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--- classes:2009:fall:phys4101.001:q_a_1104 [2009/11/03 12:51] – x500_hruby028
+++ classes:2009:fall:phys4101.001:q_a_1104 [2009/12/19 16:56] (current) – x500_sohnx020
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-===== Nov 04 (Wed)  =====
+===== Nov 04 (Wed) Laplacian in spherical coordinate, Legendre =====
 **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1102]]**\\
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 **Main class wiki page: ** [[home]]
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 ====Ekrpat 1144 12:50pm====
 A simple question about tuesday's discussion.  When solving for the eigenvector for the second and third eigenstate, I am getting
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 So what am I missing here?  Thanks.
+===Spherical Chicken ===
+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 :|  )
+===Daniel Faraday 11/3 8:30pm===
+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 Chicken ====
+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?
+=== liux0756 ===
+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.
+==== Zeno 11/4 10AM ====
+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)
+====Captain America 11/4 10:37====
+When dealing with the Hydrogen atom, we assume that the proton is a potential well for the electron.  We take electron to be a wave function and it is bound by a well, which is the proton.  Why can we take the well to be the proton?  If they are both attracting charges, should the equation work the same if the proton orbited the electron?  Would the proton then have a wave function itself, and the electron be a potential well?
+I'm also confused as to why we can call the proton a well, instead of calling it a wave function well, since in quantum mechanics all particles can be described as wave functions.  Anyone have any intuitive explanations for this?
+===prest121 11/4 12:30 ===
+When dealing with the hydrogen (or single-electron) atom, we make the assumption of a stationary nucleus.  This isn't entirely accurate, because the electron-proton system actually orbits about its center of mass.  However, since the proton is many orders of magnitude more massive than that of the electron, the center of mass of the system is very, very close to the proton.  So we take the proton to be the center of mass.
+I think this is what leads to us treating the proton as a 1/r potential well and the electron as a wave function/particle.  I think it is also related to the fact that we are primarily interested in the behavior of electrons in atoms (at the present time), not nucleons.
+===Pluto 4ever 11/5 11:11am===
+I also agree with this. We generally want to focus on the electron as opposed to the proton which we are assuming is stationary. That way in this ideal situation it becomes less complex to calculated the properties of the electron to some degree, such as such as probability of position and momentum, as well as energy and spin states.
+===Captain America 11/6 10:13am===
+But then why don't we treat the potential as a wavefunction?  Is it because the electrical charge that creates the potential doesn't behave as a wave?  I would think that the charge of the proton is distributed equally over the entire proton, and that the proton itself behaves as a wave, so the potential should not look like a pure harmonic oscillator, but a wavy harmonic oscillator instead.  Is what we do a simplification or am I over-complicating things?
+===David Hilbert's Hat 11/10 12:20pm===
+I tend to think that the effective wavefunction of a proton is very small relative to the potential it produces - for instance, whatever the effective "wavelength" of the proton's position is, it must be very small compared to how far away the coloumb potential reaches. I think finding a way to calculate these things might be difficult for any given proton (the free particle case is not easy, as we've seen) but intuitively you expect something with a charge on the order of 10^-19C to have much larger E&M properties than a quantum particle that has mass on the order of 10^-27 kg and is set at zero velocity. And very near the proton, when the quantum properties of each particle may come into play, is close to it - and the attraction between a proton and electron is repulsive, so it is unlikely that they are ever close to each other in the quantum sense.
+====Dark Helmet 11/05====
+Although i understand how to get them, what exactly is the physical interpretation of eigenstates and eigenvalues?  That still is confusing me a bit.
+=== Blackbox ===
+In quantum mechanics, operators correspond to observable variables, eigenvectors are also called eigenstates, and the eigenvalues of an operator represent those values of the corresponding variable that have non-zero probability of occurring. In other words, we can say some special wave functions are called eigenstates, and the multiples are called eigenvalues. I hope this helped.

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