Expectation Value of dynamical Quantity:

Expectation value of dynamical quantity:

In Quantum mechanics we could not pin-point the position of a particle due to uncertainty principle. We take a average value of a quantity. The average value or expectation value of a physical quantity in a state $\psi(x)$ I s defined as the average of large number of measurements taken over identical states in identical manner under identical environment.

OR

It is also defined as the average of large number of measurement over identical states $\psi(x)$ taken in identical manner in identical environment.

Mathematically;

The average or expectation value of A over $\psi(x)$ state is represented as,

$$<A>= \int_{-\infty}^\infty \psi^*(x)(\hat A\psi(x))dx\dotsm(1)$$

Where, $\psi(x)$ must be normalized and $\hat A$ should represented quantity A.

$$\hat A= Operator$$

$$A= Physical \; quantity$$

If $\psi(x)$ is not normalized then,

$$<A>=\frac{\int_{-\infty}^\infty \psi^*(x)(\hat A\psi(x))dx}{\int_{-\infty}^\infty \psi^*(x)\psi(x)dx}\dotsm(2)$$

For example:

The expectation value of position.

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

$$\Rightarrow \int_{-\infty}^\infty x\cdot [ \psi^*(x)\psi(x)]dx$$

$$\Rightarrow \int_{-\infty}^\infty x\rho(x)dx$$

$$\therefore <x>=\int_{-\infty}^\infty x\rho(x)dx$$

Where $\rho(x)$= Probability density.

Calculate:

$$<x>;<x^2>,<P_x>,<P_x>^2,\nabla x= [ <x>^2-<x>^2]^\frac12$$

And, $\nabla P_x= [ <P_x^2>-<P_x>^2]^\frac12$

And show that $\nabla x\cdot \nabla P_x=\frac{\hbar}{2}$ for state $\psi(x)= Ae^{\frac{-x^2}{2}}$ [Range -$\infty<x<+\infty$]

$\Rightarrow$ Solution: The condition of normalization is

$$\int_{-\infty}^\infty \psi^*(x)\psi(x)dx=1$$

$$or,\;\; \int_{-\infty}^\infty [ Ae^{\frac{-x^2}{2}}]^*[Ae^{\frac{-x^2}{2}}]dx=1$$

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

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

$$\therefore\;\; A= \biggl[\frac{1}{\pi}\biggr]^\frac14$$

$$\psi_N(x)=\frac{1}{\pi^{\frac14}}e^{\frac{-x^2}{2}}$$

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

$$=\int_{-\infty}^\infty \frac{1}{\pi^{\frac12}}e^{\frac{-x^2}{2}}\cdot xe^{\frac{-x^2}{2}}dx$$

$$\Rightarrow \frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty xe^{-x^2}dx=0$$

[Since the integrant is odd function of x. So integration is zero over symmetric limit.]

Again,

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

$$=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty x^2 e^{-x^2}dx$$

$$=\frac{1}{\sqrt{\pi}}I\dotsm(2)$$

Where, $I= \int_{-\infty}^\infty x^2 e^{-x^2}dx=2\int_0^\infty x^2e^{-x^2}dx$

Put$y=x^2 , x=\sqrt{y}\Rightarrow dx=\frac12\frac{1}{\sqrt{y}}dy$

$$I= 2\int_0^\infty y.e^{-y}\frac12 y^{\frac{-1}{2}}dy$$

$$=\int_0^\infty e^{-y} y^\frac12 dy$$

$$=\int_0^\infty e^{-y} y^{\frac32-1} dy$$

$$=\sqrt{\frac32}$$

$$=\frac12 \Gamma{(\frac12)}$$

$$=\frac12\sqrt{\pi}$$

From (2) $$<x>^2=\frac12\dotsm(3)$$

$$\therefore\; \nabla x= [ <x^2>-<x>^2]^\frac12$$

$$=[\frac12-0]^\frac12$$

$$=\frac{1}{\sqrt{2}}\dotsm(4)$$

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

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

$$=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-x^2}{2}}\biggl[-i\hbar\frac{\partial}{\partial x}(e^{\frac{-x^2}{2}})\biggr]dx$$

$$\frac{-\hbar}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-x^2}{2}} e^{\frac{-x^2}{2}}\biggl(\frac{-1}{2}2x\biggr)dx$$

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

$$=0$$

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

$$=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-x^2}{2}}\biggl[-\hbar^2 \frac{\partial^2}{\partial^2}(e^{\frac{-x^2}{2}})\biggr]dx$$

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

$$=\frac{\hbar^2}{\sqrt{\pi}}\int_{-\infty}^\infty e^{\frac{-x^2}{2}}\biggl[e^{\frac{-x^2}{2}} \cdot 1-xe^{\frac{-x^2}{2}} (\frac{2x}{2})\biggr]dx$$

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

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

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

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

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

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

Finally,

$$\nabla x\cdot\nabla P_x=\frac{1}{\sqrt{2}}\times \frac{\hbar}{\sqrt{2}}$$

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

$$Proved$$

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.