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Simple Harmonic Oscillator

The classical Hamiltonian of a simple harmonic oscillator is
H = \frac{p^2}{2 m} + \frac{1}{2} K x^2,
\end{displaymath} (389)

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.
\end{displaymath} (390)

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,
\end{displaymath} (391)

\epsilon = \frac{2 E}{\hbar \omega}.
\end{displaymath} (392)

Equation (390) reduces to
\frac{d^2\psi}{dy^2} - (y^2-\epsilon) \psi = 0.
\end{displaymath} (393)

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. (393) 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.
\end{displaymath} (394)

The approximate solutions to the above equation are
\psi(y) \simeq A(y) {\rm e}^{\pm y^2/2},
\end{displaymath} (395)

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},
\end{displaymath} (396)

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

Substituting Eq. (396) into Eq. (393), we obtain

\frac{d^2h}{dy^2} - 2 y \frac{dh}{dy} + (\epsilon-1) h = 0.
\end{displaymath} (397)

Let us attempt a power-law solution of the form
h(y) = \sum_{i=0}^\infty c_i y^i.
\end{displaymath} (398)

Inserting this test solution into Eq. (397), 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.
\end{displaymath} (399)

Consider the behavior of $h(y)$ in the limit $\vert y\vert\rightarrow\infty$. The above recursion relation simplifies to
c_{i+2} \simeq \frac{2}{i} c_i.
\end{displaymath} (400)

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}.
\end{displaymath} (401)

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 (398) terminate at some finite value of $i$. This implies, from the recursion relation (399), that
\epsilon = 2 n+1,
\end{displaymath} (402)

where $n$ is a non-negative integer. Note that the number of terms in the power series (398) is $n+1$. Finally, using Eq. (392), we obtain
E = (n+1/2) \hbar \omega,
\end{displaymath} (403)

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) 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}},
\end{displaymath} (404)

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

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

E_n = (n+1/2) \hbar \omega.
\end{displaymath} (405)

Assuming that the $\psi_n$ are properly normalized (and real), we have
\int_{-\infty}^\infty \psi_n \psi_m dx = \delta_{nm}.
\end{displaymath} (406)

Now, Eq. (393) can be written
\left(-\frac{d^2}{d y^2}+y^2\right)\psi_n = (2n+1) \psi_n,
\end{displaymath} (407)

where $x = d y$, and $d=\sqrt{\hbar/m \omega}$. It is helpful to define the operators
a_\pm = \frac{1}{\sqrt{2}}\left(\mp \frac{d}{dy}+y\right).
\end{displaymath} (408)

As is easily demonstrated, these operators satisfy the commutation relation
\begin{displaymath}[a_+,a_-]= -1.
\end{displaymath} (409)

Using these operators, Eq. (407) can also be written in the forms
a_+ a_- \psi_n = n \psi_n,
\end{displaymath} (410)

a_- a_+ \psi_n = (n+1) \psi_n.
\end{displaymath} (411)

The above two equations imply that
$\displaystyle a_+ \psi_n$ $\textstyle =$ $\displaystyle \sqrt{n+1} \psi_{n+1},$ (412)
$\displaystyle a_- \psi_n$ $\textstyle =$ $\displaystyle \sqrt{n} \psi_{n-1}.$ (413)

We conclude that $a_+$ and $a_-$ are raising and lowering operators, respectively, for the harmonic oscillator: i.e., 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),
\end{displaymath} (414)

from which the result
H \psi_n = (n+1/2) \hbar \omega \psi_n = E_n \psi_n
\end{displaymath} (415)

is readily deduced. Finally, Eqs. (406), (412), and (413) yield the useful expression
$\displaystyle \int_{-\infty}^\infty \psi_m x \psi_n dx$ $\textstyle =$ $\displaystyle \frac{d}{\sqrt{2}}\int_{-\infty}^{\infty}\psi_m (a_+ + a_-) \psi_n dx$ (416)
  $\textstyle =$ $\displaystyle \sqrt{\frac{\hbar}{2 m \omega}}\left(\sqrt{m} \delta_{m,n+1} + \sqrt{n} \delta_{m,n-1}\right).$  

next up previous
Next: Exercises Up: One-Dimensional Potentials Previous: Square Potential Well
Richard Fitzpatrick 2010-07-20