04. The Harmonic Oscillator

4.1 The One-Dimensional Harmonic Oscillator

Harmonic Oscillator Potential

For angular frequency $\omega$ then

Harmonic Oscillator Potential is

$$V(x)=\frac{1}{2}m{\omega }^2{x}^2$$

Proof

Consider classical particle of mass $m$ moving of one dimension under potential $V(x)$

Near a critical point $x_0$ of $V$ where $V^{\prime }(x_0)=0$ then Taylor Expansion gives

$$V(x)=V(x_0)+\frac{1}{2}{V}^{\prime \prime }(x_0)(x-x_0{)}^2+O((x-x_0{)}^3)$$

If $V^{\prime \prime }(x_0)>0$ then $x_0$ is a local minimum of potential

Without loss of generality suppose critical point $x_0$ is at origin

Thus to lowest order, potential near $x_0=0$ is

$$V(x)=V(0)+\frac{1}{2}m{\omega }^2{x}^2$$

where ${\omega }^2=\frac{V^{\prime \prime }(0)}{m}$

As additive constant $V(0)$ does not affect dynamics ($F=-V^{\prime }$) then suppose $V(0)=0$

$$V(x)=\frac{1}{2}m{\omega }^2{x}^2$$

Link to original

Parity of a Wave Function

In one dimension, stationary state wave functions satisfying

$$\psi (-x)=\pm \psi (x)$$

describes an

$$\left\{\begin{array}{c}\text{even}\\ \text{odd}\end{array}\right\}\text{ parity state}$$

Link to original

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

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

Link to original

Classical Forbidden Region

In classical system then

$$E=\frac{{\rho }^2}{2m}+V(\mathbf{x})\ge V(\mathbf{x})$$

Then

$\{x\in {\mathbb{R}}^3:E<V(\mathbf{x})\}$ is known as the classically forbidden region

Link to original

---

4.2 Higher-Dimensional Oscillators

Quantum Harmonic Oscillator in Two Dimensions

Potential for Quantum Harmonic Oscillator in Two Dimensions is

$$V(x,y)=\frac{1}{2}m({\omega }_1^2{x}^2+{\omega }_2^2{y}^2)$$

with Corresponding Stationary State Schrödinger Equation

$$-\frac{{\hslash }^2}{2m}\left(\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}\right)+V(x,y)\psi =E\psi$$

Solved by Separation of Variables $\psi (x,y)=X(x)Y(y)$ then

$$\begin{array}{r}-\frac{{\hslash }^2}{2m}\frac{d^{2}X}{d{x}^2}+\frac{1}{2}m{\omega }_1^2{x}^{2}X=E_{1}X\\ -\frac{{\hslash }^2}{2m}\frac{d^{2}Y}{d{y}^2}+\frac{1}{2}m{\omega }_2^2{y}^{2}Y=E_{2}Y\end{array}$$

where $E_1+E_2=E$

Hence by Energies and Wave Functions of Quantum Harmonic Oscillator then

$$E=E_{n_1,n_2}=\left(n_1+\frac{1}{2}\right)\hslash {\omega }_1+\left(n_2+\frac{1}{2}\right)\hslash {\omega }_2$$

where quantum numbers $n_1,n_2\in {\mathbb{Z}}^{\ge 0}$

With corresponding normalised stationary wave functions as products

$$\begin{array}{rl}{\psi }_{n_1,n_2}(x,y)& ={\psi }_{n_1}(x){\psi }_{n_2}(y)\\ & =\frac{1}{\sqrt{2^{n_1+n_2}{n}_1!n_2!}}{\left(\frac{m^2{\omega }_1{\omega }_2}{{\pi }^2{\hslash }^2}\right)}^{1/4}{H}_{n_1}\left(\sqrt{\frac{m{\omega }_1}{\hslash }}x\right)H_{n_2}\left(\sqrt{\frac{m{\omega }_2}{\hslash }}y\right)e^{-m({\omega }_1{x}^2+{\omega }_2{y}^2)/2\hslash }\end{array}$$

where ${\psi }_n$ is the normalised stationary state wave function for a one-dimensional harmonic oscillator

Link to original