Heisenberg's Uncertainty principle

1. Determine $<x>, <x^2> <P_x>, <P_x^2>$ $\nabla x$ and $\nabla P_x$ for

$$\psi_0(x)=\biggl(\frac{m\omega}{\pi \hbar}\biggr)^\frac14 e^{-\frac{m\omega}{2\hbar}}x^2$$

2. does the product $\nabla x\cdot \nabla P_x$ consistent with Heisenberg's uncertainity principle?

Reapear above calculations for \(\psi_1(x),\psi_2(x)\dotsm \psi_n(x)\

$\Rightarrow$ Given wave function,

$$\psi_0(x)= \biggl(\frac{m\omega}{\pi\hbar}\biggr)^\frac14 e^{\frac{-m\omega}{2\hbar}}\cdot x^2$$

We have,

$$\alpha= \sqrt{\frac{m\omega}{\hbar}}$$

$$So,\;\; \psi_0(x)=\biggl[\frac{\alpha}{\sqrt{\pi}}\biggr]^\frac12 e^{\frac{-\alpha^2 x^2}{2}}\dotsm(1)$$

$$<x>= \int_{-\infty}^\infty \psi_0^*(x) [\hat x \psi_0(x)] dx$$

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

$$=0$$

$$<x^2>= \int_{-\infty}^\infty \psi_0^*(x)[\hat x^2\psi_0(x)]dx$$

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

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

Put t=$\alpha^2 x^2\Rightarrow x =\frac{t H_2}{\alpha}$

$$or,\; x^2=\frac{t}{\alpha^2}$$

$$dx=\frac{1}{2\alpha}t^\frac{-1}{2} dt$$

$$<x^2>= \frac{2\alpha}{\sqrt{\pi}}\int\frac{t}{\alpha^2}e^{-t} \frac{t^{\frac12}}{2\alpha}dt$$

$$=\frac{1}{\sqrt{\pi}\alpha^2}\int_0^\infty e^{-t}t^\frac12 dt= \frac{1}{\sqrt{\pi}\alpha^2}\int_0^\infty e^{-t}t^{\frac32-1} dt$$

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

$$=\frac12\Gamma\frac12 \frac{1}{\Gamma \frac12 \alpha^2}$$

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

Again, we have, $\hat P_x= -i\frac{\hbar\partial}{\partial x}$

$$<P_x>= \int_{-\infty}^\infty \psi_0^*(x)[\hat P_x\psi_0(x)]dx$$

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

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

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

$$=0$$

and,

$$<P_x^2>=\int_{-\infty}^\infty \psi_0^*(x)[\hat P_x \psi_0(x)]dx$$

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

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

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

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

$$=\frac{\hbar^2 \alpha^3}{\sqrt{\pi}}\biggl[\frac{1}{\alpha}\sqrt{\pi}-\alpha^2 \frac{1}{2\alpha^3}\sqrt{\pi}\biggr]$$

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

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

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

Now,

$$\Delta x=[<x>^2- <x>^2]^\frac12$$

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

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

and

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

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

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

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

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

Similimaly, for $\psi_1(x),\psi_2(x)\dotsm$

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.