Trial and Error – Physicspupil

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 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 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 $a-bi$ .

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

May 25, 2021October 24, 2022

202105251602 Homework 1 (Q1)

Recall that classical wave in one spatial dimension is described by the wave equation Eq. (1):

$\displaystyle{\frac{\partial^2u}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2u}{\partial t^2}}$ .

(a) Find conditions on the constants $a$ , $f$ , $k$ and $\phi$ so that $u=a\cos (2\pi ft+kx+\phi )$ is a solution of the Eq. (1).

(b) What are the physical meanings of $a$ , $f$ , $k$ and $\phi$ ?

(c) Show that if $u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1), then so is $c_1u_1(x,t)+c_2u_2(x,t)$ where $c_1$ , $c_2$ are constants. (Mathematically, we say that the solutions form a linear space or vector space.)

(d) Consequently, $u=\sum_{i=1}^2a_i\cos (2\pi f_it+k_ix+\phi_i)$ is a solution of Eq. (1) provided that each of the two terms is also a solution of Eq. (1). What could be said about the frequency of the wave described by this linear-superpositioned solution?

Solution.

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

(a)

$\begin{aligned} \frac{\partial^2u}{\partial x^2} & = \frac{\partial}{\partial x}\bigg(\frac{\partial u}{\partial x}\bigg) \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\Big( a\cos (2\pi ft+kx+\phi )\Big)\bigg) \\& = \frac{\partial}{\partial x}\Big( -ak\sin (2\pi ft+kx+\phi) \Big) \\ & = -ak^2\cos (2\pi ft+kx+\phi) \\ \frac{\partial^2u}{\partial t^2} & = \frac{\partial}{\partial t}\bigg(\frac{\partial u}{\partial t}\bigg) \\ & = \frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\Big( a\cos (2\pi ft+kx+\phi )\Big)\bigg) \\& = \frac{\partial}{\partial t}\Big( -a(2\pi f)\sin (2\pi ft+kx+\phi) \Big) \\ & = -4a\pi^2f^2\cos (2\pi ft+kx+\phi ) \end{aligned}$

$-ak^2\cos (2\pi ft+kx+\phi )=\displaystyle{-\frac{4a\pi^2f^2}{v^2}\cos (2\pi ft+kx+\phi )}$ ,

we have $k=\pm \displaystyle{\frac{2\pi f}{v}}$ .

There are no constraints on $a$ and $\phi$ .

(b)

$a$ : amplitude
$f$ : frequency
$k$ : wave number
$\phi$ : phase shift of the wave at $x=0$ and $t=0$

(c)

Set $u(x,t)=c_1u_1(x,t)+c_2u_2(x,t)$ .

Condition $\textrm{(i)}:\enspace u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1).

$\begin{aligned} \frac{\partial^2u}{\partial x^2} & = c_1\frac{\partial^2u_1}{\partial x^2} + c_2\frac{\partial^2u_2}{\partial x^2} \\ & \stackrel{\textrm{(i)}}{=} \frac{c_1}{v^2}\frac{\partial^2u_1}{\partial t^2} + \frac{c_2}{v^2}\frac{\partial^2u_2}{\partial t^2} \\ & = \frac{1}{v^2}\bigg( c_1\frac{\partial^2u_1}{\partial t^2}+c_2\frac{\partial^2u_2}{\partial t^2} \bigg) \\ \frac{\partial^2u}{\partial t^2} & = c_1\frac{\partial^2u_1}{\partial t^2}+c_2\frac{\partial^2u_2}{\partial t^2}\\ \therefore \enspace & \frac{\partial^2u}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2u}{\partial t^2} \end{aligned}$

That $u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1) implies $c_1u_1(x,t)+c_2u_2(x,t)$ is also a solution.

(d) Frequency is not well-defined for linear-superpositioned waves.

April 14, 2021October 1, 2025

202104142106 Notes on Schrödinger equation for the delta-function barrier

The Schrödinger equation for the delta-function barrier reads

$\displaystyle{-\frac{\hbar^2}{2m}}\dfrac{\mathrm{d}^2 \varphi}{\mathrm{d}x^2}+\alpha \delta (x)\varphi =E\varphi$

It yields both bound states ( $E<0$ ) and scattering states ( $E>0$ ). For the time being we consider only scattering states.

