# Normalization constant for a 3-D wave function

## 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.)

Related Introductory Physics Homework Help News on Phys.org
Dick
Homework Helper

## 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.

PeroK
Homework Helper
Gold Member
@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:

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