Expectation value of mean momentum from ground state energy

  • #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 <x2>
c) the mean value of the momentum <p>
d) the mean value of the momentum squared <p2>

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 <p2> to <x2> 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?
 

Answers and Replies

  • #2
kuruman
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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.
 
  • #3
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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?
 
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  • #4
PeroK
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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.
 
  • #5
57
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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.
 

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