics Lecture Notes.)

Section 01 of 04

Dynamics of a particle is described by wavefunction \Psi (x,t) which is a single-valued, finite, and continuous complex function.

From statistical interpretation,

|\Psi (x,t)|^2\,\mathrm{d}x

is the probability of finding the particle in between positions x and x+\mathrm{d}x at time t.

Expectation value of position of the particle is given by

\langle \mathbf{r}\rangle \equiv \displaystyle{\iiint}\mathrm{d}V\,\mathbf{r}|\Psi (\mathbf{r},t)|^2.

Deviation.

\begin{aligned} \Delta \mathbf{r} & = \mathbf{r}-\langle\mathbf{r}\rangle \\ \langle \Delta \mathbf{r}\rangle & \equiv \langle \mathbf{r}\rangle - \langle\langle \mathbf{r}\rangle\rangle = 0 \\ \langle (\Delta \mathbf{r})^2\rangle & = \langle \mathbf{r}^2 - 2\mathbf{r}\langle\mathbf{r}\rangle + \langle \mathbf{r}\rangle^2 \rangle \\ & = \langle \mathbf{r}^2\rangle - \langle \mathbf{r}\rangle^2 \\ \end{aligned}

Dynamics of a wavefunction \Psi (\mathbf{r},t) is described by Schrödinger equation:

\displaystyle{i\hbar\frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\Psi+V(\mathbf{r},t)\Psi = \hat{H}\Psi}

Stationary states are such quantum states as

\Psi (\mathbf{r},t)=\Psi (\mathbf{r})\exp (-iEt/\hbar)

giving solutions to the time-independent Schrödinger equation (TISE):

\displaystyle{-\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi = E\Psi}.

Note that \sigma_H^2=\langle H^2\rangle - \langle H\rangle^2=0.

The general solution to Schrödinger equation is

\Psi (\mathbf{r},t)=\displaystyle{\sum_n} c_n\Psi_n(\mathbf{r})\exp (-iE_nt/\hbar )

If we have a collection of energy eigenstates \{ \Psi_1(\mathbf{r}) , \Psi_2(\mathbf{r}), \Psi_3(\mathbf{r}), \dots\} with their corresponding eigenenergy \{ E_1,E_2,E_3,\dots \}, then

\begin{aligned} \langle \hat{H}\rangle & = \int\mathrm{d}x\,\Psi^*(x,t)\hat{H}\Psi (x,t) \\ & = \sum_{n=1}^{\infty}|c_n|^2E_n \\ \end{aligned}

Note the continuity equation

\displaystyle{\frac{\partial}{\partial t}\rho = -\nabla\cdot\mathbf{j}}

such that

\rho = |\Psi (r)|^2

and

\displaystyle{\mathbf{j}\equiv -\frac{i\hbar}{2m}[\Psi^*\nabla\Psi - \Psi\nabla\Psi^*]}.

Section 02 of 04

In exactly solvable 1-D problem, we encounter bound states and scattering states defined by

\begin{cases} E<V(-\infty)\textrm{ and }V(\infty ) & \Longrightarrow \textrm{ bound state} \\ E>V(-\infty)\textrm{ or }V(\infty ) & \Longrightarrow\textrm{ scattering state} \\ \end{cases}

Boundary conditions for \Psi (x,t):

i. \Psi must be continuous;

ii. \displaystyle{\frac{\mathrm{d}\Psi}{\mathrm{d}x}} is continuous except at points where V(x) diverges; and

iii. Integrating Schrödinger equation over a region (x,x+\Delta x) and then taking the limit \displaystyle{\lim_{\Delta x\to 0}[\cdots ]} will give condition ii.

By a delta function potential it is meant that

V(x)=\alpha \delta (x).

\displaystyle{\frac{\mathrm{d}\Psi}{\mathrm{d}x}\bigg|_{x=0^+}-\frac{\mathrm{d}\Psi}{\mathrm{d}x}\bigg|_{x=0^-}=-\frac{2m\alpha}{\hbar^2}\Psi (0)}

For a scattering problem, we have

\Psi (x)=\begin{cases} Ae^{ikx}+Be^{-ikx}\quad & \qquad x<0\\ Fe^{ik'x} \quad & \qquad x>0 \\ \end{cases}

where the reflection coefficient is

\displaystyle{R=\frac{|B|^2}{|A|^2}}

and the transmission coefficient, the ratio of probability current density, is

\displaystyle{T=\frac{|F|^2k'}{|A|^2k}}.