For $x<0$ , $V=\alpha \delta (x)=0$ .

The Schrödinger equation reads

$\displaystyle{\dfrac{\mathrm{d}^2\varphi}{\mathrm{d}x^2}=-\frac{2mE}{\hbar^2}\varphi=-k^2\varphi}$ ,

where $\displaystyle{k\stackrel{\textrm{def}}{=} \frac{\sqrt{2mE}}{\hbar}}$ is real and positive.

The general solution being

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

and this time we cannot rule out either term, since neither of them blows up. Similarly, for $x>0$ ,

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

The continuity of $\varphi (x)$ at $x=0$ requires that

$F+G=A+B$

The derivatives are

$\mathrm{d}\varphi /\mathrm{d}x=ik(Fe^{ikx}-Ge^{-ikx})$ for $x>0$ ;

$\mathrm{d}\varphi /\mathrm{d}x=ik(Ae^{ikx}-Be^{-ikx})$ for $x<0$ .

Here we define

$\displaystyle{\Delta (\mathrm{d}\varphi /\mathrm{d}x)=\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg|_{0^+}-\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg|_{0^-}}$

Hence, $\Delta (d\varphi /dx)=ik(F-G-A+B)$ .

By the second boundary condition, we have

$\displaystyle{\Delta \bigg(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg)=\frac{2m}{\hbar^2}\lim_{\epsilon \to 0}\int_{-\epsilon}^{+\epsilon}V(x)\varphi (x)\,\mathrm{d}x}$ ,

which in turn gives

$\displaystyle{\Delta \bigg(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg)=\frac{2m\alpha}{\hbar^2}\varphi (0)=\frac{2m\alpha}{\hbar^2}(A+B)}$ .

Combining the results of first derivative continuity and second boundary condition, we write

$\displaystyle{ik(F-G-A+B)=\frac{2m\alpha}{\hbar^2}(A+B)}$ ,

Calculations:

Suppose the wave is coming from the left so that there will be no wave scattering from the right, i.e., $G=0$ .

Recall that $F+G=F+0=F=A+B$ by the continuity condition.

Calculations:

$\begin{aligned} F& =A+B\\ & = A-\frac{i\beta}{1+i\beta}A\\ & = \frac{1}{1+i\beta}A\\ \end{aligned}$

Reflection coefficient $R$ :

$\begin{aligned} R & \equiv \frac{|B|^2}{|A|^2}\\ & = \bigg|-\frac{i\beta}{1+i\beta}\bigg|^2\\ & = \frac{|i\beta|^2}{|1+i\beta|^2}\\ & = \frac{(\sqrt{\beta^2})^2}{(\sqrt{1^2+\beta^2})^2}\\ & = \frac{\beta^2}{1+\beta^2}. \end{aligned}$

We could find the transmission coefficient $T$ in two ways:

(1) By the formula $\displaystyle{T\equiv \frac{|F|^2}{|A|^2}}$ ; or
(2) By the fact that the sum of reflection coefficient and transmission coefficient has to be unity,

i.e., $R+T=1$ .

Calculations:

By (2):

$\begin{aligned} T & = 1-R\\ & = 1-\frac{\beta^2}{1+\beta^2}\\ & = \frac{1}{1+\beta^2} \end{aligned}$

Notice that $R$ and $T$ are functions of $\beta$ , hence:

$\displaystyle{R=\frac{1}{1+(2\hbar E/m\alpha^2)}}$ ,

$\displaystyle{T=\frac{1}{1+(m\alpha^2/2\hbar^2E)}}$ .

Readers should verify this result.

Compare this delta-function barrier with the delta-function trap, we notice that the reflection coefficient and transmission coefficient are each identical respectively in two cases.

