06 - Harmonic Oscillators

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 $V(0)=0$

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

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

Hamiltonian Operator for One-Dimensional Quantum Harmonic Oscillator

Hamiltonian Operator for One-Dimensional Quantum Harmonic Oscillator defined by

$$H=\frac{P^2}{2m}+V(X)=\frac{1}{2m}{P}^2+\frac{1}{2}m{\omega }^2{X}^2$$

Raising and Lowering Operators

Raising and Lowering Operators defined by

$$a_{\pm }\equiv \frac{1}{\sqrt{2m\omega \hslash }}(\mp iP+m\omega X)$$

As $X$ and $P$ are self-adjoint then $a_{+}=(a_{-}{)}^{\ast }$

Self-Adjoint Number Operator

Self-Adjoint Number Operator is defined as

$$N\equiv a_{+}{a}_{-}$$

Proof - Self-Adjoint

$$N^{\ast }=(a_{+}{a}_{-}{)}^{\ast }=a_{-}^{\ast }{a}_{+}^{\ast }=a_{+}{a}_{-}=N$$

Alternate Hamiltonian Form for 1d Quantum Harmonic Oscillator lemma

Alternate Hamiltonian may be defined as

$$H=\left(N+\frac{1}{2}1\right)\hslash \omega$$

where $N$ is the self-adjoint number operator

Proof

$$\begin{array}{rl}N& =a_{+}{a}_{-}=\frac{1}{2m\omega \hslash }(-iP+m\omega X)(iP+mwX)\\ & =\frac{1}{2m\omega \hslash }(P^2+m^2{\omega }^2{X}^2+im\omega [X,P])\\ & =\frac{1}{2m\omega \hslash }(P^2+m^2{\omega }^2{X}^2-m\omega \hslash 1)\\ & =\frac{1}{\hslash \omega }\left(\frac{1}{2m}{P}^2+\frac{1}{2}m{\omega }^2{X}^2\right)-\frac{1}{2}1\end{array}$$

By rearranging then get equation

Properties of Lowering and Raising Operators

1. $[a_{-},a_{+}]=1$

2. $[N,a_{\pm }]=\pm a_{\pm }$

3. $||a_{-}\psi |{|}^2=⟨\psi \mid N\psi ⟩$

4. $||a_{+}\psi |{|}^2=⟨\psi \mid (N+1)\psi ⟩$

where $\psi \in \mathcal{H}$

Proof

$(1)$

$$[a_{-},a_{+}]=\frac{1}{2m\omega \hslash }[iP+m\omega X,-iP+m\omega X]=\frac{1}{2m\omega \hslash }m\omega (-i[X,P]+i[P,X])=1$$

$(2)$

$$[N,a_{+}]=[a_{+}{a}_{-},a_{+}]=a_{+}[a_{-},a_{+}]+[a_{+},a_{+}]a_{-}=a_{+}$$

$(3)$

$$||a_{-}\psi |{|}^2=⟨a_{-}\psi \mid a_{-}\psi ⟩=⟨\psi \mid a_{+}{a}_{-}\psi ⟩=⟨\psi \mid N\psi ⟩$$

$(4)$

$$||a_{+}\psi |{|}^2=⟨a_{+}\psi \mid a_{+}\psi ⟩=⟨\psi \mid a_{-}{a}_{+}\psi ⟩=⟨\psi \mid ([a_{-},a_{+}]+a_{+}{a}_{-})\psi ⟩=⟨\psi \mid (1+N)\psi ⟩$$

Actions of Ladder Operators on Eigenstates lemma

Suppose $N\psi =\lambda \psi$ where $\lambda \in \mathbb{R}$ and $\psi \ne 0\in \mathcal{H}$ then

1. $a_{\pm }\psi$ are eigenstates of $N$ with eigenvalues $\lambda \pm 1$ (assuming non-zero)

2. $\lambda \ge 0$ with $\lambda =0$ if and only if $a_{-}\psi =0$ (ground state of $N$)

Proof

$(1)$

$$N(a_{\pm }\psi )=([N,a_{\pm }]+a_{\pm }N)\psi =(\pm a_{\pm }+\lambda a_{\pm })\psi =(\lambda \pm 1)a_{\pm }\psi$$

