The simple harmonic oscillator

The classical Hamiltonian of a simple harmonic oscillator is

$$H = \frac{p^2}{2\,m} + \frac{1}{2}\,k\,x^2,$$ (371)

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

$$\frac{d^2\psi}{dx^2} = \frac{2\,m}{\hbar^2}\left(\frac{1}{2}\,k\,x^2-E\right)\psi.$$ (372)

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

$$y = \sqrt{\frac{m\,\omega}{\hbar}}\,x,$$ (373)

and

$$\epsilon = \frac{2\,E}{\hbar\,\omega}.$$ (374)

Equation (372) reduces to

$$\frac{d^2\psi}{dy^2} - (y^2-\epsilon)\,\psi = 0.$$ (375)

We need to find solutions to the above equation which are bounded at infinity: i.e., solutions which satisfy the boundary condition $\psi\rightarrow 0$ as $\vert y\vert\rightarrow\infty$.

Consider the behavior of the solution to Eq. (375) in the limit $\vert y\vert\gg 1$. As is easily seen, in this limit, the equation simplifies somewhat to give

$$\frac{d^2\psi}{dy^2} - y^2\,\psi \simeq 0.$$ (376)

The approximate solutions to the above equation are

$$\psi(y) \simeq A(y)\,{\rm e}^{\pm y^2/2},$$ (377)

where $A(y)$ is a relatively slowly varying function of $y$. Obviously, 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

$$\psi(y) = h(y)\,{\rm e}^{-y^2/2},$$ (378)

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

Substituting Eq. (378) into Eq. (375), we obtain

$$\frac{d^2h}{dy^2} - 2\,y\,\frac{dh}{dy} + (\epsilon-1)\,h = 0.$$ (379)

Let us attempt a power-law solution of the form

$$h(y) = \sum_{i=0}^\infty a_i\,y^i.$$ (380)

Inserting this test solution into Eq. (379), and equating the coefficients of $y^i$, we obtain the recursion relation

$$a_{i+2} = \frac{(2\,i-\epsilon+1)}{(i+1)\,(i+2)}\,a_i.$$ (381)

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

$$a_{i+2} \simeq \frac{2}{i}\,a_i.$$ (382)

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

$$h(y) \sim C \sum_{j}\frac{y^{2\,j}}{j!}\sim C\,{\rm e}^{\,y^2}.$$ (383)

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, since 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 (380) terminate at some finite value of $i$. This implies, from the recursion relation (381), that

$$\epsilon = 2\,n+1,$$ (384)

where $n$ is a non-negative integer. Note that the number of terms in the power series (380) is $n+1$. Finally, using Eq. (374), we obtain

$$E = (n+1/2)\,\hbar\,\omega,$$ (385)

for $n=0,1,2,\cdots$.

Hence, we conclude that a particle moving in a harmonic potential has quantized energy levels which 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) wave-function of the lowest energy state takes the form

$$\psi_0(x) = \frac{{\rm e}^{-x^2/2\,a^2}}{\pi^{1/4}\,\sqrt{a}},$$ (386)

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

Let $\psi_n(x)$ be an energy eigenstate of the harmonic oscillator corresponding to the eigenvalue

$$E_n = (n+1/2)\,\hbar\,\omega.$$ (387)

Assuming that the $\psi_n$ are properly normalized (and real), we have

$$\int_{-\infty}^\infty \psi_n\,\psi_m\,dx = \delta_{nm}.$$ (388)

Now, Eq. (372) can be written

$$\left(-\frac{d^2}{d y^2}+y^2\right)\psi_n = (2n+1)\,\psi_n,$$ (389)

where $x = a\,y$, and $a=\sqrt{\hbar/m\,\omega}$. It is helpful to define the operators

$$a_\pm = \frac{1}{\sqrt{2}}\left(\mp \frac{d}{dy}+y\right).$$ (390)

As is easily demonstrated, these operators satisfy the commutation relation

$$[a_+,a_-]= -1.$$ (391)

Using these operators, Eq. (389) can also be written in the forms

$$a_+\,a_-\,\psi_n = n\,\psi_n,$$ (392)

or

$$a_-\,a_+\,\psi_n = (n+1)\,\psi_n.$$ (393)

The above two equations imply that

$$a_+\,\psi_n = \sqrt{n+1}\,\psi_{n+1},$$ (394)

$$a_-\,\psi_n = \sqrt{n}\,\psi_{n-1}.$$ (395)

We conclude that $a_+$ and $a_-$ are raising and lowering operators, respectively, for the harmonic oscillator: i.e., operating on the wave-function 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

$$H = \hbar\,\omega\,\left(a_+\,a_- + \frac{1}{2}\right),$$ (396)

from which the result

$$H\,\psi_n = (n+1/2)\,\hbar\,\omega\,\psi_n = E_n\,\psi_n$$ (397)

is readily deduced. Finally, Eqs. (388), (394), and (395) yield the useful expression

$$\int_{-\infty}^\infty \psi_m\,x\,\psi_n\,dx = \frac{a}{\sqrt{2}}\int_{-\infty}^{\infty}\psi_m\,(a_+ + a_-)\,\psi_n\,dx$$

$$= \sqrt{\frac{\hbar}{2\,m\,\omega}}\left(\sqrt{m}\,\delta_{m,n+1} + \sqrt{n}\,\delta_{m,n-1}\right).$$ (398)