Fourier Analysis of Angular Momentum Operator

  • Thread starter eNtRopY
  • Start date
  • #1
Okay, if I want to do a Fourier Analysis of a wavefunction, I can use the following transform pairs for real space and momentum space.

Ψ(x) = (2π hbar)^(-1/2) * ∫ dp Φ(p) exp(ipx/hbar)

Φ(p) = (2π hbar)^(-1/2) * ∫ dx Ψ(x) exp(-ipx/hbar)

So, what I want to know is what is the appropriate transform pair for angular momentum.

Let's say that our real space wavefunction is expressed in terms of cylindrical coordinates, and we are only concerned with the angular term Ψ(θ).

Do we want to transform this into angular momentum space? Is this expression correct?

Ψ(θ) = (2π hbar)^(-1/2) * ∫ dL Φ(L) exp(iLθ/hbar)

eNtRopY
 
Last edited by a moderator:

Answers and Replies

  • #2
arcnets
508
0
eNtRopY, I'm no expert, but I remember dimly that angular momentum is always quantized, so I'd expect a sum rather than an integral.
 
  • #3
Originally posted by arcnets
eNtRopY, I'm no expert, but I remember dimly that angular momentum is always quantized, so I'd expect a sum rather than an integral.

But it is only quantized because of the imposed boundary condition:

Ψ(θ=0) = Ψ(θ=2π).

Actually, that's part of what I'm trying to figure out... can we make the math tell us the integral is a sum... from momentum space?

eNtRopY
 
  • #4
arcnets
508
0
Maybe the clue is in what you call &phi(L) - angular momentum eigenstates, aren't they?

(I can't make the 'phi' show up...sorry)
 
Last edited:
  • #5
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,950
19
I think you need to step into the "why" of Fourier theory.


The reason the fourier transform works is that the set of functions

{fω(t) = eiωt | ω real}

is an orthogonal basis of the function space. Because we have a single continuous real parameter, the fourier transform takes the form of an integral over that parameter.


For angular position, we're not working with the entire space of functions; only those with a period dividing 2π. The basis for this space is only those basis vectors above that have a period dividing 2π, i.e.:

{fn(t) = e2πit/n | n a nonzero integer} U {f(t) = 1}

Here, the parameter is discrete, so the transform takes the form of a sum over n. (which is just an integeral with a different metric)

Well, to be more precise, the transform in one direction will be integrated over [0..2π] and the transform in the other direction will be a sum
 
Last edited:
  • #6
Okay, here's what I have so far.... For a 1-D ring of radius r0, we know the Schrödinger equation looks like this:

d2/dθ2 Ψ(θ) + ((2 m r02 E)/hbar2) * Ψ(θ) = 0.

Now, since I know the solution a priori, it's not a problem for me to assume the following transform pair:

Ψ(θ) = (2πhbar)-1/2 ∫ dl Φ(l) exp(ilθ)

and

Φ(l) = (2πhbar)-1/2 ∫ dθ Ψ(θ) exp(-ilθ).

So, I can use Fourier transforms to convert the differential equation to:

-l2Φ(l) + ((2 m r02 E)/hbar2) * Φ(l) = 0.

We can then easily solve for l:

l2 = ((2 m r02 E)/hbar2).

This is exactly what we wanted, but the energy is not quantized at this point because we haven't included the boundary condition.

E = (hbar2 l2)/(2 m r02)

We can clear up this problem by using the boundary condition we were given (the one in real space):

Ψ(0) = Ψ(2π).

However, I feel like there must be a way to convert the boundary condition to l-space.

Thanks.

eNtRopY
 
Last edited by a moderator:
  • #7
Tyger
398
0
In central force problems

which includes all conservative systems, and the hydrogen atom in particular, you can separate the wavefunction into radial and angular parts, and the angular part always has a discrete spectrum. There is no way to make the spectrum continuous. However in the limit of large L the system behaves like it is nearly continuous, and that, you may recall, was the basis of Bohr's Corresponence Principle.
 
Last edited:

Suggested for: Fourier Analysis of Angular Momentum Operator

Replies
1
Views
954
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
898
  • Last Post
Replies
16
Views
10K
  • Last Post
Replies
1
Views
966
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Top