Question

(a) Shallow water is non-dispersive; waves travel at a speed that is proportional to the square root of the depth. In deep water, however, the waves can’t “feel” all the way down to the bottom—they behave as though the depth were proportional to λ. (Actually, the distinction between “shallow” and “deep” itself depends on the wavelength: If the depth is less than λ, the water is “shallow”; if it is substantially greater than λ, the water is “deep.”) Show that the wave velocity of deep water waves is twice the group velocity.

(b) In quantum mechanics, a free particle of mass m traveling in the x direction is described by the wave function

$\mathbf{\psi }(\mathbf{x}\mathbf{,}\mathbf{t})\mathbf{=}{\mathbf{Ae}}^{\mathbf{i}\frac{(\mathbf{px}\mathbf{-}\mathbf{Et})}{\hslash }}$

wherep is the momentum, and $\mathbf{E}\mathbf{=}{\mathbf{p}}^2/\mathbf{2}\mathbf{m}$is the kinetic energy. Calculate the group velocity and the wave velocity. Which one corresponds to the classical speed of the particle? Note that the wave velocity is half the group velocity.

Short answer

(a) It is proved that the wave velocity of deep water waves is twice the group velocityas $v_p=2v_g$.

(b) The group velocity and the wave velocity is$\frac{\hslash k}{m}$ and the group velocity corresponds to the classical speed.

Step-by-step solution

Expression forthe wave velocity for deepwater wave and phase velocity: 

Write the expression forthe wave velocity for deepwater waves.

${\mathit{v}}_{\mathbf{p}}\mathbf{=}\mathit{\alpha }\sqrt{\mathbf{\lambda }}$ …… (1)

Here, $\alpha$is a constant and $\lambda$is the wavelength.

Write the expression for phase velocity.

${\mathit{v}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{k}}$ …… (2)

Here$\omega$, is the angular velocity and $k$is the wave number.

Determine the relation between wave velocity and group velocity:

(a)

Equate equations (1) and (2).

…… (3)

$\begin{array}{c}\frac{\omega }{k}=\alpha \sqrt{\lambda }\\ \omega =k\alpha \sqrt{\lambda }\end{array}$ …… (3)

Write the relation between the wavenumber in terms of wavelength.

$\begin{array}{c}k=\frac{2\pi }{\lambda }\\ \lambda =\frac{2\pi }{k}\end{array}$

Substitute$\lambda =\frac{2\pi }{k}$ in equation (3).

$\begin{array}{c}\omega =k\alpha \sqrt{\frac{2\pi }{k}}\\ \omega =\sqrt{2\pi }\alpha \sqrt{k}\end{array}$

Write the equation for the group velocity.

$v_g=\frac{d\omega }{dk}$

Substitute i$\omega =\sqrt{2\pi }\alpha \sqrt{k}$n the above equation.

$\begin{array}{c}{v}_g=\frac{d}{dk}\left(\sqrt{2\pi }\alpha \sqrt{k}\right)\\ v_g=\sqrt{2\pi }\alpha \frac{d}{dk}\left(\sqrt{k}\right)\\ v_g=\sqrt{2\pi }\alpha \frac{1}{2}{\left(k\right)}^{-\frac{1}{2}}\end{array}$

On further solving, the above equation becomes,

$\begin{array}{c}{v}_g=\frac{\sqrt{2\pi }\alpha }{2}\frac{1}{\sqrt{k}}\\ v_g=\frac{\alpha }{2}\sqrt{\frac{2\pi }{k}}\\ v_g=\frac{\alpha }{2}\sqrt{\lambda }\\ v_g=\frac{v_p}{2}\end{array}$

Therefore, it is proved that the wave velocity of deep water waves is twice the group velocity.

 Step 3: Determine the wave velocity and group velocity:

(b)

From the given problem, the equation is given as:

$\psi (x,t)=A{e}^{i(px-Et)/\hslash }$ …… (4)

Write the spatial representation of a wave.

$\psi (x,t)=A{e}^{i(kx-\omega t)}$ …… (5)

Equate equations (4) and (5).

$\begin{array}{l}A{e}^{i(px-Et)/\hslash }=A{e}^{i(kx-\omega t)}\\ i\frac{(px-Et)}{\hslash }=i(kx-\omega t)\\ k=\frac{p}{\hslash }\\ \omega =\frac{E}{\hslash }\end{array}$

Write the expression for the wave velocity.

$v_p=\frac{\omega }{k}$

Substitute$\omega =\frac{E}{\hslash }$ in the above expression.

$\begin{array}{c}{v}_p=\frac{(E/\hslash )}{p/\hslash }\\ v_p=\frac{E}{p}\\ v_p=\frac{p^2/2m}{p}\\ v_p=\frac{\hslash k}{2m}\end{array}$ …… (6)

Write the expression for the group velocity.

$v_g=\frac{d\omega }{dk}$

Substitute$\omega =\frac{E}{\hslash }$ in the above expression.

$\begin{array}{c}{v}_g=\frac{d}{dk}\left(\frac{E}{\hslash }\right)\\ =\frac{d}{dk}\left(\frac{p^2}{2m\hslash }\right)\\ =\frac{d}{dk}\left(\frac{{\hslash }^2{k}^2}{2m\hslash }\right)\end{array}$

On further solving, the above equation becomes,

$\begin{array}{c}{v}_g=\frac{\hslash }{2m}\frac{d}{dk}\left(k^2\right)\\ v_g=\frac{\hslash }{2m}\left(2k\right)\\ v_g=\frac{\hslash k}{m}\end{array}$ ……. (7)

From equations (6) and (7).

$\begin{array}{c}{v}_p=\frac{\hslash k}{2m}\\ v_p=\frac{v_g}{2}\end{array}$

Hence, the phase velocity is half of the group velocity.

Group velocity corresponds to the classical speed of the particle. This can be understood as follows.

Write the expression for the classical speed.

$v_c=\frac{p}{m}$

Substitute$p=\hslash k$ in the above expression.

$\begin{array}{c}{v}_c=\frac{\hslash k}{m}\\ v_c=v_g\end{array}$

Hence, the from the equation the wave velocity do not represent the classical speed of the particle.

Therefore, the group velocity and the wave velocity is $\frac{\hslash k}{m}$and the group velocity corresponds to the classical speed.