Chapter 28, Problem 47E

(a)

To determine

The boundary conditions for the Schrödinger wave equation.

Answer to Problem 47E

The boundary conditions are $\Psi \left(x=a\right)=0$ and $\Psi \left(x=-a\right)=0$.

Explanation of Solution

The particle moves between $x=a$ and $x=-a$. So, the corresponding wave function also exists between these boundaries.

Beyond these boundaries, there is no existence of the particle.

Conclusion:

Thus, the boundary conditions are $\Psi \left(x=a\right)=0$ and $\Psi \left(x=-a\right)=0$.

(b)

To determine

Whether both sine and cosine solutions are acceptable wave functions.

Answer to Problem 47E

Both sine and cosine solutions are acceptable wave functions.

Explanation of Solution

Write the expression for the Schrӧdinger equation for the particle.

    $\frac{d^2\Psi \left(x\right)}{d{x}^2}+\frac{2m}{{\hslash }^2}\left[E-V\left(x\right)\right]\Psi \left(x\right)=0$        (I)

Here, $\Psi \left(x\right)$ is the wave function corresponding to the particle, $h$ is Planck’s constant, $E$ is the total energy of the particle and $V\left(x\right)$ is the potential energy of the particle.

Conclusion:

Substittue $0$ for $V\left(x\right)$ in equation (I).

  $\frac{d^2\Psi \left(x\right)}{d{x}^2}+\left(\frac{2m}{{\hslash }^2}E\right)\Psi \left(x\right)=0$

Substitute $k$ for $\sqrt{\frac{2m}{{\hslash }^2}E}$ in above equation.

  $\frac{d^2\Psi \left(x\right)}{d{x}^2}+k^2\Psi \left(x\right)=0$

Write the solution for $\Psi \left(x\right)$.

    $\Psi \left(x\right)=ASin\left(kx\right)+BCos\left(kx\right)$

Apply the first boundary condition $\Psi \left(x=a\right)=0$ to the solution for $\Psi \left(x\right)$.

    $ASin\left(ka\right)+BCos\left(ka\right)=0$

Solve the above equation.

  $BCos\left(ka\right)=0$

Apply the second boundary condition $\Psi \left(x=-a\right)=0$ to the solution for $\Psi \left(x\right)$.

    $-ASin\left(ka\right)+BCos\left(ka\right)=0$

Solve the above equation.

    $ASin\left(ka\right)=0$

Write the solution for $\Psi \left(x\right)$ corresponding to the first boundary condition.

    ${\Psi }_n\left(x\right)=BCos\left(\frac{n\pi x}{2a}\right),n=1,3,\mathrm{5......}$

Write the solution for $\Psi \left(x\right)$ corresponding to the second boundary condition.

  ${\Psi }_n\left(x\right)=ASin\left(\frac{n\pi x}{2a}\right),n=2,4,\mathrm{6......}$

Thus, both sine and cosine solutions are acceptable wave functions.

(c)

To determine

The normalized wave functions and the corresponding energies.

Answer to Problem 47E

The normalized wave functions are ${\Psi }_n\left(x\right)=\sqrt{\frac{1}{a}}Sin\left(\frac{n\pi x}{2a}\right)$ and ${\Psi }_n\left(x\right)=\sqrt{\frac{1}{a}}\text{Cos}\left(\frac{n\pi x}{2a}\right)$ and the energies are $\frac{n^2{\pi }^2{h}^2}{8m{a}^2}$.

Explanation of Solution

Write the expression for the normalization condition of a wave function.

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

Here, $\Psi \left(x\right)$ is the wave function.

Write the expression for the energy in $n^{\text{th}}$ state.

    $E_n=\frac{{\hslash }^2{k}_n^2}{2m}$        (III)

Here, $E_n$ is the energy, $h$ is Planck’s constant, $k_n$ is the wave vector for $n^{\text{th}}$ state and $m$ is the mass of the particle.

Conclusion:

Substitute $ASin\left(\frac{n\pi x}{2a}\right)$ for $\Psi \left(x\right)$ in equation (II).

    ${\left|A\right|}^2{\displaystyle \underset{0}{\overset{a}{\int }}Si{n}^2\left(\frac{n\pi x}{2a}\right)dx}=1$

Simplify the above equation.

    $A=\sqrt{\frac{1}{a}}$

Substitute $BCos\left(\frac{n\pi x}{2a}\right)$ for $\Psi \left(x\right)$ in equation (II).

  ${\left|B\right|}^2{\displaystyle \underset{0}{\overset{a}{\int }}Co{s}^2\left(\frac{n\pi x}{2a}\right)dx}=1$

Simplify the above equation.

    $B=\sqrt{\frac{1}{a}}$

The solution for $ka$ are $\left(n\pi /2\right)$.

Substitute $\frac{n\pi }{2a}$ for $k$ in equation (III).

    $E_n=\frac{n^2{\pi }^2{h}^2}{8m{a}^2},n=1,2,3,\mathrm{......}$

Thus, the normalized wave functions are ${\Psi }_n\left(x\right)=\sqrt{\frac{1}{a}}Sin\left(\frac{n\pi x}{2a}\right)$ and ${\Psi }_n\left(x\right)=\sqrt{\frac{1}{a}}\text{Cos}\left(\frac{n\pi x}{2a}\right)$ and the energies are $\frac{n^2{\pi }^2{h}^2}{8m{a}^2}$.