Quantum Mechanics * – Page 2

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

July 26, 2019July 28, 2023

201907260745 Homework 1 (Q1)

Consider a probability density of the Gaussian distribution

$|\Psi |^2=\rho (x)=Ae^{-\lambda(x-a)^2}$

where $A$ , $a$ , and $\lambda$ are constants.

You probably wish to know .

Determine $A$ according to the normalizing rules.
Find $\langle x\rangle$ , $\langle x^2\rangle$ , and $\sigma$ , the standard deviation of $x$ .
Sketch the graph of $\rho (x)$ .

Solution:

Roughwork.

$\begin{aligned}1 & = \int_{-\infty}^{+\infty} |\Psi |^2 \mathrm{d}x \\1& = \int_{-\infty}^{+\infty} Ae^{-\lambda (x-a)^2}\mathrm{d}x \\\frac{1}{A} & = \int_{-\infty}^{+\infty} e^{-\big(\lambda^{\frac{1}{2}}(x-a)\big)^2} \mathrm{d}x \\\dots & \Bigg( \because \enspace \frac{\mathrm{d}\big( \lambda^{\frac{1}{2}}(x-a) \big)}{\mathrm{d}x} = \sqrt{\lambda} \Bigg) \dots \\\frac{1}{A} & = \frac{1}{\sqrt{\lambda}} \int_{-\infty}^{+\infty} e^{-\big(\lambda^{\frac{1}{2}}(x-a)\big)^2} \mathrm{d}\big( \lambda^{\frac{1}{2}}(x-a) \big) \\\frac{1}{A} & =\sqrt{\frac{\pi}{\lambda}} \\A & = \sqrt{\frac{\lambda}{\pi}}\end{aligned}$
Recall , Eq. (1.17) in Griffith's Introduction to Quantum Mechanics

First,

$\begin{aligned} \langle x\rangle & = A\int_{-\infty}^{+\infty} xe^{-\lambda (x-a)^2}\mathrm{d}x \\ & = A\int_{-\infty}^{+\infty} (u+a)e^{-\lambda u^2}\mathrm{d}u \\ & = A \int_{-\infty}^{+\infty} u\,\mathrm{d}u + Aa\int_{-\infty}^{+\infty}e^{-\lambda u^2}\,\mathrm{d}u \\ & = 0+Aa\sqrt{\frac{\pi}{\lambda}} \\ & = a \end{aligned}$

The section below sets a bad example of computation, the second equality sign being wishful thinking, and what follows thence is incorrect.

$\begin{aligned} \langle x^2\rangle & = A\int_{-\infty}^{+\infty} x^2e^{-\lambda (x-a)^2}\mathrm{d}x \\ & = A \int_{-\infty}^{+\infty} x^2e^{-\lambda x^2}\mathrm{d}x + A \int_{-\infty}^{+\infty} x^2e^{2\lambda ax}\mathrm{d}x + A \int_{-\infty}^{+\infty} x^2e^{-\lambda a^2}\mathrm{d}x \end{aligned}$
Recall the formula for integration by parts is , or, .

Evaluate term-by-term, the first term is

$\begin{aligned} &\quad A\int_{-\infty}^{+\infty} x^2e^{-\lambda x^2}\,\mathrm{d}x\\ & = \frac{A}{2} \int_{-\infty}^{+\infty} xe^{-\lambda x^2}\,\mathrm{d}(x^2) \\ & =\frac{A}{2} \Bigg\{ \bigg[ \frac{-xe^{-\lambda x^2}}{\lambda}\bigg]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty} \frac{-e^{-\lambda x^2}}{\lambda}\,\mathrm{d}(x^2) \Bigg\} \\ \dots & \Bigg( \quad \mathrm{d}(\sqrt{\lambda}x^2) = \frac{1}{2\sqrt{\lambda}}\,\mathrm{d}(x^2) \quad \Bigg) \dots \\ & =\frac{A}{2} \Bigg\{ \bigg[ \frac{-xe^{-\lambda x^2}}{\lambda}\bigg]_{-\infty}^{+\infty} +\frac{2\sqrt{\lambda}}{\lambda}\int_{-\infty}^{+\infty}e^{-(\sqrt{\lambda} x)^2} \mathrm{d}(\sqrt{\lambda}x^2) \Bigg\} \\ & = \frac{A}{2} \bigg( 0+2\sqrt{\frac{\pi}{\lambda}}\bigg) \\ & = \frac{\sqrt{\frac{\lambda}{\pi}}}{2} \bigg( 2\sqrt{\frac{\pi}{\lambda}}\bigg) \\ & = 1 \end{aligned}$

The terrible blunder ends here.

