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classes:2009:fall:phys4101.001:q_a_1111

Nov 11 (Wed) Radial Wave Function (finite well, hydrogen)

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spillane

Could someone help clarify what the diff. is between j0 and n0 as discussed in todays lecture on pg 143 ?

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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classes/2009/fall/phys4101.001/q_a_1111.txt · Last modified: 2009/11/16 22:20 by yk