Probability of finding one dimensional Simple Harmonic Oscillator

Probability of finding one dimensional S.H.O outside classical limit at ground state.

\(\Rightarrow\) The ground state wave function for one dimensional simple harmonic oscillator is,

$$\psi_0(x)=\biggl[\frac{\alpha}{\sqrt{\pi}}\biggr]^\frac12 e^{\frac{-\alpha^2 x^2}{2}}\dotsm(1)$$

Where, \(\alpha= \sqrt{\frac{m\omega}{\hbar}}\)

The corresponding Probability density,

$$\rho_0(x)= \psi_0^*(x)\psi_0(x)$$

$$or,\;\; \rho_0(x)=\frac{\alpha}{\sqrt{\pi}}e^{-\alpha^2 x^2}\dotsm(2)$$

The probability of finding particle within distance x to x+dx from equilibrium position (x=0) is given by,

$$dp(x)= \rho_0(x)dx$$

$$or,\; dp(x)=\frac{\alpha}{\sqrt{\pi}}e^{-\alpha^2 x^2} dx\dotsm(3)$$

ground state energy of 1- Dimensional Simple harmonic oscillator is

$$E_0= \frac{\hbar \omega}{2}\dotsm(4)$$

The corresponding energy in classical mechanics is,

$$E_0= \frac12 m\omega^2 A^2\dot(5)$$

Where, A is amplitude.

From (4) and (5)

$$or,\;\; \frac{\hbar\omega}{2}= \frac12 m\omega^2 A^2$$

$$or,\;\; A=\pm\sqrt{\frac{\hbar}{m\omega}}$$

$$or,\;\; A= \pm \frac12\dotsm(6)$$

The classical limits of 1-Dimensional simple harmonic oscillation \(\frac{-1}{\alpha}\) to \(+\frac{1}{\alpha}\).

The probability of finding quantum harmonic oscillator at ground state with classical limits ( From \(\frac{-1}{2}\) to \(\frac{+a}{\alpha}\)) is given by,

$$P_{withinclassicallimit}= \int_{\frac{-1}{\alpha}}^{\frac{+1}{\alpha}} \frac{\alpha}{\sqrt{\pi}} e^{-\alpha^2 x^2} dx$$

$$=\frac{\alpha}{\sqrt{\pi}}\int_{\frac{-1}{\alpha}}^{\frac{+1}{\alpha}} e^{-\alpha^2 x^2 } dx$$

$$=\frac{2\alpha}{\sqrt{\pi}}\int_0^{\frac{1}{\alpha}} e^{-\alpha^2 x^2}dx$$

Let \(\alpha x= t\Rightarrow dx=\frac{dt}{\alpha}\)

When x=0\(\Rightarrow t=0\)

$$x=\frac{1}{\alpha}\Rightarrow t=1$$

So, $$P=\frac{2\alpha}{\sqrt{\pi}}\int_0^1 e^{-t^2}\frac{dt}{\alpha}$$

$$=\frac{2}{\sqrt{\pi}}\int_0^1 \biggl[\sum_{n=0}^\infty \frac{(-t^2)^n}{n!}\biggr]dt$$

$$\frac{2}{\sqrt{\pi}}\int_0^1 \biggl[ 1-\frac{t^2}{1!}+\frac{t^4}{2!}-\frac{t^6}{3!}+\frac{t^8}{4!}-\frac{t^{10}}{5!}\dotsm\biggr]dt$$

$$=\frac{2}{\sqrt{\pi}}\biggl[t-\frac{t^3}{3\times 1!}+\frac{t^5}{5\times 2!}-\frac{t^7}{7\times 3!}+\frac{t^9}{9\times 4!}-\frac{t^{11}}{11\times 5!}\dotsm\biggr]_0^1$$

$$=\frac{2}{\pi}\biggl[1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}-\frac{1}{216}\dotsm\biggr]$$

$$\approx 0.84$$

$$=84%$$

Hence the probability of finding one-D simple harmonic oscillator within classical limit is 0.84 or 84% at ground state, Hence probability of finding oscillator outside classical limits is given by,

$$P_{outside}=1- P_{within}$$

$$=1-0.84$$

$$\approx 0.16$$

$$=16%$$

This value of probability outside classical limits is due to wave mechanical behavior of particle, which can not be explained in classical mechanics,

Question:

Probability of finding 1-D simple harmonic oscillator outside classical limit at first excited state.

$$OR$$

Write Schrodinger equation for 1-D unsymmetric simple harmonic oscillator for potential of the form

$$V(x)= \frac12 kx^2 \; for \; x>0$$

$$=\infty \; for\; x\leq 0$$

Also find energy eigen function, energy eigen value and the first or ground state wave function.

\(\Rightarrow\)

Ground state \(\psi_1(x)\) (n=1)

$$E_1=\frac32 \hbar \omega$$

first excited state (n=3)

$$E_3= \frac{7\hbar \omega}{2}$$

All even number does not vanish at x=0. So take only odd numbers.

$$i.e,\; For, n=1$$

$$\psi_1(x)=\biggl[\frac{\alpha}{2^1\sqrt{\pi}1!}\biggr]^\frac12 H_1(\alpha x)e^{-\alpha^2 \frac{x^2}{2}}$$

$$=\biggl[\frac{\alpha}{2\sqrt{\pi}}\biggr]^\frac12 2\alpha\cdot x e^{-\alpha^2 \frac{x^2}{2}}$$

$$For n=3,$$

$$\psi_3(x)=\biggl[ \frac{\alpha}{2^3 \sqrt{\pi}8!}\biggr]^\frac12 H_3(\alpha x)e^{-\frac{\alpha^2 x^2}{2}}$$

$$=\biggl[\frac{\alpha}{48\sqrt{\pi}}\biggr]^\frac12 (8\alpha^3 x^3-12\alpha x)e^{-\frac{\alpha^2 x^2 }{2}}$$

$$For\; n=0$$

$$\psi_0(x)=\biggl[\frac{\alpha}{\sqrt{\pi}}\biggr]^\frac12 e^{-\frac{\alpha^2 x^2}{2}}$$

$$For,\; n=0,$$

$$\psi_0(x)\ne 0$$

$$For\; n=2$$

$$\psi(x)= \biggl[\frac{\alpha}{8\sqrt{\pi}}\biggr]^\frac12 (4\alpha^2 x^2 - 2) e^{-\alpha^2 \frac{x^2}{2}}$$

$$For x=0\Rightarrow \psi_2(x)\ne 0$$

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.