April 14, 2021July 25, 2022

202104142057 Sidenote on “deriving momentum operator”

$\begin{aligned} \langle x\rangle &:= \int_{-\infty}^{+\infty}x|\Psi (x,t)|^2\,\mathrm{d}x.\\ \frac{\mathrm{d}\langle x\rangle}{\mathrm{d}t} & = \int_{-\infty}^{+\infty}x\frac{\partial}{\partial t}|\Psi (x,t)|^2\,\mathrm{d}x.\\ & = \int_{-\infty}^{+\infty}x\frac{i\hbar}{2m}\Bigr(\Psi^*\frac{\partial^2\Psi}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi \Bigr)\,\mathrm{d}x\\ & = \frac{i\hbar}{2m}\int_{-\infty}^{+\infty}(x)\Bigr(\Psi^*\frac{\partial^2\Psi}{\partial x^2}+ \frac{\partial \Psi \partial \Psi^*}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi -\frac{\partial \Psi \partial \Psi^*}{\partial x^2}\Bigr)\,\mathrm{d}x\\ & =\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}(x)\Bigr[\frac{\partial}{\partial x}\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}\Bigr)- \frac{\partial}{\partial x}\Bigr(\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\Bigr]\,\mathrm{d}x\\ &= \frac{i\hbar}{2m}\int_{-\infty}^{+\infty}\frac{\partial}{\partial x}(x)\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\,\mathrm{d}x\\ &= -\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\,\mathrm{d}x\\ &= -\frac{i\hbar}{2m}\Bigr[\int_{-\infty}^{+\infty}\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr)-\int_{-\infty}^{+\infty}\Bigr(\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\,\mathrm{d}x\Bigr]\\ &= -\frac{i\hbar}{2m}\Bigr[\Bigr(\Psi^*\Psi \Bigr|^{+\infty}_{-\infty}-\int_{-\infty}^{+\infty}\Psi \frac{\partial \Psi^*}{\partial x}\,\mathrm{d}x\Bigr)-\Bigr(\Psi \Psi^* \Bigr|^{+\infty}_{-\infty}-\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr) \Bigr]\\ & = -\frac{i\hbar}{2m}\Bigr( 0-\int_{-\infty}^{+\infty}\Psi \frac{\partial \Psi^*}{\partial x}\,\mathrm{d}x-0+\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr) \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}\langle x\rangle}{\,\mathrm{d}t} & = -\frac{i\hbar}{2m}\Bigr[-\Bigr(\Psi \Psi^* \Bigr|^{+\infty}_{-\infty}-\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr)+\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr]\\ &= -\frac{i\hbar}{2m}\Bigr(2 \int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr)\\ &= -\frac{i\hbar}{m}\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\\ &= \frac{\hbar}{im}\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\\ \langle p\rangle & := m\frac{\mathrm{d}\langle x\rangle}{\,\mathrm{d}t}\\ & = \int_{-\infty}^{+\infty}\Psi^* \Bigr(\frac{\hbar}{i}\frac{\partial}{\partial x}\Bigr)\Psi \,\mathrm{d}x \end{aligned}$

April 14, 2021October 24, 2022

202104161839 Homework 1 (Q2)

At time $t=0$ a particle is represented by the wave function

$\Psi (x) = \begin{cases} Ax/a, & \textrm{if }0\leq x\leq a \\ A\big( (x-a)^2+1\big), &\textrm{if }a\leq x\leq 2a \\ A(a^2+1)\displaystyle{\frac{3a-x}{a}},&\textrm{if }2a\leq x\leq 3a\\ 0, & \textrm{otherwise} \end{cases}$

where $A$ , $x$ are constants.

(a) Normalize $\Psi$ (that is, find $A$ in terms of $a$ ).
(b) Sketch $\Psi (x,0)$ as a function of $x$ .
(c) Where is the particle most likely to be found, at $t=0$ ?
(d) What is the probability of finding the particle to the left of $a$ .
(e) What is the expectation value of $x$ ?

(a)

