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A simple question about tuesday's discussion. When solving for the eigenvector for the second and third eigenstate, I am getting

<math>|s_2\rangle = |s_3\rangle = \begin{bmatrix} 0
1
1 \end{bmatrix}</math>

but the answers we arrived at in discussion were

<math>|s_2\rangle = \begin{bmatrix} 0
1
0 \end{bmatrix}</math> and <math>|s_3\rangle = \begin{bmatrix} 0
0
1 \end{bmatrix}</math>

So what am I missing here? Thanks.

I assume you're talking about the eigenstate of H, because as I remember, off the top of my head, those were the solutions I got for that set. The reason we divided the eigenstate into two was because even though (0,1,1) were technically the values that would be allowable, we could also have (0,3,1) or (0,1,5) etc etc… the two values of 1 can change independent of each other, so we have to represent them independently. this way, we can describe all possible eigenvalues of H as scalar products of (0,1,0) and (0,0,1), which cannot be done if you have only (0,1,1). (this is kind of a round about way to answer this – but that's the idea, I believe, without technical terms/proofs :| )

I think also that the eigenvectors must be orthonormal. So, in this problem, the eigenvalue 2 has a degeneracy of 2, so if you pick (0,1,1) to be one of the eigenvectors for the eigenvalue 2, you have to pick a second eigenvector which is orthogonal to (0,1,1), and still satisfies the conditions for that eigenvalue. Does that make sense? I hope so, I am tired.

Since the conditions of the problem allow you to choose your orthonormal basis vectors for this eigenvalue to be (0,0,1) and (0,1,0), we can just choose that to make things easy.

Spherical coord. are helpful when dealing with mostly spherical volumes – but if we were only interested in volumes… for the problems we're dealing with, would it really change much to approximate the volume as a small box, not a sphere? We've not really dealt with the different spin and momentum volumes l,m, etc so obviously not there.. but if we weren't concerned with r, couldn't we still keep the coordinates rectangular?

I think why we use spherical coordinates depends on the potential profile. For example, in hydrogen atom the potential is spherical-symmetric, so it is convenient to use spherical coordinates. Otherwise we may use rectangular coordinates, for example 3D infinite square well.

On pg 149-150 Griffiths develops and discusses Bohr's allowed energy formula and says that he obtained it “by a serendipitous mixture of inapplicable classical physics and premature quantum theory.” I don't understand how classical arguments could lead to such an anti intuitive result (classically). Or is this how it was proved that there are discrete energy levels so the electron would always be in the same phase in its 'orbit,' not interfering with itself and wouldn't emit photons and spiral into the nucleus? (which I guess would be a fine classical argument with premature quantum theory that still describes the electron as a wave)

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