Chapter 2, Problem 2.42P

(a)

To determine

The normalized wave function.

Answer to Problem 2.42P

The normalized wave function is $\underset{\_}{\Psi \left(x,0\right)={\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}{e}^{ilx}}$.

Explanation of Solution

Given that the wave function is;

  $\Psi \left(x,0\right)={\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}{e}^{ilx}$        (I)

Normalize the wave function.

  $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\Psi \right|}^{2}dx}=1\\ {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|A{e}^{-a{x}^2}{e}^{ilx}\right|}^{2}dx}=1\\ A^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-2a{x}^2}dx}=1\\ A^2\sqrt{\frac{\pi }{2a}}=1\\ A={\left(\frac{2a}{\pi }\right)}^{1/4}\end{array}$        (II)

Use the normalization constant to rewrite the wave function.

  $\Psi \left(x,0\right)={\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}{e}^{ilx}$        (III)

Conclusion:

Therefore, the normalized wave function is $\underset{\_}{\Psi \left(x,0\right)={\left(\frac{2a}{\pi }\right)}^{1/4}{e}^{-a{x}^2}{e}^{ilx}}$.

(b)

To determine

The wave function at time $t$, the speed of the Gaussian envelop and the speed of the wave.

Answer to Problem 2.42P

The wave function at time $t$ is $\underset{\_}{\Psi \left(x,t\right)={\left(\frac{2a}{\pi }\right)}^{1/4}\frac{1}{\gamma }{e}^{-a{\left(x-\frac{\hslash lt}{m}\right)}^2/{\gamma }^2}{e}^{il\left(x-\frac{\hslash lt}{2m}\right)}}$. The speed of the Gaussian envelope is $\underset{\_}{\hslash l/m}$ and the speed of the wave is $\underset{\_}{\hslash l/2m}$.

Explanation of Solution

Write the solution to the generic quantum problem, for a free particle;

  $\varphi \left(k\right)=\frac{1}{\sqrt{2\pi }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\Psi \left(x,0\right)e^{-ikx}}dx$        (IV)

Use equation (III) in (IV).

  $\varphi \left(k\right)=\frac{1}{\sqrt{2\pi }}{\left(\frac{2a}{\pi }\right)}^{1/4}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{x}^2}{e}^{ilx}{e}^{-ikx}}dx$        (V)

Thus, the wave function at time $t$ can be represented as;

  $\Psi \left(x,t\right)=\frac{1}{\sqrt{2\pi }}\frac{1}{{\left(2\pi a\right)}^{1/4}}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-\frac{{\left(k-l\right)}^2}{4a}}{e}^{i\left(kx-\frac{\hslash k^{2}t}{2m}\right)}dk}$        (VI)

Let $u\equiv k-l$, so $k=u+l$ and $dk=du$ (since $l$ is constant). Now equation (VI) can be written as;

  $\begin{array}{c}\Psi \left(x,t\right)=\frac{1}{\sqrt{2\pi }}\frac{1}{{\left(2\pi a\right)}^{1/4}}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-\frac{u^2}{4a}}{e}^{i\left[ux+lx-\left(\hslash t/2m\right)\left(u^2+2ul+l^2\right)\right]}du}\\ =\frac{1}{\sqrt{2\pi }}\frac{1}{{\left(2\pi a\right)}^{1/4}}{e}^{il\left(x-\frac{\hslash lt}{2m}\right)}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-u^2\left(\frac{1}{4a}+i\frac{\hslash t}{2m}\right)+iu\left(x-\frac{\hslash lt}{m}\right)}du}\end{array}$        (VII)

The integral in equation (VII) becomes;

  $\frac{1}{\sqrt{\frac{1}{4a}+i\frac{\hslash t}{2m}}}{e}^{{\left(x-\frac{\hslash lt}{m}\right)}^2/4\left(\frac{1}{4a}+i\frac{\hslash t}{2m}\right)}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-y^2}dy}=\frac{2\sqrt{a}}{\gamma }{e}^{-a{\left(x-\frac{\hslash lt}{m}\right)}^2/{\gamma }^2}\sqrt{\pi }$

Thus, the wave function at time $t$ can be expressed as;

  $\Psi \left(x,t\right)={\left(\frac{2a}{\pi }\right)}^{1/4}\frac{1}{\gamma }{e}^{-a{\left(x-\frac{\hslash lt}{m}\right)}^2/{\gamma }^2}{e}^{il\left(x-\frac{\hslash lt}{2m}\right)}$        (VIII)

In the above expression, the first exponential term which represent the Gaussian envelop travels at speed $\hslash l/m$, and the second exponential term which represent the sinusoidal wave travel at speed $\hslash l/2m$.

