Particles wave packet spreads in time

Particles wave packet spreads in time

Show that width of wave packet of free particle ( matter ) changes with time during its propagation.

The form of time dependent wave packet for any particle is given by

$$\psi (x,t)= \int_{-\infty}^\infty A(k)e^{i(kx-\omega(k)t}dk\dotsm(1)$$

Where, A(k) is amplitude of wave packet taken as Gaussian.

$$A(k)= e^{-\sigma (k-k_0)^2}\dotsm(2)$$

To observe effect of propagation we must expand \(\omega(k)\) around \(k_0\). For simplicity of calculation we use Tyler series expansion.

$$\omega(k)= \omega (k_0)+ \frac{(k-k_0)'}{1!}\frac{d\omega}{dk}|_{k=k_0}+\frac{(k-k_0)^2}{2!}\frac{d^2\omega}{dk^2}|_{k=k_0}+\dotsm$$

$$=\omega(k_0)+ (k-k_0)\alpha+ (k-k_0)^2\beta+\dotsm$$

Where, \(\alpha= \frac{d\omega}{dk}|_{k=k_0}\) and \(\beta=\frac{d^2\omega}{2dk^2}|_{k=k_0}\)

In case of photon

$$E=\hbar\omega$$

$$=\frac{h}{2\pi}2\pi\nu$$

$$=\frac{h}{2\pi} 2\pi \frac{c}{\lambda}$$

$$=\frac{hck}{2\pi}$$

$$\omega= \frac{hck}{2\pi \hbar}$$

$$=\frac{hck}{2\pi \frac{h}{2\pi}}$$

$$\therefore \omega= ck\dotsm(4)$$

In case of photon \(\omega\propto k\) ( \(\omega\) is linear function of k)

$$\alpha=\frac{d\omega}{dk}|_{k=k_0}$$

$$or,\;\; \frac{d(ck)}{dk}\biggl|_{k=k_0}= c$$

\(\alpha= c = V_g=\) Group velocity

$$\beta=\frac{d^2\omega}{dk^2}\biggl|_{k=k_0}$$

$$=\frac{d}{dk}\biggl[\frac{d}{dk}(ck)\biggr]\biggl|_{k=k_0}$$

$$=\frac{d}{dk}(c)\biggl|_{k=k_0}$$

$$\therefore \beta=0$$

For photon equation (3) becomes,

$$\omega(k)= \omega(k_0)+ (k-k_0) V_g\dotsm(5)$$

Using (2) and (5) in equation (1)

$$\psi(x,t)= \int_{-\infty}^\infty e^{-\sigma(k-k_0)^2} e^{i(kx-\omega(k_0)t-(k-k_0)tv_g}dk$$

$$=\int_{-\infty}^\infty e^{-\sigma (k-k_0)^2} e^{i(kx-\omega t v_g)} e^{i(k_0tv_g-\omega (k_0 t))}dk$$

At t=0

$$\psi (x,0)= \int_{-\infty}^\infty e^{-\sigma(k-k_0)^2 e^{ikx}}dk$$

For \(t\ne 0\)

\(\psi(x,t)=\psi(x-V_gt, t)= \psi (x- (t,t))\)

$$x=v_g\cdot t$$

From above equation it has been seen that form of wave packet for photon remains unchanged.

i.e The peak of wave packet travels a distance \(x= V_g\cdot t\)

There is no any dispersion of wave packet for photon.

In case of free particle (non -relativistic ):

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

$$=\frac{hbar^2 k^2}{2m}$$

$$or, \; \; \hbar \omega= \frac{\hbar^2 k^2}{2m}$$

$$\therefore \omega= \frac{\hbar k^2}{2m}\dotsm(4)$$

$$\alpha=\frac{d\omega}{dk}\biggl|_{k=k_0}$$

$$=\hbar \frac{2k}{2m}\biggl|_{k=k_0}$$

$$=\frac{\hbar k_0}{m}$$

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

$$=V_g$$

IN this case \(\alpha \) is also equals to group velocity.

$$\beta=\frac12 \frac{d^2\omega}{dk^2}\biggl|_{k=k_0}$$

$$=\frac12 \frac{d^2}{dk^2}\biggl(\frac{\hbar k^2}{2m}\biggr)$$

$$=\frac12 \frac{d}{dk}\biggl(\frac{\hbar 2k}{2m}\biggr)$$

$$=\frac{\hbar}{2m}\ne 0$$

Now equation (5) in this case become not equal to zero

$$\omega(k)=\omega(k_0)+(k-k_0)V_g+(k-k_0)^2\beta\dotsm(5)$$

Substituting (2) and (5) in equation (1)

$$\psi(x,t)=\int_{-\infty}^\infty e^{-\sigma(k-k_0)^2 } e^{i[ kx-(\omega(k_0)+(k-k_0)v_g+(k-k_0)^2\beta)t]}dk$$

$$\int_{-\infty}^{\infty} e^{-\sigma(k-k_0)^2}e^{i[ kx-\omega(k_0)t-(k-k_0)v_gt-(k-k_0)^2\beta t]}dk$$

Put \(k-k_0=k'\)

$$\psi(x,t)=\int_{-\infty}^\infty e^{-\sigma k'^2 } e^{i[k'x+k_0 x- \omega (k_0)t-k'v_g t- k'^2\beta t]}dk'$$

$$=\int_{-\infty}^\infty e^{-(\sigma+i\beta t)k'^2+k'[i(x-v_g t)]} e^{i[k_0 x- \omega (k_0)t]}dk'$$

$$=e^{i(k_0 x-\omega(k_0)t} \int_{-\infty}^\infty e^{-(\sigma+\beta i t)k'^2+ik'(x-v_g t)}dk'$$

The form of above integral is same as

$$\int_{-\infty}^\infty e^{-\sigma k'^2+ik'x} dx'=\sqrt{\frac{\pi}{\sigma}}$$

$$\therefore \psi(x,t)=\sqrt{\frac{\pi}{\sigma+i\beta t}}\cdot e^{i(k_0x-\omega(x_0)t)} e^{\frac{-(x-v_g t)^2}{4(\sigma+ i\beta t)}}$$

The probability density corresponding to above wave packet is

$$|\psi(x,t)|^2=\psi (x,t)^* \psi(x,t)$$

$$=\sqrt{\frac{\pi^2}{(\sigma - i\beta t)(\sigma+ i\beta t)}} e^0 e^{-\biggl[\frac{(x-v_g t)^2}{4(\sigma+ i\beta t)}\times \frac{(x-V_g t)^2}{4(\sigma-i\beta t)}\biggr]}$$

$$=\sqrt{\frac{\pi^2}{\sigma^2+\beta^2 t^2}} e^{-\frac{(x-v_gt)^2}{4}\biggl[\frac{\sigma+i\beta+\sigma- i\beta t}{\sigma^2+\beta^2t^2}\biggr]}$$

$$=\sqrt{\frac{\pi^2}{\sigma^2+\beta^2 t^2}} e^{-\frac{(x-v_g t)^2 2\sigma}{4(\sigma^2+\beta^2t^2})}$$

in this case \(\Delta x= 2\sqrt{2 \sqrt{\sigma^2 + \beta^2 t^2}}\)

For free particle the width of wave packet increases with time

$$\Delta x= 2\sigma ^\frac12 \sqrt{2\sqrt{\frac{1+\beta^2 t^2}{\sigma^2}}}$$

i.e Vacuum acts as dispersive medium of propagation of wave pack for free particle .

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.