How to Normalize and Integrate a Wave Function in the Range 0 to L?

In summary: L).In summary, to normalize a wave function, you must integrate the square of the function and set it equal to one, then solve for A. This can be done by using trigonometric identities and substitution to simplify the integrand. However, integrating the statement cos ((2*pi*x^2)/L^2) dx may require using more complex methods such as the Fresnel C Integral.
  • #1
theblender
3
0
Pretty basic question here, but I was wondering if someone could explain to me how to normalize a wave function. Specifically:

ψn(x) = A * √x * sin (n∏x2/L2), n = 1, 2, 3, ...

Normalized in the range 0 to L.

Thanks for the help, a little bit of a walk through would be much appreciated.
 
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  • #2
What have you tried so far, and where are you stuck?
 
  • #3
Well, I'm working it now, its been shed under some new light, so what I am doing is just integrating the square of psi and setting it equal to one, then solving for A. Which is what I tried initially, but I just got confused with multiplying the original by the complex conjugate.
 
  • #4
Also having trouble integrating the statement cos ((2*pi*x^2)/L^2) dx.
 
  • #5
theblender said:
Also having trouble integrating the statement cos ((2*pi*x^2)/L^2) dx.

Hint: you need to substitute x for some variable y so that you end up with cos(y^2) in the integrand. Then you can try messing around with trig, for example cos(y^2) = cos(y*y) = ?

You realize there is no easy way to expand that. In fact, wolframalpha gives:

http://www.wolframalpha.com/input/?i=cos(x^2)

which uses the "Fresnel C Integral" which I haven't even come across until this example.

So the integral is not representable by standard elementary functions...Edit: However integrating the square of your original wavefunction is quite straightforward once you calculate the integral of x(sin^2)(x^2).

Use the fact that (sin^2)(y) = (1/2)(1-cos(y))
 
Last edited:

What is normalizing a wave function?

Normalizing a wave function is the process of adjusting the amplitude of a wave function so that the total probability of finding the particle is equal to 1. This ensures that the wave function is a valid probability distribution.

Why is normalizing a wave function important?

Normalizing a wave function is important because it allows us to make meaningful predictions about the location and behavior of a particle. Without normalization, the total probability of finding the particle could be greater than 1, making it impossible to accurately predict its behavior.

How is a wave function normalized?

A wave function is normalized by dividing the entire function by the square root of the integral of the squared function over all space. This process ensures that the total probability of finding the particle is equal to 1.

What happens if a wave function is not normalized?

If a wave function is not normalized, it means that the total probability of finding the particle is greater than 1. This makes it impossible to accurately predict the behavior of the particle using the wave function, and the results would not align with experimental observations.

Can a wave function be normalized to a value other than 1?

Yes, a wave function can be normalized to a value other than 1. In some cases, it may be more convenient to normalize the wave function to a different value, such as 2 or 10. However, normalizing to a value other than 1 may make it more difficult to interpret the results of the wave function.

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