- #1

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**1. The problem statement**

Consider a particle of mass m under the action of the one-dimensional harmonic oscillator potential. The Hamiltonian is given by

[tex] H = \frac{p^2}{2m} + \frac{m \omega ^2 x^2}{2} [/tex]

Knowing that the ground state of the particle at a certain instant is described by the wave function

[tex] \psi(x) = (\frac{m \omega}{\pi \hbar})^\frac{1}{4} e^{-\frac{m \omega x^2}{2 \hbar}} [/tex]

calculate (for the ground state)

a) the mean value of the position <x>

b) the mean value of the position squared <x

^{2}>

c) the mean value of the momentum <p>

d) the mean value of the momentum squared <p

^{2}>

## Homework Equations

[tex] H = \frac{p^2}{2m} + \frac{m \omega ^2 x^2}{2} [/tex]

[tex] <x^2> = \int_{-\infty}^{\infty} \psi^*(x) x^2 \psi(x) dx [/tex]

[tex] <p^2> = - \hbar ^2 \int_{-\infty}^{\infty} \psi^*(x) \frac{\partial^2 \psi(x)}{\partial x^2} dx [/tex]

## The Attempt at a Solution

My question relates to the fourth part of the question so I'll give the first three answers here.

[tex] a) <x> = 0 [/tex]

[tex] b) <x^2> = \frac{\hbar}{2m \omega} [/tex]

[tex] c) <p> = 0 [/tex]

I attempted to relate <p

^{2}> to <x

^{2}> by assuming that

[tex] H = E = E_k + V = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 [/tex] and using the ground state energy of a harmonic oscillator:

[tex] E = \frac{1}{2} \hbar \omega [/tex]

I then took the expectation value of both sides

[tex] <E> = \frac{1}{2} \hbar \omega = \frac{1}{2m} <p^2> + \frac{1}{2} m \omega^2 <x^2> [/tex]

[tex] \frac{1}{2} \hbar \omega = \frac{1}{2m}<p^2>+\frac{1}{2}m \omega^2 \frac{\hbar}{2m\omega}[/tex]

[tex] \frac{1}{2} \hbar \omega = \frac{1}{2m}<p^2> + \frac{1}{4}\hbar \omega [/tex]

[tex] <p^2> = \frac{1}{2} m \omega \hbar [/tex]

I'm just not sure if I'm correct in my assumptions. Can I assume H = E in this situation?