For a harmonic oscillator we write

\begin{aligned} \hat{H} & = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2x^2 \\ \hat{a}_+ & = \frac{1}{\sqrt{2\hbar m\omega}}(-i\hat{p}_x+m\omega x) \\ \hat{a}_- & = \frac{1}{\sqrt{2\hbar m\omega}}(i\hat{p}_x+m\omega x) \\ \end{aligned}

then

\begin{aligned} \hat{H} & = \hbar\omega \bigg(\hat{a}_-\hat{a}_+-\frac{1}{2}\bigg)\\ & = \hbar\omega \bigg(\hat{a}_+\hat{a}_- +[\hat{a}_-,\hat{a}_+]-\frac{1}{2}\bigg) \\ E_n & = \bigg(n+\frac{1}{2}\bigg)\hbar\omega \\ \hat{a}_+\Psi_n & = \alpha_n\Psi_{n+1} \\ \hat{a}_-\Psi_n & = \beta_n\Psi_{n-1} \\ \alpha_n & = \sqrt{n+1} \\ \beta_n & = \sqrt{n} \\ \end{aligned}

Section 03 of 04

Hermitian conjugates are such that:

\langle\Psi |\hat{Q}\Psi\rangle = \langle\hat{Q}^\dagger\Psi |\Psi\rangle

Hermitian operators are such that:

\hat{Q}=\hat{Q}^\dagger

We introduce some principles below:

Theorem 1. Eigenvalues of Hermitian operators are real.

Theorem 2. Eigenfunctions corresponding to different eigenvalues are orthogonal.

Axiom. The eigenfunctions of an Hermitian observable operator are complete: any function in the Hilbert space can be expressed as a linear combination of them. This complete set of eigenfunctions can be further transformed to a complete orthogonal set by the Gram-Schmidt orthogonalization procedure, i.e., \{\Psi_1,\Psi_2,\dots ,\Psi_n\} where \langle \Psi_n|\Psi_m\rangle =\delta_{nm}.

Generalized statistical interpretation If you measure a physical observable (described by the Hermitian operator \hat{Q}), you are certain to get one of the eigenvalues of \hat{Q}. If \hat{Q} has a discrete spectrum, the probability of getting the particular eigenvalue q_n associated with the orthonormalized eigenfunction f_n(x) is |c_n|^2 where c_n=\langle f_n|\Psi\rangle. Since we have a complete and orthonormalized eigenfunctions, an arbitrary wavefunction is expandable:

\Psi (x)=\displaystyle{\sum_n c_nf_n(x)}.

\begin{aligned} 1 & = \langle \Psi |\Psi\rangle \\ & = \int\mathrm{d}x\sum_nc_n^*f_n^*(x)\sum_{n'}c_{n'}f_{n'}(x) \\ & = \sum_{n}\sum_{n'} c_n^*c_{n'}\underbrace{\int\mathrm{d}x\, f_{n}^*(x)f_{n'}(x)}_{\delta_{nn'}} \\ & = \sum_{n}|c_n|^2 \\ \end{aligned}

The expectation value of \hat{Q}:

\begin{aligned} \langle \hat{Q}\rangle & = \langle\Psi |\hat{Q}\Psi\rangle \\ & = \int\mathrm{d}x\sum_{n}c_n^*f_n^*(x)\hat{Q}\sum_{n'}c_{n'}f_{n'}(x) \\ & = \sum_{n}\sum_{n'}c_{n}^*c_{n'}\int\mathrm{d}x\, f_{n}^*(x)\hat{Q}f_{n'}(x) \\ & = \sum_n\sum_{n'}c_n^*c_{n'}q_{n'}\delta_{nn'} \\ & = \sum_n|c_n|^2q_n \\ \end{aligned}

If \hat{Q} has a continuous spectrum

i.e., q takes continuous values in \hat{Q}f_q(x)=qf_q(x);

and \Psi_q(x) is Dirac-orthonormalized

i.e., \langle f_q|f_{q'}\rangle =\delta (q-q'),

then the probability of getting the outcome for \hat{Q} measurement in the range (q,q+\mathrm{d}q) is |c(q)|^2\,\mathrm{d}q where c(q)=\langle f_q|\Psi\rangle.

