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

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SUMMARY

The discussion focuses on 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. The normalization constant is defined as A = (m*omega_o/(h_bar*pi))^1/4. The user seeks clarification on whether integrating the wave function provides Xav and how to properly compute these values using the integral of the wave function.

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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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