Trouble with normalizing a wave function

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Homework Help Overview

The discussion revolves around normalizing a wave function for a quantum particle described by the state ψ(x,t) = Ae-a[mx2/h-bar)+it]. Participants are exploring the normalization condition and the implications of the integration bounds in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the normalization condition ∫ψ² = 1 and questioning the origin of a factor of 2 in the solution provided by the book. They discuss the integration bounds and the implications of symmetry in the wave function.

Discussion Status

Some participants have provided insights regarding the integration bounds and the reasoning behind the factor of 2, indicating a productive exploration of the topic. However, there is no explicit consensus on the appropriateness of the chosen bounds or the normalization method.

Contextual Notes

Participants are considering the domain of the wave function as all x and t > 0, and the bounds of the integral from 0 to ∞, which raises questions about the typical integration limits used in Gaussian integrals.

marineric
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Homework Statement


A particle of mass m is in the state ψ(x,t) = Ae-a[mx2/h-bar)+it]

Find A

Homework Equations


I know that to normalize a wave function I should use ∫ψ2 = 1


The Attempt at a Solution


The book gives the solution as 1 = 2abs(A)2∫ e-2amx2/h-bar) dx

My question is where does the "2" factor in front of the A2 come from?
 
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Hmm...could you perhaps provide a little more information on
(i) the domain of the wavefunction
(ii) the bounds of that integral
 
i) the domain is for all x and t > 0?

ii) the bounds of the integral are from 0 to ∞
 
I think I got it. The solutions manual integrated from 0 to ∞, and multiplied by 2 because of symmetry (to get the -∞ to 0 part)
 
marineric said:
I think I got it. The solutions manual integrated from 0 to ∞, and multiplied by 2 because of symmetry (to get the -∞ to 0 part)
Yup, seems like it. It is rather odd of them to do that though; usually we like -∞ to ∞ bounds because they allow us to use the Gaussian integration formula.
 

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