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--- classes:2009:fall:phys4101.001:q_a_1111 [2009/11/09 17:11] – x500_spil0049
+++ classes:2009:fall:phys4101.001:q_a_1111 [2009/11/16 22:20] (current) – yk
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-===== Nov 11 (Wed)  =====
+===== Nov 11 (Wed) Radial Wave Function (finite well, hydrogen) =====
 **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1109]]**\\
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 I know it was a ? in class, but iam not clear on what the answer was.
+== Mercury 11/10/2009 7:30pm ==
+j0 are the Bessel functions and n0 are the Neumann functions. We only deal with Bessel functions in physics, because the Neumann functions blow up at the origin and therefore have no physical significance. It's just like with the Legendre polynomials that we discussed in class--Legendre functions are second order equations, meaning that there are two solutions. The first kind are the ones we use P(z), but there is a second kind Q(z) that we do not discuss because they diverge at z=1.
+==== Schrodinger's Dog 11/10/2009 8:00pm ====
+Why is it so important to find simultaneous eigenstates, when we have compatible operators. Griffths does with the angular momentum quantities, but I don't see why it is so important. What does it mean to be simultaneous? Does it just mean that you can measure both quantities at the same time or are they other implications that go along with this?
+Thanks!
+=== liux0756 11/11/2009 4:00pm ===
+Yes, simultaneous eigenstates are important. Generally speaking if two operators do not have simultaneous eigenstates, then they can't be measured at the same time. For example, x and p do not have simultaneous eigenstates, so we can't get the exact value of x and p at the same time: <math>\delta x \delta p \ge \hbar/2</math> But angular momentum <math>l^2</math> and <math>l_z</math> have common eigenstates and in those states both of them can have exact measurement.
+====Dagny====
+Basic question: How do we go (what's the process) from dxdydz to <math>r^2sin\theta</math><math>dr</math><math>d\theta</math><math>d\phi</math>
+===Blackbox 11/12/2009 12:15pm===
+The process is simple, dxdydz=dv for the cartesian coordinate system. In similar way, dv for the spherical coordinate system is <math>dr*rd\theta*rsin\theta d\phi</math>. It may be easily understood if you look at the sherical coordinate system plot from any EM refernece book.
+===Daniel Faraday 11/12 10:30pm===
+Wait a second: the sin term isn't squared in dV, is it? I thought dV was just
+<math>r^2sin\theta dr d\theta d\phi</math>  ??
+===Dagny===
+Correct!
+====Blackbox 11/12/2009 11:50am====
+I just forgot,, Could anyone tell me about the unit of <math> k=\frac{sqrt{2mE}}{\hbar} </math>? Thanks,
+===liux0756===
+The unit of k is [length]^-1. Because momentum <math>p=\hbar k</math>, angular momentum <math> L= n \hbar = momentum*length </math>, so k is [length]^-1.
+====Dagny====
+What is the very detailed process for the transformation of the gradient operator from cartesian to spherical coordinates? Just one operator term is fine, like del/delx. (Because I'm just interested in the process and also how we come about this process.) I've found a couple of website examples on how to do it, but nothing that provides a clear explanation.
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 **Q&A for the previous lecture: [[Q_A_1109]]**\\
 **Q&A for the next lecture: [[Q_A_1113]]**

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