Quantum Mechanics * – Physicspupil

May 8, 2024May 8, 2024

202405081402 Pastime Exercise 010

About the graph below, tell some stories as probable as probable can be.

Roughwork.

We naturally assume there are no forms of negative energy. That is, kinetic energy $\textrm{KE: } K\geqslant 0$ , potential energy $\textrm{PE: }U\geqslant 0$ , and (total) mechanical energy $K+U=E=\textrm{Const.}\geqslant 0$ .

The graph is divided into Zone ①, Zone ②, and Zone ③.

$\begin{aligned} V(x) & = \begin{cases} +\infty & \textrm{for }x\leqslant 1 \\ 2 & \textrm{for }1<x\leqslant 2 \\ -2|x-3|+4& \textrm{for }2\leqslant x\leqslant 5 \\ \frac{1}{2}(x^2-10x+25) & \textrm{for }5\leqslant x\\ \end{cases} \\ \end{aligned}$

For any conservative system, total energy $E$ takes the form of a horizontal line $y=\textrm{Const.}$ as in a graph. Take $E=2$ as an example:

$\begin{aligned} & \qquad E: y=2 \\ & \Longrightarrow \begin{cases} (E-U=)K> 0\Rightarrow x\in (4,7) \\ (E-U=)K = 0\Rightarrow x \in (1,2]\cup\{ 4\}\cup\{ 7\} \\ (E-U=)K< 0\Rightarrow x\in (-\infty ,1]\cup (2,4)\cup (7,\infty) \\ \end{cases} \\ \end{aligned}$

$\begin{aligned} & \qquad K=\frac{1}{2}mv^2 = \frac{p^2}{2m} \\ & \Longrightarrow \begin{cases} K(x)>0\Rightarrow \textrm{a particle moves at }x \\ K(x)=0\Rightarrow\textrm{a particle stays at }x \\ K(x)<0\Rightarrow \textrm{no particle exists at }x \\ \end{cases} \\ \end{aligned}$

(to be continued)

May 30, 2023May 30, 2023

202305301123 Exercise 5.1

The operators $\hat{Q}_1$ , $\hat{Q}_2$ , and $\hat{Q}_3$ are defined to act as follow upon a (well-behaved) function $f(x)$ :

$\hat{Q}_1$ squares the first derivative of $f(x)$ ;
$\hat{Q}_2$ differentiates $f(x)$ twice;
$\hat{Q}_3$ multiplies $f(x)$ by $x^4$ .

(a) For each of these operators, write down an explicit expression for $\hat{Q}_i\,f(x)$ .
(b) Simplify the operators expression $\hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2$ , writing your result in terms of the variable $x$ . Do $\hat{Q}_2$ and $\hat{Q}_3$ commute?
(c) Derive an expression for $\hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2$
in terms of the operators $\hat{x}$ and $\hat{p}$ .
(d) An operator $\hat{Q}$ is linear if, for arbitrary well-behaved functions $f_1(x)$ and $f_2(x)$ , the following property holds:

$\hat{Q}\,[c_1f_1(x)+c_2f_2(x)]=c_1\,\hat{Q}\,f_1(x)+c_2\,\hat{Q}\,f_2(x)$ .

Prove whether or not each of the three operators defined above is linear.

_{Extracted from M. A. Morrison. (1990). Understanding Quantum Physics A User’s Manual.}

Roughwork.

(a)

$\begin{aligned} \hat{Q}_1\,f(x) & =\bigg[\frac{\mathrm{d}}{\mathrm{d}x}f(x)\bigg]^2 \\ \hat{Q}_2\,f(x) & = \frac{\mathrm{d}^2}{\mathrm{d}x^2}f(x) \\ \hat{Q}_3\,f(x) & = x^4f(x) \\ \end{aligned}$

(b)

If $\hat{Q}_1\hat{Q}_2=\hat{Q}_2\hat{Q}_1$ , we say $\hat{Q}_1$ and $\hat{Q}_2$ are commuting operators to each other.

