01 - Energies and Wave Functions of Quantum Harmonic Oscillator

Energies and Wave Functions of Quantum Harmonic Oscillator

Consider One-Dimensional Quantum Harmonic Oscillator of angular frequency $\omega$ then

Energies

$$E=E_n=\left(n+\frac{1}{2}\right)\hslash \omega$$

for $n\in {\mathbb{Z}}^{\ge 0}$ a non-negative integer

Corresponding Normalised Stationary State Wave Functions

$${\psi }_n(x)=\frac{1}{\sqrt{2^{n}n!}}{\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}{H}_n(\xi )e^{-{\xi }^2/2}$$

where
$\xi =\sqrt{\frac{m\omega }{\hslash }}x$
$H_n$ is the $n$th Hermite polynomial with closed form

$$H_n(\xi )=e^{{\xi }^2/2}{\left(\xi -\frac{d}{d\xi }\right)}^n{e}^{-{\xi }^2/2}$$

Proof

Consider Stationary State Schrödinger Equation of energy $E$ then

$$-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+\frac{1}{2}m{\omega }^2{x}^2\psi =E\psi$$

Remove physical constants by using dimensionless variables

$$ϵ\equiv \frac{2E}{\hslash \omega },\xi \equiv \sqrt{\frac{m\omega }{\hslash }}x$$

Hence equation becomes

$$-\frac{d^2\chi }{d{\xi }^2}+{\xi }^2\chi =ϵ\chi$$

where

$$\psi (x)\equiv \chi (\xi )=\chi \left(\sqrt{\frac{m\omega }{\hslash }}x\right)$$

One solution is $\chi (\xi )=e^{\mp {\xi }^2/2}$ with $ϵ=\pm 1$ as

$$\frac{d}{d\xi }(e^{\mp {\xi }^2/2})=\mp \xi e^{\mp {\xi }^2/2},\frac{d^2}{d{\xi }^2}(e^{\mp {\xi }^2/2})=({\xi }^2\mp 1)e^{\mp {\xi }^2/2}$$

Need normalisable solutions to Schrödinger Equation hence

$${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx<\infty$$

Hence through change of variables then for $\chi (\xi )=e^{\mp {\xi }^2/2}$ require

$$\sqrt{\frac{\hslash }{m\omega }}{\int }_{-\infty }^{\infty }{e}^{\mp {\xi }^2}d\xi <\infty$$

which is finite only for minus sign

Hence take solution

$$\chi (\xi )=e^{-{\xi }^2/2}\text{ and }ϵ=1$$

Consider change of variable

$$\chi (\xi )\equiv f(\xi )e^{-{\xi }^2/2}$$

Hence

$$\begin{array}{rl}\frac{d\chi }{d\xi }& =\left(\frac{df}{d\xi }-\xi f\right)e^{-{\xi }^2/2}\\ \frac{d^2\chi }{d{\xi }^2}& =\left[\frac{d^{2}f}{d{\xi }^2}-\xi \frac{df}{d\xi }-f-\xi \left(\frac{df}{d\xi }-\xi f\right)\right]e^{-{\xi }^2/2}\end{array}$$

So differential equation becomes

$$\frac{d^{2}f}{d{\xi }^2}-2\xi \frac{df}{d\xi }+(ϵ-1)f=0$$

Suppose $f(\xi )$ is a power series solution

$$f(\xi )=\sum _{k=0}^{\infty }{a}_k{\xi }^k$$

Hence

$$\xi \frac{df}{d\xi }=\sum _{k=0}^{\infty }k{a}_k{\xi }^k\text{ and }\frac{d^{2}f}{d{\xi }^2}=\sum _{k=0}^{\infty }(k+1)(k+2)a_{k+2}{\xi }^k$$

Substituting then

$$\sum _{k=0}^{\infty }[(k+1)(k+2)a_{k+2}-2k{a}_k+(ϵ-1)a_k]{\xi }^k=0$$

Hence as each power of $\xi$ is zero then

$$a_{k+2}=\frac{2k+1-ϵ}{(k+1)(k+2)}{a}_k$$

Thus even and odd powers are separated, so gives two linearly independent series solutions

$$\begin{array}{rl}{f}_{\text{even}}(\xi )& =a_0\left[1+\frac{(1-ϵ)}{2!}{\xi }^2+\frac{(5-ϵ)(1-ϵ)}{4!}{\xi }^4+\cdots \right]\\ f_{\text{odd}}(\xi )& =a_0\left[\xi +\frac{(3-ϵ)}{3!}{\xi }^3+\frac{(7-ϵ)(3-ϵ)}{5!}{\xi }^5+\cdots \right]\end{array}$$

For normalisable solutions then series must terminate (skipped proof)
Suppose $n\ge 0$ is the least integer such that $a_{n+2}=0$ then

$$2n+1-ϵ=0$$

As $ϵ=\frac{2E}{\hslash \omega }$ then

$$E=E_n=\left(n+\frac{1}{2}\right)\hslash \omega$$

Ground State is when $f\equiv 1$ and $ϵ=1$ with $n=0$ hence

$${\psi }_0(x)=a_0{e}^{-m\omega x^2/2\hslash }$$

Normalising ground state wave function then

$$1={\int }_{-\infty }^{\infty }|{\psi }_0(x){|}^{2}dx=|a_0{|}^2{\int }_{-\infty }^{\infty }{e}^{-m\omega x^2/\hslash }dx$$

Hence using Gaussian Integral for a Normal Distribution of Variance ${\sigma }^2=\frac{\hslash }{2m\omega }$ then

$$|a_0{|}^2=\sqrt{\frac{m\omega }{\pi \hslash }}$$

Thus full normalised time-dependent ground state wave function is

$${\Psi }_0(x,t)={\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}{e}^{-(m\omega x^2+i\hslash \omega t)/2\hslash }$$

Generally Stationary State Wave Functions are

$${\chi }_n(\xi )=f_n(\xi )e^{-{\xi }^2/2}$$

where $f_n(\xi )$ is an even/odd polynomial in $\xi =\sqrt{\frac{m\omega }{\hslash }}x$ of degree $n$ for $n$ even/odd respectively

Polynomials $f_n$ can be determined explicitly by setting $ϵ-1=2n$ in recurrence relation
Before normalisation

$$f_n(\xi )\equiv H_n(\xi )\text{is the }n\text{th}\text{ Hermite polynomial}$$