# 4.7: Simple Harmonic Oscillator


The classical Hamiltonian of a simple harmonic oscillator is $H = \frac{p^{\,2}}{2\,m} + \frac{1}{2}\,K\,x^{\,2},$ where $$K>0$$ is the so-called force constant of the oscillator. Assuming that the quantum mechanical Hamiltonian has the same form as the classical Hamiltonian, the time-independent Schrödinger equation for a particle of mass $$m$$ and energy $$E$$ moving in a simple harmonic potential becomes $\label{e5.90} \frac{d^{\,2}\psi}{dx^{\,2}} = \frac{2\,m}{\hbar^{\,2}}\left(\frac{1}{2}\,K\,x^{\,2}-E\right)\psi.$ Let $$\omega = \sqrt{K/m}$$, where $$\omega$$ is the oscillator’s classical angular frequency of oscillation. Furthermore, let $y = \sqrt{\frac{m\,\omega}{\hbar}}\,x,$ and $\label{e5.92} \epsilon = \frac{2\,E}{\hbar\,\omega}.$ Equation ([e5.90]) reduces to $\label{e5.93} \frac{d^{\,2}\psi}{dy^{\,2}} - (y^{\,2}-\epsilon)\,\psi = 0.$ We need to find solutions to the previous equation which are bounded at infinity: that is, solutions which satisfy the boundary condition $$\psi\rightarrow 0$$ as $$|y|\rightarrow\infty$$.

Consider the behavior of the solution to Equation ([e5.93]) in the limit $$|y|\gg 1$$. As is easily seen, in this limit the equation simplifies somewhat to give $\frac{d^{\,2}\psi}{dy^{\,2}} - y^{\,2}\,\psi \simeq 0.$ The approximate solutions to the previous equation are $\psi(y) \simeq A(y)\,{\rm e}^{\pm y^{\,2}/2},$ where $$A(y)$$ is a relatively slowly varying function of $$y$$. Clearly, if $$\psi(y)$$ is to remain bounded as $$|y|\rightarrow\infty$$ then we must chose the exponentially decaying solution. This suggests that we should write $\label{e5.96} \psi(y) = h(y)\,{\rm e}^{-y^{\,2}/2},$ where we would expect $$h(y)$$ to be an algebraic, rather than an exponential, function of $$y$$.

Substituting Equation ([e5.96]) into Equation ([e5.93]), we obtain $\label{e5.97} \frac{d^{\,2}h}{dy^{\,2}} - 2\,y\,\frac{dh}{dy} + (\epsilon-1)\,h = 0.$ Let us attempt a power-law solution of the form $\label{e5.98} h(y) = \sum_{i=0,\infty} c_i\,y^{\,i}.$ Inserting this test solution into Equation ([e5.97]), and equating the coefficients of $$y^{\,i}$$, we obtain the recursion relation $\label{e5.99} c_{i+2} = \frac{(2\,i-\epsilon+1)}{(i+1)\,(i+2)}\,c_i.$ Consider the behavior of $$h(y)$$ in the limit $$|y|\rightarrow\infty$$. The previous recursion relation simplifies to $c_{i+2} \simeq \frac{2}{i}\,c_i.$ Hence, at large $$|y|$$, when the higher powers of $$y$$ dominate, we have $h(y) \sim C \sum_{j}\frac{y^{\,2 j}}{j!}\sim C\,{\rm e}^{\,y^{\,2}}.$ It follows that $$\psi(y) = h(y)\,\exp(-y^{\,2}/2)$$ varies as $$\exp(\,y^{\,2}/2)$$ as $$|y|\rightarrow\infty$$. This behavior is unacceptable, because it does not satisfy the boundary condition $$\psi\rightarrow 0$$ as $$|y|\rightarrow\infty$$. The only way in which we can prevent $$\psi$$ from blowing up as $$|y|\rightarrow\infty$$ is to demand that the power series ([e5.98]) terminate at some finite value of $$i$$. This implies, from the recursion relation ([e5.99]), that $\epsilon = 2\,n+1,$ where $$n$$ is a non-negative integer. Note that the number of terms in the power series ([e5.98]) is $$n+1$$. Finally, using Equation ([e5.92]), we obtain $E = (n+1/2)\,\hbar\,\omega,$ for $$n=0,1,2,\cdots$$.

Hence, we conclude that a particle moving in a harmonic potential has quantized energy levels that are equally spaced. The spacing between successive energy levels is $$\hbar\,\omega$$, where $$\omega$$ is the classical oscillation frequency. Furthermore, the lowest energy state ($$n=0$$) possesses the finite energy $$(1/2)\,\hbar\,\omega$$. This is sometimes called zero-point energy. It is easily demonstrated that the (normalized) wavefunction of the lowest energy state takes the form $$\psi_{0}(x)=\frac{\mathrm{e}^{-x^{2} / 2 d^{2}}}{\pi^{1 / 4} \sqrt{d}}$$

Let $$\psi_n(x)$$ be an energy eigenstate of the harmonic oscillator corresponding to the eigenvalue $E_n = (n+1/2)\,\hbar\,\omega.$ Assuming that the $$\psi_n$$ are properly normalized (and real), we have $\label{e5.107} \int_{-\infty}^\infty \psi_n\,\psi_m\,dx = \delta_{nm}.$ Now, Equation ([e5.93]) can be written $\label{e5.108} \left(-\frac{d^{\,2}}{d y^{\,2}}+y^{\,2}\right)\psi_n = (2n+1)\,\psi_n,$ where $$x = d\,y$$, and $$d=\sqrt{\hbar/m\,\omega}$$. It is helpful to define the operators $\label{e5.109} a_\pm = \frac{1}{\sqrt{2}}\left(\mp \frac{d}{dy}+y\right).$ As is easily demonstrated, these operators satisfy the commutation relation $[a_+,a_-] = -1.$ Using these operators, Equation ([e5.108]) can also be written in the forms $a_+\,a_-\,\psi_n = n\,\psi_n,$ or $a_-\,a_+\,\psi_n = (n+1)\,\psi_n.$ The previous two equations imply that \begin{aligned} \label{e5.113} a_+\,\psi_n &= \sqrt{n+1}\,\psi_{n+1},\\[0.5ex] a_-\,\psi_n &=\sqrt{n}\,\psi_{n-1}.\label{e5.114}\end{aligned} We conclude that $$a_+$$ and $$a_-$$ are raising and lowering operators, respectively, for the harmonic oscillator: that is, operating on the wavefunction with $$a_+$$ causes the quantum number $$n$$ to increase by unity, and vice versa. The Hamiltonian for the harmonic oscillator can be written in the form $H = \hbar\,\omega\,\left(a_+\,a_- + \frac{1}{2}\right),$ from which the result $H\,\psi_n = (n+1/2)\,\hbar\,\omega\,\psi_n = E_n\,\psi_n$ is readily deduced. Finally, Equations ([e5.107]), ([e5.113]), and ([e5.114]) yield the useful expression \begin{aligned} \label{e5.xxx} \int_{-\infty}^\infty \psi_m\,x\,\psi_n\,dx = \frac{d}{\sqrt{2}}\int_{-\infty}^{\infty}\psi_m\,(a_+ + a_-)\,\psi_n\,dx= \sqrt{\frac{\hbar}{2\,m\,\omega}}\left(\sqrt{m}\,\delta_{m,n+1} + \sqrt{n}\,\delta_{m,n-1}\right).\end{aligned}