Recall addition and subtraction of operators:

$\begin{aligned} (\hat{Q}_1\pm\hat{Q}_2)\,f(x) & = \hat{Q}_1\,f(x)\pm\hat{Q}_2\,f(x) \\ & = \pm\hat{Q}_2\,f(x)+\hat{Q}_1f(x) \\ \end{aligned}$

so as to write

$\begin{aligned} & \quad\enspace (\hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2)\,f(x)\\ & = (\hat{Q}_2\hat{Q}_3)\,f(x)-(\hat{Q}_3\hat{Q}_2)\,f(x) \\ & = \hat{Q}_2\,(\hat{Q}_3\,f(x))-\hat{Q}_3\,(\hat{Q}_2\,f(x)) \\ & = \hat{Q}_2\,(x^4f(x)) -\hat{Q}_3\,(f''(x)) \\ & = (x^4f''(x)+12x^2f(x)) - x^4f''(x) \\ & = 12x^2f(x) \\ & \neq 0 \\ \end{aligned}$

Therefore $\hat{Q}_2$ and $\hat{Q}_3$ do not commute.

(c)

The position operator, $\hat{x}$ , in the $x$ -representation, multiplies $f(x)$ by $x$ ; and the momentum operator, $\hat{p}_x$ , by $\displaystyle{-i\hbar\frac{\partial}{\partial x}}$ :

$\begin{aligned} |f\rangle & \stackrel{|x\rangle}{\longrightarrow} \langle x|f\rangle = f(x) \\ \hat{x}|f\rangle & \stackrel{|x\rangle}{\longrightarrow} \langle x|\hat{x}|f\rangle = xf(x) \\ \hat{p}|f\rangle & \stackrel{|x\rangle}{\longrightarrow} \langle x|\hat{p}|f\rangle = -i\hbar\nabla f\\ \end{aligned}$

$\hat{Q}_2\hat{Q}_3-\hat{Q}_3\hat{Q}_2=\textrm{I don't know?}$

Note the symmetry between the $x$ and the $p$ representations and a one-to-one mapping onto each other.

(d) Not to be attempted.

March 17, 2022January 2, 2023

202203171422 Half Picture of Quantum Mechanics 01B

(This half picture is based on the second half of manuscript of Chapter One, 2015-2016 PHYS3351 Quantum Mechanics Lecture Notes.)

Section 01

Momentum

As the wavefunction describes the state of a quantum system, how do we find the information of velocity or momentum?

Expectation value $\langle\mathbf{r}\rangle$ of position $\mathbf{r}$

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

is a function of time $t$ .

$\begin{aligned} \frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t} & = \iiint\mathrm{d}V\,\mathbf{r}\frac{\partial}{\partial t}|\Psi |^2 \\ & = \frac{i\hbar}{2m}\iiint\mathrm{d}V\,\mathbf{r}\nabla\cdot [\Psi^*\nabla\Psi - (\nabla\Psi^* )\Psi ] \\ \dots &\textrm{ integration by parts }\dots \\ \frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t} & = -\frac{i\hbar}{2m}\iiint [\Psi^*\nabla\Psi - (\nabla\Psi^* )\Psi ]\,\mathrm{d}V \\ \dots &\textrm{ integration by parts }\dots \\ \underbrace{\frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t}}_{\textrm{velocity}} & =\iiint\mathrm{d}V\,\Psi^*\bigg( -\frac{i\hbar}{m}\nabla\bigg)\Psi \\ \end{aligned}$

$\displaystyle{\frac{\mathrm{d}\langle \mathbf{r}\rangle}{\mathrm{d}t}}$ is the “velocity” of the central position of the particle “cloud”; whereas

$\displaystyle{m\frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t}=\iiint\mathrm{d}V\,\Psi^*(-i\hbar\nabla )\Psi}$

is a very suggestive form for us to introduce the momentum operator:

$\displaystyle{\hat{p}=\frac{\hbar}{i}\nabla}$ .

Expectation value:

$\begin{aligned} \langle\hat{p}\rangle & = -i\hbar\iiint\mathrm{d}V\,\Psi^*\nabla\Psi \\ & = m\frac{\mathrm{d}\langle \mathbf{r}\rangle}{\mathrm{d}t} \\ \end{aligned}$

Kinetic energy:

$\hat{T}=\displaystyle{\frac{\hat{p}}{2m}=-\frac{\hbar^2}{2m}\nabla^2}$

Angular momentum:

$\hat{L}=\mathbf{r}\times \hat{p} = \displaystyle{\mathbf{r}\times \bigg(\frac{\hbar}{i}\nabla\bigg)}$ .

