Schrodinger equation for rigid rotator.

Schrodinger equation for the rigid Rotator:

We have the time independent Schrodinger equation in spherical polar co-ordinate is given by,

$$\nabla^2\psi(r,\theta,\phi)+\frac{2\mu}{\hbar^2}[E-v(r)]\psi(r,\theta, \phi)=0$$

$$or,\; \frac{1}{r^2}\frac{\partial}{\partial r}\biggl(r^2\frac{\partial\psi(r,\theta,\phi)}{\partial r}\biggr)+\frac{1}{r^2sin\theta}\frac{\partial}{\partial \theta}\biggl(sin\theta\frac{\partial\psi(r,\theta,\phi)}{\partial \theta}\biggr)+\frac{1}{r^2sin^2\theta}\frac{\partial\psi(r,\theta,\phi)}{\partial\phi^2}$$

$$+\frac{2y}{\hbar^2}[E-v(r)]\psi(r,\theta,\phi)=0\dotsm(1)$$

Since, $$V(r)=0$$

$$\frac{\partial\psi(r)}{\partial r}=0$$

The \(\psi(r,\theta,\phi)=\psi(theta,\phi)\)

Now the equation (1) becomes

$$\frac{1}{r^2}\frac{\partial }{\partial r}\biggl(r^2\frac{\partial \psi(\theta,\phi)}{\partial r}\biggr)+\frac{1}{r^2sin\theta}\frac{\partial}{\partial \theta}\biggl(sin\theta\frac{\partial}{\partial \theta}\psi(\theta,\phi)biggr)+\frac{1}{r^2sin^2\theta}\frac{\partial^2}{\partial \phi}[\psi(\theta,\phi)$$

$$+\frac{2\mu}{\hbar^2}E\psi(\theta,\phi)=0$$

$$or,\; \frac{1}{r^2sin\theta}\frac{\partial}{\partial\theta}\biggl(sin\theta\frac{\partial\psi(\theta,\phi)}{\partial \theta}\biggr)+\frac{1}{r^2sin^2\theta}\frac{\partial^2\psi(\theta,\phi)}{\partial\phi^2}+\frac{2\mu E\psi(\theta,\psi)}{\hbar^2}=0$$

$$or,\;\; \frac{1}{sin\theta}\frac{\partial}{\partial\theta}\biggl(sin\theta\frac{\partial\psi(\theta,phy)}{\partial\theta}\biggr)+\frac{1}{sin^2\theta}\frac{\partial^2\psi(\theta,\phi)}{\partial\phi^2}+\frac{2\mu r^2 E}{\hbar^2}\psi(\theta,\phi)=0$$

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

This is the Schrodinger equation for the rigid rotator.

Solution of wave function of Rigid Rotator.

To solve Schrodinger equation of rigid rotator, let us consider the separation of variables as,

$$\psi(\theta,\phi)= Y_l^m (\theta,\phi)= \Theta (\theta)\Phi(\phi)\dotsm(3)$$

Now, equation (2) can be written as,

$$\frac{\Theta}{sin\theta}\biggl(\frac{d}{d\theta}sin\theta \frac{d \Theta}{d\theta}\biggr)+\frac{\Theta}{sin^2\theta}\frac{d^2\Phi}{d\phi^2}+ \frac{2IE}{\hbar^2}\Theta \Phi=0$$

$$\frac{sin\theta}{\Theta}\biggl(\frac{d}{d\theta}sin\theta\frac{d\Theta}{d\theta}\biggr)+\frac{2IE}{\hbar^2}sin^2\theta= -\frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2}=m^2(say)\dotsm(2)$$

Where, \(m^2= Separation constant\)

Taking 2nd and 3rd part of equation (4), we get

$$\frac{d^2\Phi}{d\phi^2}+ \Phi m^2=0\dotsm(5)$$

The solution of equation (5) can be written as,

$$\Phi(\phi)= Ae^{\pm im\phi}\dotsm(6)$$

Using periodic boundry condition we get,

$$Ae^{-im\phi}= Ae^{\pm in\phi}e^{\pm im\pi}$$

$$or,\; cos 2\pi m+ i sin2\pi m=1$$

Equation the real part we get,

$$cos(2\pi m)=1$$

$$or,\; cos(2\pi m)= cos (2\pi n)$$

Where, \(n=0 ,\pm 1, \pm 2, ..\)

\(or, \; m-n, \; m=0, \pm 1, \pm , ..\) is the magnetic quantum number.

the \(\Phi\) equation of rigid rotator is,

$$\Phi(\phi)= Ae ^{im\phi}\dotsm(7)$$

Using the condition of normalization. For equation (7) we get,

$$\int_0^{2\pi} [ \Phi(\phi)]^* [ \Phi(\phi)]d\phi=1$$

$$or,\;\; |A|^2 \int_0^{2\pi} d\phi=1$$

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

\(\therefore\) The normalized \(\Phi\) equation for rigid rotator is given by

$$\Phi(\phi)= \frac{1}{\sqrt{2\pi}}e^{im\phi}\dotsm(8)$$

This is the normalized \(\Phi-\) equation of rigid rotation.

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.