Question

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

Answer 1

Answer on Question #85011 Physics / Quantum Mechanics

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

Solution:

In the ground state the wave function and energy of a simple harmonic oscillator

$$\psi_ {0} (x) = \left(\frac {m \omega}{\pi \hbar}\right) ^ {\frac {1}{4}} \exp \left(- \frac {m \omega x ^ {2}}{2 \hbar}\right)$$ $$E _ {0} = \frac {\hbar \omega}{2}$$

The mean kinetic energy

$$\begin{aligned} \langle K \rangle &= \int_{-\infty}^{\infty} \psi_ {0} (x) \left(- \frac {\hbar^ {2}}{2 m} \frac {d ^ {2}}{d x ^ {2}}\right) \psi_ {0} (x) d x \\ &= \frac {\hbar^ {2}}{2 m} \left(\frac {m \omega}{\pi \hbar}\right) ^ {\frac {1}{2}} \int_{-\infty}^{\infty} \exp \left(- \frac {m \omega x ^ {2}}{\hbar}\right) \left(\frac {m \omega}{\hbar} - \frac {m ^ {2} \omega ^ {2}}{\hbar^ {2}} x ^ {2}\right) d x \\ &= \frac {\hbar^ {2}}{2 m} \left(\frac {m \omega}{\pi \hbar}\right) ^ {\frac {1}{2}} \left(\frac {m \omega}{\hbar} \sqrt {\frac {\pi \hbar}{m \omega}} - \frac {m ^ {2} \omega^ {2}}{\hbar^ {2}} \frac {\hbar}{2 m \omega} \sqrt {\frac {\pi \hbar}{m \omega}}\right) = \frac {\hbar \omega}{4} \end{aligned}$$

The mean potential energy

$$\langle V \rangle = E _ {0} - \langle K \rangle = \frac {\hbar \omega}{2} - \frac {\hbar \omega}{4} = \frac {\hbar \omega}{4}$$

Answer: $\langle K\rangle = \langle V\rangle = \frac{\hbar\omega}{4}$

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