$\begin{aligned} \langle\hat{T}\rangle & = -\frac{\hbar^2}{2m}\iiint\mathrm{d}V\,\Psi^*\nabla^2\Psi \\ \langle \hat{L}\rangle & = \iiint\mathrm{d}V\,\Psi^*\bigg(\mathbf{r}\times\frac{\hbar}{i}\nabla \bigg)\Psi \\ \end{aligned}$

Summary.

In quantum mechanics, all observable quantities, such as momentum, position, or energy, are described by operators. When we describe a particle by the wavefunction $\Psi (\mathbf{r},t)$ , we are looking at the probability distribution at the position representation. In this representation the position operator takes the special form of a number: $\mathbf{r}$ ; whereas other observables such as momentum and energy involve the differential operator $\nabla$ . For a particle described by a wavefunction, the expectation value of the observed quantity is given by

$\langle\hat{S}\rangle = \displaystyle{\iiint\mathrm{d}V\,\Psi^*\hat{S}\Psi}$

where $\hat{S}$ denotes the operator form of the observable.

Section 02

Uncertainty principle

Like the standard deviation of position distribution, we can also evaluate the standard deviation of momentum:

$\sigma_{p_x}^2=\langle\hat{p}_x^2\rangle - \langle\hat{p}_x\rangle^2$ .

For any possible wavefunction, one finds that

$\sigma_x\sigma_{p_x}\geqslant\displaystyle{\frac{\hbar}{2}}$ .

Example: Gaussian wavepacket

$\begin{aligned} \Psi (x,0) & = A\exp \bigg(-\frac{x^2}{4\sigma^2}+iKx\bigg) \\ A & = (2\pi\sigma^2)^{-1/4} \\ \hat{p}_x & = \frac{\hbar}{i}\frac{\partial}{\partial x} \\ \Psi (x,0) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi (k)e^{ikx}\,\mathrm{d}k \\ \phi (k) & = (2\sigma^2/\pi )^{1/4}\exp [-\sigma^2(k-K)^2] \\ \end{aligned}$

$\begin{aligned} \langle \hat{p}_x^n\rangle & = \int_{-\infty}^{\infty}\bigg[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi^*(k)e^{-ikx}\,\mathrm{d}k\bigg](\hat{p}_x)^n\bigg[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi (k')e^{ik'x}\,\mathrm{d}k'\bigg] \\ & = \int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi^* (k)e^{-ikx}\,\mathrm{d}k\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi (k')(\hbar k')^ne^{ik'x}\,\mathrm{d}k' \\ & = \int_{-\infty}^{\infty}\mathrm{d}k\,\phi^* (k)\int_{-\infty}^{\infty}\phi (k')\delta (k-k')(\hbar k')^n\,\mathrm{d}k' \\ & = \int_{-\infty}^{\infty}(\hbar k)^n|\phi (k)|^2\,\mathrm{d}k \\ \end{aligned}$

$\begin{aligned} \langle\hat{p}_x\rangle & = \int\mathrm{d}x\,\Psi^*\bigg( \frac{\hbar}{i}\frac{\partial}{\partial x}\bigg)\Psi = \hbar K \\ \Delta p_x & = \langle \hat{p}_x^2\rangle - \langle \hat{p}_x\rangle^2 = \frac{\hbar}{2\sigma} \\ \dots \textrm{ because }&\Delta x=\sigma \\ \Delta x\cdot\Delta p_x & = \frac{\hbar}{2} \\ \end{aligned}$

$\therefore$ Gaussian wavepackets satisfy the uncertainty relation. This relation tells us that the position and the momentum of a particle can $\textrm{\scriptsize{NEVER}}$ be simultaneously determined with precision. This also shows the significance of the constant $\hbar$ . Only when we talk about quantities with this scale, quantum effect is significant, that is, things cannot be described by classical mechanics. But if we cannot measure things with such precision, you won’t “notice” the quantum effect: that is why our daily experience is classical.

Von Neumann projection postulate:

Measurement collapses the wavefunction.

E.g., A position measurement with the outcome $\mathbf{r}_0$ collapses the wavefunction from $\Psi (\mathbf{r},t)$ to $\delta (\mathbf{r}-\mathbf{r}_0)$ . So a second measurement immediately following will also find the particle at $\mathbf{r}_0$ .

Note the subtle difference of expectation value from mean value in the usual sense.

(to be continued)

March 17, 2022October 24, 2022

202203170840 Half Picture of Quantum Mechanics 01A

(This half picture is based on the first half of manuscript of Chapter One, 2015-2016 PHYS3351 Quantum Mechanics Lecture Notes.)

