Harmonic oscillator – position, momentum and energy

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.11.

The lowest two states for the harmonic oscillator are the ground state ${\psi_{0}}$ and first excited state ${\psi_{1}}$ . We can work out some mean values of various quantities using explicit integration.

Consider first the ground state ${\psi_{0}}$ . Using the substitutions ${\xi\equiv\sqrt{m\omega/\hbar}x}$ and ${\alpha\equiv(m\omega/\pi\hbar)^{1/4}}$ we have ${\psi_{0}=\alpha e^{-\xi^{2}/2}}$ . We can simplify the operations considerably if we note the even and odd natures of some of the functions to be integrated. Since ${\psi_{0}(x)}$ is even, ${x\psi_{0}^{2}(x)}$ is odd, so ${\langle x\rangle=0}$ . To calculate ${\langle p\rangle}$ , since the operator ${p=(\hbar/i)d/dx}$ , we need the derivative ${d\psi_{0}/dx=\sqrt{\hbar/m\omega}d\psi_{0}/d\xi=\sqrt{\hbar/m\omega}(-2\xi)\psi_{0}}$ . This is again an odd function, so ${\langle p\rangle=0}$ as well.

For the mean square values, we do need to do some integrals (I’ve used software for this).

$\displaystyle \langle x^{2}\rangle$	$\displaystyle =$	$\displaystyle \frac{\hbar}{m\omega}\langle\xi^{2}\rangle$
$\displaystyle$	$\displaystyle =$	$\displaystyle \left(\frac{\hbar}{m\omega}\right)^{3/2}\alpha^{2}\int_{-\infty}^{\infty}\xi^{2}e^{-\xi^{2}}d\xi$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{\hbar}{2m\omega}$

Here we used ${d\xi\equiv\sqrt{m\omega/\hbar}dx}$ to convert the integration variable from ${x}$ to ${\xi}$ in the second line.

$\displaystyle \langle p^{2}\rangle$	$\displaystyle =$	$\displaystyle \sqrt{m\omega}\hbar^{3/2}\alpha^{2}\int_{-\infty}^{\infty}(1-\xi^{2}e^{-\xi^{2}/2})e^{-\xi^{2}/2}d\xi$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{1}{2}m\omega\hbar$

For ${\psi_{1}}$ we have ${\psi_{1}(\xi)=\sqrt{\frac{2\pi\hbar}{m\omega}}\alpha^{3}\xi e^{-\xi^{2}/2}}$ which is an odd function. The square of an odd function is an even function, so ${\psi_{1}^{2}(\xi)}$ is even, which means that the function to be integrated to find ${\langle x\rangle}$ is again the product of an even function and an odd function, so ${\langle x\rangle=0}$ here as well.

Considering ${\langle p\rangle}$ , we calculate the derivative of ${\psi_{1}(\xi),}$ which is of form ${K(1-\xi^{2})e^{-\xi^{2}/2}}$ for a constant ${K}$ , which is even. Thus to obtain ${\langle p\rangle}$ we must integrate this even function multiplied by the odd function ${\psi_{1}}$ so the result is ${\langle p\rangle=0}$ .

To get the mean square values, we do the integrals:

$\displaystyle \langle x^{2}\rangle$	$\displaystyle =$	$\displaystyle \frac{\hbar}{m\omega}\langle\xi^{2}\rangle$
$\displaystyle$	$\displaystyle =$	$\displaystyle 2\pi\left(\frac{\hbar}{m\omega}\right)^{5/2}\alpha^{6}\int_{-\infty}^{\infty}\xi^{4}e^{-\xi^{2}}d\xi$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{3}{2}\frac{\hbar}{m\omega}$

And for the momentum:

$\displaystyle \langle p^{2}\rangle$	$\displaystyle =$	$\displaystyle -2\pi\frac{\hbar^{5/3}}{\sqrt{m\omega}}\alpha^{6}\int_{-\infty}^{\infty}\xi e^{-\xi^{2}/2}\frac{d^{2}}{d\xi^{2}}(\xi e^{-\xi^{2}/2})d\xi$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{3}{2}\hbar m\omega$

For ${\psi_{0}}$ using the results above, the uncertainty principle here comes out to

$\displaystyle \sigma_{p}\sigma_{x}=\sqrt{\langle x^{2}\rangle\langle p^{2}\rangle}=\frac{\hbar}{2}$

For ${\psi_{1}}$ , we have

$\displaystyle \sigma_{p}\sigma_{x}=\sqrt{\langle x^{2}\rangle\langle p^{2}\rangle}=\frac{3\hbar}{2}$

The mean kinetic and potential energies can be worked out from the above results without doing any more integration. We get ${\langle T\rangle=\frac{\langle p^{2}\rangle}{2m}=\hbar\omega/4}$ for ${\psi_{0}}$ and ${\langle T\rangle=\frac{\langle p^{2}\rangle}{2m}=3\hbar\omega/4}$ for ${\psi_{1}}$ .

${\langle V\rangle=\frac{k\langle x^{2}\rangle}{2}=\hbar\omega/4}$ for ${\psi_{0}}$ and ${\langle V\rangle=\frac{k\langle x^{2}\rangle}{2}=3\hbar\omega/4}$ for ${\psi_{1}.}$ Adding these together to get the total energy ${E}$ gives ${\hbar\omega/2}$ for ${\psi_{0}}$ and ${3\hbar\omega/2}$ for ${\psi_{1}}$ as it should.

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By Virial theorem « Physics tutorials on Tuesday, 9 October 2012 at 17:38

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