Probability of finding a particle at x=0 for a simple HO

1. Jan 7, 2015

rwooduk

1. The problem statement, all variables and given/known data
What is the probability of finding the particle at x=0 for a simple harmonic oscillator.

2. Relevant equations
$$\Psi _{0} = (\frac{m\omega}{\pi\hbar})^{\frac{1}{4}} exp (\frac{-m\omega x^{2}}{2\hbar})$$

3. The attempt at a solution
Going back to basics with this one, too much revision, cant think straight! Please could someone confirm that the probability is:

$$P(x=0)= |\Psi (0)|^{2} = (\frac{m\omega}{\pi \hbar})^{\frac{1}{2}}$$

dont know why I'm getting confused with this, feel as though it should be zero at x=0 and the only way to get this would be integrating the term also.

2. Jan 7, 2015

Fightfish

Why do you think it should be zero at x = 0?
Also, just to be pedantic, it is not right to speak of the probability finding a particle at a particular position, because position is a continuous observable. You probably mean the probability per unit length of finding the particle about x = 0.

Last edited: Jan 7, 2015
3. Jan 7, 2015

rwooduk

Good point! there is no reason why it would be zero, i was thinking of an infinite well for some reason. And yes it's a bit of a funny question asking for a specfic position. Many thanks for your help!

4. Jan 7, 2015

5. Jan 7, 2015

rwooduk

I'm not back at University so can't access this paper, but I would be interested to know what this means to the following statement:

and inparticular how it applies to the ground state wave function of an infinite well at the boundaries.

6. Jan 7, 2015

BvU

IS the ground state explicitly mentioned in the exercise ?

7. Jan 8, 2015

rwooduk

Hi, yes the ground state wave function is given.

8. Jan 8, 2015

BvU

Well, then the probability density follows from the wave function. You've done that already in post 1. Seems the best answer to me.
Happily sitting there with a minimum energy that still satisfies the uncertainty principle.

Asking for the probability at x=0 is unphysical :) (post #2)
Expectation value for x in ground state is 0, though.

Re infinite well: if it's a square well, continuity of the wave function forces $\Psi = 0$ at the walls.

9. Jan 8, 2015

rwooduk

excellent thanks for the additional info!

10. Jan 8, 2015

Fightfish

This probably strays away from the original question already, but since we're on the topic of wavefunction "nodes", the vanishing of the wavefunction at the boundaries of an infinite square well potential are not considered as "nodes".

In the wavefunction picture, a node occurs when the wavefunction "changes sign" and crosses the axis. In terms of the probability density picture, a node would manifest itself as a stationary point (one that touches the axis, of course). There are some interesting theorems concerning the number of nodes and the level of the energy eigenfunctions if you like to investigate further.