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

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To normalize the wave function ψn(x) = A * √x * sin(nπx²/L²) in the range 0 to L, one must integrate the square of the wave function and set the result equal to one to solve for the normalization constant A. The integration involves handling the term cos((2πx²)/L²), which can be simplified by substituting variables to yield cos(y²). However, this integral does not have a solution in terms of standard elementary functions, as indicated by the use of the Fresnel C Integral. Despite this complexity, integrating the square of the original wave function can be simplified using the identity sin²(y) = (1/2)(1 - cos(y)). The discussion highlights the challenges and strategies involved in normalizing wave functions in quantum mechanics.
theblender
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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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What have you tried so far, and where are you stuck?
 
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.
 
Also having trouble integrating the statement cos ((2*pi*x^2)/L^2) dx.
 
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))
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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