Example of Normalization

Q.1 Normalize: \(\psi(x)=e^{\frac{-x^2}{2}}; -b\leq x\leq b\).

Here, \(\psi(x)= e^{\frac{-x^2}{2}}; -b\leq x\leq b\)

Let \(\psi_N(x)= A\psi(x)= Ae^{\frac{-x^2}{2}}; -b\leq x\leq b\)

We have, $$\int_{-b}^b \psi_N^*(x) \psi_N(x)dx=1$$

$$|A|^2\int_{-b}^b e^{-x^2} dx=1$$

Case (I)

$$If \; b\longrightarrow \infty \; then$$

$$|A|^2 \int_{-\infty}^\infty e^{-x^2} dx=1$$

$$or,\;\; |A|^2 \sqrt{\pi}=1$$

$$A= \biggl(\frac{1}{\pi}\biggr)^\frac14$$

Case(II)

Using $$e^{-x^2}=1+\frac{-x^2}{1!} + \frac{(-x^2)^2}{2!}+\dotsm$$

$$=1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}+\dotsm$$

Only when |x|<b

$$or,\; |A|^2 \int_{-b}^b\biggl[ 1-\frac{x^2}{1!}+ \frac{x^4}{2!}+\dotsm\biggr]dx=1$$

$$or,\; |A|^2 \biggl[|x|_{-b}^b -\frac{1}{1!} \biggl|\frac{x^3}{3}\biggr|_{-b}^b +\frac{1}{2!} \biggl|\frac{x^5}{5}\biggr|_{-b}^b\dotsm\biggr]=1$$

$$or,\; |A|^2 \biggl[2b-\frac{2b^3}{3\times 1}+\frac{2b^5}{5\times 2!}\dotsm \biggr]=1$$

$$\therefore\; A= \frac{1}{\sqrt{2(b-\frac{b^3}{3}+\frac{b^5}{10}-\dotsm)}}$$

Hence, the normalized wave function is,

$$\psi_N(x)=\biggl(\frac{1}{\pi}\biggr)^\frac14 e^{\frac{-x^2}{2}}; -b\leq x\leq b, b\longrightarrow \infty$$

$$=\frac{e^{\frac{-x^2}{2}}}{\sqrt{2(b-\frac{b^3}{3}+\frac{b^5}{10}\dotsm)}}; -b\leq x\leq b, |x|<b.$$

Q.2

1. Normalize, \(\psi(x)= Ae^{\frac{-\alpha^2 x^2}{2}}\) over \(-\infty\; to\; +\infty\)
2. Find \(<x>. <x^2> \)and \(\Delta x=[<x^2>-<x>^2]^\frac12\)

iii. Find\( <P_x>, <P_x^2>\) and \(\Delta P_x=[<P_x^2>- <P_x>^2]^\frac12\)

1. Does this wave packet is in consistent with, Heisenberg's uncertainity principle?

\(\Rightarrow\) $$ A=\biggl(\frac{\alpha}{\sqrt{\pi}}\biggr)^\frac12$$

Normalized wave function

$$\psi_N(x)=\biggl(\frac{\alpha}{\sqrt{\pi}}\biggr)^\frac12 e^{\frac{-\alpha^2 x^2}{2}}$$

Now,

$$<x>=\int_{-\infty}^\infty \psi_N^*(x)[\hat x \psi_N(x)]dx$$

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

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

$$\therefore\; <x>=0$$

Again, $$<x^2>=\int_{-\infty}^\infty \psi_N^*x)[\hat x^2 \psi_N(x)]dx$$

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

$$=\frac{2\alpha}{\sqrt{\pi}}\int_0^\infty x^2 e^{-\alpha^2 x^2}dx$$

\(Put\; y= \alpha^2 x^2\Rightarrow \frac{y}{\alpha^2}= x^2\)

