Physics Friday 58 « Twisted One 151’s Weblog

Physics Friday 58

By twistedone151

In a previous Friday post, I demonstrated one method of determining the energy eigenvalues for the one-dimensional quantum harmonic oscillator. In particular, we took the (time-independent) Schrödinger equation $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{m\omega^2}{2}x^2\psi=E\psi$ , and by defining dimensionless parameters $\epsilon\equiv\frac{E}{\hbar\omega}$ , $\xi\equiv\sqrt{\frac{m\omega}{\hbar}}x$ , and then attempting a solution of the form $\psi=u(\xi)e^{-\xi^2/2}$ , we derived the differential equation $u''-2\xi{u'}+(2\epsilon-1)u{\xi}={0}$ , where the series solution $u(\xi)=\sum_{n=0}^{\infty}a_n\xi^n$ has recursion relation $a_{n+2}=\frac{2n+1-2\epsilon}{(n+2)(n+1)}a_n$ . Then, the requirement that the wavefunction be normalizable requires that the series solution terminate after finitely many terms, requiring that $\epsilon=n+\frac{1}{2}$ ⇒ $E_n=\hbar\omega(n+\frac{1}{2})$ , for n a non-negative integer.

Now, let us consider the ground state case: n=0. Then $\epsilon=\frac{1}{2}$ , and the series recursion relation becomes $a_{n+2}=\frac{2n}{(n+2)(n+1)}a_n$ , so that $a_2=\frac{2\cdot{0}}{(0+2)(0+1)}a_0=0$ , and the terminating solution is that u(ξ) is a constant (a₀). Then $\psi_0(\xi)=Ce^{-\xi^2/2}$ , and so
$\psi_0(x)=Ce^{-\frac{m\omega}{2\hbar}x^2}$ .
Normalizing this wavefunction,
$\begin{eqnarray}1&=&\int_{-\infty}^{\infty}\left|\psi_0(x)\right|^2\,dx\\&=&\int_{-\infty}^{\infty}C^2e^{-\frac{m\omega}{\hbar}x^2}\,dx\\&=&C^2\int_{-\infty}^{\infty}e^{-\frac{m\omega}{\hbar}x^2}\,dx\\1&=&C^2\sqrt{\frac{\pi\hbar}{m\omega}}\\C^2&=&\sqrt{\frac{m\omega}{\pi\hbar}}\\C&=&\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\end{eqnarray}$
and
$\psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}e^{-\frac{m\omega}{2\hbar}x^2}$ .
So the ground state wavefunction of the 1-dimensional quantum harmonic oscillator is a gaussian.

Now, let us examine the uncertainties of position and momentum. The Heisenberg Uncertainty Principle tells us that $\Delta{x}\Delta{p}\ge\frac{\hbar}{2}$ (see here). For our wavefunction, we see
$\begin{eqnarray}\langle{x}\rangle&=&\int_{-\infty}^{\infty}x\left|\psi_0(x)\right|^2\,dx\\&=&\sqrt{\frac{m\omega}{\pi\hbar}}\int_{-\infty}^{\infty}xe^{-\frac{m\omega}{\hbar}x^2}\,dx\\&=&{0}\end{eqnarray}$
and
$\begin{eqnarray}\langle{p}\rangle&=&\int_{-\infty}^{\infty}\psi_0^*(x)\frac{\hbar}{\imath}\frac{d}{dx}\psi_0(x)\,dx\\\imath\sqrt{\frac{m^3\omega^3}{\pi\hbar}}\int_{-\infty}^{\infty}xe^{-\frac{m\omega}{\hbar}x^2}\,dx\\&=&{0}\end{eqnarray}$
as in both cases the integrand is odd (see here).
Thus
$\begin{eqnarray}(\Delta{x})^2&=&\left\langle\left(x-\langle{x}\rangle\right)^2\right\rangle\\&=&\left\langle{x^2}\right\rangle\\&=&\int_{-\infty}^{\infty}x^2\left|\psi_0(x)\right|^2\,dx\\&=&\sqrt{\frac{m\omega}{\pi\hbar}}\int_{-\infty}^{\infty}x^2e^{-\frac{m\omega}{\hbar}x^2}\,dx\\(\Delta{x})^2&=&\frac{\hbar}{m\omega\sqrt{\pi}}\int_{-\infty}^{\infty}u^2e^{-u^2}\,du\end{eqnarray}$
Using $\int_{0}^{\infty}u^ne^{-u^2}\,du=\frac{1}{2}\operatorname{\Gamma}\left(\frac{n+1}{2}\right)$ and $\operatorname{\Gamma}\left(\frac{3}{2}\right)=\frac{1}{2}\operatorname{\Gamma}\left(\frac{1}{2}\right)=\frac{\sqrt\pi}{2}$ , we see that
$\int_{-\infty}^{\infty}u^2e^{-u^2}\,du=\frac{\sqrt\pi}{2}$ , and so
$\begin{eqnarray}(\Delta{x})^2&=&\frac{\hbar}{m\omega\sqrt{\pi}}\frac{\sqrt\pi}{2}\\&=&\frac{\hbar}{2m\omega}\\\Delta{x}&=&\sqrt{\frac{\hbar}{2m\omega}}\end{eqnarray}$ .
Similarly,
$\begin{eqnarray}(\Delta{p})^2&=&\left\langle\left(p-\langle{p}\rangle\right)^2\right\rangle\\&=&\left\langle{p^2}\right\rangle\\&=&\int_{-\infty}^{\infty}\psi_0^*(x)(-\hbar)\frac{d^2}{dx^2}\psi_0(x)\,dx\\&=&\sqrt{\frac{m^3\hbar\omega^3}{\pi}}\int_{-\infty}^{\infty}\left(1-\frac{m\omega}{\hbar}x^2\right)e^{-\frac{m\omega}{\hbar}x^2}\,dx\\&=&\sqrt{\frac{m^3\hbar\omega^3}{\pi}}\int_{-\infty}^{\infty}\left(1-\frac{m\omega}{\hbar}x^2\right)e^{-\frac{m\omega}{\hbar}x^2}\,dx\\&=&\frac{m\hbar\omega}{\sqrt{\pi}}\int_{-\infty}^{\infty}(1-u^2)e^{-u^2}\,du\end{eqnarray}$ .
Now,
$\int_{-\infty}^{\infty}(1-u^2)e^{-u^2}\,du=\sqrt\pi-\frac{\sqrt\pi}{2}=\frac{\sqrt\pi}{2}$ ,
so
$\begin{eqnarray}(\Delta{p})^2&=&\frac{m\hbar\omega}{\sqrt{\pi}}\frac{\sqrt\pi}{2}\\&=&\frac{m\hbar\omega}{2}\\\Delta{p}&=&\sqrt{\frac{m\hbar\omega}{2}}\end{eqnarray}$
Taking the product,
$\Delta{x}\Delta{p}=\sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{m\hbar\omega}{2}}=\frac{\hbar}{2}$ ,
which is exactly the lower limit allowed by the uncertainty principle. Thus, the non-zero value of the ground-state energy (zero-point energy) can be seen as being a result of the Heisenberg Uncertainty Principle.

Tags: Friday Physics, Gaussian Integral, Ground State, Ground State Energy, Harmonic Oscillator, Heisenberg, physics, Quantum Harmonic Oscillator, Quantum Mechanics, Uncertainty Principle, Zero-Point Energy

This entry was posted on February 6, 2009 at 12:01 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.

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Physics Friday 58

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