Section 01

Given a particle of mass $m$ and a force $F(x,t)$ , we classify

$\begin{cases}\textrm{conservative }F =\displaystyle{-\frac{\partial V(x)}{\partial x}}\enspace\big( V(x)\textrm{: potential energy}\big)\\ \textrm{nonconservative} \\ \end{cases}$

In classical mechanics, description of dynamics is given by

i. trajectory (i.e., position as a function of time):

$x(t)$

ii. velocity:

$\mathbf{v}=\displaystyle{\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}}$

iii. momentum:

$p=m\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}t}}$

iv. kinetic energy:

$T=\displaystyle{\frac{1}{2}m\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2}$

v. solving from Newton’s second law $\textrm{Net }F=ma$ :

$\displaystyle{F=ma\qquad\Longrightarrow\qquad m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=F=-\frac{\partial V}{\partial x}}$

In quantum mechanics, description of dynamics is given by wavefunction $\Psi (x,t)$ which is a single-valued, finite, and continuous complex function.

Statistical interpretation.

$|\Psi (x,t)|^2\,\mathrm{d}x$ : the probability of particle being between $x$ and $x+\mathrm{d}x$ at time $t$

In 3D case, $|\Psi (x,y,z,t)|^2\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ : the probability of particle being in the volume between $x$ and $x+\mathrm{d}x$ , $y$ and $y+\mathrm{d}y$ , and $z$ and $z+\mathrm{d}z$ .

$\Psi (x)$ is also known as the probability amplitude.

Imagine the particle as a “cloud” which is moving and changing its shape, instead of a point object following a trajectory curve in space. The particle has a larger probability to be at the denser point of the “cloud”. Even if you know everything about the particle dynamics, that is, the wavefunction $\Psi (\mathbf{r},t)$ , you cannot predict with certainty the outcome of an experiment to measure its position. Instead, you can say the probability you will detect the particle in the volume between $x$ and $x+\mathrm{d}x$ , $y$ and $y+\mathrm{d}y$ , and $z$ and $z+\mathrm{d}z$ is

$|\Psi (x,y,z,t)|^2\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ .

Normalization.

$\displaystyle{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\, |\Psi (x,y,z,t)|^2=1}$

Q. Which one of the following is $\textrm{\scriptsize{NOT}}$ a possible wavefunction in the region $x\geqslant 0$ for $\beta >0$ :

(I) $Axe^{\beta x}$ ; (II) $Axe^{-\beta x}$ ; (III) $Ae^{-\beta x}$ .

A. (I)

Important characteristics of a wavefunction

A discrete example

There are a number of positions $x_1$ , $x_2$ , … , $x_{10}$ on a 1D line, at which the particle can appear with probability $\rho_1$ , $\rho_2$ , … , $\rho_{10}$ .

Normalization: $\rho_1 + \rho_2+\cdots +\rho_{10}=1$

If we throw dice for $N\gg 1$ times, each time we get a particle at one of the positions:

$x(j)\in\{ x_1, x_2, \dots ,x_{10}\}$ for $j=1,2,\dots ,N$

the times we get the particle at position $x_i$ is given by $N\rho_i$ ,

i.e., the number of $x(j)\big|_{j=1,\dots ,N}=x_i$ is $N\rho_i$ .

Average position:

$\begin{aligned} \langle x\rangle & = \sum_{j=1}^{N}x(j)\frac{1}{N}\\ & = \sum_{i=1}^{10}\frac{x_iN\rho_i}{N} \\ & = \sum_{i=1}^{10}x_i\rho_i \\ \end{aligned}$

Numerical measure of the amount of spread:

$\begin{aligned} \Delta x(j) & \equiv x(j)-\langle x\rangle \\ \sum_{j}\Delta x(j) & = 0 \\ \langle (\Delta x)^2\rangle & \equiv \sum_{j}\frac{1}{N}(\Delta x(j))^2 \\ & = \sum_{j}\frac{1}{N}(x^2(j)+\langle x\rangle^2-2x(j)\langle x\rangle )\\ & = \sum_{j}\frac{1}{N}x^2(j) - \langle x\rangle^2 \\ & = \langle x^2\rangle -\langle x\rangle^2 \textrm{ (always +ve)} \end{aligned}$

Owing to the statistical description of quantum mechanics, we can no longer speak of the precise position of a particle, nevertheless, we still care about two quantities: (I) the expectation value of position; and (II) the uncertainty in position.