$$\Rightarrow x= \frac{y^\frac12}{\alpha}$$

$$\Rightarrow dx= \frac{1}{2\alpha}y^{\frac{-1}{2}}dy$$

$$<x^2>=\frac{2\alpha}{\sqrt{\pi}}\int_0^\infty \frac{y}{\alpha^2} e^{-y}\frac{y^\frac{-1}{2}}{2\alpha}dy$$

$$=\frac{2\cdot 1}{2\sqrt{\pi}\alpha^2}\int_0^\infty e^{-y}y^\frac12 dy$$

$$=\frac{1}{\sqrt{\pi}\alpha^2}\gamma{\frac32}$$

$$\frac12 \Gamma{\frac12} \frac{1}{\sqrt{\pi}\alpha^2}$$

$$=\frac{1}{2\alpha^2}$$

$$\therefore\; \Delta x= [ <x^2>- <x>^2]^\frac12= \biggl[\frac{1}{2\alpha^2}-0\biggr]^\frac12$$

$$=\biggr[\frac{1}{2\alpha^2}\biggr]^\frac12$$

$$=\frac{1}{\alpha\sqrt{2}}$$

Again, we have \(\hat P_x= -\hbar\frac{\partial}{\partial x}\)

$$<P_x>=\int_{-\infty}^\infty \psi^*(x)[\hat P_x \psi(x)]dx$$

$$\frac{\alpha}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-\alpha^2 x^2}{2}}\biggl[-i\hbar\frac{\partial}{\partial x}\biggl(e^{\frac{-\alpha^2 x^2}{2}}\biggr)\biggr]dx$$

$$=(-i\hbar)\frac{\alpha}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-\alpha^2 x^2}{2}}e^{\frac{-\alpha^2 x^2 }{2}}\biggl(\frac{-2x\alpha^2}{2}\biggr)dx$$

$$=i\hbar\frac{\alpha^3}{\sqrt{\pi}}\int_{-\infty}^\infty xe^{-\alpha^2 x^2}dx$$

$$=0$$

And,$$<P_x^2>=\int_{-\infty}^\infty \psi^*(x)[\hat P_x^2\psi(x)]dx$$

$$=\frac{\alpha}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-\alpha^2 x^2}{2}}\biggl[-\hbar^2\frac{\partial^2}{\partial x^2}(e^{\frac{-\alpha^2 x^2}{2}})\biggr]dx$$

$$=\frac{-\hbar^2\alpha}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-\alpha^2 x^2}{2}} \frac{\partial}{partial x}\biggl[e^{\frac{-\alpha^2 x^2}{2}}\bigg(\frac{-2x\alpha^2}{2}\biggr)\biggr]dx$$

$$=\frac{\hbar^2\alpha^3}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-\alpha^2x^2}{2}}\biggl(e^{\frac{-\alpha^2x^2}{2}}-\frac{x2x\alpha^2}{2}e^{\frac{-\alpha^2 x^2}{2}}\biggr)dx$$

$$=\frac{\hbar^2\alpha^3}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-\alpha^2 x^2} dx-\frac{\hbar^2 \alpha^5}{\sqrt{\pi}}\int_{-\infty}^\infty x^2 e^{-\alpha^2 x^2 } dx$$

$$=\frac{\hbar^2\alpha^3}{\sqrt{\pi}}\frac{\sqrt{\pi}}{\alpha}-\frac12 \frac{\sqrt{\pi}}{\alpha^2}\frac{\hbar^2\alpha^5}{\sqrt{\pi}}$$

$$=\frac{\hbar^2\alpha^2}{2}$$

Now, $$\Delta P_x=[<P_x^>-<P_x>^2]^\frac12$$

$$[\frac{\hbar^2\alpha^2}{2}-0]^\frac12$$

$$=\frac{\hbar\alpha}{\sqrt{2}}$$

Then, $$\Delta x\cdot \Delta P_x=\frac{1}{\alpha\sqrt{2}}\frac{\hbar\alpha}{\sqrt{2}}$$

$$=\frac{\hbar}{2}$$

$$\therefore\;\; \Delta x\cdot \Delta P_x=\frac{\hbar}{2}$$

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.