Energy eigen value of one-dimensional simple harmonic oscillator.

Energy eigen value of one-dimensional Simple Harmonic Oscillator.

Asymptotic solution ( very large vale of y)

Equation (10) becomes,

$$\frac{d^2\psi}{dy^2}+ (-y^2)\psi=0$$

$$\frac{d^2\psi}{dy^2}-y^2\psi=0\dotsm(11)$$

The form of solution of equation (11) is

$$\psi(y)\propto e^{\pm\frac{y^2}{2}}$$

One of the form of solution. i.e.

$$e^{+\frac{y^2}{2}}$$is divergent.

At \(x= \pm \infty\) i.e \(e^({\frac{\pm \infty)^2}{2}}=\infty\)

So it is not physically acceptable solution.

$$\therefore \psi(y)\propto e^{-\frac{y^2}{2}}$$

For finite value of y( finite distance)

The solution of equation (10) can be written as,

$$\psi(y)= e^{\frac{-y^2}{2}}H(y)\dotsm(1)$$

Where, H(y) is finite polynomial of y.

$$\frac{d\psi}{dy}= \frac{dH}{dy} e^{-\frac{y^2}{2}}+ H e^{-\frac{y^2}{2}} d(-\frac{y^2}{2})$$

$$=\frac{dH}{dy}e^{-\frac{y^2}{2}}- H\frac{2y}{2} e^{\frac{-y^2}{2}}$$

$$=\frac{dH}{dy}e^{-\frac{y^2}{2}}- y\cdot H\cdot e^{\frac{-y^2}{2}}$$

Again,

$$\frac{d^2\psi}{dy^2}= \frac{d^2H}{dy^2} e^{\frac{-y^2}{2}}+ \frac{dH}{dy}e^{\frac{-y^2}{2}}(\frac{-2y}{2})- y\frac{dH}{dy}e^{\frac{-y^2}{2}}- He^{\frac{-y^2}{2}}- yHe^{\frac{-y^2}{2}}(\frac{-2y}{2})$$

$$or,\;\; \frac{d^2\psi}{dy^2}= \frac{d^2 H}{dy^2}e^{\frac{-y^2}{2}}- 2y\frac{dH}{dy}e^{-\frac{y^2}{2}}+ (y^2-1)He^{\frac{-y^2}{2}}$$

$$\therefore \;\; \frac{d^2\psi}{dy^2}= \biggl[\frac{d^2H}{dy^2}- 2y\frac{dH}{dy}+ (y^2-1)H e^{\frac{-y^2}{2}}\dotsm(13)$$

From equation (10) we get,

$$\frac{d^2H}{dy^2}- 2y\frac{dH}{dy}+ (y^2-1)H+ (\lambda- y^2)H e^{\frac{-y^2}{2}}=0$$

$$or,\;\; \frac{d^2H}{dy^2}- 2y\frac{dH}{dy}+ (\lambda-1)H=0\dotsm(14)$$

As \(e^{\frac{-y^2}{2}}\ne 0 \) for finite value of y

From series solution

$$H(y)= \sum_{r=0}^\infty a_r y^{k+4}\dotsm(15)$$

Where \(a_0=0\)

$$\frac{dH}{dy}= \sum_{r=0}^\infty a_r(k+r)y^{k+r-1}$$

$$\frac{d^2H}{dy^2}=\sum_{r=0}^\infty a_r(k+r)(k+r-1)y^{k+r-2}\dotsm(16)$$

Substituting (15) and (16) in equation (14), we get,

$$or,\;\; \sum_{r=0}^\infty a_r (k+r)(k+r-1)y^{k+r-2}-2\sum_{r=0}^\infty a_r(k+r)y^{k+r}+(\lambda-1)\sum_{r=0}^\infty a_r y^{k+4}=0$$

$$or,\;\; \sum_{r=0}^\infty a_r(k+r)(k+r-1) y^{k+r-2}- \sum_{r=0}^\infty a_r[ 2k+ 2r-\lambda+1]y^{k+r}=0\dotsm(17)$$

We are going to determine the value of \(\lambda\) for finite terminating series solution.

Equation the coefficient of \(y^{k-2}\) to zero,

$$a_0(k+0)(k+0-1)=0$$

$$a_0\ne 0\Rightarrow k(k-1)=0$$

$$Either\; k=0\; or\; k=1$$

Equating the coefficient of general term [ i.e. the coefficient if \(y^{k+r}\) to zero.]

$$a_{r+2}(k+r+2)(k+r+1)-a_r(2k+2r-\lambda+1)=0$$

$$or,\;\; a_{r+2}=\frac{a_r[2k+2r-\lambda+1]}{(k+r+2)(k+r+1)}$$

$$or,\;\; \frac{a_{r+2}}{a_r}= \frac{2k+2r-\lambda +1}{(k+r+2)(k+r+1)}$$

For, k=0

$$\frac{a_{r+2}}{a_r}= \frac{2r-\lambda+1}{(r+2)(r+1)}\dotsm(18)$$

For terminating series solution \(a_r\ne 0\), but \(a_{r+2}\)=0\)

The series terminate at rth term all the coeff. After rth is zero.

$$[\lambda= 2r+1= 2n+1]\dotsm(19)$$

Where, r=0,1,2,…,r=n

Substituting value of \(\lambda\) in equation (9), we get,

$$E= \frac{\hbar\omega}{2}\cdot (2n+1)$$

$$\therefore\;\; E_n= \biggl(n+\frac12\biggr) \hbar \omega\dotsm(20)$$

Where, n= 0,1,2,3,…

Equation (20) gives the energy eigen value of one-dimensional simple harmonic oscillator.

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.