Expectation value is the weighted average of all possible values that a physical observable can take.

Expectation value of position:

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

Deviation.

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

For the $x$ -, $y$ -, $z$ -components of position, observe that

$\begin{aligned} \langle (\Delta x)^2\rangle & = \langle x^2\rangle - \langle x\rangle^2 \\ \langle (\Delta y)^2\rangle & = \langle y^2\rangle - \langle y\rangle^2 \\ \langle (\Delta z)^2\rangle & = \langle z^2\rangle - \langle z\rangle^2 \\ \langle (\Delta\mathbf{r})^2\rangle & = \langle (\Delta x)^2\rangle + \langle (\Delta y)^2\rangle + \langle (\Delta z)^2\rangle \\ \end{aligned}$

Standard deviation in position (aka uncertainty)

$\sigma$ is defined by $\sigma^2=\langle (\Delta\mathbf{r})^2\rangle$ ;

The expectation value of position $\langle \mathbf{r}\rangle$ gives the central position of the particle “cloud”, and $\sigma$ tells how much the “cloud” spreads out in space. Roughly speaking, you will find the particle in the space interval of

$(\langle x\rangle\pm\sigma_x, \langle y\rangle\pm\sigma_y, \langle z\rangle\pm\sigma_z)$

Example (Gaussian wavepacket)

Consider a wavefunction

$\Psi (x,0)=A\exp \bigg( \displaystyle{-\frac{x^2}{4\sigma^2}+iKx} \bigg)$ .

By the normalization requirement,

$\begin{aligned} \int\mathrm{d}x\, |\Psi (x,0)|^2 & = 1 \\ |A|^2\int\mathrm{d}x\,\exp\bigg( -\frac{x^2}{2\sigma^2}\bigg) & = 1 \\ A & = (2\pi\sigma^2)^{1/4} \end{aligned}$

$\begin{aligned} \langle x\rangle & = \int_{-\infty}^{\infty}\mathrm{d}x\, x(2\pi\sigma^2)^{-1/2}\exp \bigg( -\frac{x^2}{2\sigma^2}\bigg) = 0 \\ \langle x^2\rangle & = \int_{-\infty}^{\infty}\mathrm{d}x\, x^2(2\pi\sigma^2)^{-1/2}\exp \bigg( -\frac{x^2}{2\sigma^2}\bigg) \\ & = \int_{-\infty}^{\infty}\frac{1}{2}\,\mathrm{d}x^2\, x(2\pi\sigma^2)^{-1/2}\exp \bigg( -\frac{x^2}{2\sigma^2}\bigg) \\ & = \int_{-\infty}^{\infty}(-\sigma^2)\,\mathrm{d}\bigg(\exp\bigg( -\frac{x^2}{2\sigma^2} \bigg)\bigg)x(2\pi\sigma^2)^{-1/2} \\ & = \int_{-\infty}^{\infty}\sigma^2(2\pi\sigma^2)^{-1/2}\exp\bigg( -\frac{x^2}{2\sigma^2} \bigg)\,\mathrm{d}x \\ & = \sigma^2 \\ \Rightarrow \langle (\Delta x)^2\rangle & = \sigma^2 \\ \end{aligned}$

Section 02

The dynamics of the wavefunction $\Psi (\mathbf{r},t)$ is described by Schrödinger equation:

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

Hamiltonian operator:

$\hat{H}=\hat{T}+\hat{V}$

Kinetic energy operator:

$\displaystyle{\hat{T}=-\frac{\hbar^2}{2m}\nabla^2}$

Potential energy operator:

$\hat{V}=V(\mathbf{r},t)$

Planck constant:

$\hbar =1.0546\times 10^{-34}\,\mathrm{J\, s}$

Given the wavefunction $\Psi (\mathbf{r},t_0)$ at time $t_0$ , the Schrödinger equation determines $\Psi (\mathbf{r},t)$ for all future time.

What is quantum mechanics? The simplest answer is wavefunction (and its statistical interpretation) plus Schrödinger equation.

In integral form, Schrödinger equation reads:

$\Psi (x,t)=\Psi (x,t_0)\underbrace{\exp \bigg(-i\frac{\hat{H}}{\hbar}(t-t_0) \bigg)}_{\textrm{evolution operator}}$

Probability density and flux

According to the probability interpretation, wavefunction has to be normalized. In a 1D example:

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

Motivation.

