Schrodinger time-independent equation in spherical polar co-ordinate system [ continue ]

Cont..

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

Let \(\psi(\theta,\phi)= H (\theta)\cdot \Phi (\phi)\dotsm(7)\)

Be the form of solution of equation (6)

Substituting equation (7) in equation (8) we get,

$$\frac{\Phi}{sin\theta}\frac{d}{d\theta}\biggl[sin\theta\frac{dH}{d\theta}\biggr]+\frac{H}{sin^2\theta}\frac{d^2\Phi}{d\phi^2}+\frac{2IE}{\hbar^2}H\Phi=0$$

Dividing both sides by \(\phi H we get\)

$$or,\;\; \frac{1}{sin\theta}\frac{d}{d\theta}\biggl[sin\theta\frac{dH}{d\theta}\biggr]+\frac{1}{\Phi sin^2\theta}\frac{d^2\Phi}{d\phi^2}+\frac{2IE}{\hbar^2}=0$$

Multiplying both sides by \(sin^2\theta\) we get,

$$or, \; sin^2\theta\biggl[\frac{1}{H sin\theta}\frac{d}{d\theta}\biggl(sin\theta\frac{dH}{d\theta}\biggr)+\frac{2IE}{\hbar^2}\biggr]+\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}= m^2(say)$$

$$or,\;\; sin^2\theta\biggl[\frac{1}{H sin\theta}\frac{d}{d\theta}\biggl(sin\theta\frac{d}{d\theta}\biggr)+\frac{2IE}{\hbar^2}\biggr]=\frac{-1}{\Phi}\frac{d^2\Phi}{d\phi^2}= m^2$$

Where, \(m^2 \) is separation constant. It’s value is to be determined.

Taking 2^nd and 3^rd term of equation (8), we get

$$\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}+ m^2=0$$

$$\frac{d^2\Phi}{d\phi^2}+ m^2\Phi=0\dotsm(9)$$

Equation (9) is called \(\Phi-\) equation for rigid rotator.

Solution of equation (9) is

$$\Phi(\phi)= Ae^{\pm im\phi}\dotsm(10)$$

Here, A and m are constant to be determined.

Using periodic boundary condition, i.e. value of \(\Phi(\phi)\) remains same over same angular position.

$$or, \;\; \Phi(\phi)= \Phi(\phi+2\pi)$$

Here, \(2\pi\)= angular period

$$Ae^{\pm im\phi}= Ae^{\pm im ( \phi+ 2\pi)}$$

$$or,\;\; 1= e^{\pm 2im\pi}$$

$$or,\;\; 1= cos(2m\pi)+ isin(2m\pi)$$

Equation real parts on both sides,

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

Where, \(n=0,\pm 1, \pm 2,\pm 3,…\)

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

$$m=n =0, \pm 1, \pm 2,\pm 3, …\dotsm(11)$$

Here, m is called azimuthal quantum number.

Substituting value of m from (11) in (10) we get,

$$\Phi(\phi)=Ae^{im \phi}\dotsm(12)$$

Where, \(m=0,\pm1, \pm2,…\)

To determine A, we use the condition of Normalization.

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

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

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

$$or,\;\; A=\frac{1}{\sqrt{2\pi}}\dotsm(13)$$

Normalization constant of \(\Phi\)-wave function.

$$\Phi(\phi)= \frac{1}{\sqrt{2\pi}}e^{im\phi}\dotsm(14)$$

With \(m=0,\pm1,\pm2,…\)

Here, +ve and –ve value of m indicates anticlockwise and clockwise rotation around circular orbit.

Taking 1^st and last part of equation (8) we get,

$$\frac{1}{sin\theta}\frac{d}{d\theta}\biggl[sin\theta\frac{d H}{d\theta}\biggr]+\biggl[\frac{2IE}{\hbar^2}-\frac{m^2}{sin^2\theta}\biggr]H=0$$

$$or,\;\; \frac{1}{sin\theta}\frac{d}{d\theta}\biggl[sin\theta\frac{dH}{d\theta}\biggr]+\biggl[\beta-\frac{m^2}{sin^2\theta}\biggr]H=0\dotsm(15)$$

Where, $$\beta=\frac{2IE}{\hbar^2}\dotsm(16)$$

Equation (15) is called as \(\Theta\)- part of schrodinger equation for rigid rotator.

Reference:

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.