Introduction

Harmonic Oscillator:

If a particle experienced a force which is directly proportional to displacement of particle and directed opposite to displacement from mean position ( Equilibrium position). Then the particle is said to be simple harmonic oscillator i.e.

$$\vec F= -k\vec k$$

$$\therefore\; F= -kx$$

Where k is force constant.

$$or,\;\; \frac{-\partial u(x)}{\partial x}= -kx$$

$$or,\;\; \int \partial u(x)= \int kx \partial x$$

$$or,\;\; \biggl[u(x)= \frac{kx^2}{2}\biggr]$$

$$or,\;\;\biggl[ V(x)=\frac12 kx^w\biggr]\dotsm(1)$$

Equation (1) gives the potential energy of a simple harmonic oscillator it is also called parabolic potential well as shown in figure.

The Hamiltonian of one dimensional simple harmonic Oscillators is,

$$\hat H(x)= \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+ \hat V(x)$$

$$=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+\frac12 kx^2\dotsm(2)$$

The one dimensional Schrodinger equation for simple harmonic oscillator is,

$$\hat H \psi(x)= E\psi(x)$$

Where,

E= energy of oscillator

m = mass of oscillator.

$$or,\;\; \frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+\frac12 kx^2\psi(x)= E\psi(x)$$

$$or,\;\; \frac{d^2\psi(x)}{dx^2}+ \frac{2m}{\hbar^2}\biggl[E-\frac12 kx^2\biggr]\psi(x)=0$$

Since,

$$\omega= \sqrt{\frac{k}{m}}= angular\; frequency$$

$$\therefore\; k= m\omega^2$$

$$or,\; \frac{d^2\psi(x)}{dx^2}+\frac{2m}{\hbar^2}[E-\frac12 m\omega^2x^2]\psi(x)=0$$

By solving equation (3), we have to determine energy eigen values of simple harmonic oscillator and the energy eigen functions of the oscillator.

$$or,\;\; \frac{d^2\psi(x)}{dx^2}+\biggl[\frac{2mE}{\hbar^2}-\frac{m^2\omega^2}{\hbar^2}x^2\biggr]\psi(x)=0\dotsm(4)$$

To solve above equation we changed variable in such a way that the selected variable is a dimension less parameter.

$$i.e, \;\; y=\alpha x\dotsm(5)[\; dimensionless \;variable\;]$$

$$\alpha^4= \frac{m^2\omega^2}{\hbar}\Rightarrow \alpha=\sqrt{\frac{m\omega}{\hbar}}$$

Now,

$$\frac{d\psi}{dx}=\frac{d\psi}{dy}\frac{dy}{dx}$$

$$=\frac{d\psi}{dy}\cdot \frac{d(\alpha x}{dx}$$

$$=\alpha\frac{d\psi}{dy}$$

Again, \frac{d}{dx}\biggl[\frac{d\psi}{dx}\biggr]=\frac{d^2\psi}{dx^2}$$

$$=\frac{d}{dy}\biggl[\alpha\frac{d\psi}{dY}\biggr]\cdot \frac{dy}{dx}$$

$$\therefore \; \frac{d^2\psi}{dx^2}=\alpha^2\\frac{d^2\psi}{dy^2}\dotsm(6)$$

Substituting equation (5) and (6) in equation (4), We get

$$\alpha^2\frac{d^2\psi}{dy^2}+\biggl[\frac{2mE}{\hbar^2}-\frac{m^2\omega^2}{\hbar^2}\frac{y^2}{\alpha^2}\biggr]\psi=0$$

$$or,\;\; \frac{d^2\psi}{dy^2}+\biggl[\frac{2mE}{\hbar^2\alpha^2}-\frac{m^2\omega^2}{\hbar^2}\frac{1}{\alpha^4}y^2\biggr]\psi=0\dotsm(7)$$

We select \(\alpha\) in such a way that \(\frac{m^2\omega^2}{\hbar^2}\cdot\frac{1}{\alpha^4}=1\)

$$\therefore \alpha^4=\frac{m^2\omega^2}{\hbar^2}$$

Let, \(\frac{2mE}{\hbar^2\alpha^2}=\lambda\dotsm(3)\)

$$or, \;\; E=\frac{\hbar^2}{2m}\alpha^2\cdot \lambda$$

$$or,\;\; E=\frac{\hbar^2}{2m}\cdot \frac{m\omega}{\hbar}\cdot \lambda$$

$$\therefore\; \; E= \frac{\hbar \omega}{2}\cdot \lambda\dotsm(9)$$

By determining value of \(\lambda\), we can determine eigen value of S.H.O

Substituting equation (8) and value of \(\alpha\) in equation (7) we get,

$$\frac{d^2\psi}{dy^2}+[\lambda-y^2]\psi=0\dotsm(10)$$

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.