$\begin{aligned} 1 & = \int_{-\infty}^{+\infty}|\Psi |^2\,\mathrm{d}x \\ & = \int_{0}^{a}\frac{|A|^2x^2}{a^2}\,\mathrm{d}x + \int_{a}^{2a}|A|^2[(x-a)+1]^2\,\mathrm{d}x+\int_{2a}^{3a}|A|^2(a+1)^2\bigg( \frac{3a-x}{a}\bigg)^2\,\mathrm{d}x \\ & = \frac{|A|^2}{a^2}\bigg[\frac{x^3}{3}\bigg]\bigg|^a_{0} + |A|^2\bigg[\frac{\big( (x-a)+1\big)^3}{3}\bigg]\bigg|_{1}^{a+1} + |A|^2\bigg(\frac{a+1}{a}\bigg)^2\bigg[\frac{(x-3a)^3}{3}\bigg]\bigg|_{-a}^{0} \\ & = \frac{|A|^2a}{3} + |A|^2\bigg(\frac{(2)^3}{3} - \frac{(2-a)^3}{3} \bigg) + |A|^2\bigg( \frac{a+1}{a} \bigg)^2\bigg( \frac{(-3a)^3}{3} - \frac{(-4a)^3}{3} \bigg) \\ \dots \enspace & \textrm{ (to be continued) }\dots \end{aligned}$

Roughwork.

Simplifying the second term,

$\begin{aligned} &\quad |A|^2\bigg(\frac{(2)^3}{3} - \frac{(2-a)^3}{3} \bigg) \\ & = |A|^2\bigg[ \bigg(\frac{8}{3}\bigg) -\bigg( \frac{8-12a+6a^2-a^3}{3}\bigg) \bigg] \\ & = |A|^2\bigg( \frac{4}{3}a - 2a^2 + \frac{a^3}{3}\bigg) \end{aligned}$

Simplifying the third term,

$\begin{aligned} & \quad |A|^2\bigg( \frac{a+1}{a} \bigg)^2\bigg( \frac{(-3a)^3}{3} - \frac{(-4a)^3}{3} \bigg) \\ & = |A|^2\bigg( 1+\frac{2}{a}+\frac{1}{a^2}\bigg) \bigg( \frac{-27a^3}{3} + \frac{64a^3}{3} \bigg)\\ & = |A|^2\bigg( 1+\frac{2}{a}+\frac{1}{a^2}\bigg) \bigg( \frac{37a^3}{3}\bigg) \\ & = |A|^2\bigg( \frac{37}{3}a^3+\frac{74}{3}a^2 + \frac{37}{3}a \bigg) \end{aligned}$

$\begin{aligned} \dots\enspace &\textrm{ (continue) }\dots \\ 1 & = |A|^2 \bigg[ \bigg( \frac{1}{3} + \frac{4}{3} + \frac{37}{3} \bigg) a + \bigg( -2+\frac{74}{3} \bigg) a^2 + \bigg( \frac{1}{3}+\frac{37}{3} \bigg) a^3 \bigg] \\ 1 & = |A|^2 \bigg( 14a + \frac{68}{3}a^2 + \frac{38}{3}a^3 \bigg) \\ A & = \frac{1}{\sqrt{14a+\frac{68}{3}a^2+\frac{38}{3}a^3}} \end{aligned}$

(b)

(c)

at $x=2a$ where $|\Psi |^2$ attains its maximum value.

(d)

$\begin{aligned} P(x\leqslant a) & =\int_{0}^{a}|\Psi |^2\,\mathrm{d}x \\ & = \int_{0}^{a}\bigg|\frac{Ax}{a}\bigg|^2\,\mathrm{d}x \\ & = |A|^2\frac{a^2}{3} \end{aligned}$

(e)

$\begin{aligned} \langle x\rangle & = \int_{0}^{3a}x|\Psi |^2\,\mathrm{d}x \\ & = |A|^2\bigg( \int_{0}^{a}\frac{x^3}{a^2}\,\mathrm{d}x + \int_{a}^{2a}[(x-a)^2+1]^2x\,\mathrm{d}x + \int_{2a}^{3a}(a^2+1)^2\bigg(\frac{3a-x}{a}\bigg)^2 x\,\mathrm{d}x \bigg) \\ \dots\enspace & \textrm{ (to be continued) }\dots \end{aligned}$

November 11, 2020July 25, 2022

202011110657 Sidenote of Separation of Variables

Two-particle system: $\displaystyle{i\hbar\frac{\partial \Psi}{\partial t}=H\Psi}$

in which $\displaystyle{H=-\frac{\hbar^2}{2m_1}\nabla_1^2-\frac{\hbar^2}{2m_2}\nabla_2^2+V(\mathbf{r_1},\mathbf{r_2},t)}$ .

Assuming that the potential is time-independent.

