Why are Hydrogen Orbitals Asymmetrical Despite a Spherically Symmetric Problem?

[LeonhardEuler]

I was thinking back to my QM class last semester. My teacher derived what the orbitals of the hydrogen atom should look like. I just realized something a little strange about the result that it did not occur to me at the time to ask: How is it that orbitals resulted from a spherically symetric problem which were not themselves sherically symetric? Unfortunately I can not find the whole derivation, but from what I remember, lobes appeared that were either aligned with, or perpendicular to, the polar axis that was used for spherical coordinates. Since the choice of this axis was arbitrary, how did it end up affecting the result? Or maybe it didn't and I'm missing something, or remembering something incorrectly?

 

[Hurkyl]

Consider solutions to the equation x + y = 5.

This equation is symmetric in interchanging x and y.
What can you say about the solutions?

 

[LeonhardEuler]

That for any solution (a,b), there is a solution (b,a). But in the case of the hydrogen atom, you would expect that symetry would imply that the value of $\Psi$ would be the same at all points at the same distance, which isn't the case. I'm sorry, I don't see the connection yet.

 

[ZapperZ]

I was thinking back to my QM class last semester. My teacher derived what the orbitals of the hydrogen atom should look like. I just realized something a little strange about the result that it did not occur to me at the time to ask: How is it that orbitals resulted from a spherically symetric problem which were not themselves sherically symetric?

But is this really that strange? Consider the classical central force problem. You can have, as one solution, an elliptical orbit with the source of the central force at one of the focus of the ellipse. This is not symmetric about that central force.

Zz.

 

[LeonhardEuler]

True, but that is only for asymetric initial conditions. Is this also the case with the orbitals: that initial conditions determine the position of the lobes?

 

[ZapperZ]

True, but that is only for asymetric initial conditions. Is this also the case with the orbitals: that initial conditions determine the position of the lobes?

Er..no. It's a POSSIBLE SOLUTION. It means that just because the problem is symmetric, it doesn't mean ALL possible "stationary" solutions have to be.

Zz.

 

[LeonhardEuler]

Then is it the case that there are actually infinitely many possible solutions with the lobes pointing in every possible direction?

 

[ZapperZ]

Then is it the case that there are actually infinitely many possible solutions with the lobes pointing in every possible direction?

Why not? Let the principle quantum number be n=100. How many l,m solutions can support that in the orbital part of the wavefuntion?

Zz.

 

[LeonhardEuler]

A lot, although I forget how to calculate exactly how many. But suppose n=3. You would still not expect the choice of the polar axis to influence the solution for any particular value of n because it is just a mathematical abstraction chosen arbitrarily. Are there possible solutions for n=3 with the lobes pointing in any direction? (The reason I am restricting it to n=3 is because from what I remember, an increase in n means an increase in energy, so an n=100 orbital could not be the same as an n=3 orbital, and you would expect that for every energy, the solutions should be symetric. Am I remembering that correctly? Because if I'm not then I see how my question is answered.)

 

[reilly]

Here the invariance of the potential, and thus of the Hamiltonian under rotations comes from the spherical symmetry of the potential and the KE. So angular momentum is conserved (true with spin, but let's save that for later) In fact, in spherical coordinates, the KE (Laplacian) has a term proportional to L**2, where l is orbital angular momentum. See any book on basic QM and or classical mechanics for the details.

The idea is, sort of like: a normal 3D vector maps into another 3D vector under rotations, and keeps its norm. Angular mometum states have the property that the value of L**2 or l*(1+1) is preserved under rotations, and for a given l, the new eigenstate is a superposition of all the Lz states, all of which are degenerate. (Technically, the states for a given l form a finite representation of the rotation group. Books on angular momentum, and, indeed, group theory talk a lot about these matters.)

Regards,
Reilly Atkinson

 

[DaTario]

Quantum Physics of Atoms, Molecules, Solids, Nuclei and Partilcles,
Eisberg Resnik

after figure 7.9.
You are right, in order to observe this z direction explicitly, one would have to provide an experimental set up which unavoidably disturbs the simetry of the potential (Hamiltonian). The degeneracy in energy between spherical and not spherical eigenstates helps maintaining these "oriented" states of the free atom hidden.

Best Regards

DaTario