QM harmonic oscillator - integrating over a gaussian?

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Homework Help Overview

The discussion revolves around calculating the probability of finding a particle in the first excited state of a quantum harmonic oscillator within a specific range of position, -0.2 to 0.2. The original poster attempts to integrate the square of the wavefunction, which is a Gaussian function, but encounters challenges due to the limits of integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration of the Gaussian function and question how to handle the limits of integration. Some suggest looking into the error function for assistance, while others propose numerical methods or series expansion as potential approaches.

Discussion Status

The discussion is ongoing, with participants providing various suggestions for tackling the integration problem. There is no explicit consensus on a single method, but multiple approaches are being explored, including numerical integration and series expansion.

Contextual Notes

Participants are considering the use of calculators or software for numerical integration, and there is mention of constraints regarding the use of specific algorithms or tools. The original poster's understanding of Gaussian integrals and their limits is also under examination.

tarkin
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Homework Statement


[/B]
For the first excited state of a Q.H.O., what is the probability of finding the particle in -0.2 < x < 0.2

Homework Equations



Wavefunction for first excited state: Ψ= (√2) y e-y2/2
hj055m

hj02cd


where:
hj02ns

hj05fr

The Attempt at a Solution



To find the probability, I tried the integral of : |Ψ|2

but this gives the integral of gaussian. From what I've read, the integral of a gaussian can only be solved from -infinity to infinity. So how can I find it from -0.2 to 0.2?
 
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Look up error function, then use a canned algorithm such as on a spreadsheet to find its value.
 
Use that ##e^x=\sum\limits_{n=0}^{\infty}{\frac{x^n}{n!}}## so that ##e^{-\frac{x^2}{2}}=\sum\limits_{n=0}^{\infty}{(-1)^n\frac{x^{2n}}{2^nn!}}##.

So you can integrate like it is a polynomial with infinite terms. You can choose up to which term of n to keep but I think for your value of x between 0.2 and -0.2 , the first three or four terms of integration are enough. You ll probably have to use a computer program or at least a calculator if you choose a very high value for n like the first 10 terms or more.
 
Last edited:

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