Relation between group velocity and particle velocity

Show that group velocity is always equal to particle velocity.

Case I: For free non-relativistic particle

The total energy of non-relativistic free particle is equal to it's K.E only.

$$K.E=\frac12 mv^2=\frac{P^2}{2m}$$

$$E= \frac{p^2}{2m}\dotsm(1)$$

Where,

m= mass of particle

p=mv= linear momentum of particle

v= particle velcoity.

We have, group velocity of the associated wave packet is,

$$V_g=\frac{dE}{dp}\dotsm(2)$$

Using (1) and (2)

$$V_g=\frac{d}{dp}\biggl(\frac{p^2}{2m}\biggr)$$

$$=\frac{2p}{2m}$$

$$=\frac{2\times m\times v}{2\times m}$$

$$\therefore V_g= V\dotsm(3)$$

Case 2: For bounded (non-relativistic )

For bounded particle the total energy is the sum of K.E and potential energy.

$$E=\frac{p^2}{2m}+V(r)\dotsm(4)$$

We have, the group velocity of the associated wave packet is ,

$$V_g= \frac{dE}{dP}\dotsm(*)$$

Using equation (4) and (*)

$$V_g= \frac{d}{dp}\biggl[\frac{p^2}{2m}+V(v)\biggr]$$

$$=\frac{2p}{2m}+0$$

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

$$\therefore\; V_g= V\dotsm(5)$$

Case 3: For relativistic particle:

The total energy of relativistic particle in terms of relativistic momentum p

$$i.e P= mv=\frac{m_0 V}{\sqrt{1-\frac{v^2}{c^2}}} $$

Is given by

$$E=\sqrt{p^2c^2+m_0^2 c^4}\dotsm(6)$$

We have the group velocity of the associated wave packet is

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

$$=\frac{d(p^2c^2+m_0^2c^4)^\frac12}{dp}$$

$$=\frac12\times \sqrt{p^2 c^2 +m_0^2 c^4}\times (2pc^2+0)$$

$$=\frac{p^2c^2}{E}$$

$$=\frac{mvc^2}{mc^2}$$

$$\therefore\; V_g= V$$

Hence group velocity is always equal to the particle velocity. The information associated with particle travels with group velocity not phase velocity.

Show that \(\Delta x\cdot \Delta P_x= h<\frac{hbar}{2}\) from the concept of wave packet.

Where,

\(\Delta x\) = Uncertainity in the measurement of position.

\(\Delta P_x\)= Corresponding in the measurement of linear momentum.

Consider the superposition of two sine wave of having same amplitude 'a' and ave number k,k\(\Delta k\) and angular frequencies \(\omega\) and \(\omega+\Delta\omega\) 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 particle represented by wave packet must lies between any two successive node. The width of wave packet is also known as uncertainity in the measurement of position of particle represented by wave packet.

Here, $$\Delta x= x_2- x_1\dotsm(5)$$

Where,

\(x_1\)= position of first node

\(x_2\)= position of 2^nd node

At the position of node amplitude must be equal to zero.

$$2acos\biggl[\frac{\Delta k}{2} x- \frac{\Delta \omega }{2} t\biggr]=0$$

$$\Rightarrow 2a\ne 0$$

$$\Rightarrow cos\biggl(\frac{\Delta k}{2}x-\frac{\Delta \omega}{2}t\biggr)=0$$

$$\Rightarrow cos\biggl(\frac{\Delta k}{2} x-\frac{\Delta \omega}{2}t\biggr)= cos\biggl[(2n+1)\frac{\pi}{2}\biggr]$$

Where, n=0,1,2,3,4,….

$$\Rightarrow \frac{\Delta k}{2} x-\frac{\Delta \omega}{2} t= (2n+1)\frac{\pi}{2}\dotsm(6)$$

For the position of first node at any time t

Put n=0

$$\frac{\Delta k}{2}x-\frac{\Delta \omega}{2}=\frac{\pi}{2}\dotsm(7)$$

For the position of 2^nd node at any time t

Put n=1

$$\frac{\Delta k}{2}x-\frac{\Delta \omega 2}{2}t=\frac{3\pi}{2}\dotsm(8)$$

Subtracting equation (7) and (8) we get,

$$\frac{\Delta k}{2}(x_2-x_1)=\pi$$

$$or,\;\; \Delta k\cdot \Delta x= 2\pi\dotsm(9)$$

We have,

$$k=\frac{2\pi}{\lambda}=\frac{2\pi}{h}p$$

$$\therefore k=\frac{P_x}{\hbar}$$

$$\Rightarrow \Delta k=\frac{\Delta P_x}{\hbar}\dotsm(10)$$

From equation (9) and (10)

$$\Delta x\cdot \Delta P_x= 2\pi\cdot \hbar>\frac{\hbar}{2}$$

$$or,\; \Delta x\cdot \Delta P_x=2\pi\cdot\frac{h}{2\pi}$$

$$or,\; \Delta x\cdot \Delta P_x= h> \frac{\hbar }{2}\dotsm(11)$$

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.