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

$$\begin{equation}\psi_{0}(x)=\frac{\mathrm{e}^{-x^{2} / 2 d^{2}}}{\pi^{1 / 4} \sqrt{d}}\end{equation}$$

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 $$\begin{aligned} \label{e5.xxx} \int_{-\infty}^\infty \psi_m\,x\,\psi_n\,dx = \frac{d}{\sqrt{2}}\int_{-\infty}^{\infty}\psi_m\,(a_+ + a_-)\,\psi_n\,dx= \sqrt{\frac{\hbar}{2\,m\,\omega}}\left(\sqrt{m}\,\delta_{m,n+1} + \sqrt{n}\,\delta_{m,n-1}\right).\end{aligned}$$

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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