Correction.

$\begin{aligned} \langle x^2 \rangle & = \int_{-\infty}^{+\infty} x^2Ae^{-\lambda (x-a)^2}\mathrm{d}x \\ & = A\int_{-\infty}^{+\infty} (u+a)^2 e^{\lambda u^2}\mathrm{d}u \\ & = A\int_{-\infty}^{+\infty} u^2e^{-\lambda u^2}\mathrm{d}u + 2Aa\int_{-\infty}^{+\infty} ue^{-\lambda u^2}\mathrm{d}u + Aa^2\int_{-\infty}^{+\infty}e^{-\lambda u^2}\mathrm{d}u \end{aligned}$

Step back to look closer, the third term is the easiest to compute:

$\begin{aligned} & \quad Aa^2\int_{-\infty}^{+\infty}e^{-\lambda u^2}\mathrm{d}u \\ & = \frac{Aa^2}{\sqrt{\lambda}}\int_{-\infty}^{+\infty}e^{-(\sqrt{\lambda}u)^2}\mathrm{d}(\sqrt{\lambda}u) \\ & =Aa^2\sqrt{\frac{\pi}{\lambda}}\\ & = \sqrt{\frac{\lambda}{\pi}} a^2\sqrt{\frac{\pi}{\lambda}}\\ & = a^2 \end{aligned}$

The first two terms might need to be evaluated using integration by parts. On the other hand, from the angle of parity, in the second term

$2Aa\displaystyle{\int_{-\infty}^{+\infty}} f(u)\,\mathrm{d}u$

where $f(u)\stackrel{\mathrm{def}}{=}ue^{-\lambda u^2}$

$f(-u)=(-u)e^{-\lambda(-u)^2}=-f(u)$

is an odd function. The definite integral upon evaluation will be nought:

$2Aa\displaystyle{\int_{-\infty}^{+\infty}} ue^{-\lambda u^2}\,\mathrm{d}u \equiv 0$ .

The first term can be checked

$A \displaystyle{\int}g(u)\,\mathrm{d}u$

where $g(u) \stackrel{\mathrm{def}}{=}u^2e^{-\lambda u^2}$

that $g(-u)=(-u)^2e^{-\lambda (-u)^2}=g(u)$ is an even function.

Upon evaluation the definite integral will have the property that

$A \displaystyle{\int_{-\infty}^{+\infty}}u^2e^{-\lambda u^2}\,\mathrm{d}u = 2A \displaystyle{\int_{0}^{+\infty}}u^2e^{-\lambda u^2}\,\mathrm{d}u$ ,

though it seems not useful here. Doing integration by parts,

$\begin{aligned} & \quad A \displaystyle{\int_{-\infty}^{+\infty}}u^2e^{-\lambda u^2}\mathrm{d}u \\ & = \frac{A}{\sqrt{\lambda}} \int_{-\infty}^{+\infty} u^2e^{-(\sqrt{\lambda}u)^2}\mathrm{d}(\sqrt{\lambda}u) \\ & = \frac{A}{\sqrt{\lambda}} \Bigg\{ \bigg[ \frac{u^2e^{-(\sqrt{\lambda}u)^2}}{-2(\sqrt{\lambda}u)} \bigg]\bigg|_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty} \frac{e^{-(\sqrt{\lambda}u)^2}}{-2(\sqrt{\lambda}u)} \Big( \frac{2u}{\sqrt{\lambda}} \Big) \mathrm{d}(\sqrt{\lambda}u) \Bigg\} \\ & = \frac{A}{\sqrt{\lambda}} \Bigg\{ 0+ \frac{1}{\lambda}\int_{-\infty}^{+\infty} e^{-(\sqrt{\lambda}u)^2} \mathrm{d}(\sqrt{\lambda}u) \Bigg\} \\ & = A\frac{1}{\sqrt{\lambda}}\frac{1}{\lambda}\sqrt{\pi} \\ & = \sqrt{\frac{\lambda}{\pi}} \frac{1}{\sqrt{\lambda}}\frac{1}{\lambda}\sqrt{\pi} \\ & = \frac{1}{\lambda} \end{aligned}$

Thus $\langle x^2\rangle = \frac{1}{\lambda} + a^2$

As $\sigma^2=\langle x^2\rangle -\langle x\rangle^2$ ,

$\sigma^2 = \frac{1}{\lambda} + a^2 - a^2 = \frac{1}{\lambda}$ .