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classes:2009:fall:phys4101.001:q_a_1123 [2009/11/22 22:29] – gebrehiwetclasses:2009:fall:phys4101.001:q_a_1123 [2009/11/28 22:18] (current) – ely
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-===== Nov 23 (Mon)  =====+===== Nov 23 (Mon) Spin operators, probabilities, Larmor precession, S-G experiment =====
 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1120]]**\\ **Q&A for the previous lecture: [[Q_A_1120]]**\\
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 **Main class wiki page: ** [[home]] **Main class wiki page: ** [[home]]
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 ====Daniel Faraday 11/22 6pm==== ====Daniel Faraday 11/22 6pm====
 In the proof that [L^2,Lx] = 0 on page 161, how does Griffiths get from the first line to the second line? In the proof that [L^2,Lx] = 0 on page 161, how does Griffiths get from the first line to the second line?
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 ===Schrodinger's Dog 11/22 7:50pm=== ===Schrodinger's Dog 11/22 7:50pm===
 He uses the commutation relation [AB,C] = A[B,C] + [A,C]B. The first term [Lx^2,Lx], is zero since Lx commutes with itself. Then using the commutation relation I introduced you expands [Ly^2,Lx] and [Lz^2,Lx]. Lets just expand [Ly^2,Lx], just to illustrate how the commutation relation works. In this case, A=Ly, B=Ly, and C=Lx. Using the relation we have [Ly*Ly,Lx] = Ly[Ly,Lx] + [Ly,Lx]Ly. He similar does this with [Lz^2,Lx].  He uses the commutation relation [AB,C] = A[B,C] + [A,C]B. The first term [Lx^2,Lx], is zero since Lx commutes with itself. Then using the commutation relation I introduced you expands [Ly^2,Lx] and [Lz^2,Lx]. Lets just expand [Ly^2,Lx], just to illustrate how the commutation relation works. In this case, A=Ly, B=Ly, and C=Lx. Using the relation we have [Ly*Ly,Lx] = Ly[Ly,Lx] + [Ly,Lx]Ly. He similar does this with [Lz^2,Lx]. 
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 ====Daniel Faraday 11/22 8:30pm==== ====Daniel Faraday 11/22 8:30pm====
 Where does the minus sign come from when you operate Sz on <math>\chi-</math> in eq. 4.144 (pg 174)? Where does the minus sign come from when you operate Sz on <math>\chi-</math> in eq. 4.144 (pg 174)?
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 When you multiply the definition matrices for each value you have: When you multiply the definition matrices for each value you have:
 <math> S_z \chi_- = \frac{\hbar}{2} \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left(\begin{array}{c} 0 \\ 1 \end{array} \right) = \frac{\hbar}{2} \left(\begin{array}{c} 1*0+0*1 \\ 0*0+(-1)*1 \end{array} \right) = \frac{\hbar}{2} \left(\begin{array}{c} 0 \\ -1 \end{array} \right) = \frac{-\hbar}{2} \left(\begin{array}{c} 0 \\ 1 \end{array} \right) = \frac{-\hbar}{2}\chi_-</math> <math> S_z \chi_- = \frac{\hbar}{2} \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left(\begin{array}{c} 0 \\ 1 \end{array} \right) = \frac{\hbar}{2} \left(\begin{array}{c} 1*0+0*1 \\ 0*0+(-1)*1 \end{array} \right) = \frac{\hbar}{2} \left(\begin{array}{c} 0 \\ -1 \end{array} \right) = \frac{-\hbar}{2} \left(\begin{array}{c} 0 \\ 1 \end{array} \right) = \frac{-\hbar}{2}\chi_-</math>
 +
 +=== Daniel Faraday 11/23 8am ===
 +But aren't the 'definition matrices' that you used above derived from eq 4.144 in the first place? I think there's something kind of obvious in eq. 4.144 that I'm not seeing that gives the minus sign.
 +
 +=== Blackbox 11/23/10:30am ===
 +The minus sign comes from "m". Since electrons have spin 1/2, "m" has 1/2 and -1/2. 
 +Therefore <math>S_z\chi_+ = \frac{\hbar}{2}\chi_+ </math>, <math> S_z\chi_- = -\frac{\hbar}{2} \chi_- </math>.
  
 ==== Zeno 11/22 9:30pm ==== ==== Zeno 11/22 9:30pm ====
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 ==Schrodinger's Dog 11/22 10:27pm== ==Schrodinger's Dog 11/22 10:27pm==
 I find myself answering my own question a lot, while I write on here to.  I find myself answering my own question a lot, while I write on here to. 
 +
 +
 +==== Zeno 11/23 10:15am ====
 +This is kind of a tangential question regarding the Stern-Gerlach experiment, but how would one produce an electron beam? And how would the medium through which the beam passes affect the precision? It seems like electrons would be diffracted and the beam would be diffused relatively quickly in air, or even in a decent vacuum chamber that you could reasonably hope to create in a lab. Any thoughts?
 +
 +===chavez 11/23 10:56am===
 +Cathode ray tubes (http://en.wikipedia.org/wiki/Cathode_ray) are pretty standard for creating electron beams. There is plenty of information out there on them if you are curious.
 +
 +== Zeno 11/25 9am ==
 +I knew there was something obvious that I was overlooking. Thanks!
 +
 +===David Hilbert's Hat 11/25 12:30pm===
 +As far as I know they actually cooked silver atoms from a furnace. Wikipedia mentions that you need to have particles with a total electric charge of zero because otherwise they would deflect under the influence of a magnetic field before coming out of the apparatus - http://en.wikipedia.org/wiki/Stern-Gerlach seems to suggest that anything with a total neutral charge and unpaired electrons orbiting it can be used. If I recall correctly there is a pretty lengthy description of it in the 2000 level quantum book. 
 +
 +===David Hilbert's Hat 11/25 1pm===
 +Also http://hyperphysics.phy-astr.gsu.edu/Hbase/spin.html seems to have a few extra details about the silver atoms. 
 +
 +===Dark Helmet 11/28 10:15===
 +I don't know if it is exactly the same thing, but a free-electron laser would be a beam of electrons.  Or is that a laser made from accelerating electrons?  Now i can't remember.  Well, anyway, an electron beam wouldn't be perfectly precise even in a perfect vaccuum due to the uncertainty in the beam width.  Another one of those uncertaintly principle relations
 +
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 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
classes/2009/fall/phys4101.001/q_a_1123.1258950551.txt.gz · Last modified: 2009/11/22 22:29 by gebrehiwet

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