Let $\Psi (\mathbf{r}_1,\mathbf{r}_2,t)=\psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)$ ,

$\begin{aligned} & i\hbar \frac{\partial \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)}{\partial t} \\ = & -\frac{\hbar^2}{2m_1}\nabla_1^2\bigg( \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)\bigg) -\frac{\hbar^2}{2m_2}\nabla_2^2\bigg( \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t) \bigg) \\ &\qquad\enspace +V(\mathbf{r_1},\mathbf{r_2})\psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t) \\ & \\ & \textrm{\scriptsize OR} \\ & \\ & i\hbar\Big[ \frac{\partial \psi (\mathbf{r}_1,\mathbf{r}_2)}{\partial t}\varphi (t)+\psi (\mathbf{r}_1,\mathbf{r}_2)\frac{\partial \varphi (t)}{\partial t}\Big]\\ = & -\frac{\hbar^2}{2m_1}\varphi (t)\nabla_1^2\psi (\mathbf{r}_1,\mathbf{r}_2)-\frac{\hbar^2}{2m_2}\varphi (t)\nabla_2^2\psi (\mathbf{r}_1,\mathbf{r}_2) +V\psi \varphi \\ \end{aligned}$

April 24, 2020October 24, 2022

202004240406 Homework 1 (Q6)

Derive the Planck function $B_\nu$ .

HINTS: You can make the following steps.

(1) Show that the density of states for photons is given by

$g(E)=\displaystyle{\frac{\mathrm{d}N}{\mathrm{d}E}=\frac{8\pi VE^2}{h^3c^3}}$ ,

(2) the photon density per unit frequency is given by

$\displaystyle{\frac{\mathrm{d}n}{\mathrm{d}\nu}}=\displaystyle{\frac{8\pi \nu^2}{c^3(e^{h\nu /kT+1})}}$ , and

(3) the definition of the specific intensity $B_\nu =I_\nu$ is the energy flux per unit frequency per unit solid angle.

Note: the energy and momentum of photons are related as

$E=h\nu =pc$ ,

the energy flux per unit frequency is given by

$h\nu \displaystyle{\frac{\mathrm{d}n}{\mathrm{d}\nu}}c$ ,

and total solid angle for isotropic emission is $4\pi$ .

Attempts.

In attempting to derive the Planck function, extensive reference was made to the following several sources found on the Internet:

i. G. B. Rybicki & A. P. Lightman. (1979). Radiative Processes in Astrophysics. John Wiley & Sons, Inc.

ii. https://edisciplinas.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank(DOT)pdf

iii. https://en.wikipedia(DOT)org/wiki/Planck\%27s_law#Derivation

iv. http://web.phys.ntnu.no/~stovneng/TFY4165_2013/BlackbodyRadiation(DOT)pdf

Here I follow Rybicki and Lightman’s derivation (pp. 20–22, 1979) direct, and have changed only some wordings.

Solution.

The wave vector of the photon of frequency $\nu$ propagating in direction $\mathbf{n}$ is $\mathbf{k}=(2\pi /\lambda )\,\mathbf{n}=(2\pi\nu /c)\,\mathbf{n}$ . For a photon gas in a box of dimension $(L_x,L_y,L_z)$ , the number of nodes of the standing wave is $n_i=k_iL_i/2\pi$ for each direction $i\in\{ x,y,z\}$ and for some wave number $k_i$ . For all directions, write the variation of the number of nodes with the wave number

$\Delta n_i=\displaystyle{\frac{L_i\Delta k_i}{2\pi}}$

The three-dimensional wave vector element $\Delta k_x\Delta k_y\Delta k_z\equiv \mathrm{d}^3k$ has the number of states to be

$\Delta N=\Delta n_x\Delta n_y\Delta n_z=\displaystyle{\frac{L_xL_yL_z\,\mathrm{d}^3k}{(2\pi )^3}}$

By the fact that $L_xL_yL_z=V$ is the volume of the box and the fact that photons have two independent polarizations (i.e., two states per wave vector $\mathbf{k}$ ), for each 3D-wave vector, the number of states for every unit volume is

$\displaystyle{\frac{\Delta N}{V\,\mathrm{d}^3k}}=2/(2\pi )^3$ .