Does $\Psi (x,t)$ remain normalized in the evolution according to Schrödinger equation? I.e.,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x\stackrel{?}{=}0}$

The answer to this question is affirmative.

Proof.

To begin with,

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x & = \int_{-\infty}^{\infty}\bigg[ \frac{\partial}{\partial t}|\Psi (x,t)|^2\bigg]\,\mathrm{d}x \end{aligned}$

evaluating the square-bracketed term in the integrand,

$\displaystyle{\frac{\partial}{\partial t}|\Psi (x,t)|^2=\Psi^*\frac{\partial \Psi}{\partial t}+\frac{\partial\Psi^*}{\partial t}\Psi}$

by identities of Schrödinger’s

$\begin{aligned} \frac{\partial \Psi}{\partial t} & = \frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi\\ \frac{\partial \Psi^*}{\partial t} & = -\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2} +\frac{i}{\hbar}V\Psi^*\\ \end{aligned}$

it follows that

$\begin{aligned} \frac{\partial}{\partial t}|\Psi |^2 & = \frac{i\hbar}{2m}\bigg( \Psi^*\frac{\partial^2\Psi}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi \bigg) \\ & = \frac{\partial}{\partial x}\bigg[ \frac{i\hbar}{2m}\bigg( \Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*}{\partial x}\Psi \bigg) \bigg]\\ \end{aligned}$

and

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x = \frac{i\hbar}{2m}\bigg( \Psi^*\frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi\bigg)\bigg|_{-\infty}^{\infty}}$

since $\Psi (x,t)$ is normalized, i.e.,

$\Psi (x=\pm\infty ,t)=0$ ,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x=0}$ .

$\therefore$ The Schrödinger equation guarantees the wavefunctions remain normalized.

QED

Defining the probability density by:

$\rho (x,t)\equiv |\Psi (x,t)|^2$

and the probability current by:

$\displaystyle{j=-\frac{i\hbar}{2m}\bigg(\Psi^*\frac{\partial \Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi\bigg)}$

we write the continuity equation:

$\displaystyle{\frac{\partial}{\partial t}\rho =-\frac{\partial}{\partial x}j}$ .

Note that

$\underbrace{\frac{\partial}{\partial t}\int_{a}^{b}\rho\,\mathrm{d}x}_{\textrm{(A)}}= \underbrace{-j(b)}_{\textrm{(B)}} +\underbrace{j(a)}_{\textrm{(C)}}$

Physical meanings of (A), (B), and (C):

$\textrm{(A)}$ : change in the probability of being between $a$ and $b$ ;
$\textrm{(B)}/\textrm{(C)}$ : positive (/negative) value means inward (/outward) flux into the region between $a$ and $b$ .

In 3D case:

$\begin{aligned} \mathbf{j}(\mathbf{r}) & =-\frac{i\hbar}{2m}[\Psi^*\nabla\Psi - (\nabla\Psi^*)\Psi ] \\ \frac{\partial}{\partial t}|\Psi |^2 & = -\nabla\cdot\mathbf{j} \\ \end{aligned}$

Probability flux through a surface area is $\mathbf{j}\cdot\mathrm{d}\mathbf{s}$ .

(to be continued)

March 16, 2022May 4, 2023

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,

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

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

March 15, 2022October 24, 2022

202203151002 Exercise 3.3

Starting from

$\langle x\rangle =\displaystyle{\int \psi^*(x)x\psi (x)\,\mathrm{d}x}$

and using equation (3)

$\displaystyle{\psi (x)=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\phi (p)e^{ipx/\hbar}\,\mathrm{d}p}$

to express $\psi (x)$ in terms of $\phi (p)$ , deduce equation (33):

$\displaystyle{x=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}}$ .

_{Extracted from D. S. Saxon. (2012). Elementary Quantum Mechanics.}

Some useful formulae. (Relationship between wave functions in configuration space $\psi (x)$ and in momentum space $\phi (p)$ )

In differential forms,

$\displaystyle{x\exp (ipx/\hbar )=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\exp (ipx/\hbar)}$

$\displaystyle{p\exp (-ipx/\hbar)=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}x}\exp (-ipx/\hbar )}$

or, in integral forms,

$\psi (x)=\displaystyle{\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi (p)\exp (ipx/\hbar )\,\mathrm{d}p}$

$\phi (p)=\displaystyle{\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\psi (x)\exp (ipx/\hbar )\,\mathrm{d}x}$

Derivation.

