Chapter 28, Problem 54P

(a)

To determine

To show that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a freely moving particle with the wave function $\Psi \left(x\right)=A{e}^{ikx}$.

Answer to Problem 54P

It is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a freely moving particle with the wave function $\Psi \left(x\right)=A{e}^{ikx}$.

Explanation of Solution

Write the Schrodinger’s equation.

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}+U\Psi =E\Psi$        (I)

Here, $\hslash$ is the reduced Planck’s constant, $m$ is the mass of the particle, $\Psi$ is the given wave function, $U$ is the potential energy and $E$ is the total energy of the particle.

Write the statement to be proved.

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=K\Psi$        (II)

Here, $K$ is the kinetic energy of the particle.

Write the expression of the given wavefunction.

  $\Psi \left(x\right)=A{e}^{ikx}$        (III)

Here, $A$ is the normalization constant and $k$ is the propagation constant.

Put equation (III) in equation (II).

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=KA{e}^{ikx}$        (IV)

Take the derivative equation (III) with respect to $x$ .

  $\begin{array}{c}\frac{d\Psi }{dx}=A{e}^{ikx}\left(ik\right)\\ =ikA{e}^{ikx}\end{array}$

Take the derivative of the above equation with respect to $x$ .

  $\begin{array}{c}\frac{d^2\Psi }{d^{2}x}=\frac{d}{dx}\left(ikA{e}^{ikx}\right)\\ =ikA{e}^{ikx}\left(ik\right)\\ =i^2{k}^{2}A{e}^{ikx}\\ =-k^{2}A{e}^{ikx}\end{array}$        (V)

Put equations (V) in the left-hand side of equation (II) and rearrange it.

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=-\frac{{\hslash }^2}{2m}\left(-k^{2}A{e}^{ikx}\right)\\ =\frac{{\hslash }^2{k}^2}{2m}\left(A{e}^{ikx}\right)\end{array}$        (VI)

Write the equation for the reduced Planck’s constant.

  $\hslash =\frac{h}{2\pi }$        (VII)

Here, $h$ is the Planck’s constant.

Write the equation for the wave vector.

  $k=\frac{2\pi }{\lambda }$        (VIII)

Here, $\lambda$ is the wavelength of the particle.

Put equation (VII) and (VIII) in (VI).

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=\frac{{\left(\frac{h}{2\pi }\right)}^2{\left(\frac{2\pi }{\lambda }\right)}^2}{2m}\left(A{e}^{ikx}\right)\\ =\frac{4{\pi }^2{h}^2}{8{\pi }^{2}m{\lambda }^2}\left(A{e}^{ikx}\right)\\ =\frac{1}{2m}{\left(\frac{h}{\lambda }\right)}^2\left(A{e}^{ikx}\right)\end{array}$        (IX)

Write the equation for the de Broglie wavelength.

  $\lambda =\frac{h}{p}$

Here, $p$ is the momentum of the particle.

Rewrite the above equation for $p$ .

  $p=\frac{h}{\lambda }$        (X)

Put the above equation in equation (IX).

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=\frac{1}{2m}{p}^2\left(A{e}^{ikx}\right)\\ =\frac{p^2}{2m}\left(A{e}^{ikx}\right)\end{array}$        (XI)

Write the equation for kinetic energy.

  $K=\frac{p^2}{2m}$        (XII)

Put the above equation in equation (XI).

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=K\left(A{e}^{ikx}\right)$        (XIII)

Conclusion:

Equation (XIII) is exactly the same as equation (IV) which has to be proved.

Thus, it is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a freely moving particle with the wave function $\Psi \left(x\right)=A{e}^{ikx}$.

(b)

To determine

To show that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a particle in a box with the wave function $\Psi \left(x\right)=A\sin kx$.

Answer to Problem 54P

It is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a particle in a box with the wave function $\Psi \left(x\right)=A\sin kx$.

Explanation of Solution

Write the expression of the given wavefunction.

  $\Psi \left(x\right)=A\sin kx$        (XIV)

Put equation (XIV) in equation (II).

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=K\left(A\sin kx\right)$        (XV)

Take the derivative equation (XIV) with respect to $x$ .

  $\begin{array}{c}\frac{d\Psi }{dx}=A\cos kx\left(k\right)\\ =Ak\cos kx\end{array}$

Take the derivative of the above equation with respect to $x$ .

  $\begin{array}{c}\frac{d^2\Psi }{d^{2}x}=\frac{d}{dx}\left(Ak\cos kx\right)\\ =Ak\left(-\sin kx\right)\left(k\right)\\ =-A{k}^2\sin kx\end{array}$

Put the above equation in the left-hand side of equation (XV) and rearrange it.

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=-\frac{{\hslash }^2}{2m}\left(-A{k}^2\sin kx\right)\\ =\frac{{\hslash }^2{k}^2}{2m}\left(A\sin kx\right)\end{array}$

Put equation (VII) and (VIII) in the above equation.

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=\frac{{\left(\frac{h}{2\pi }\right)}^2{\left(\frac{2\pi }{\lambda }\right)}^2}{2m}\left(A\sin kx\right)\\ =\frac{4{\pi }^2{h}^2}{8{\pi }^{2}m{\lambda }^2}\left(A\sin kx\right)\\ =\frac{1}{2m}{\left(\frac{h}{\lambda }\right)}^2\left(A\sin kx\right)\end{array}$

Put equation (X) in the above equation.

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=\frac{1}{2m}{p}^2\left(A\sin kx\right)\\ =\frac{p^2}{2m}\left(A\sin kx\right)\end{array}$

Put equation (XII) in the above equation.

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}=K\left(A\sin kx\right)$        (XVI)

Conclusion:

Equation (XVI) is exactly the same as equation (XV) which has to be proved.

Thus, it is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a particle in a box with the wave function $\Psi \left(x\right)=A\sin kx$.