Conclusion:

Therefore, the wave function at time $t$ is $\underset{\_}{\Psi \left(x,t\right)={\left(\frac{2a}{\pi }\right)}^{1/4}\frac{1}{\gamma }{e}^{-a{\left(x-\frac{\hslash lt}{m}\right)}^2/{\gamma }^2}{e}^{il\left(x-\frac{\hslash lt}{2m}\right)}}$. The speed of the Gaussian envelope is $\underset{\_}{\hslash l/m}$ and the speed of the wave is $\underset{\_}{\hslash l/2m}$.

(c)

To determine

The probability density ${\left|\Psi \left(x,t\right)\right|}^2$ and to plot the same.

Answer to Problem 2.42P

The probability density is $\underset{\_}{{\left|\Psi \left(x,t\right)\right|}^2=\sqrt{\frac{2}{\pi }}\omega e^{-2{\omega }^2{\left(x-\frac{\hslash lt}{m}\right)}^2}}$ where $\omega =\sqrt{\frac{a}{1+{\theta }^2}}$. The plot of ${\left|\Psi \left(x,t\right)\right|}^2$ is shown in Figure 1.

Explanation of Solution

The wave function is obtained from part (b) is given in equation (VIII).

  $\Psi \left(x,t\right)={\left(\frac{2a}{\pi }\right)}^{1/4}\frac{1}{\gamma }{e}^{-a{\left(x-\frac{\hslash lt}{m}\right)}^2/{\gamma }^2}{e}^{il\left(x-\frac{\hslash lt}{2m}\right)}$

The probability density can be expressed as;

  ${\left|\Psi \left(x,t\right)\right|}^2={\left(\frac{2a}{\pi }\right)}^{1/2}\frac{1}{{\left|\gamma \right|}^2}{e}^{a{\left(x-\frac{\hslash lt}{m}\right)}^2\left[\frac{1}{{\gamma }^2}+\frac{1}{{\left({\gamma }^{*}\right)}^2}\right]}$        (IX)

Simplify the term in the square bracket.

  $\begin{array}{c}\left[\frac{1}{{\gamma }^2}+\frac{1}{{\left({\gamma }^{*}\right)}^2}\right]=\frac{1}{{\left|\gamma \right|}^4}\left[{\left({\gamma }^{*}\right)}^2+{\gamma }^2\right]\\ =\frac{1}{{\left|\gamma \right|}^4}\left(1-\frac{2i\hslash t}{m}+1+\frac{2i\hslash t}{m}\right)\\ =\frac{2}{{\left|\gamma \right|}^4}\end{array}$        (X)

The term ${\left|\gamma \right|}^2$ can be expressed as;

  $\begin{array}{c}{\left|\gamma \right|}^2=\sqrt{\left(1+\frac{2ia\hslash t}{m}\right)\left(1-\frac{2ia\hslash t}{m}\right)}\\ =\sqrt{1+{\theta }^2}\end{array}$        (XI)

Where, $\theta \equiv \frac{2a\hslash t}{m}$.

Thus, equation (IX) can be modified as;

  $\begin{array}{c}{\left|\Psi \left(x,t\right)\right|}^2={\left(\frac{2a}{\pi }\right)}^{1/2}\frac{1}{\sqrt{1+{\theta }^2}}{e}^{2a{\left(x-\frac{\hslash lt}{m}\right)}^2/\left(1+{\theta }^2\right)}\\ =\sqrt{\frac{2}{\pi }}\omega e^{-2{\omega }^2{\left(x-\frac{\hslash lt}{m}\right)}^2}\end{array}$        (XI)

Where $\omega =\sqrt{\frac{a}{1+{\theta }^2}}$.with $\theta =\frac{2\hslash at}{m}$.

Thus, the graph of ${\left|\Psi \left(x,t\right)\right|}^2$ will be flattening Gaussian as shown in Figure 1.

Conclusion:

Therefore, the probability density is $\underset{\_}{{\left|\Psi \left(x,t\right)\right|}^2=\sqrt{\frac{2}{\pi }}\omega e^{-2{\omega }^2{\left(x-\frac{\hslash lt}{m}\right)}^2}}$ where $\omega =\sqrt{\frac{a}{1+{\theta }^2}}$. The plot of ${\left|\Psi \left(x,t\right)\right|}^2$ is shown in Figure 1.