$\begin{aligned} \langle x\rangle & = \int \psi^*x\psi (x)\,\mathrm{d}x \\ & = \frac{1}{2\pi\hbar}\iiint\mathrm{d}x\,\mathrm{d}p\,\mathrm{d}p'\,\big( \phi^*(p')\exp (ip'x/\hbar )\big) x\big( \phi (p)\exp (ipx/\hbar )\big)\\ \dots &\enspace \textrm{by the fact }x\exp (ipx/\hbar )=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\exp (ipx/\hbar )\enspace \dots \\ \langle x\rangle & = \frac{1}{2\pi\hbar}\iiint\mathrm{d}x\,\mathrm{d}p\,\mathrm{d}p'\phi^*(p')\exp (ip'x/\hbar )\phi (p)\bigg( -\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\exp (ipx/\hbar ) \bigg) \\ \dots &\enspace \textrm{integrating wrt }x\textrm{ by parts }\enspace \dots \\ \langle x\rangle & = \frac{1}{2\pi\hbar}\iiint\mathrm{d}x\,\mathrm{d}p\,\mathrm{d}p'\,\phi^*(p')\exp [ix(p'-p)/\hbar ]\,\frac{\hbar}{i}\frac{\mathrm{d}\phi (p)}{\mathrm{d}p} \\ \dots &\enspace \textrm{doing }x\textrm{ integration }\enspace\dots \\ & = \iint \mathrm{d}p\,\mathrm{d}p'\,\phi^*(p')\frac{\hbar}{i}\frac{\mathrm{d}\phi (p)}{\mathrm{d}p}\delta (p'-p) \\ & = \int\mathrm{d}p\,\phi^*(x)\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\phi (p) \\ \end{aligned}$

Afterword. (Using bra-ket/Dirac notation)

_{Reference: Question 86824 answered by joshphysics on Nov 17, 2013 (AT)physics.stackexchange(DOT)com}

For any physically admissible state functions , which necessarily vanish at infinity, we see that

$\begin{aligned} & \quad \langle p|[\hat{x},\hat{p}]|\psi\rangle \\ & = \langle p|\hat{x}\hat{p}-\hat{p}\hat{x}|\psi\rangle \\ & = \langle p|\hat{x}\hat{p}|\psi\rangle - \langle p|\hat{p}\hat{x}|\psi\rangle \\ & = \langle p|\hat{x}\hat{p}|\psi\rangle - p\langle p|\hat{x}|\psi\rangle \\ \end{aligned}$

By the canonical commutation relation between and :

$[\hat{x},\hat{p}]=i\hbar\hat{I}$

where is the identity operator, we have

$\begin{aligned} \langle p|i\hbar I|\psi\rangle & = \langle p|\hat{x}\hat{p}|\psi\rangle - p\langle p|\hat{x}|\psi\rangle \\ p\langle p|\hat{x}|\psi\rangle & = \langle p|\hat{x}\hat{p}|\psi\rangle - i\hbar\langle p|\psi\rangle \\ \end{aligned}$

Manipulating the first term on RHS:

$\begin{aligned} &\quad \langle p|\hat{x}\hat{p}|\psi \rangle \\ & = \int\mathrm{d}x\,\langle p|\hat{x}|x\rangle \langle x|\hat{p}|\psi\rangle \\ & = \int \mathrm{d}x\, x\exp (ipx/\hbar )\bigg( -i\hbar\frac{\mathrm{d}}{\mathrm{d}x}\psi (x)\bigg) \\ & = i\hbar\int\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\big( (x\exp (ipx/\hbar ))\psi (x) \big) \\ & = i\hbar\int\mathrm{d}x\exp (ipx/\hbar )\psi (x)+i\hbar\int\mathrm{d}x\, x\bigg(\frac{ip}{\hbar}\bigg) \exp (ipx/\hbar )\psi (x) \\ & = i\hbar\psi (p)-p\int\mathrm{d}x\, x\exp (ipx/\hbar )\psi (x) \\ & = i\hbar\psi (p)+i\hbar p\frac{\mathrm{d}}{\mathrm{d}p}\int\mathrm{d}x\, \exp (ipx/\hbar )\psi (x)\\ & = i\hbar\psi (p)+i\hbar p\frac{\mathrm{d}}{\mathrm{d}p}\psi (p) \end{aligned}$

thus the result

$\displaystyle{p\langle p|\hat{x}|\psi\rangle = i\hbar p\frac{\mathrm{d}}{\mathrm{d}p}\psi (p)}$ .

