Fourier Analysis of Angular Momentum Operator


by eNtRopY
Tags: analysis, angular, fourier, momentum, operator
eNtRopY
#1
Jun29-03, 02:27 PM
P: n/a
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
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
arcnets
arcnets is offline
#2
Jun29-03, 02:50 PM
P: 513
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.
eNtRopY
#3
Jun29-03, 03:40 PM
P: n/a
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

arcnets
arcnets is offline
#4
Jun29-03, 03:46 PM
P: 513

Fourier Analysis of Angular Momentum Operator


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)
Hurkyl
Hurkyl is offline
#5
Jun29-03, 04:00 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
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
eNtRopY
#6
Jun29-03, 04:06 PM
P: n/a
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
Tyger
Tyger is offline
#7
Jun30-03, 03:45 PM
Tyger's Avatar
P: 402
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.


Register to reply

Related Discussions
Angular Momentum Raising Operator Advanced Physics Homework 4
matrix of angular momentum operator Quantum Physics 4
Fourier Integral involving a position operator Quantum Physics 27
Angular momentum and orbital angular momentum problems Introductory Physics Homework 3