Schrodinger time-independent equation in spherical polar co-ordinate system.

Rigid Rotator:

It is a system of two points particles separated by a fixed distance, for example: at low temperature diatomic molecules behaves as a rigid rotator.

Eg, \( co, H_2, O_2, N_2, Hcl , NO \) etc are diatomic molecules.

Fig: Rigid rotator

Consider, two particles having masses \(m_1\) and \(m_2\) having position vectors \(r_1\) and \(r_2\) respectively. With respect to centre of mass.

The distance between two particle is constant. i.e.

$$r= r_1+r_2\dotsm(1)$$

and sum of moment of mass about C.M is zero. i.e.

$$m_1\vec r_1+ m_2\vec r_2=0$$

$$\therefore\;\; m_1 r_1= m_2 r_2 \; [ Taking\; magnitude \; only]$$

The moment of inertia of rigid rotator ( diatomic molecule ) is given by

$$I= m_1 r_1^2+ m_2 r_2^2\dotsm(2)$$

We substitute \(r_1\) and \(r_2\) in terms of r and \(m_1, m_2\) in equation (2) as,

$$r_1=\frac{m_2 r}{m_1+m_2}$$

$$r_2= \frac{m_1 r}{m_1+ m_2}\dotsm(3)$$

$$I= \biggl[m_1\frac{m_2^2}{(m_1+m_2)^2}+m_2\frac{m_1^2}{(m_1+ m_2)^2}\biggr]\cdot r^2$$

$$=m_1 m_2\biggl[\frac{(m_2+m_1)}{(m_1+m_2)^2}\biggr]\cdot r^2$$

$$\frac{m_1m_2}{m_1+ m_2}\cdot r^2$$

$$I=\mu\cdot r^2\dotsm(4)$$

Where, \(\mu=\frac{m_1m_2}{m_1+m_2}\), reduced mass of rotator.

Here a rigid rotator can be treated as a point particle of mass \(\mu\) moving on the surface of sphere of constant radius 'r' ( dist. between two particles)

We write time independent Schrodinger equation in spherical polar co-ordinate system and solve this equation to explain quantum mechanical behavior of rigid rotator.

Fig: Rigid Rotator ( spherical polar co-ordinate system)

The Schrodinger equation for this problem is,

$$\nabla^2\psi+ \frac{2m}{\hbar}[E-v(r)]\psi=0\dotsm(5)$$

Where, V(r)= potential between \(m_1\) and \(m_2\)

= 0 ( The reference potential is zero)

= Constant

$$\nabla^2=\frac{1}{r^2} \frac{\partial }{\partial r}\biggl[r^2\frac{\partial}{\partial r}\biggr]+\frac{1}{r^2 sin\theta}\frac{\partial }{\partial \theta}\biggl[sin\theta \frac{\partial}{\partial \theta}\biggl[sin\theta\frac{\partial}{\partial \theta}\biggr]+\frac{1}{r^2sin^2\theta}\frac{\partial^2}{\partial\phi^2}$$

We replace m by reduced mass \(\mu\):

So equation (5) becomes,

$$\nabla^2=\frac{1}{r^2} \frac{\partial }{\partial r}\biggl[r^2\frac{\partial}{\partial r}\biggr]+\frac{1}{r^2 sin\theta}\frac{\partial }{\partial \theta}\biggl[sin\theta \frac{\partial}{\partial \theta}\biggl[sin\theta\frac{\partial}{\partial \theta}\biggr]+\frac{1}{r^2sin^2\theta}\frac{\partial^2}{\partial\phi^2}+\frac{2me E}{\hbar^2}=0$$

In this problem,, r= constant

$$\frac{1}{r^2}\frac{\partial}{\partial r}\biggl[r^2\frac{\partial \psi}{\partial r}\biggr]=0$$

Then above equation become, \([\psi(r,\theta,\phi)=\psi(\theta,\phi)]\)

$$\frac{1}{sin\theta}\frac{\partial}{\partial \theta}\biggl[sin\theta\frac{\partial \psi}{\partial \theta}\biggr]+\frac{1}{sin^2\theta}\frac{\partial^2\psi}{\psi\phi^2}+\frac{2(\mu r^2)}{\hbar^2}\psi=0$$

Since, \(\mu r^2= I\)

$$or,\;\; \frac{1}{sin\theta}\frac{\partial}{\partial\theta}\biggl[sin\theta\frac{\partial\psi(\theta, \phi)}{\partial \theta}\biggr]+\frac{1}{sin^2\theta}\frac{\partial^2\psi(\theta,\phi)}{\partial \phi^2}+\frac{2IE}{\hbar^2}\psi(\theta,\phi)=0\dotsm(6)$$

The equation (6) is the time independent Schrodinger equation for rigid rotator.

We solve equation (6) by the method of separation of variable.

To be continue...

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.