April 14, 2021 – Physicspupil

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)

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