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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

but the answers we arrived at in discussion were

and

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 :| )

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.

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Q&A for the previous lecture: Nov 2 (Mon) Chap 4, finally!
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