Answer to Question #86637 in Atomic and Nuclear Physics for Anand

Question

Calculate the mean kinetic and potential energies of a simple harmonic oscillator
which is in its ground state.

Answer 1

The eigenvalues and eigenfunctions for a simple 1-D harmonic oscillator in the coordinate representation have the following form:

$$E_n = \hbar \omega (n + \frac{1}{2}), \\ \psi_n (\xi)= \frac{1}{\sqrt{n! 2^n \sqrt{\pi}}} H_n (\xi) \exp{(-\frac{\xi^2}{2})},$$

where

$$\xi = x \sqrt{\frac{m \omega}{\hbar}}$$

and H_n are Hermite polynomials.

The mean potential energy of the system can be calculated as follows:

$$<U>_n = \int_{-\infty}^{\infty} \psi_n^2 (\xi) \frac{k x^2}{2} d\xi = \frac{k \hbar}{2 m \omega} \int_{-\infty}^{\infty} \psi_n^2 (\xi) \xi^2 d\xi$$

Hermite polynomials satisfy the following relation:

$$\xi H_n(\xi) = n H_{n-1} (\xi) + \frac{1}{2} H_{n+1} (\xi)$$

Applying this twice and taking into account the explicit expression for the eigenfunctions, we obtain:

$$\xi^2 \psi_n(\xi) = \frac{1}{2} \sqrt{n(n-1)} \psi_{n-2} + (n+\frac{1}{2})\psi_n + \frac{1}{2} \sqrt{(n+1)(n+2)} \psi_{n+2}$$

Substituting this result into the integral for <U> and taking into account the orthonormality of the eigenfunctions, we obtain

$$<U>_n = \frac{k \hbar}{2 m \omega} (n + \frac{1}{2}) = \frac{\hbar \omega}{2} (n + \frac{1}{2})$$

where we substitute

$$k = m \omega^2$$

For the ground state n = 0, hence,

$$<U>_0 = \frac{\hbar \omega}{4}$$

The total energy (eigenenergy) of a harmonic oscillator is integral of motion (i.e., it conserves). Hence, applying the averaging procedure, we obtain

$$<E>=E=<T+U>=<T>+<U>,\\ <T> = E - <U>$$

Substituting the values for the ground state, we get

$$<T>_0 = \frac{\hbar \omega}{2} - \frac{\hbar \omega}{4} = \frac{\hbar \omega}{4}.$$

Answer:

$$<T>_0 = <U>_0 = \frac{\hbar \omega}{4}$$