Normalization of wavefunctions

[dyn]

Hi
I have been working my way through some past papers and then checking the solutions but I am confused about the following. One question asked for the normalisation constant for the following wavefunction ψ( r, θ , ø ) = Aexp(-r/R) where R is a constant. The solution requires a triple volume integral using dV=r^2sinθdrdθdø. The 2nd question asks to normalise exp(imø) but this time the integral is just done over dø from 0 to 2∏.
The 1st wavefunction only involves r but I have to do a triple integral. The 2nd only involves ø but I just need a single integral. I'm confused !

 

[strangerep]

You have to integrate over the domain on which the wavefunction is defined. In your 1st case, it's a 3D domain. In the 2nd case it's a 1D domain (actually, just a circle).

 

[dyn]

In the first case I can see that it is a 3-D domain as the wavefunction is given as ψ ( r , θ , ø ) even though the function only depends on r. But in the 2nd case the function is just given as exp(imø) not in the form ψ ( ) so how would know if the domain is 3-D or 1-D ?

 

[strangerep]

In the first case I can see that it is a 3-D domain as the wavefunction is given as ψ ( r , θ , ø ) even though the function only depends on r. But in the 2nd case the function is just given as exp(imø) not in the form ψ ( ) so how would know if the domain is 3-D or 1-D ?

If the question is well-posed, then it should be clear from the context. If it just says $e^{im\phi}$, with no other information, then it's reasonable to assume it depends only on $\phi$. I can't say for sure without seeing the wording (and context) of the question(s).

Also, if you try normalizing $e^{im\phi}$ over an unrestricted 3D domain, you won't get a sensible answer since the integral diverges.

 

[dyn]

I've just done another question which asked to determine the constant A in the wavefunction
ψ ( r ) = Aexp(-r/a). The solution involved the triple integral over r ,θ , ø. But if the function is given as a function of only one variable how do I know to do the triple integral ? Whereas when given exp(imø) which again is only one variable I only do a single integral ?

 

[strangerep]

I've just done another question which asked to determine the constant A in the wavefunction
ψ ( r ) = Aexp(-r/a). The solution involved the triple integral over r ,θ , ø. But if the function is given as a function of only one variable how do I know to do the triple integral ? Whereas when given exp(imø) which again is only one variable I only do a single integral ?

I can't offer any further answer without seeing the full questions. Can you post a link to them?

 

[dyn]

Hi,
I can't post a link to them but I will give the question as best I can.( I don't know how to post complicated equations , sorry )

The question starts by giving the time-independent Schrodinger equation for a particle in 3-D where ψ = ψ ( vector r ). The question then asks to show that for an electron in a hydrogen atom , the spherically symmetric solutions satisfy an equation which only involves the r coordinate. ( the Laplacian operator in spherical polar coordinates is given as a hint ).
The next part then asks to show that the wavefunction ψ ( scalar r ) = A exp(-r/a) is a solution to the equation I had previously calculated.
I don't understand how to know whether to do the triple or single integral as the function only involves r ; but with the exp(imø) it was only a single integral.
Sorry if I'm not making much sense.

 

[PeterDonis]

( I don't know how to post complicated equations , sorry )

See here:

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[jtbell]

A hydrogen atom is a three-dimensional object, so it is described by a three-dimensional wave function. The fact that the wave function depends only on r (for the specific state that you're dealing with) means that the probability distribution is spherically symmetric; it doesn't turn the atom into a one-dimensional object.

As a classical analogy, consider a solid sphere of uniform density. All you need to describe it is the density ρ and the radius R. But if you want to find the mass of the sphere, you need to integrate the density over a three-dimensional volume. This would give you the result M = (4/3)πR³ρ, i.e. the volume of a sphere of radius R, times the density.

As for the $e^{im\phi}$, does the problem statement not give any context at all for it? What is it supposed to be describing?

One possibility that comes to my mind is that the three-dimensional hydrogen wave function is a product of three one-dimensional functions: $\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$. When you normalize it, the three-dimensional integral separates into a product of three one-dimensional integrals. It's possible to normalize each of them individually so the overall integral is 1·1·1 = 1.

 

[dyn]

Thanks for that. The exp(imø) problem is for the eigenfunction of the z-component of angular momentum with eigenvalue mh/2∏. The normalization integral is a single integral from 0 to 2∏. I understand that this would be impossible to normalise with r ranging from 0 to ∞ but it don't understand why this is a single integral when the other ones are triple integrals.

 

[dyn]

A hydrogen atom is a three-dimensional object, so it is described by a three-dimensional wave function. The fact that the wave function depends only on r (for the specific state that you're dealing with) means that the probability distribution is spherically symmetric; it doesn't turn the atom into a one-dimensional object.

As a classical analogy, consider a solid sphere of uniform density. All you need to describe it is the density ρ and the radius R. But if you want to find the mass of the sphere, you need to integrate the density over a three-dimensional volume. This would give you the result M = (4/3)πR³ρ, i.e. the volume of a sphere of radius R, times the density.

As for the $e^{im\phi}$, does the problem statement not give any context at all for it? What is it supposed to be describing?

One possibility that comes to my mind is that the three-dimensional hydrogen wave function is a product of three one-dimensional functions: $\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$. When you normalize it, the three-dimensional integral separates into a product of three one-dimensional integrals. It's possible to normalize each of them individually so the overall integral is 1·1·1 = 1.

The $e^{im\phi}$ I mentioned is the ##\Phi(\phi) you mention above so as you say it would only involve a 1-D integral so would the R(r) integral. But there lies my confusion ! Both ψ (r) and R(r) are functions of a single coordinate but the ψ (r ) is a 3-D integral and R(r) is a 1-D integral

 

[dyn]

Sorry the above post didn't come out right. I was trying LateX for the first time. Only got it half right !