Using solid angle as in spherical coordinates, rewrite

$\mathrm{d}^3k=k^2\,\mathrm{d}k\,\mathrm{d}\Omega =\displaystyle{\frac{(2\pi )^3\nu^2\,\mathrm{d}\nu\,\mathrm{d}\Omega}{c^3}}$ .

Then, the density of states, i.e., the number of states per solid angle per volume per frequency, is given by

$\rho_s\stackrel{\textrm{def}}{=}\displaystyle{\frac{\mathrm{d}N}{\mathrm{d}\Omega\,\mathrm{d}V\,\mathrm{d}\nu}}=\displaystyle{\frac{2\nu^2}{c^3}}$ .

As each state of $n$ photons each of energy $h\nu$ is of energy $E_n=nh\nu$ , and according to statistical mechanics the probability of a state of energy $E_n$ is proportional to $\mathrm{exp}(-\beta E_n)$ (where $\beta =1/kT$ and $k=\textrm{ the Boltzmann's constant}$ ), the average energy of each state is

$\overline{E}=\displaystyle{\frac{\sum_{n=0}^\infty E_n\mathrm{exp}(-\beta E_n)}{\sum_{n=0}^\infty \mathrm{exp}(-\beta E_n)}}=-\displaystyle{\frac{\mathrm{\partial}}{\mathrm{\partial}\beta}\ln \bigg( \sum_{n=0}^\infty \mathrm{exp}(-\beta E_n)\bigg)}$ .

Recall the formula for the sum of a geometric series:

$\displaystyle{\sum_{n=0}^\infty}\mathrm{exp}(-\beta E_n)=\displaystyle{\sum_{n=0}^\infty}\mathrm{exp}(-nh\nu\beta )=(1-\mathrm{exp}(-\beta h\nu ))^{-1}$ ,

one has therefore the result (in Bose-Einstein statistics)

$\overline{E}=\displaystyle{\frac{h\nu\,\mathrm{exp}(-\beta h\nu )}{1-\mathrm{exp}(-\beta h\nu )}=\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}$ .

where it can be seen that the occupation number

$n_\nu =\bigg[ \mathrm{exp}\bigg( \displaystyle{\frac{h\nu}{kT}}\bigg) -1\bigg]^{-1}$

means the average number of photons in some frequency $\nu$ , as one energy $h\nu$ corresponds to one frequency $\nu$ .

The energy per solid angle per volume per frequency is the product of i. $\overline{E}$ , the average energy of each state, and ii. $\rho_s$ , the density of states.

I.e., $\overline{E}\cdot \rho_s = \bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}$

Define the specific energy density $u_\nu$ the energy per unit volume per unit frequency range. Then the energy density per unit solid angle can be expressed as

$\begin{aligned} u_\nu (\Omega ) & =\displaystyle{\frac{\mathrm{d}E}{\mathrm{d}V\,\mathrm{d}\Omega\,\mathrm{d}\nu}} \\ \dots \textrm{arranging } & \textrm{and equating the previous two equations}\dots \\ u_\nu (\Omega )\,\mathrm{d}V\,\mathrm{d}\nu\,\mathrm{d}\Omega & =\bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}\,\mathrm{d}V\,\mathrm{d}\nu\,\mathrm{d}\Omega \\ \dots \textrm{comparing} & \dots \\ u_\nu (\Omega ) & = \bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}} \end{aligned}$

Observe that the specific energy density $u_\nu$ and the specific intensity (or brightness) $I_\nu$ are related by

$u_\nu (\Omega )=\displaystyle{\frac{I_\nu}{c}}$ .

Now that the specific intensity, $I_\nu$ , is the same as the source function (of specific emission mechanism), $B_\nu$ .
The frequency distribution is said to be a blackbody form, i.e.,

The Planck function is expressed as

$\boxed{I_\nu =B_\nu (T)=\displaystyle{\frac{2h\nu^3}{c^2}}\bigg[ \mathrm{exp}\bigg( \displaystyle{\frac{h\nu}{kT}} \bigg) -1\bigg]^{-1}}$

(Units: $\mathrm{erg\cdot s^{-1}\cdot Hz^{-1}\cdot cm^{-2}\cdot ster^{-1}}$ )