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,

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