Expectation value of mean momentum from ground state energy

[TheBigDig]

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
$$H = \frac{p^2}{2m} + \frac{m \omega ^2 x^2}{2}$$
Knowing that the ground state of the particle at a certain instant is described by the wave function
$$\psi(x) = (\frac{m \omega}{\pi \hbar})^\frac{1}{4} e^{-\frac{m \omega x^2}{2 \hbar}}$$
calculate (for the ground state)
a) the mean value of the position <x>
b) the mean value of the position squared <x²>
c) the mean value of the momentum <p>
d) the mean value of the momentum squared <p²>

Homework Equations

$$H = \frac{p^2}{2m} + \frac{m \omega ^2 x^2}{2}$$
$$<x^2> = \int_{-\infty}^{\infty} \psi^*(x) x^2 \psi(x) dx$$
$$<p^2> = - \hbar ^2 \int_{-\infty}^{\infty} \psi^*(x) \frac{\partial^2 \psi(x)}{\partial x^2} dx$$

The Attempt at a Solution

My question relates to the fourth part of the question so I'll give the first three answers here.
$$a) <x> = 0$$
$$b) <x^2> = \frac{\hbar}{2m \omega}$$
$$c) <p> = 0$$

I attempted to relate <p²> to <x²> by assuming that
$$H = E = E_k + V = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ and using the ground state energy of a harmonic oscillator:
$$E = \frac{1}{2} \hbar \omega$$
I then took the expectation value of both sides
$$<E> = \frac{1}{2} \hbar \omega = \frac{1}{2m} <p^2> + \frac{1}{2} m \omega^2 <x^2>$$
$$\frac{1}{2} \hbar \omega = \frac{1}{2m}<p^2>+\frac{1}{2}m \omega^2 \frac{\hbar}{2m\omega}$$
$$\frac{1}{2} \hbar \omega = \frac{1}{2m}<p^2> + \frac{1}{4}\hbar \omega$$
$$<p^2> = \frac{1}{2} m \omega \hbar$$

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

 

[kuruman]

Can I assume H = E in this situation?

No you cannot. H is an operator, E is an energy (scalar). However, you can assume that the expectation value <H> = E₁ because $\psi(x)$ is the ground eigenfunction.

 

[TheBigDig]

No you cannot. H is an operator, E is an energy (scalar). However, you can assume that the expectation value <H> = E₁ because $\psi(x)$ is the ground eigenfunction.

So then my answer is correct, I just made the wrong assumption?

 

[PeroK]

So then my answer is correct, I just made the wrong assumption?

Yes, that's a neat shortcut and better than calculating $\langle p^2 \rangle$ by hand.

 

[TheBigDig]

Yes, that's a neat shortcut and better than calculating $\langle p^2 \rangle$ by hand.

Excellent! Thanks for the confirmation. A few of my classmates have tried out the integral by hand and gotten the same answer. I'm just really lazy.