Solving for A: Integral Help in Quantum Mechanics

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SUMMARY

The discussion focuses on solving the integral of the function \(\int_{-\infty}^{\infty} Ae^{-\lambda(x - a)^2} dx\) where A, \(\lambda\), and a are positive constants. The integral is set to equal 1, leading to the conclusion that \(A = \sqrt{\frac{\lambda}{\pi}}\). The solution utilizes the known result \(\int_{-\infty}^{\infty} e^{-a x^2} dx = \sqrt{\frac{\pi}{a}}\) from Griffiths' "Quantum Mechanics" Chapter 1. A substitution \(u = x - a\) is suggested to simplify the integration process.

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Hi,
I'm having problems integrating the following function:
[tex]\int^\infty_{-\infty}Ae^{-\lambda(x - a)^2}[/tex]
Where A, [itex]\lambda[/itex], and a are positive, real constants. If anyone could point me in the right direction I'd really appreciate it.

The integral is supposed to be equal to 1 and using that fact I'm supposed to solve for A. This is from Griffiths' Quantum Mechanics Chapter 1.
 
Last edited:
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You can do the substitution u=x-a (you shouldn't actually have to do that, just think about what it would do), and then use the fact that:

[tex]\int_{-\infty}^{\infty} e^{-a x^2} dx = \sqrt{\frac{\pi}{a}}}[/tex]

That's something worth memorizing, but you can derive it using the technique shown on http://en.wikipedia.org/wiki/Gaussian_function" page.
 
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Thank you! So assuming I made the correct substitutions and such I end up with the integral being equal to:

[tex]A\sqrt{\frac{\pi}{\lambda}}[/tex]

And since the integral is equal to 1 (it's a probability thing)

[tex]A = \sqrt{\frac{\lambda}{\pi}}[/tex]
 

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