Schrodinger equation for rigid rotator. [ continue ]

\(\Theta\) equation for rigid rotator:

Now, taking first and third part of equation (4) we get,

$$\frac{sin\theta}{\Theta}\frac{d}{d\theta}\biggl(sin\theta\frac{d\Theta}{d\theta}\biggr)+\frac{2IE}{\hbar^2}=\frac{m^2}{sin^2\theta}$$

$$or\;\; \frac{1}{\Theta sin\theta}\frac{d}{d\theta}sin\theta\frac{d\Theta}{d\theta}+\biggl(B-\frac{m^2}{sin^2\theta}\biggr)\Theta=0\dotsm(9)$$

this is the \(\theta\) equation for rigid rotator.

Where, $$B=\frac{2IE}{\hbar^2}\dotsm(10$$

To sove the equation (9)

Let, \(x= cos\theta\Rightarrow sin^2\theta= t-x^2\)

$$or,\;\; \frac{dx}{d\theta}=-sin\theta$$

$$or,\; \frac{d\Theta}{d\theta}=\frac{d\Theta}{dx}\cdot \frac{dx}{d\theta}$$

$$=-sin\theta\frac{d\Theta}{dx}$$

$$\Rightarrow \; \frac{1}{sin\theta}\frac{d}{d\theta}=\frac{-d}{dx}$$

Substituting all these quantities in equation (9) we get,

$$-\frac{d}{dx}\biggl[sin\theta(-sin\theta\frac{d\Theta}{dx})\biggr]+(B-\frac{m^2}{1-\alpha^2})\Theta=0$$

$$or,\; \frac{d}{dx}\biggl[(1-x^2)\frac{d\Theta}{dx}+\biggl(B-\frac{m^2}{1-x^2}\biggr)\Theta=0\dotsm(11)$$

This is the form of associated Legendre polynomial whose solution can be written as,

$$\Theta(\theta)= BP_l^m (x)\dotsm(12)$$

Where,

$$P_l^m= (1-x^2)^\frac m2 \frac{d^m}{dx^m}P_l(x)$$

Where, \(P_l(x)\) is the legendre polynomial of first order. The equation (12) in terms of \(\theta\) can be written as,

$$\Theta(\theta)= BP__l^m(cos\theta)\dotsm(13)$$

Now, the normalized equation can be obtained by using the property of orthogonality; we have,

$$\int_0^\pi [ \Theta (\theta)]^* [ \Theta(\theta0]sin\theta d\theta=1$$

$$or,\;\; |B|^2 \int_0^\pi |P_l^m (cos\theta)|^2 sin\theta d\theta=1$$

$$or,\;\; |B|^2\biggl[\frac{2}{2l+1}\cdot \frac{(l+m)!}{(l-m)!}\biggr]^\frac12\dotsm(14)$$

So, the normalized \(\Theta\) equation for rigid rotator is,

$$\Theta(\theta)= \biggl[\frac{2l+1}{2}\cdot \frac{l-m}{2}\biggr]^\frac12 P_l^m (cos\theta)\dotsm(15)$$

This is the required expression for \(\Theta\)- equation of rigid rotator.

Complete wave function for rigid rotator:

The complete wave function for rigid rotator is given by

$$\psi(\theta,\phi)= Y_l^m (\theta, \phi)= \Theta (\theta)\cdot \Phi(\phi)$$

$$\therefore\;\; Y_l^m (\theta,\phi)= \frac{1}{\sqrt{2\pi}}e^{im\phi}\biggl[\frac{2l+1}{2}\frac{(l-m)!}{(l+m)!}\biggr]^\frac12 P_l^m (cos\theta)$$

$$i.e.\;\; Y_l^m (\theta,\phi)= \biggl[\frac{2l+1}{4\pi}\cdot \frac{(l-m)!}{(l+m)!}\biggr]^\frac12 P_l^m (cos\theta) e^{im\phi}$$

Energy level Diagram for rigid rotator:

We have for rigid rotation,

$$B=\frac{2IE}{\hbar^2}\; and\; B= l(l+1)$$

So,

$$\frac{2IE}{\hbar^2}= l(l+1)$$

$$or,\; E= \frac{\hbar^2 l(l+1)}{2I}$$

$$or,\; E= Bl(l+1)$$

Where, \(B= \frac{\hbar^2}{2I}= \) rotational constant.

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.