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classes:2009:fall:phys4101.001:q_a_1111

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classes:2009:fall:phys4101.001:q_a_1111 [2009/11/12 12:18] – x500_sohnx020classes: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]]**\\ **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1109]]**\\ **Q&A for the previous lecture: [[Q_A_1109]]**\\
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 === liux0756 11/11/2009 4:00pm === === 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. 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.
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 ====Dagny==== ====Dagny====
-Basic question: How do we go (what's the process) from dxdydz to <math>r^2sin^2\theta</math><math>dr</math><math>d\theta</math><math>d\phi</math>+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=== ===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. 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.
  
-====Blackbox 11/12/2009 11:50pm====+===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, 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 previous lecture: [[Q_A_1109]]**\\
 **Q&A for the next lecture: [[Q_A_1113]]** **Q&A for the next lecture: [[Q_A_1113]]**
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classes/2009/fall/phys4101.001/q_a_1111.1258049928.txt.gz · Last modified: 2009/11/12 12:18 by x500_sohnx020

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