Superposition of plane waves

Show that \(\Delta x\cdot \Delta k> 1 \) from the superposition of infinite number of plane waves.

For the construction of wave packet having a single peak we take large number of waves having different amplitude, frequency and wave number

According to principle of superposition

$$\psi (x,t)= \sum_{i=1}^\infty A_i e^{i(k_i x- \omega_i t)}\dotsm(1)$$

For chorent superposition of infinite wave. We write

$$\psi (x,t)= \int A(k) e^{i(kx-\omega t)}dk\dotsm(2)$$

At t=0

$$\psi (x,0)= \int_{-\infty}^\infty A(x)e^{ikx} dk\dotsm(3)$$

We select A(k) in the form of Gaussian distribution

$$A(k)= e^{-\sigma(k-k_0)^2}\dotsm(4)$$

Where, \(\sigma\) is the parameter, k is variable and \(k_0\) is constant. The shape of A(k) is symmetric.

Substituting equation (4) to (3)

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

Put ,

$$(k-k_0)= k'$$

$$\Rightarrow dk= dk'$$

$$k= k_0+ k'$$

When,

$$k=-\infty\Rightarrow k'= -\infty$$

$$k=+\infty \Rightarrow k'= +\infty$$

Now,

$$\psi (x,0)= \int_{-\infty}^{\infty} e^{-\sigma k'^2} e^{ik_0 x} e^{ik'x} \cdot dk'$$

$$=e^{ik_0 x} \int_{-\infty}^\infty e^{-\sigma k'^2+ ik'x} dk'$$

$$\psi(x,0)= e^{ik_0 x} \int_{-\infty}^{\infty} e^{-\sigma \biggl[ k'^2-2k'\cdot \frac{ix}{2\sigma}+(\frac{ix}{2\sigma})^2\biggr]}+ e^{\sigma\biggl(\frac{ix}{2\sigma}\biggr)^2} dk'$$

$$=e^{ik_0 x} e^{\frac{-x^2}{4\sigma}} \int_{-\infty}^\infty e^{-\sigma \biggl( k'-\frac{ix}{2\sigma}\biggr)^2}dk'$$

We have from standard integral,

$$\int_{-infty}^\infty e^{-\alpha x^2}=\sqrt{\frac{\pi}{\alpha}}$$

$$\therefore \; \psi (x,0) = \sqrt{\frac{\pi}{\sigma}} e^{\frac{-x^2}{4\sigma} }e^{ik_0 x}\dotsm(5)$$

Now, probability density function of wave packet is

$$|\psi(x,0)|^2= \psi_{(x,0)}^* \psi_{(x,0)}$$

$$= \biggl(\sqrt{\frac{\pi}{\sigma}}\biggr)^2\biggl(e^{\frac{-x^2}{2\sigma}}\biggr)^2 e^{-ik_0 x} e^{ik_0 x}$$

$$|\psi (x,0)|^2= \frac{\pi}{\sigma}e^{-\frac{x^2}{2\sigma}}\dotsm(6)$$

$$|A(k)|^2= e^{-2\sigma (k- k_0)^2}\dotsm(7)$$

The width of wave packet in position space corresponds to value of x at which \(|\psi(x,0)^2|^2\) becomes \(\frac1e\times\) its maximum value.

$$\frac{1}{e}\times \frac{\pi}{\sigma}=|\psi(x,0)|^2$$

$$or, \; e^{-1} \frac{\pi}{\sigma}= \frac{\pi}{\sigma} e^{-x_1^2}{2\sigma}$$

$$or,\;\; x_1=\pm \sqrt{ 2\sigma}$$

$$\therefore \Delta x= 2x_1= 2\sqrt{2\sigma}\dotsm(8)$$

For the value of \(\Delta k\)

$$\frac1e \times 1= e^{-2\sigma\cdot k_1^2}$$

Where, \(k_1= (k-k_0)\)

$$\Rightarrow k_1= \pm \frac{1}{\sqrt{2\sigma}}$$

$$\therefore \Delta k= \frac{2}{\sqrt{2\sigma}}\dotsm(9)$$

$$\Delta x\cdot \Delta k= 2\sqrt{2\sigma}\cdot \frac{2}{\sqrt{2\sigma}}$$

$$\therefore \Delta x\cdot \Delta k=4> 1$$

We have,

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

$$or,\;\; \Delta x\cdot \Delta P_x = 4\hbar > \frac{\hbar}{2}$$

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.