Importance of normalization

Importance of Normalization:

1. If \(\psi(r,t)\) is the solution of equation

$$i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar}{2m}\nabla^2\psi+V\psi$$

then \(A\psi(r,t); B\psi(r,t)\) are also the solution of same equation.

Here A and B are arbitrary constant.

2. To represent particle by wave packet, the amplitude of wave function should be fixed. ( By the process of normalization) if wave function is normalized then only we can calculate the probability of finding particle in certain region of space.

3. The process of fixing the amplitude of wave function is known as Normalization.

Properties of valid wave function:

1. A wave factor or wavepacket should be finite. Because probability dennsity should be finite ( measurable ).

2. Wave function must be single valued. Because the probability of finding particle is maximum at a point in one instant of time.

3. Wave function must be continuous.

4. Wave function should be square integrable. i.e.

$$\int_{-\infty}^\infty |\psi(x)|^2 dx= N<\infty$$

Where N is a finite.

\(\Rightarrow\) The valid wavefunction must be zero at x=\(\pm \infty\) ( infinity ).

Wave function [ Physical significance of \(\psi\)]

\(\Rightarrow\) \(\psi(r,t) is the solution of Schrodinger wave-equation it describes the de-Broglie wave so it is called wave function. It may be real or complex. It describe the behavior of single particle or photon. i.e. Quantum Mechanical description of particle.

According to max-Born \(\psi\) can be regarded as the probability of finding a particle in a region of space at a time. Since the probability is non-negative and real quantity. So \(\psi*\psi\) is the probability density.

$$dp(x)=\psi^*\psi=|\psi|^2$$

\(\int_{-\infty}^\infty \psi^*\psi dxdydz \) gives probability of finding particle in certain region.

Limitation of Wave function (\(\psi\):-

- It should be finite for all values of x,y,z and t.

- It should be single valued [ i.e \(\psi(x)=\sqrt{x^2}=\pm x\)] not allowed.

- It should be continuous in every region expect at the region of infinite potential.

- It should be analytic. [ i.e. it possesses continuous first order derivative ].

- The wave function is vanishes at boundaries ( infinity).

Normalized wave function:

As we know, the probability of finding particle in the volume element \(d\tau\) is given by \(\psi\psi^* d\tau\) or \(|\psi|^2d\tau\). Where \(d\tau= dxdydz\). Now, the total probability of finding particle in entire space must be unity. i.e.

$$ \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\psi^*(r,t)\psi(r,t)dxdydz=1\dotsm(1)$$

Determine the probability current density corresponding to

$$\psi(x)= Ae^{ikx}+ Be^{-ikx}$$

\(\Rightarrow\) Solution:

$$\psi(x)= Ae^{ikx}+ be^{-ikx}$$

$$\psi^*(x)= Ae^{-ikx}+ Be^{ikx}$$

$$\frac{d\psi(x)}{dx}= (ik)[Ae^{ikx}-Be^{-ikx}]$$

$$\frac{d\psi^*(x)}{dx}= (-ik)[A^* e^{-kx}-B^* e^{ikx}]$$

Now, probability current density is,

$$J(x)=\frac{\hbar}{2im}[\psi^*\frac{d\psi}{dx}-\psi\frac{d\psi^*}{dx}]$$

$$=\frac{\hbar}{2im}\biggl[(A^*e^{-ikx}+B^*e^{ikx})(ik)(Ae^{ikx}-Be^{-ikx})-(Ae^{ikx}+Be^{-ikx})(-ik)(A^*e^{-ikx}-B^*e^{ikx})\biggr]$$

$$=\frac{\hbar}{2im}(ik)\biggl[|A|^2-A^*Be^{-2kx}+AB^*e^{2ikx}-|B|^2+|A|^2-AB^* e^{2kx}+A^*Be^{-2ikx}-|B|^2\biggr]$$

$$=\frac{\hbar}{2m}k[2|A|^2-2|B|^2]$$

$$=\frac{\hbar}{m}k[|A|^2-|B|^2]$$

$$=\frac{\hbar k}{m}|A|^2+\biggl[\frac{-\hbar k}{m}|B|^2\biggr]$$

$$=\vec J_{inc}+ \vec J_{ref}$$

Where, \(\vec J_{incident}= \frac{\hbar k}{m}|A|^2\) and \(\vec J_{reflected}= \frac{\hbar k}{m}|B|^2\)

Here, probability current density is equal to product of probability density (\(|A|^2\)) and group velocity \(\biggl(\frac{\hbar k}{m}\biggr)\)

q. If \(\psi(x)=\frac{e^{ikx}}{x}\)

Find the probability density and probability current density. Also show that, J= pro.density times group velocity.

\(\Rightarrow\) Solution:

$$\psi(x)= \frac{e^{ikx}}{x}$$

Probability density \(\rho= \psi\psi^*=\frac{e^{ikx}}{x}\frac{e^{-ikx}}{x}\)

$$\frac{1}{x^2}$$

Now, $$\psi^*(x)= \frac{e^{ikx}}{x}$$

$$\frac{d\psi^*}{dx}=\frac{(-ik)e^{-ikx} x- e^{-ikx}}{x^2}$$

$$=\frac{-e^{-ikx}(ikx+1)}{x^2}$$

$$\frac{d\psi}{dx}=\frac{ikxe^{ikx}- e^{ikx}}{x^2}$$

$$=\frac{e^{ikx}(ikx-1)}{x^2}$$

Then the probability current density is,

$$\vec J= \frac{\hbar}{2im}[\psi^*\frac{d\psi}{dx}-\frac{\psi d\psi^*}{dx}]$$

$$=\frac{\hbar}{2im}\biggl[\frac{e^{-ikx}}{x}\biggl(\frac{e^{ikx}(ikx-1)}{x^2}\biggr)-\frac{e^{ikx}}{x}\biggl(\frac{-e^{ikx}(ikx+1)}{x^2}\biggr)\biggr]$$

$$=\frac{\hbar}{2m}\frac{1}{x^3}(ikx-1+ikx+1)$$

$$=\frac{\hbar 2ikx}{2im x^3}$$

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

$$=\frac{1}{x^2}\biggl(\frac{\hbar k}{m}\biggr)$$

$$= Probability \;density\; times \;group \;velocity $$

$$Proved$$

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.