(d)

To determine

The expectation values $\langle x\rangle$, $\langle p\rangle$, $\langle x^2\rangle$, $\langle p^2\rangle$ and ${\sigma }_x$ and ${\sigma }_p$

Answer to Problem 2.42P

The expectation values and $\sigma$ values are obtained as; $\langle x\rangle =\frac{\hslash l}{m}t$, $\langle x^2\rangle =\frac{1}{4{\omega }^2}+{\left(\frac{\hslash lt}{m}\right)}^2$, $\langle p\rangle =\hslash l$, $\langle p^2\rangle ={\hslash }^2\left(a+l^2\right)$, ${\sigma }_x=\frac{1}{2\omega }$, and ${\sigma }_p=\hslash \sqrt{a}$.

Explanation of Solution

Write the expression for the expectation value of $x$.

  $\langle x\rangle ={\displaystyle {\int }_{-\infty }^{\infty }x{\left|\Psi \left(x,t\right)\right|}^{2}dx}$        (XII)

Use equation (XI) in (IX) and solve the integral.

  $\langle x\rangle =\sqrt{\frac{2}{\pi }}{\displaystyle {\int }_{-\infty }^{\infty }x\omega e^{-2{\omega }^2{\left(x-\frac{\hslash lt}{m}\right)}^2}dx}$

Let $y\equiv x-vt$, so $x=y+vt$.

  $\begin{array}{c}\langle x\rangle =\sqrt{\frac{2}{\pi }}{\displaystyle {\int }_{-\infty }^{\infty }\left(y+vt\right)\omega e^{-2{\omega }^2{y}^2}dy}\\ \langle x\rangle =vt\\ \langle x\rangle =\frac{\hslash l}{m}t\end{array}$

Here, the first integral is trivially zero; and the second is $1$ by normalization.

The expectation value of $p$ can be found as;

  $\langle p\rangle =m\frac{d\langle x\rangle }{dt}$        (XIII)

Substitute $\langle x\rangle$ in (XIII).

  $\begin{array}{c}\langle p\rangle =m\frac{d}{dt}\left(\frac{\hslash l}{m}t\right)\\ \langle p\rangle =\hslash l\end{array}$

The expectation value $\langle x^2\rangle$ can be determined as;

  $\begin{array}{c}\langle x^2\rangle =\sqrt{\frac{2}{\omega }}{\displaystyle {\int }_{-\infty }^{\infty }{\left(y+vt\right)}^2}\omega e^{-2{\omega }^2{y}^2}dy\\ \langle x^2\rangle =\frac{1}{4{\omega }^2}+0+{\left(vt\right)}^2\\ \langle x^2\rangle =\frac{1}{4{\omega }^2}+{\left(\frac{\hslash lt}{m}\right)}^2\end{array}$

The expectation value $\langle p^2\rangle$ can be found out as;

  $\langle p^2\rangle =-{\hslash }^2{\displaystyle {\int }_{-\infty }^{\infty }{\Psi }^{*}\frac{{\partial }^2\Psi }{\partial x^2}dx}$        (XIV)

  $\begin{array}{c}\frac{\partial \Psi }{\partial x}=\left[-\frac{2a}{{\gamma }^2}\left(x-\frac{\hslash lt}{m}\right)+il\right]\Psi \\ \frac{{\partial }^2\Psi }{\partial x^2}=-\frac{2a}{{\gamma }^2}\Psi +{\left[-\frac{2a}{{\gamma }^2}\left(x-\frac{\hslash lt}{m}\right)+il\right]}^2\Psi \\ \frac{{\partial }^2\Psi }{\partial x^2}=\left(A{x}^2+Bx+C\right)\Psi \end{array}$

Where, $A$, $B$, and $C$ are defined as;

  $\begin{array}{c}A\equiv \left(\frac{2a}{{\gamma }^2}\right)\\ B\equiv -{\left(\frac{2a}{{\gamma }^2}\right)}^2\frac{2\hslash lt}{m}-\frac{4ial}{{\gamma }^2}\\ B\equiv -\frac{4ial}{{\gamma }^2}\end{array}$

  $\begin{array}{c}C\equiv -\frac{2a}{{\gamma }^2}+{\left(\frac{2a}{{\gamma }^2}\right)}^2{\left(\frac{\hslash lt}{m}\right)}^2+\left(\frac{4ial}{{\gamma }^2}\right)\left(\frac{\hslash lt}{m}\right)-l^2\\ C\equiv -\frac{1}{{\gamma }^4}\left(2a{\gamma }^2+l^2\right)\end{array}$

Write $\langle p^2\rangle$ in terms of $A$, $B$, and $C$.

