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

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Homework Help Overview

The discussion revolves around the probability of finding a particle at a specific position, x=0, for a simple harmonic oscillator. The original poster presents a wave function and expresses confusion regarding the probability at that point, questioning whether it should be zero.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the wave function at x=0 and question the validity of interpreting probability at a continuous position. There is a discussion about nodes in the probability distribution and how they relate to the ground state of the system.

Discussion Status

Participants are actively engaging with the concepts, with some providing clarifications on the nature of probability in quantum mechanics. There is an acknowledgment of the original poster's confusion, and various interpretations of the problem are being explored without reaching a consensus.

Contextual Notes

There is a mention of the ground state wave function and its implications for the probability density, as well as references to the infinite square well potential and its boundary conditions. The original poster's question is noted as potentially unphysical, prompting further examination of the assumptions involved.

rwooduk
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Homework Statement


What is the probability of finding the particle at x=0 for a simple harmonic oscillator.

Homework Equations


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

The Attempt at a Solution


Going back to basics with this one, too much revision, can't 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.

thanks again for any advice
 
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rwooduk said:
dont know why I'm getting confused with this, feel as though it should be zero at x=0
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.
 
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Fightfish said:
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.
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!
 
Fightfish said:
In fact, it cannot be zero - there cannot be any nodes in the probability distribution for a particle in the ground state:
http://scitation.aip.org/content/aapt/journal/ajp/70/8/10.1119/1.1479742

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:

A wave function node occurs at points where the wave function is zero and changes signs. The electron has zero probability of being located at a node.

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

thanks again for the reply!
 
IS the ground state explicitly mentioned in the exercise ?
 
BvU said:
IS the ground state explicitly mentioned in the exercise ?

Hi, yes the ground state wave function is given.
 
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.
 
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BvU said:
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.

excellent thanks for the additional info!
 
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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.
 
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