# Hamiltonian operator in 3d

1. Apr 3, 2012

### fluidistic

If I consider the problem of for example the hydrogen atom. I.e. a central force problem with an effective potential V(r) that depends only of r, the distance between the positively charged nucleous and the negatively charged electron.
In the Schrödinger's equation, one considers the Hamiltonian operator as $-\frac{\hbar }{2m} \nabla ^2 +V (\vec r, t )$. From classical mechanics we know that in a central force problem, the motion is constrained into a plane. My question is thus: why is the Laplacian taken in spherical coordinates (this assumes a 3d motion) instead of polar coordinates (this assumes a motion constrained into 2 dimensions)?

2. Apr 3, 2012

### M Quack

The reason is purely mathematical.

In spherical coordinates, you can write the wave functoin as a neat product of a radial wave function that contains the central potential and depends only on r, and the angular part that depends on theta and phi, and which is nicely solved by spherical harmonics.

3. Apr 3, 2012

### fluidistic

I know this.
What I don't understand is that from classical mechanics we know that a central force problem simplifies the problem because it implies that the motion is in a plane rather than a 3 dimensional space.
So why can't I simply take the Laplacian in polar coordinates, which would result in much simpler algebra/arithmetics than in the spherical coordinates?

4. Apr 3, 2012

### The_Duck

But this is not classical mechanics. In quantum mechanics, the fact that the force is central does not imply that motion is confined to a plane. If the electron's position was known to lie in a certain plane, the uncertainty principle would tell us that there was infinite uncertainty in the component of momentum perpendicular to that plane, so that the electron must shortly leave the plane. If you've seen pictures of electron orbitals, you'll know that they are not planar, but rather three dimensional probability distributions.

5. Apr 3, 2012

### fluidistic

I see, thanks.
I still don't really get it. In classical mechanics we didn't assume the motion would be in either 1d, 2d or 3d. The mathematics shows that it's in 2d (I think a cross product shows that $\vec r$ lies in a plane orthogonal to $\vec L$). While in quantum mechanics it seems we simply assumes that the motion of an electron in a central force will be in 3d without demonstrating it. If we had to assume something, I think that'd be a motion in 2d as in classical mechanics. I never read any comment on this assumption yet; maybe I lack the book(s). But that deserves an explanation in my opinion.
Edit: In classical mechanics a motion in 3d would mean a non central force is acting on the particle. If we assume the electron has a 3d motion, then it's hard to believe that the force acting on it is simply a central force. Unless of course that the maths shows it.

6. Apr 3, 2012

### The_Duck

You're right--initially we need make no assumptions about whether the motion will be confined to some smaller number of dimensions than three. In classical or quantum mechanics, we should first write down the full 3D equations of motion, and if we find that those equations say that motion is confined to some particular smaller-dimensional subspace, we shall count ourselves lucky that the problem is simpler than it initially appeared.

Thus, in classical mechanics you can write down the equations of motion for a point particle in three dimensions moving in a central potential, and you find that the particle stays in one 2D plane within that 3D space. Having discovered this, you may reorient your coordinate system so that, say, the z axis points perpendicularly to that plane, and then drop the equation of motion for the z coordinate, since it is trivial. Thus you've reduced the 3D problem to a 2D problem. This is how we justify solving a 2D problem instead of a 3D problem in classical mechanics.

In quantum mechanics you can write down the Schrodinger equation for the wave function of a particle moving in a 3D potential. But now, even if in the initial state the wave function is only nonzero within a certain plane, you find that it will quickly evolve to a state where the wave function is nonzero outside that plane. Thus we can't simplify to a 2D problem.

7. Apr 3, 2012

### fluidistic

Thanks, flawless logical argument.

8. Apr 3, 2012

### tom.stoer

One should mention one major difference between the orbit and the Hamiltonian; the Hamiltonian defines the problem whereas the orbit is just one single solution; there may be others constrained to a different plane.

So whereas one single solution is 2d, the space of all solutions has again full 3d rotational symmetry, w/o any preferred plane or axis.