The wavefunction is expressed in

\Psi (x)=\displaystyle{\int\mathrm{d}q\, c(q)f_q(x)}.

Uncertainty principle in general,

\sigma_{A}^2\sigma_{B}^2\geqslant \bigg( \displaystyle{\frac{1}{2i}}\langle [\hat{A},\hat{B}]\rangle \bigg)^2;

in particular, due to Heisenberg, from [\hat{x},\hat{p}]=i\hbar, we have

\displaystyle{\sigma_x\sigma_p\geqslant \frac{\hbar}{2}}

In Dirac notation, the coordinate space wavefunction \Psi (x,t) is the representation of |\Psi (t)\rangle in the basis of position eigenfunctions:

\begin{aligned} \Psi (x,t) & = \langle x|\Psi (t)\rangle \\ \hat{Q}|q\rangle & = q|q\rangle \\ \end{aligned}

After choosing a complete set of orthonormal bases (aka representation), i.e., \{ |e_n\rangle\} such that \langle e_n|e_m\rangle =\delta_{nm}, a state vector can be expressed as a column vector

|\alpha\rangle \Rightarrow\begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n \end{bmatrix}

where \alpha_n=\langle e_n|\alpha\rangle or |\alpha\rangle = \sum_n\alpha_n|e_n\rangle.

In this matrix notation, the inner product of two state vectors is

\begin{aligned} \langle \beta |\alpha \rangle & = \begin{bmatrix}\beta_1^* & \beta_2^* & \cdots & \beta_n^* \end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\\vdots \\ \alpha_n\end{bmatrix} \\ & = \sum_n\beta_n^*\alpha_n \\ \end{aligned}

Besides, the operation |\beta\rangle =\hat{Q}|\alpha\rangle takes the matrix product form:

\begin{bmatrix}\beta_1 \\ \beta_2 \\ \vdots \\ \beta_n\end{bmatrix} = \begin{bmatrix} Q_{11} & Q_{12} & \cdots & Q_{1n} \\ Q_{21} & Q_{22} & \cdots & Q_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ Q_{n1} & Q_{n2} & \cdots & Q_{nn}\end{bmatrix}\begin{bmatrix}\alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n\end{bmatrix}

The Schrödinger equation in bra-ket/Dirac notation:

\displaystyle{i\hbar\frac{\mathrm{d}}{\mathrm{d}t}|\Psi\rangle = \hat{H}|\Psi\rangle}.

Basic matrix algebra:

\begin{aligned} \langle \Psi |\hat{Q}\Phi\rangle & = \langle\Psi |\cdot \hat{Q}\cdot |\Phi\rangle \\ (|\hat{Q}\Phi\rangle )^\dagger & = (\hat{Q}\cdot |\Phi\rangle )^\dagger \\ & = (|\Phi\rangle )^\dagger \cdot \hat{Q}^\dagger \\ & = \langle\Phi |\cdot\hat{Q}^\dagger \\ & = \langle \hat{Q}\Phi | \\ \end{aligned}

Section 04 of 04

Hydrogen atom and its angular momentum

\begin{aligned} \hat{L}_{x} & = \hat{y}\hat{p}_{z} - \hat{z}\hat{p}_{y} \\ \hat{L}_{y} & = \hat{z}\hat{p}_{x} - \hat{x}\hat{p}_{x} \\ \hat{L}_{z} & = \hat{x}\hat{p}_{y} - \hat{y}\hat{p}_{x} \\ \end{aligned}

\begin{aligned} [\hat{L}_x,\hat{L}_y] & = i\hbar (\hat{x}\hat{p}_y-\hat{y}\hat{p}_x) \\ & = i\hbar\hat{L}_z \\ [\hat{L}_y,\hat{L}_z] & = i\hbar\hat{L}_x \\ [\hat{L}_z,\hat{L}_x] & = i\hbar\hat{L}_y \\ [\hat{L}^2,\hat{L}_y] & = 0 \\ [\hat{L}^2,\hat{L}_z] & = 0 \\ \end{aligned}

