How Do You Normalize a Function to Unity?

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To normalize a function to unity, one must define the norm using the integral of the modulus squared of the wavefunction. Specifically, for a function like (2/L)^(1/2) sin(n*pi*x/L), the normalization condition is expressed as the integral of |ψ(x)|² over the relevant range equaling one. In the case of an infinite square well from x=0 to x=L, this condition simplifies to the integral from 0 to L. The result ensures that the probability density integrates to one, confirming the function is normalized. This approach is essential in quantum mechanics for ensuring accurate probability interpretations.
ehrenfest
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Homework Statement



I understand how you normalize vectors to unity, but how would you normalize a function to unity?

For example, how would you show that the function (2/L)^(1/2) sin(n*pi*x/L) is normalized to unity? You cannot just just take its modulus and set it to one since because x is variable...right?

Homework Equations





The Attempt at a Solution

 
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Define the norm. Looks like QM, so you want L^2. The integral over the range of x of psi(x)*conjugate(psi(x))=1 defines a normalized function.
 
ehrenfest said:

Homework Statement



I understand how you normalize vectors to unity, but how would you normalize a function to unity?

For example, how would you show that the function (2/L)^(1/2) sin(n*pi*x/L) is normalized to unity? You cannot just just take its modulus and set it to one since because x is variable...right?

Homework Equations





The Attempt at a Solution



It's important to realize that the modulus squared (not just the modulus), | \psi(x)|^2 of a wavefunction is a probability density , not a probability. This means that the quantity | \psi(x)|^2 dx is aprobability, which represents the probability of finding the particle in the infinitesimal interval (x,x+dx). What must be normalized to one is therefore

\int_{- \infty}^{\infty} | \psi(x)|^2 dx =1

It seems as if you are considering an infinite squqre well located from x=0 to x=L. In that case, the wavefunction is zero outside of the well and the above condition reduces to


\int_{0}^{L} | \psi(x)|^2 dx =1


hope this helps

Patrick
 
I see. Thanks.
 

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