Simple Harmonic Oscillator

The classical Hamiltonian of a simple harmonic oscillator is

$$H = \frac{p^{ 2}}{2 m} + \frac{1}{2} K x^{ 2},$$ (C.106)

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

$$\frac{d^{ 2}\psi}{dx^{ 2}} = \frac{2 m}{\hbar^{ 2}}\left(\frac{1}{2} K x^{ 2}-E\right)\psi.$$ (C.107)

Let $\omega = \sqrt{K/m}$ , where $\omega$ is the oscillator's classical angular frequency of oscillation. Furthermore, let

$$y = \sqrt{\frac{m \omega}{\hbar}} x,$$ (C.108)

and

$$\epsilon = \frac{2 E}{\hbar \omega}.$$ (C.109)

Equation (C.107) reduces to

$$\frac{d^{ 2}\psi}{dy^{ 2}} - \left(y^{ 2}-\epsilon\right)\psi = 0.$$ (C.110)

We need to find solutions to the previous equation that are bounded at infinity. In other words, solutions that satisfy the boundary condition $\psi\rightarrow 0$ as $\vert y\vert\rightarrow\infty$ .

Consider the behavior of the solution to Equation (C.110) in the limit $\vert y\vert\gg 1$ . As is easily seen, in this limit, the equation simplifies somewhat to give

$$\frac{d^{ 2}\psi}{dy^{ 2}} - y^{ 2} \psi \simeq 0.$$ (C.111)

The approximate solutions to the previous equation are

$$\psi(y) \simeq A(y) {\rm e}^{ \pm y^{ 2}/2},$$ (C.112)

where $A(y)$ is a relatively slowly-varying function of $y$ . Clearly, if $\psi(y)$ is to remain bounded as $\vert y\vert\rightarrow\infty$ then we must chose the exponentially decaying solution. This suggests that we should write

$$\psi(y) = h(y) {\rm e}^{-y^{ 2}/2},$$ (C.113)

where we would expect $h(y)$ to be an algebraic, rather than an exponential, function of $y$ .

Substituting Equation (C.113) into Equation (C.110), we obtain

$$\frac{d^{ 2}h}{dy^{ 2}} - 2 y \frac{dh}{dy} + (\epsilon-1) h = 0.$$ (C.114)

Let us attempt a power-law solution of the form

$$h(y) = \sum_{i=0,\infty} c_i y^{ i}.$$ (C.115)

Inserting this test solution into Equation (C.114), and equating the coefficients of $y^{ i}$ , we obtain the recursion relation

$$c_{i+2} = \frac{(2 i-\epsilon+1)}{(i+1) (i+2)} c_i.$$ (C.116)

Consider the behavior of $h(y)$ in the limit $\vert y\vert\rightarrow\infty$ . The previous recursion relation simplifies to

$$c_{i+2} \simeq \frac{2}{i} c_i.$$ (C.117)

Hence, at large $\vert y\vert$ , when the higher powers of $y$ dominate, we have

$$h(y) \sim C \sum_{j}\frac{y^{ 2 j}}{j!}\sim C \exp\left(+y^{ 2}\right).$$ (C.118)

It follows that $\psi(y) = h(y) \exp\left(-y^{ 2}/2\right)$ varies as $\exp\left(+y^{ 2}/2\right)$ as $\vert y\vert\rightarrow\infty$ . This behavior is unacceptable, because it does not satisfy the boundary condition $\psi\rightarrow 0$ as $\vert y\vert\rightarrow\infty$ . The only way in which we can prevent $\psi$ from blowing up as $\vert y\vert\rightarrow\infty$ is to demand that the power series (C.115) terminate at some finite value of $i$ . This implies, from the recursion relation (C.116), that

$$\epsilon = 2 n+1,$$ (C.119)

where $n$ is a non-negative integer. Note that the number of terms in the power series (C.115) is $n+1$ . Finally, using Equation (C.109), we obtain

$$E = (n+1/2) \hbar \omega,$$ (C.120)

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

$$\psi_0(x) = \frac{{\rm e}^{-x^{ 2}/2 d^{ 2}}}{\pi^{1/4}\sqrt{d}},$$ (C.121)

where $d=\sqrt{\hbar/m \omega}$ .