$(2)$

$$N\psi =\lambda \psi ⟹\lambda ||\psi |{|}^2=⟨\psi \mid N\psi ⟩=||a_{-}\psi |{|}^2\ge 0$$

Hence $\lambda \ge 0$ with equality if and only if $a_{-}\psi =0$

Eigenvalues of Number Operator

Eigenvalues of $N$ are $n\in {\mathbb{Z}}_{\ge 0}$

If there is a unique ground state ${\psi }_0$ of $N$ up to normalisation then there is
Eigenstate with eigenvalue $n$ proportional to $a_{+}^n{\psi }_0$

Proof

Let $\psi$ be an eigenstate of $N$ with eigenvalue $\mu$

Suppose for all $n\ge 0$ then

$$a_{-}^n\psi \ne 0\in \mathcal{H}$$

Inductively $a_{-}^n\psi$ are eigenstates of $N$ with eigenvalues $\mu -n$

For sufficiently large $n$ then
Eigenvalue $\lambda =\mu -n<0$ contradicting $(2)$ of Actions of ladder operators on eigenstates

Hence exists smallest $n\ge 0$ where $a_{-}^n\psi \ne 0$ but $a_{-}^{n+1}\psi =0$ then

$$a_{-}^n\psi \text{ has eigenvalue }\lambda =\mu -n$$

So $\mu =n$
Hence spectrum of $N$ is subset of ${\mathbb{Z}}_{\ge 0}$

Suppose ${\psi }_0$ is a ground state of $N$

By induction then $a_{+}^n{\psi }_0$ is an eigenstate of $N$ with eigenvalues $n\in {\mathbb{Z}}_{\ge 0}$
States are non-zero by Properties of lowering and raising operators so

$$||a_{+}^{n+1}{\psi }_0||=⟨a_{+}^n{\psi }_0\mid (N+1)a_{+}^n{\psi }_{0}z⟩=(n+1)||a_{+}^n{\psi }_0|{|}^2>0$$

where $||{\psi }_0|{|}^2>0$ so holds by induction
Hence spectrum of $N$ is ${\mathbb{Z}}_{\ge 0}$

Suppose $\psi$ is any state with eigenvalue $n>1$
Hence $a_{-}^n\psi$ has eigenvalue zero under $N$ hence so does $a_{-}^n(a_{+}^n{\psi }_0)$

Both are hence ground states hence if ground state is unique up to normalisation then

$$a_{-}^n\psi =\kappa a_{-}^n{a}_{+}^n{\psi }_0\text{ where }\kappa \in \mathbb{C}$$

Let

$$\tilde{\psi }\equiv \psi -\kappa a_{+}^n{\psi }_0$$

Hence $N\tilde{\psi }=n\tilde{\psi }$ as eigenvalue $n$ but $a_{-}^n\tilde{\psi }=0$
Thus inductively then

$$\begin{array}{rl}||\tilde{\psi }|{|}^2& =\frac{1}{n}⟨\tilde{\psi }\mid N\tilde{\psi }⟩=\frac{1}{n}||a_{-}\tilde{\psi }|{|}^2=\frac{1}{n(n-1)}⟨a_{-}\tilde{\psi }\mid N{a}_{-}\tilde{\psi }⟩=\frac{1}{n(n-1)}||a_{-}^2\tilde{\psi }|{|}^2\\ & =\cdots =\frac{1}{n!}||a_{-}^n\tilde{\psi }|{|}^2=0\end{array}$$

Hence $\tilde{\psi }=0$ so eigenstates with eigenvalue $n$ are unique up to normalisation

Energies of One-Dimensional Quantum Harmonic Oscillator corollary

Consider One-Dimensional Quantum Harmonic Oscillator of Frequency $\omega$

Energy is

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

with corresponding stationary states $a_{+}^n{\psi }_0$
where ${\psi }_0$ is ground state

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

where $a_0\in \mathbb{C}$ is a integration constant which is normalised for $a_0={\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}$

Normalised Stationary States of Quantum Harmonic Oscillator

Normalised Stationary States of Quantum Harmonic Oscillator are

$${\psi }_n\equiv \frac{1}{\sqrt{n!}}{a}_{+}^n{\psi }_0$$

as with normalisation constant

with orthonormal property

$$⟨{\psi }_m\mid {\psi }_n⟩={\delta }_{mn}$$