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\ \ \ \ \ (1)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\hbar}{m\omega}\right)^{3/2}\alpha^{2}\int_{-\infty}^{\infty}\xi^{2}e^{-\xi^{2}}d\xi\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar}{2m\omega} \ \ \ \ \ (3)

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}\left(1-\xi^{2}\right)e^{-\xi^{2}}d\xi\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}m\omega\hbar \ \ \ \ \ (5)

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\ \ \ \ \ (6)
\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\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{3}{2}\frac{\hbar}{m\omega} \ \ \ \ \ (8)

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\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{3}{2}\hbar m\omega \ \ \ \ \ (10)

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} \ \ \ \ \ (11)

For {\psi_{1}}, we have

\displaystyle  \sigma_{p}\sigma_{x}=\sqrt{\langle x^{2}\rangle\langle p^{2}\rangle}=\frac{3\hbar}{2} \ \ \ \ \ (12)

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.

7 thoughts on “Harmonic oscillator – position, momentum and energy

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