Chapter 6, Problem 37P

(a)

To determine

Verify that given value of $C$ normalizes the wave function.

Answer to Problem 37P

The value of $C$ that normalizes the wave is $1/\sqrt{2}$ .

Explanation of Solution

Write the expression for normalization constant.

  $1={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\text{ψ}\left(\text{x}\right)\right|}^{2}dx}$        (I)

Substitute $C\left[{\psi }_1\left(x\right)+{\psi }_2\left(x\right)\right]$ for $\text{ψ}\left(\text{x}\right)$ in equation (I).

  $\begin{array}{c}1=C^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\left[{\psi }_1\left(x\right)+{\psi }_2\left(x\right)\right]\right|}^{2}dx}\\ 1=C^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left\{{\psi }_1^{*}+{\psi }_2^{*}\right\}\left\{{\psi }_1+{\psi }_2\right\}dx}\end{array}$

Expand the above expression.

  $1=C^2\left\{{\displaystyle \int {\left|{\psi }_1\right|}^{2}dx}+{\displaystyle \int {\left|{\psi }_2\right|}^{2}dx}+{\displaystyle \int {\psi }_2^{*}{\psi }_{1}dx}+{\displaystyle \int {\psi }_1^{*}{\psi }_{2}dx}\right\}$        (II)

Conclusion:

${\psi }_1\left(x\right)$ and ${\psi }_2\left(x\right)$ are normalized function; so, the value of first two integrals is equal to 1.

For last two integrals, the integral value of both is same as the functions ${\psi }_1\left(x\right)$ and ${\psi }_2\left(x\right)$ is real.

Solve the last integral part using waveform.

  $\begin{array}{c}{\displaystyle \int {\psi }_2{\psi }_{1}dx}=\frac{2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\sin \left(\frac{\pi x}{L}\right)\sin \left(\frac{2\pi x}{L}\right)dx}\\ =\frac{1}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\left\{\cos \left(\frac{\pi x}{L}\right)-\cos \left(\frac{3\pi x}{L}\right)\right\}dx}\\ =\frac{1}{L}{\left[\frac{L}{\pi }\sin \left(\frac{\pi x}{L}\right)-\frac{L}{3\pi }\sin \left(\frac{3\pi x}{L}\right)\right]}_0^L\\ =0\end{array}$

Substitute 1 for ${\displaystyle \int {\left|{\psi }_1\right|}^{2}dx}$ , 1 for ${\displaystyle \int {\left|{\psi }_2\right|}^{2}dx}$ , 0 for ${\displaystyle \int {\psi }_2^{*}{\psi }_{1}dx}$ and 0 for ${\displaystyle \int {\psi }_1^{*}{\psi }_{2}dx}$ in equation (II).

  $\begin{array}{c}1=C^2\left\{1+1+0+0\right\}\\ 1=2C^2\\ C=\frac{1}{\sqrt{2}}\end{array}$

Thus, the value of $C$ that normalizes the wave is $1/\sqrt{2}$ .

(b)

To determine

The value of function at any later time $t$ .

Answer to Problem 37P

The function at any later time $t$  is $\text{ψ}\left(x,t\right)=C\left\{{\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}$ .

Explanation of Solution

Write the expression for time dependent wave function.

  $\text{ψ}\left(x,t\right)=C\left\{{\psi }_1\left(x\right)e^{-i{\omega }_{1}t}+{\psi }_2\left(x\right)e^{-i{\omega }_{2}t}\right\}$

Substitute $E/\hslash$ for $\omega$ in above expression.

  $\text{ψ}\left(x,t\right)=C\left\{{\psi }_1\left(x\right)e^{-i{E}_{1}t/\hslash }+{\psi }_2\left(x\right)e^{-i{E}_{2}t/\hslash }\right\}$

Conclusion:

Thus, the function at any later time $t$  is $\text{ψ}\left(x,t\right)=C\left\{{\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}$ .

(c)

To determine

Verify that average energy in the superposition is the arithmetic mean of ground and first excited state energies.

Answer to Problem 37P

Superposition is not the stationary state but average energy in the superposition is the arithmetic mean of ground and first excited state energies.

Explanation of Solution

The stationary state of any function is defined by the Eigen function. A wave is in stationary state when it is an Eigen function of energy operator.

Write the expression for action of energy operator on function.

  $\left[E\right]\text{ψ}=i\hslash \frac{\partial }{\partial t}\left\{\text{ψ}\left(x,t\right)\right\}$        (III)

Write the expression for average energy.

  $\langle E\rangle ={\displaystyle \int {\Psi }^{*}\left[E\right]\Psi dx}$        (IV)

Conclusion:

Substitute $C\left\{{\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}$ $\text{ψ}\left(x,t\right)$ in equation (III).

  $\begin{array}{c}\left[E\right]\text{ψ}=i\hslash \frac{\partial }{\partial t}\left[C\left\{{\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}\right]\\ =Ci\hslash \left\{\left(\frac{-i{E}_1}{\hslash }\right){\psi }_1{e}^{-i{E}_{1}t/\hslash }+\left(\frac{-i{E}_2}{\hslash }\right){\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}\\ =C\left\{E_1{\psi }_1{e}^{-i{E}_{1}t/\hslash }+E_2{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}\end{array}$

Here, $E_1$ and $E_2$ are not equal; hence the operator does not return the multiples of original function. Thus, the function is not in stationary state.

Substitute $C\left\{{\psi }_1^{*}{e}^{+i{E}_{1}t/\hslash }+{\psi }_2^{*}{e}^{+i{E}_{2}t/\hslash }\right\}$ for ${\Psi }^{*}$ and $C\left\{E_1{\psi }_1{e}^{-i{E}_{1}t/\hslash }+E_2{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}$ for $\left[E\right]\text{ψ}$ in equation (IV).

  $\begin{array}{c}\langle E\rangle =C^2{\displaystyle \int \left\{\left({\psi }_1^{*}{e}^{+i{E}_{1}t/\hslash }+{\psi }_2^{*}{e}^{+i{E}_{2}t/\hslash }\right)\left(E_1{\psi }_1{e}^{-i{E}_{1}t/\hslash }+E_2{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right)\right\}dx}\\ =C^2\left\{E_1{\displaystyle \int {\left|{\psi }_1\right|}^{2}dx+E_2{\displaystyle \int {\left|{\psi }_2\right|}^{2}dx}}\right\}\\ ={\left(\frac{1}{\sqrt{2}}\right)}^2\left\{E_1\left(1\right)+E_2\left(1\right)\right\}\\ =\frac{E_1+E_2}{2}\end{array}$

Thus, superposition is not the stationary state but average energy in the superposition is the arithmetic mean of ground and first excited state energies.