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Simple Harmonic Oscillator

The classical Hamiltonian of a simple harmonic oscillator is

$\displaystyle H = \frac{p^{ 2}}{2 m} + \frac{1}{2} K x^{ 2},$ (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

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

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

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

and

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

Equation (C.107) reduces to

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

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

The approximate solutions to the previous equation are

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

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

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

Let us attempt a power-law solution of the form

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

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

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

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

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

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

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

$\displaystyle \epsilon = 2 n+1,$ (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

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

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

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

next up previous
Next: Angular Momentum Up: Wave Mechanics Previous: Three-Dimensional Wave Mechanics
Richard Fitzpatrick 2016-01-25