How Do You Calculate Xav, (X^2)av, and deltaX for a Simple Harmonic Oscillator?

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Homework Help Overview

The discussion revolves around calculating the average position (Xav), the average of the square of the position ((X^2)av), and the uncertainty in position (deltaX) for a simple harmonic oscillator using its ground-state wave function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the normalization constant and the wave function to perform integrals but questions whether this approach will yield Xav. Participants suggest that inserting x into the integral is necessary for finding Xav.

Discussion Status

Participants are engaging in clarifying the steps needed to compute the required averages and are discussing the normalization of the wave function. There is a focus on ensuring the correct application of the wave function in the integrals.

Contextual Notes

There is an emphasis on the normalization of the wave function and the need for clarity in the mathematical steps involved in the calculations. The original poster expresses uncertainty about their approach.

GreenLRan
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Homework Statement



Use the ground-state wave function of the simple harmonic oscillator to find: Xav, (X^2)av and deltaX. Use the normalization constant A= (m*omegao/(h_bar*pi))^1/4.

Homework Equations



deltaX=sqrt((X^2)av-(Xav)^2)

wavefunc=A*e^(-ax^2) ?

The Attempt at a Solution



I'm not sure if I'm on the right path, but I started out by plugging in A and doing the integral of the wavefunction. My question is... does doing this give me Xav? If not, how would I go about solving this problem?
 
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If you plug the right A in and do the integral of \psi^* \psi you should get 1. It's normalized. Getting <x>=Xav? requires inserting an x into the integral.

&lt;f(x)&gt;=\int \psi^*(x) f(x) \psi(x) dx
 
Last edited:
Of course, you know you need to use Psi^2.
 
Meir Achuz said:
Of course, you know you need to use Psi^2.

I said that in a pretty sloppy way. I've edited the post to clarify.
 

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