Analytical Method for Solving the Simple Harmonic Oscillator

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Quantum Mechanics A
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian \mathcal{H}, it describes how a state |\Psi\rangle evolves in time.
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Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
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Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
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Oscillation Theorem
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Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
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Charged Particles in an Electromagnetic Field
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The Heisenberg Picture: Equations of Motion for Operators
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Angular Momentum as a Generator of Rotations in 3D
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Propagator for the Harmonic Oscillator
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Coulomb Potential Scattering

In addition to the elegant operator-based method described previously to solve the harmonic oscillator, one may also write down and solve the Schrödinger equation for the system. The equation is

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}m\omega^2x^2\psi=E\psi.

If we now introduce the dimensionless position, \xi=\sqrt{\frac{m\omega}{\hbar}}x, and the dimensionless energy, \epsilon=\frac{E}{\hbar\omega}, then this reduces to the much simpler form,

\frac{d^2\psi}{d\xi^2}+(2\epsilon-\xi^2)\psi=0.

Let us now look at the asymptotic behavior of the wave function. For large values of \xi,\! our equation may be approximated as

\frac{d^2\psi}{d\xi^2}\approx\xi^2\psi.

We now make the following ansatz for its solution:

\psi(\xi) = e^{k\xi^2}

Substituting this in the asymptotic equation, we get

4k^2\xi^2 + 2k = \xi^2\!

or, in the limit of large \xi\!,

k = \pm\tfrac{1}{2}.

with this value of k,

\psi(\xi)=Ae^{-\xi^2/2}+Be^{\xi^2/2}.

For ψ(ξ) to remain finite at the origin, we must set B = 0.\! Therefore, the long-distance behavior of the wave function is given by \psi(\xi)\approx Ae^{-\xi^2/2}.

Now that we have separated out the asymptotic behavior, we now postulate that the complete solution, valid everywhere, can be written as

\psi(\xi)=e^{-\xi^2/2}h(\xi),

where h(ξ) is a function that diverges more slowly than e^{\xi^2/2} for large ξ. The first and second derivatives of this form are

\psi'(\xi) = [h'(\xi) - \xi h(\xi)]e^{-\xi^2/2}

and

\psi''(\xi) = [h''(\xi)-2\xi h'(\xi) + (\xi^2-1)h(\xi)]e^{-\xi^2/2}

Substituting these into the differential equation, we get

h''(\xi) - 2\xi h'(\xi) + (2\epsilon-1)h(\xi) = 0.\!

Let us try a series solution for h(\xi):\!

h(\xi) = \sum_{j=0}^{\infty}a_j \xi^j

Substituting this into the differential equation and collecting like powers yields

\sum_{j=0}^{\infty}[(j+1)(j+2)a_{j+2} - (2j-2\epsilon+1)a_j]\xi^j = 0.

Equating the coefficients of each power of \xi\! on both sides, we obtain the recursion relation,

a_{j+2} = \frac{2j-2\epsilon+1}{(j+1)(j+2)}a_j.

For large values of j,\! this may be approximated as

a_{j+2}\approx\frac{2}{j+2}a_j.

If we assume that a_j\! is only finite for even j,\! then this is satisfied by

a_{2n}=\frac{2^n}{(2n)!!}a_0=\frac{1}{n!}a_0.

This yields the power series for e^{\xi^2}, which diverges more rapidly for large \xi\! than e^{-\xi^2/2}. Taking only odd coefficients yields a similar result, namely the power series for

\frac{\sqrt{\pi}}{2}e^{\xi^2}\text{erf}(x),

where erf(x) is the error function. This means that, to obtain a wave function that goes to zero as \xi\rightarrow\pm\infty,\! we must truncate the power series at a finite order; i.e., h(\xi)\! must be a polynomial. To truncate the series to order n,\! where n\! is a non-negative integer, we must set

2\epsilon-1=2n,\!

and take only those coefficients a_j\! for which j\! has the same "parity" (odd or even) as n.\! The fact that the resulting polynomial will either be an even or odd function could have been seen from the fact that the system under investigation has an even potential. The above condition yields the allowed energy levels for the system,

E_n=\left (n+\tfrac{1}{2}\right )\hbar\omega.

With this constraint on ε, the recursion relation for the power series coefficients becomes

a_{j+2} = \frac{2(j-n)}{(j+1)(j+2)}a_j.

This recursion relation will yield the Hermite polynomial, H_n(\xi).\!

The properly normalized eigenfunctions are

\psi_n(x)=\frac{1}{\sqrt{2^n n!}}\left (\frac{m\omega}{\pi\hbar} \right )^{1/4}e^{-{m\omega x^2/2\hbar}}H_n\left (\sqrt{\frac{m\omega}{\hbar}}x\right ),

as we found in the previous section.

Problem

(Schaum, Theory and Problems of Quantum Mechanics, Chapter 5)

Consider a particle with charge e\! in a three-dimensional isotropic harmonic potential,

V(r)=\frac{1}{2}m{\omega }^2{r}^2,

and in the presence of a constant electric field, \mathbf{E}(\mathbf{r})=E_0\hat{\mathbf{x}}.\! Find the energy levels and eigenstates of the patricle.

Solution

Reference

Introduction to Quantum Mechanics, 2nd ed. , by D. J. Griffiths

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