  $\begin{array}{c}\langle p^2\rangle =-{\hslash }^2{\displaystyle {\int }_{-\infty }^{\infty }{\Psi }^{*}\left[A{x}^2+Bx+C\right]}\Psi dx\\ =-{\hslash }^2\left[A\langle x^2\rangle +B\langle x\rangle +C\right]\end{array}$        (XV)

Use $\langle x^2\rangle$ and $\langle x\rangle$ in equation (XV).

  $\begin{array}{c}\langle p^2\rangle =-\frac{{\hslash }^2}{{\gamma }^4}\left[4a^2\left(\frac{1}{4{\omega }^2}+{\left(\frac{\hslash lt}{m}\right)}^2\right)-4ial\left(\frac{\hslash lt}{m}\right)-\left(2a{\gamma }^2+l^2\right)\right]\\ \langle p^2\rangle =-\frac{{\hslash }^2}{{\gamma }^4}\left\{\left[a+a{\left(\frac{2\hslash at}{m}\right)}^2-2a-\frac{4i{a}^2{\hslash }^{2}t}{m}\right]+l^2\left[-1-\frac{4ia\hslash t}{m}+4{\left(\frac{\hslash at}{m}\right)}^2\right]\right\}\\ \langle p^2\rangle =-\frac{{\hslash }^2}{{\gamma }^4}\left(-a{\gamma }^4-l^2{\gamma }^4\right)\\ \langle p^2\rangle ={\hslash }^2\left(a+l^2\right)\end{array}$

The values of ${\sigma }_x$ and ${\sigma }_p$ can be found as follows.

  $\begin{array}{c}{\sigma }_x^2=\langle x^2\rangle -{\langle x\rangle }^2\\ {\sigma }_x^2=\frac{1}{4{\omega }^2}+{\left(\frac{\hslash lt}{m}\right)}^2-{\left(\frac{\hslash lt}{m}\right)}^2\\ {\sigma }_x=\sqrt{\frac{1}{4{\omega }^2}}\\ {\sigma }_x=\frac{1}{2\omega }\end{array}$

  $\begin{array}{c}{\sigma }_p^2=\langle p^2\rangle -{\langle p\rangle }^2\\ {\sigma }_p^2={\hslash }^{2}a+{\hslash }^2{l}^2-{\hslash }^2{l}^2\\ {\sigma }_p=\sqrt{{\hslash }^{2}a}\\ {\sigma }_p=\hslash \sqrt{a}\end{array}$

Conclusion:

Therefore, the expectation values and $\sigma$ values are obtained as; $\langle x\rangle =\frac{\hslash l}{m}t$, $\langle x^2\rangle =\frac{1}{4{\omega }^2}+{\left(\frac{\hslash lt}{m}\right)}^2$, $\langle p\rangle =\hslash l$, $\langle p^2\rangle ={\hslash }^2\left(a+l^2\right)$, ${\sigma }_x=\frac{1}{2\omega }$, and ${\sigma }_p=\hslash \sqrt{a}$.

(e)

To determine

Whether the uncertainty principle holds or not.

Answer to Problem 2.42P

The uncertainty principle holds for the given case.

Explanation of Solution

The values of ${\sigma }_x$ and ${\sigma }_p$ are obtained as;

  ${\sigma }_x=\frac{1}{2\omega }$

  ${\sigma }_p=\hslash \sqrt{a}$

The product of  ${\sigma }_x$ and ${\sigma }_p$ is obtained as;

  ${\sigma }_x{\sigma }_p=\frac{1}{2\omega }\hslash \sqrt{a}$

Use the expression for $\omega$ in the uncertainty product.

  $\begin{array}{c}{\sigma }_x{\sigma }_p=\frac{\hslash \sqrt{a}}{2\left(\sqrt{a/1+{\left(\frac{2\hslash at}{m}\right)}^2}\right)}\\ {\sigma }_x{\sigma }_p\ge \frac{\hslash }{2}\end{array}$

Thus, uncertainty principle holds for this case.

Conclusion:

Therefore, the uncertainty principle holds for the given case.