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## Main Question or Discussion Point

When we solve the Schrodinger equation for a central potential, such as the coulomb potential for the hydrogen atom, we get eigenstates which are not, in general, spherically symmetric. So they depend on the choice of coordinates.

If we can get two distinct solutions, just by choosing different axes, how do we know which solution is the correct one?

My prof says that they are the same solution, no matter which way they are oriented, and I disagree with him. If we solve the angular dependence using [itex] \phi [/itex] and [itex]\theta[/itex] in the usual definition, we get solutions (of |Psi|^2) with cylindrical symmetry about the z-axis, but if we use angles [itex]\phi ' [/itex] and [itex]\theta ' [/itex] such that [itex]\theta ' [/itex] is the polar angle from the y-axis and [itex] \phi ' [/itex] is the azimuthal angle around from the z-axis toward the x-axis, we get solutions with cylindrical symmetry around the y-axis. And in general [itex] Y_l^m(\theta, \phi) \neq Y_l^m(\theta ', \phi ') [/itex]

The only solution I can think of is if you can make the eigenstates that you get using one set of coordinates, from linear combinations of the eigenstates you get using any other set of coordinates.

Anyone know the answer? Are they in fact the same solution, and I'm confusing myself by adding orientation to the problem, when it doesn't matter? Or can you make the eigenstates you find from one set of coordinates from the eigenstates from another set? Or is there some other solution that I can't see.

If we can get two distinct solutions, just by choosing different axes, how do we know which solution is the correct one?

My prof says that they are the same solution, no matter which way they are oriented, and I disagree with him. If we solve the angular dependence using [itex] \phi [/itex] and [itex]\theta[/itex] in the usual definition, we get solutions (of |Psi|^2) with cylindrical symmetry about the z-axis, but if we use angles [itex]\phi ' [/itex] and [itex]\theta ' [/itex] such that [itex]\theta ' [/itex] is the polar angle from the y-axis and [itex] \phi ' [/itex] is the azimuthal angle around from the z-axis toward the x-axis, we get solutions with cylindrical symmetry around the y-axis. And in general [itex] Y_l^m(\theta, \phi) \neq Y_l^m(\theta ', \phi ') [/itex]

The only solution I can think of is if you can make the eigenstates that you get using one set of coordinates, from linear combinations of the eigenstates you get using any other set of coordinates.

Anyone know the answer? Are they in fact the same solution, and I'm confusing myself by adding orientation to the problem, when it doesn't matter? Or can you make the eigenstates you find from one set of coordinates from the eigenstates from another set? Or is there some other solution that I can't see.

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