Question

For the ground state of linear harmonic oscillator calculate average value of "x" and "x^2".

Answer 1

Answer on Question #44133-Physics-Quantum Mechanics

For the ground state of linear harmonic oscillator calculate average value of "x" and "x^2".

Solution

For the ground state of the harmonic oscillator, the expectation value of the position operator $x$ is given by

$$< x >_0 = \int \psi_0^*(x) \cdot x \cdot \psi_0(x) \, dx,$$

where

$$\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}} e^{-\frac{m\omega}{2\hbar}x^2}.$$ $$< x >_0 = \int_{-\infty}^{\infty} \sqrt{\frac{m\omega}{\pi\hbar}} \cdot x \cdot e^{-\frac{m\omega}{\hbar}x^2} \, dx = 0.$$

The fact that this expression vanishes can be seen either by brute force calculation, or by symmetry (the integrand is odd and the limits of integration are symmetric).

The expectation value of $x^2$, however, doesn't vanish:

$$< x^2 >_0 = \int \psi_0^*(x) \cdot x^2 \cdot \psi_0(x) \, dx = \int_{-\infty}^{\infty} \sqrt{\frac{m\omega}{\pi\hbar}} \cdot x^2 \cdot e^{-\frac{m\omega}{\hbar}x^2} \, dx.$$

Here is a useful trick: knowing the fundamental integral $\int_{-\infty}^{\infty} e^{-cx^2} \, dx = \sqrt{\frac{\pi}{c}}$, we can take a derivative with respect to $c$ to obtain

$$\int_{-\infty}^{\infty} x^2 e^{-cx^2} \, dx = -\frac{d}{dc} \int_{-\infty}^{\infty} e^{-cx^2} \, dx = -\frac{d}{dc} \left(\sqrt{\frac{\pi}{c}}\right) = \frac{\sqrt{\pi}}{2c^2}.$$

Here $c = \frac{m\omega}{\hbar}$, and thus

$$< x^2 >_0 = \int_{-\infty}^{\infty} \sqrt{\frac{m\omega}{\pi\hbar}} \cdot x^2 \cdot e^{-\frac{m\omega}{\hbar}x^2} \, dx = \sqrt{\frac{m\omega}{\pi\hbar}} \frac{1}{2} \sqrt{\frac{\pi\hbar^3}{m^3\omega^3}} = \frac{\hbar}{2m\omega}.$$

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