Simple Harmonic Oscillator

Consider the motion of a particle of mass $m$ in the simple harmonic oscillator potential

$$U(x)= \frac{1}{2}\,\kappa\,x^{\,2},$$ (4.165)

where $\kappa>0$ is the so-called force constant of the oscillator. According to classical physics, a particle trapped in this potential executes simple harmonic motion at the angular frequency $\omega=\!\sqrt{\kappa/m}$. (See Section 1.3.6.) The time-independent Schrödinger equation for a particle of mass $m$ and energy $E$ moving in a simple harmonic potential becomes

$$\frac{d^{2}\psi}{dx^{2}} = \frac{2\,m}{\hbar^{2}}\left(\frac{1}{2}\,\kappa\,x^{2}-E\right)\psi.$$ (4.166)

[See Equation (4.71).] Let

$$y = \sqrt{\frac{m\,\omega}{\hbar}}\,x,$$ (4.167)

and

$$\epsilon = \frac{2\,E}{\hbar\,\omega}.$$ (4.168)

Equation (4.166) reduces to

$$\frac{d^{2}\psi}{dy^{2}} - (y^{2}-\epsilon)\,\psi = 0.$$ (4.169)

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

Consider the behavior of the solution to Equation (4.169) in the limit $\vert y\vert\gg 1$. As is easily seen, in this limit the equation simplifies somewhat to give

$$\frac{d^{2}\psi}{dy^{2}} - y^{2}\,\psi \simeq 0.$$ (4.170)

The approximate solutions to the previous equation are

$$\psi(y) \simeq A(y)\,{\rm e}^{\pm y^{2}/2},$$ (4.171)

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

$$\psi(y) = h(y)\,{\rm e}^{-y^{2}/2},$$ (4.172)

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

Substituting Equation (4.172) into Equation (4.169), we obtain

$$\frac{d^{2}h}{dy^{2}} - 2\,y\,\frac{dh}{dy} + (\epsilon-1)\,h = 0.$$ (4.173)

Let us attempt a power-law solution of the form

$$h(y) = \sum_{i=0,\infty} c_i\,y^{\,i}.$$ (4.174)

Inserting this test solution into Equation (4.173), and equating the coefficients of $y^{\,i}$, we obtain the recursion relation

$$c_{i+2} = \frac{(2\,i-\epsilon+1)}{(i+1)\,(i+2)}\,c_i.$$ (4.175)

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

$$c_{i+2} \simeq \frac{2}{i}\,c_i.$$ (4.176)

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

$$h(y) \sim C \sum_{j}\frac{y^{\,2 j}}{j!}\sim C\,{\rm e}^{\,y^{2}}.$$ (4.177)

It follows that $\psi(y) = h(y)\,\exp(-y^{2}/2)$ varies as $\exp(\,y^{2}/2)$ 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 (4.174) terminate at some finite value of $i$. This implies, from the recursion relation (4.175), that

$$\epsilon = 2\,n+1,$$ (4.178)

where $n$ is a non-negative integer. Note that the number of terms in the power series (4.174) is $n+1$. Finally, using Equation (4.168), we obtain

$$E = \left(\frac{1}{2}+n\right)\hbar\,\omega,$$ (4.179)

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 another example of zero-point energy. (See Section 4.3.1.) It is easily demonstrated that the (normalized) wavefunction of the lowest energy state takes the form

$$\psi_0(x) = \frac{{\rm e}^{-x^{2}/2\,d^{2}}}{\pi^{1/4}\!\sqrt{d}},$$ (4.180)

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