Physics Friday 55 « Twisted One 151’s Weblog

Physics Friday 55

By twistedone151

Classically, a one-dimensional harmonic oscillator is a system with a mass under a restoring force proportional to displacement from the equilibrium position: F=-kx. The energy is $H=T+V=\frac{p^2}{2m}+\frac{k}{2}x^2=\frac{m}{2}\dot{x}^2+\frac{k}{2}x^2$ , and the equation of motion $\ddot{x}=-\frac{k}{m}x$ has solution $x(t)=A\cos(\omega{x}+\phi)$ , where $\omega\equiv\sqrt{\frac{k}{m}}$ .

Analogously, a one-dimensional quantum harmonic oscillator is a particle with Hamiltonian $H=\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2$ , where here p is the quantum momentum operator, and x the position operator. In the position basis, this is then
$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{m\omega^2}{2}x^2$ .
A cursory examination of the expectation for energy
$\langle{H}\rangle=\int_{-\infty}^{\infty}\psi^{*}(x)H\psi(x)\,dx$
shows that we can expect our energies to be non-negative.

Most quantum mechanics books and courses I have encountered address the energy eigenstates by means of the raising and lowering operators. Instead, here we will use an analytical method of finding the eigenvalues to the time-independent Schrödinger equation
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{m\omega^2}{2}x^2\psi=E\psi$ .
Let us replace x and E with corresponding dimensionless variables: we define the dimensionless variables $\epsilon=\frac{E}{\hbar\omega}$ , $\xi=\sqrt{\frac{m\omega}{\hbar}}x$ . Then $\frac{m\omega^2}{2}x^2=\frac{\hbar\omega}{2}\xi^2$ , $\frac{d^2\psi}{dx^2}=\frac{m\omega}{\hbar}\frac{d^2\psi}{d\xi^2}$
the Schrödinger equation becomes
$\begin{eqnarray}-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{m\omega^2}{2}x^2\psi&=&E\psi\\-\frac{\hbar^2}{2m}\frac{m\omega}{\hbar}\frac{d^2\psi}{d\xi^2}+\frac{\hbar\omega}{2}\xi^2\psi&=&\hbar\omega\epsilon\psi\\-\frac{\hbar\omega}{2}\frac{d^2\psi}{d\xi^2}+\frac{\hbar\omega}{2}\xi^2\psi&=&\hbar\omega\epsilon\psi\\\frac{d^2\psi}{d\xi^2}+(2\epsilon-\xi^2)\psi&=&{0}\end{eqnarray}$ .

To find a solution to this differential equation, we consider asymptotic behavior. Namely, when the energy is small, the ξ² dominates over ε, and our equation approaches $\psi''=\xi^2\psi$ , which hints at a solution that behaves like a Gaussian. Adopting, then, a test form
$\psi=u(\xi)e^{-\xi^2/2}$ , and plugging this into our differential equation $\frac{d^2\psi}{d\xi^2}+(2\epsilon-\xi^2}\psi={0}$ , we get
$u''e^{-\xi^2/2}-2\xi{u'}e^{-\xi^2/2}+(2\epsilon-1)u(\xi)e^{-\xi^2/2}={0}$ ,
or, eliminating the common Gaussian factor,
$u''-2\xi{u'}+(2\epsilon-1)u(\xi)={0}$
Applying the power series method, we plug in $u(\xi)=\sum_{n=0}^{\infty}a_n\xi^n$ to find
$\sum_{n=0}^{\infty}C_n\left[n(n-1)\xi^{n-2}-2n\xi^n+(2\epsilon-1)\xi^n\right]={0}$
Which gives recursion relation $C_{n+2}=\frac{2n+1-2\epsilon}{(n+2)(n+1)}C_n$ .

Now, we return to the physics of the problem, namely that the wavefunctions have to be normalized. This means that ψ→0 as x→±∞. Now, considering that
$\lim_{n\to\infty}\frac{C_{n+2}}{C_n}=\lim_{n\to\infty}\frac{2n+1-2\epsilon}{(n+2)(n+1)}=\frac{2}{n}$ , and comparing to the series expansion of the gaussian, we see our series diverges too fast for ψ to be normalized, unless the series terminates. This requires $2n+1-2\epsilon=0$ for some n, which gives $\epsilon=n+\frac{1}{2}$ ⇒ $E_n=\hbar\omega(n+\frac{1}{2})$ , and as E≥0, we see that the ground state has energy $E_0=\frac{1}{2}\hbar\omega\gt{0}$ ; this is known as the ground state energy, or zero-point energy.

Solving for the wavefunctions themselves requires more mathematics, and gives solutions involving the Hermite polynomials.

Tags: Friday Physics, physics, Quantum Mechanics, Schrödinger Equation, Harmonic Oscillator, Quantum Harmonic Oscillator, Ground State, Ground State Energy, Zero-Point Energy, Hermite Polynomials

This entry was posted on January 16, 2009 at 5:21 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Physics Friday 55”

Monday Math 56 « Twisted One 151’s Weblog Says:
January 26, 2009 at 10:35 am | Reply
[...] includes a number of notable polynomial sequences (including the Hermite polynomials mentioned in a recent Physics Friday post). Note that the sequence Pn(x)=xn is a trivial example. Now, we can obtain from the relation , or [...]
Physics Friday 58 « Twisted One 151’s Weblog Says:
February 6, 2009 at 12:02 am | Reply
[...] Friday 58 By twistedone151 In a previous Friday post, I demonstrated one method of determining the energy eigenvalues for the one-dimensional quantum [...]

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Physics Friday 55

2 Responses to “Physics Friday 55”

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