Normalization constant for a 3-D wave function

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SUMMARY

The normalized wave function for a particle in a three-dimensional box with dimensions a, b, and c is given by Ψ(x,y,z) = √(8/abc) * sin(nxπx/a) * sin(nyπy/b) * sin(nzπz/c). The normalization condition requires that the integral ∫0^a dx ∫0^b dy ∫0^c dz Ψ*(x,y,z)Ψ(x,y,z) = 1. The integrand Ψ*(x,y,z)Ψ(x,y,z) can be separated into three distinct integrals, each corresponding to the respective dimensions of the box, which simplifies the normalization process.

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  • Familiarity with normalization conditions in quantum mechanics
  • Knowledge of multivariable calculus and integration techniques
  • Experience with trigonometric functions and their properties
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  • Learn about the implications of normalization in quantum systems
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Students and professionals in physics, particularly those focusing on quantum mechanics, as well as educators teaching wave functions and normalization in three-dimensional systems.

Valeria
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Homework Statement


Show that the normalized wave function for a particle in a three-dimensional box with sides of length a, b, and c is:
Ψ(x,y,z) = √(8/abc) * sin(nxπx/a)* sin(nyπy/b)* sin(nzπz/c).

Homework Equations


Condition for the normalization:
0adx ∫0bdy ∫0cdz Ψ*(x,y,z)Ψ(x,y,z) = 1.

The Attempt at a Solution


From the 1-D case I know that I should arrive to this:
0adx ∫0bdy ∫0cdz Ψ*(x,y,z)Ψ(x,y,z) = 1 = (AxAyAz)20asin2(nxπx/a)dx ∫0bsin2(nyπy/b)dy ∫0csin2(nzπz/c)dz

However, I do not understand why Ψ*(x,y,z)Ψ(x,y,z) (unlike a 1-D case) is outside of the integrals. How exactly Ψ*(x,y,z)Ψ(x,y,z) is related to ∫0adx ∫0bdy ∫0cdz then? I do not understand how it is split between 3 integrals to give the formula above.

(I know how to proceed with the solution once I can arrive to this formula, so actually finding the normalization constant for me is not important.)
 
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Valeria said:

Homework Statement


Show that the normalized wave function for a particle in a three-dimensional box with sides of length a, b, and c is:
Ψ(x,y,z) = √(8/abc) * sin(nxπx/a)* sin(nyπy/b)* sin(nzπz/c).

Homework Equations


Condition for the normalization:
0adx ∫0bdy ∫0cdz Ψ*(x,y,z)Ψ(x,y,z) = 1.

The Attempt at a Solution


From the 1-D case I know that I should arrive to this:
0adx ∫0bdy ∫0cdz Ψ*(x,y,z)Ψ(x,y,z) = 1 = (AxAyAz)20asin2(nxπx/a)dx ∫0bsin2(nyπy/b)dy ∫0csin2(nzπz/c)dz

However, I do not understand why Ψ*(x,y,z)Ψ(x,y,z) (unlike a 1-D case) is outside of the integrals. How exactly Ψ*(x,y,z)Ψ(x,y,z) is related to ∫0adx ∫0bdy ∫0cdz then? I do not understand how it is split between 3 integrals to give the formula above.

(I know how to proceed with the solution once I can arrive to this formula, so actually finding the normalization constant for me is not important.)

In multiple integrals like this they sometime write all of the integration variables and limits on the left and the integrand on the right. This is not supposed to mean that the integrand is 'outside' of the integral or anything. It's just supposed to make it a little easier to read. I think you know exactly what to do. Just ignore that writing convention and put the integrand where you know it should be.
 
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@Valeria There are two ways to write an integral, whether is has one variable or several

##\int f(x)dx##

Might be considered the normal mathematician's convention.

Or:

##\int dx f(x)##

which physicists may prefer.

They both mean the same. When it comes to multivariable integrals, I prefer the former. For example:

##\int \int \int f(x,y,z) dx dy dz##
 
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