Symmetry of Hamiltonian and eigenstates

[Sum Guy]

Suppose we have an electron in a hydrogen atom that satisfies the time-independent Schrodinger equation:
$$-\frac{\hbar ^{2}}{2m}\nabla ^{2}\psi - \frac{e^{2}}{4\pi \epsilon_{0}r}\psi = E\psi$$

How can it be that the Hamiltonian is spherically-symmetric when the energy eigenstate isn't? I was thinking along the lines of rotations with angular momentum operators but I'm not sure I can come up with a nice explanation. Can someone help me see why this is the case?

 

[A. Neumaier]

It just means that the eigenspaces are invariant under the symmetry group. Since symmetry groups of interest rarely have a nontrivial 1-dimensional representation, this means in practice that most energy eigenstates states are degenerate. One can easily verify this in the hydrogen atom, or in a pair of equal harmonic oscillators.

 

[Strilanc]

Solving a differential equation tends to exponentiate things. Consider the complex circle $2 e^{i \theta}$. What happens when we exponentiate it? Make a parametric plot of $e^{2 e^{i \theta}}$ and you get this:

Which is not circularly symmetric like its input was.

That's not a full explanation, but I hope it gives an intuition for why the spherical symmetry could be lost.

(Actually, because the Hamiltonian is scaled by $i$, exponentiating tends to introduce cycling. But the cycles are in the phases of amplitudes, not in physical space.)

 

[Sum Guy]

Solving a differential equation tends to exponentiate things. Consider the complex circle $2 e^{i \theta}$. What happens when we exponentiate it? Make a parametric plot of $e^{2 e^{i \theta}}$ and you get this:

View attachment 98168

Which is not circularly symmetric like its input was.

That's not a full explanation, but I hope it gives an intuition for why the spherical symmetry could be lost.

(Actually, because the Hamiltonian is scaled by $i$, exponentiating tends to introduce cycling. But the cycles are in the phases of amplitudes, not in physical space.)

Thank you for this - it's a nice thought. Is there any way you could apply this mode of thinking to the situation where $\psi = R(r)Y(\theta, \phi) = R(r)cos(\theta)$ say?

 

[Strilanc]

Thank you for this - it's a nice thought. Is there any way you could apply this mode of thinking to the situation where $\psi = R(r)Y(\theta, \phi) = R(r)cos(\theta)$ say?

I actually don't know enough about it to do that, unfortunately. Actually I'm half-expecting the next poster to complain that it's a really misleading way to think about a Hamiltonian, so we'll see.

 

[secur]

How can it be that the Hamiltonian is spherically-symmetric when the energy eigenstate isn't?

To see this intuitively consider Newtonian orbital dynamics. The hamiltonian is very similar, with a central 1/r^2 force and a kinetic energy term, mv^2/2 - or, if written as in Schroedinger's, p^2/2m. So ask the same question about planetary orbits: the Hamiltonian is spherically-symmetric but orbits don't have to be, why?

The reason is, the kinetic energy term is formally (as written) symmetric but not when we plug in actual values for a planet's initial position and velocity. If those initial conditions are just right we get a spherically-symmetric orbit (circle) but much more likely it will be an ellipse (or hyperbola, but that's irrelevant here). True the angular momentum is always symmetric, meaning: at any angle it's the same. But position and velocity (or, momentum) are not.

The hydrogen atom case is, roughly, the same idea, but it's more difficult because the electron's a wave function not a solid body like the Earth. You get complicated standing wave patterns instead of simple ellipses. But it's still true that the position and momentum (NOT angular) operators will give non-spherically-symmetric values in a given eigenstate (except s-orbitals).

I may be wrong on some detail or terminology, but I hope this intuitive picture helps

 

[Strilanc]

The reason is, the kinetic energy term is formally (as written) symmetric but not when we plug in actual values for a planet's initial position and velocity. If those initial conditions are just right we get a spherically-symmetric orbit (circle) but much more likely it will be [...]

Nit: a circular orbit also isn't spherically-symmetric. Good example.

 

[secur]

Good catch! But actually I did cover that objection:

I may be wrong