Derivation for Group velocity

Derivation for the group velocity:

For simplicity of calculation we take two sine-wave having same amplitude 'a', wave number k and k+\(\Delta\)k and angular frequency \(\omega\) and \(\omega+\Delta\omega\) respectively travelling along x-axis. i.e.

$$\psi_k (x,t)= a sin(kx-\omega t)\dotsm(1)$$

$$\psi_{k+\Delta k}(x,t)= asin [ (k+\Delta k)x- (\omega+\Delta \omega)t]\dotsm(2)$$

According to the principle of superposition the resultant wave is given by

$$\psi (x,t)=\psi_k(x,t)+\psi_{k+\Delta k}(x,t)$$

$$=a\biggl[ 2cos (\frac{\Delta k}{2} x-\frac{\Delta \omega}{2} t) sin\biggl((k+\frac{\Delta k}{2})x-(\omega t+\frac{\Delta \omega}{2})t\biggr)\biggr]$$

$$= A sin\biggl[\biggl(k+\frac{\Delta k}{2}\biggr)x-\biggl(\omega+\frac{\Delta \omega}{2}\biggr)t\biggr]\dotsm(3)$$

Where,

$$A= 2a cos\biggl(\frac{\Delta k}{2}x-\frac{\Delta\omega}{2}t\biggr)\dotsm(4)$$

Equation (3) gives resultant wave. Equation (4) gives modulated amplitude of resultant wave.

The phase velocity of modulated amplitude is given by,

$$V_g= \frac{\frac{\Delta \omega}{2}}{\frac{\Delta k}{2}}$$

$$=\frac{\Delta\omega}{\Delta k}\dotsm(5)$$

If we add infinite number of wave for the complete construction of wave packet then the group velocity is written as,

$$V_g= \lim_{\Delta k\to 0} \frac{\Delta\omega}{\Delta k}$$

$$V_g= \frac{d\omega}{dk}\dotsm(6)$$

Group velocity can also be defined as the rate of change of angular frequency with the wave number.

Derivation of other form of group velocity :

We have, $$v_g=\frac{d\omega}{dk}$$

$$=\frac{d(\hbar \omega)}{d(\hbar k)}$$

$$=\frac{dE}{dp}$$

Where,\(\hbar \omega= E\) and \(\hbar k= p\)

$$\therefore\; V_g= \frac{dE}{dP}\dotsm(7)$$

Again, we have

$$V_{ph}= \frac{\omega}{k}$$

$$\omega= V_{ph}\cdot k\dotsm(8)$$

Using (8) in equation (6)

$$V_g= \frac{d}{dk}(V_{ph}\cdot k)$$

$$= V_{ph} \frac{dk}{dk}+k\cdot \frac{dV_{ph}}{dk}$$

$$=V_{ph}+k\frac{dV_{ph}}{dk}\dotsm(9)$$

Again,

$$P= \hbar k\Rightarrow dp= \hbar dk$$

$$k=\frac{p}{\hbar}\Rightarrow dk= \frac{dp}{\hbar}$$

Substituting in equation (9)

$$V_g= V_{ph} + \frac{p}{\hbar}\frac{dV_{ph}}{\frac{dp}{\hbar}}$$

$$V_g= V_{ph}+ p \frac{d V_{ph}}{dp}\dotsm(10)$$

Again we have

$$k=\frac{2\pi}{\lambda}=2\pi (\lambda^{-1})$$

$$\frac{dk}{d\lambda}= (2\pi) \frac{d\lambda^{-1}}{d\lambda}$$

$$\Rightarrow dk= \frac{-2\pi}{\lambda^2}d\lambda $$

Substituting above in equation (9)

$$ V_g= V_{ph} + \biggl(\frac{2\pi}{\lambda}\biggr)\biggl[\frac{dV_{ph}}{\frac{-2\pi}{\lambda^2}d\lambda}\biggr]$$

$$V_g= V_{ph}-\lambda \frac{dV_{ph}}{d\lambda}\dotsm(11)$$

Equation (11) is the required [removed] relation ) between phase velocity, group velocity and wave length.

Case 1 : For non-dispersive medium.

$$\frac{dV_{ph}}{d\lambda}=0$$

\(V_{ph}\) independent of \(\lambda\)

$$\therefore V_g= V_{ph}$$

In this case group velocity is equal to phase velocity.

Example; Sped of photon in vaccum

$$V_g=V_{ph}=C= Particle velocity$$

Case 2: For normal dispersion

$$\frac{dV_{ph}}{d\lambda}>0$$

In this case group velocity is less than phase velocity.

Case 3: for anomalous dispersion :

$$\frac{dV_{ph}}{d\lambda}<0$$

in this case group velocity is greater than phase velocity.

Reference:

1. Mathews, P.M and K Venkatesan. A Text Book of Quantum Mechanics. New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E. Quantum Mechanics . New York: John Wiley, 1969.
3. Prakash, S and S Salauja. Quantum Mechanics. Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh. Quantum Mechanics. Chand & Company Ltd., 2002.