Raising operator and lowering operator:

\hat{L}_{\pm}\equiv \hat{L}_x\pm i\hat{L}_y

Theorem.

\begin{cases}\hat{L}^2Y =\lambda Y \\ \hat{L}_zY = MY\end{cases}\Longrightarrow\begin{cases}\hat{L}^2(\hat{L}\pm Y)=\lambda (\hat{L}\pm Y)\\\hat{L}_z(\hat{L}\pm Y)=(M\pm\hbar )(\hat{L}\pm Y)\end{cases}

\langle \hat{L}^2\rangle\geqslant \langle \hat{L}_z^2\rangle

\Longrightarrow \begin{cases} \hat{L}^2Y_l^m & = \hbar(l+1)lY_l^m \\ \hat{L}_zY_l^m & = \hbar mY_l^m \\ \hat{L}_\pm Y_l^m & = \hbar\sqrt{(l\mp m)(l\pm m+1)}Y_l^{m\pm 1} \\ \end{cases}

\boxed{l = 0, 1, 2,\dots }
\boxed{m=-l,-l+1,\dots, l-1,l}

Half-integers (e.g. l=\frac{1}{2},\frac{3}{2}) are excluded because Y_l^m(\theta,\phi ) is a function in real space such that Y_l^m(\theta, \phi+2\pi )=Y_l^m(\theta ,\phi ).

Spin, an intrinsic angular momentum, has no wavefunction in real space.

Assumptions.

\begin{aligned} [\hat{S}_{x},\hat{S}_{y}] & = i\hbar \hat{S}_{z} \\ [\hat{S}_{y},\hat{S}_{z}] & = i\hbar \hat{S}_{x} \\ [\hat{S}_{z},\hat{S}_{x}] & = i\hbar \hat{S}_{y} \\ \hat{S}^2|s, m\rangle & =s(s+1)\hbar^2|s, m\rangle \\ \hat{s}_z|s, m\rangle & = m\hbar|s, m\rangle \\ \hat{s}_{\pm}|s, m\rangle & = \sqrt{(s\mp m)(s\pm m+1)}\,\hbar|s, m\pm 1\rangle \\ \textrm{(where }& m=-s,-s+1,\dots ,s\textrm{ )} \\ \end{aligned}

The spin number s can be integer (if bosons) or half-integer (if fermions).

Larmor Precession. (spin in a magnetic field)

\hat{H}=-\gamma \mathbf{B}\cdot\hat{\mathbf{S}}=-\gamma (B_x\hat{S}_x+B_{y}\hat{S}_y+B_{z}\hat{S}_z)

For hydrogen atom,

\begin{aligned} \hat{H}_0 & = \frac{\hat{p}^2}{2m}-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r} \\ \hat{p}^2 & = -\hbar^2\nabla^2 \\ \nabla^2 & = \frac{1}{r^2}\frac{\partial}{\partial r}\bigg( r^2\frac{\partial}{\partial r}\bigg) - \frac{1}{\hbar^2r^2}\hat{L}^2\\ \end{aligned}

\begin{aligned} \hat{H}_0\Psi_{n,l,m} & =E_n\Psi_{n,l,m} \\ \Psi_{n,l,m} & = R_{n,l}(r)Y_l^m(\theta ,\phi ) \\ \textrm{where }n & = 1,2,3,\dots\\ l&= 0,1,\dots ,n-1 \\ m & = -l,-l+1,\dots ,l \\ E_n & = -\frac{1}{n^2}R_y \\ \textrm{where }R_y & = \bigg[ \frac{m}{2\hbar^2}\bigg( \frac{e^2}{4\pi\epsilon_0}\bigg)^2\bigg] =13.6\,\mathrm{eV} \\ \end{aligned}