December 4, 2021June 6, 2023

202112041209 Homework 1 (Q5)

Suppose that light of intensity $10^{-9}\,\mathrm{W/m^2}$ normally shines on a metal surface. The metal is made up of a simple cubic lattice of atoms with lattice spacing $0.3\,\mathrm{nm}$ . Each atom has one free electron. The binding energy at the metal surface is $8\,\mathrm{eV}$ . Suppose further that the light is uniformly distributed over the surface and all its energy is absorbed by the surface electrons.

(a) If the incident radiation were well described by classical physics, how long would one have to wait after switching on the light source until an electron gains enough energy to be released as a photoelectron?

(b) In reality, how long is this time duration? Explain briefly.

Reading Comprehension.

Highlighting some keyword(s) will help doing the question:

(S1) […] light […] normally shines on a […] surface […] ;

(S2) […] metal is made up of a simple cubic lattice of atoms […] ;

(S3) […] Each atom has one free electron […] .

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2265 Modern Physics Homework 1 Solution.)

(a)

There are eight atoms each lattice and one electron each atom. Hence,

$\begin{aligned} t & = \frac{8 \times (1.602\times 10^{-19})}{10^{-9} \times (0.3\times 10^{-9})^2} \\ & = 1.424\times 10^{10}\,\mathrm{s} \\ \end{aligned}$

(b)

No electron can be released as a photoelectron, for no photon has energy greater than $8\,\mathrm{eV}$ . (Why?)

September 29, 2021August 11, 2022

202109291137 SQL Table 003

CREATE TABLE SQL_Table_003

(ID int,

Formulations_of_quantum_mechanics varchar(255),

Proposer varchar(255),

Time_(Yr) varchar(255)

);

INSERT INTO SQL_Table_003

(Formulations_of_quantum_mechanics, Proposer, Time_(Yr))

VALUES

('transformation theory', 'Paul Dirac', '~1927 ')

('matrix mechanics', 'Werner Heisenberg, Max Born, Pascual Jordan', '~1925')

('wave mechanics', 'Erwin Schrödinger', '~1925')

Output.

+------+-------------------------------------+--------------------------------+-------------+

+------+-------------------------------------+--------------------------------+-------------+

| 1 | transformation theory | Paul Dirac | ~1927 |

| 2 | matrix mechanics | Werner Heisenberg, Max Born, | ~1925 |

| 3 | wave mechanics | Erwin Schrödinger | ~1925 |

| 4 | path integral formulation | Richard Feynman | ~1948 |

+------+-------------------------------------+--------------------------------+-------------+

September 28, 2021November 29, 2021

202109281720 SQL Table 002

CREATE TABLE SQL_Table_002

(ID int,

Velocity varchar(255),

Nonquantum varchar(255),

Quantum varchar(255)

);

INSERT INTO SQL_Table_002

(Velocity, Nonquantum, Quantum)

VALUES

('v << c', 'Classical Mechanics (Newton)', 'Quantum Mechanics (Bohr, Heisenberg, Schrödinger, et al.)')

('0 < |v-c| << 1', 'Special Relativity (Einstein)', 'Quantum Field Theory (Dirac, Pauli, Feynman, Schwinger, et al.)')

Output.

+------+------------------+-----------------------+----------------------------+

+------+------------------+-----------------------+----------------------------+

| 1 | v << c | Classical Mechanics | Quantum Mechanics |

| 2 | 0 < |v-c| << 1 | Special Relativity | Quantum Field Theory |

+------+------------------+-----------------------+----------------------------+

(D. J. Griffiths on Introduction to Electrodynamics (1999), pg. xi)

September 15, 2021October 24, 2022

202109151516 CYU 7.1

If $a=3+4i$ , what is the product $a^*a$ ?

Background. (Complex conjugate)

The complex conjugate of $a+bi$ is equal to $a-bi$ .

$\begin{aligned} a^*a & = (3-4i)(3+4i) \\ & = (3)(3)+(3)(4i)+(-4i)(3)+(-4i)(4i) \\ & = 9 - 16i^2 \\ & = 9 - 16(-1)\\ & = 9+16 \\ & = 25 \\ \end{aligned}$