Decompose wave packet into eigenvalues of L2, Lz and k

[diegzumillo]

Homework Statement

The free particle wave packet in question is $$\psi=ce^{-(r/r_0)^2}$$

Homework Equations

The Attempt at a Solution

I've been going through books and class notes but I really have no idea where this came from. I'm thinking that if I can decompose this in plane waves I could also write this in terms of Bessel functions and Legendre polynomials, where some $l$s would appear (eigenvalues of $L^2$. But other than that, I'm totally stumped!

 

[king vitamin]

I've been going through books and class notes but I really have no idea where this came from. I'm thinking that if I can decompose this in plane waves I could also write this in terms of Bessel functions and Legendre polynomials, where some $l$s would appear (eigenvalues of $L^2$. But other than that, I'm totally stumped!

I think you have the right idea. The decomposition will involve spherical Bessel functions and spherical harmonics, but notice that your problem has a very simple angular dependence which will make your particular problem relatively easy. You can start by writing down the free-particle eigenfunctions with the same spherical symmetry as your wavepacket, and then concentrate on writing your wavepacket in terms of these eigenfunctions.

 

[diegzumillo]

Well, I settled with a solution. Possibly wrong or, at least, incomplete. I'm writing this wave function as an integral over all space of it times a delta
$$e^{-(r/r_0)²}=\int_{-\infty}^{\infty}dr′e^{-((r-r′)/r_0)²}\delta (r′)$$
and then I use
$$\delta (r′)=(1/(2π))\int_{-\infty}^{\infty}dke^{-ir′k}$$
there. Finally I rewrite this plane wave exponential as the expansion in spherical harmonics and bessel function (Rayleigh relation, I think it's called). It looks horrible! I'm not going to write that down here. There might be a way to make it more compact, solving one of the integrals, maybe. But it's late and I'm calling it done :)

Thanks for the input, King Vitamin.

 

[king vitamin]

Indeed, your resulting integral over r' will kill almost all of the terms in your resulting expression.

Since I don't want you to give up, I'll give you big hint. Your wave packet has l = 0, so the only expansion will be in terms of l = 0 eigenfunctions of the free Schrödinger equations. The l=0 Schrödinger equation is extremely simple, and the solution is a familiar function which you will have an easy time working with.

 

[diegzumillo]

OH! it's spherically symmetric. I totally neglected that information. So l=0 and my summation over l becomes a single term... Or are you suggesting I ditch this delta approach completely?

On second thought. The summation over l that appears in my expression comes from the expansion of a general plane wave that I got from my delta. I don't think I can simply cancel the terms of it. So the delta approach doesn't work then. I solved the Schrodinger equation for l=0 and the result are plane waves $$e^{\pm ax}$$ How do I write a gaussian as a combination of plane waves? that's my initial problem that I tried to solve with the insertion of a delta.

 

[king vitamin]

In your delta function approach, the nonzero l terms need to somehow not contribute. Like I said above, I'm pretty sure the integrals over r' will be zero for all terms except l = 0. I think I've seen this expression you're referring to, but don't remember it off the top of my head. I do think that solving the spherical coordinate Schrodinger is easier - you already know that the angular solutions are spherical harmonics which have zero overlap with your wave packet, so you only need to solve the l=0 case.

I don't quite agree with your plane-wave solution either. I'm assuming you're working with the reduced function $u(r) = rR(r)$, which satisfies the "centrifugal" equation

$$-\frac{\hbar^2}{2m}\frac{d^2u(r)}{dr^2} + \left( V(r) + \frac{\hbar^2 l (l+1)}{2mr} \right) u(r) = E u(r).$$

I assume that you found that for l=0, the function u(r) must be plane waves, $u(r) = Ae^{i k r} + Be^{-i k r}$ where $k = \sqrt{2mE}/\hbar$. However, importantly, the function $R(r) = u(r)/r$ must be finite at $r = 0$. It turns out that there is only one possible combination of A and B which satisfies this (up to overall normalization).

Notice that this decomposition will be far easier since the original wave packet and these waves are both in the radial direction.

 

[diegzumillo]

Interesting, I probably made some mistake. Following your logic the solutions will be linear combinations of $\pm \frac{sin(r)}{r}$ then. Which is pretty close to what I got, actually.
The relation we are talking about is this
$$e^{-irkcos\theta}=\sum_{l=0}^{\infty}(i)^{l}(2l+1)P_{l}(cos\theta)j_{l}(kr)$$
(there's another version with spherical harmonics as well) And my final equation looks like this
$$e^{-(r/r_0)²}=\frac{1}{2\pi}\int \int dkdr′e^{-((r-r′)/r_0)²}\frac{sin(kr′)}{kr′}$$
Strange that I have two integrals still though. But they are exactly equal, maybe there's something to be done with that.

 

[king vitamin]

You need to be much more careful with your method. The expression

$$e^{-irkcos\theta}=\sum_{l=0}^{\infty}(i)^{l}(2l+1)P_{l}(cos\theta)j_{l}(kr)$$

only really makes sense for $r>0$, whereas your integral is over negative and positive r (otherwise converting the delta function into a Fourier integral isn't valid), so your expression isn't right. If you really wanted to use your method carefully, you should use three-dimensional delta functions:

$$e^{-(r/r_0)^2} = \int d^3x' \delta^3(\mathbf{x'}) e^{-((|x| - |x'|)/r_0)^2} = \int d^3x' \frac{d^3k}{(2 \pi)^3} e^{-i \mathbf{k} \cdot \mathbf{x}'} e^{-((|x| - |x'|)/r_0)^2}.$$

Now you would like to write $e^{-i \mathbf{k} \cdot \mathbf{x}'} = e^{-ikr\cos\theta}$, but the angle here depends on both the vector $\mathbf{k}$ AND $\mathbf{x}'$. The resulting integral is far harder (I can't see how to do it by eye).

In contrast, if you simply use the fact that your expression needs to take the form
$$e^{-(r/r_0)^2} = \int_0^{\infty} dk c(k) \frac{\sin(kr)}{kr}$$
and use usual methods to find the function c(k), you should find your answer pretty quickly.

 

[diegzumillo]

I see now. Thanks for all the help. I won't be able to continue solving this problem (new homework, new headaches) but it's pretty clear what's going on now, and that's the whole point of these problems.