Question

Determine the expectation value of the position of a harmonic oscillator in its ground state.

Short answer

The expectation value of harmonic oscillator in ground state is 0.

Step-by-step solution

Expectation Value

Theprobabilistic value of an experiment in the quantum mechanics is called the expectation value. It is different from the most probable value of wave function.

Determination of expectation value of the position of a harmonic oscillator

The wave function of harmonic oscillator is given as:

${\psi }_x={\left(\frac{m\omega }{\sqrt{\pi }\hslash }\right)}^{1/2}{e}^{-\left(\frac{m\omega }{2\hslash }\right)x^2}$

The expectation value of harmonic oscillator in ground state is given as:

localid="1660044825093" $\overline{x}\left(x\right)=\int {x\left|{\psi }_x\right|}^{2}dx$

Substitute all the values in the above equation.

localid="1660044830148" $\overline{x}\left(x\right)=\int x{\left|\left({\left(\frac{m\omega }{\sqrt{\pi }\hslash }\right)}^{1/2}{e}^{-\left(\frac{m\omega }{2\hslash }\right)x^2}\right)\right|}^{2}dx$

Substitute x = 0 for ground state

localid="1660044834568" $$\begin{gathered} \overline{x}\left(0\right)=\int \left(0\right){\left|\left({\left(\frac{m\omega }{\sqrt{\pi }\hslash }\right)}^{1/2}{e}^{-\left(\frac{m\omega }{2\hslash }\right){\left(0\right)}^2}\right)\right|}^{2}dx \\ \overline{x}\left(0\right)=0 \end{gathered}$$

Therefore, the expectation value of harmonic oscillator in ground state is 0.