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classes:2009:fall:phys4101.001:q_a_1118 [2009/11/17 22:18] – gebrehiwetclasses:2009:fall:phys4101.001:q_a_1118 [2009/11/30 09:00] (current) – x500_bast0052
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 ===Esquire (Age of No Ideas)=== ===Esquire (Age of No Ideas)===
 I have no idea what an eigenspinor physically represents. I have no idea what an eigenspinor physically represents.
 +
 +==Devlin==
 +Neither do I.
 +
 ===Green Suit 11/17=== ===Green Suit 11/17===
 This is what I found on Wikipedia: This is what I found on Wikipedia:
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 I looked at how Griffths calculated the probability when given a state, but how do you figure out the coefficients of psi+ and psi-? Without this, I am at a lost of calculated probability for certain spin states.   I looked at how Griffths calculated the probability when given a state, but how do you figure out the coefficients of psi+ and psi-? Without this, I am at a lost of calculated probability for certain spin states.  
  
 +===David Hilbert's Hat 11/18 2pm===
 +Do you mean ψ or χ?
 +
 +===David Hilbert's Hat 11/18 2pm===
 +As far as I can tell, a and b are always given or just some constants, so you can use [4.139] as well as the corresponding eigenvalue for whatever operator you're looking at. It is done in example 4.2 in the book for <math> S_{z} </math> and <math> S_{x} </math>.
 +
 +====Schrodinger's Dog 11/18====
 +χ
 +
 +====Captain America 11/18 10:26 ====
 +I'm looking for a better way to conceptualize what we are doing in class, can anyone help me understand how a particle can have 1/2 of a spin?  I know "spin" is a more or less a made up term when it comes to electrons or gravitons and the like, but why don't we just multiply all the spins by 2 and make it easier to understand classically (which is the reason for calling it spin in the first place, if I'm not mistaken)?  This is confusing because how can something have half a spin?  I guess it's all relative because gravitons have spin of 2, which also makes no sense. Is there a specific reason for there being 1/2 spins instead of just calling everything by twice that?
 +
 +===David Hilbert's Hat 11/18 2pm===
 +If you look at equation 4.135, it seems to say that for those operators the observable values must be given in terms of s and <math> \hbar </math>. So you can't really arbitrarily change the eigenvalues/observables for spin because they must obey this relation with the respective operators. 
 +
 +Your question seems to be similar to one thing I was thinking of, which is how are these operators and observables grounded to reality? If you're given some operator, which you apply in a lab setting by taking a measurement, the observable value is the eigenvalue - so I guess you can't just set all spins to be integers instead of half-integers, because integer values might not properly satisfy these eigenvalue equations or you leave out observable quantities but excluding half-integers. 
 +
 +====Zeno 11/18 10:45====
 +
 +This is a question relative to the discussion and problem 2.38, back when we were working with the infinite square well. If you recall, that was the problem where an infinite square well doubled in size, and the probability of measuring the same energy as before the expansion was 1/2 (I think). That problem was a mathematical curiosity, but our discussion problem last week with the cubical well approximated the FCC system we were describing fairly accurately. My questions regarding this concept are:
 +
 +1. If you expanded the dimensions, or deformed them like in the extra-credit problem, would you get a similar result in terms of having a non-unity probability of measuring the same energy? 
 +
 +2. Would the energy change between the expectation value of the first and second measurement be the energy required to deform the infinite cubic well (i.e., the FCC structure)?
 +
 +3. Does the validity of our approximation of the structure as an infinite cubic well break down if it's not cubic or if it's deformed much? I guess the extra credit problem proved that is does deviate to some extent. 
 +
 +I may have to work this out if I have some extra time. I was just wondering about it because we can deform these structures (at least slightly) and the 1-D infinite square well expansion led to an anti-intuitive result. 
 +
 +=== Yuichi ===
 +I like this question.
 +
 +
 +====Pluto 4ever 11/18 5:30PM====
 +Does the value of the quantum number s depend on m, or is it completely independent of the other terms and deals with purely spin alone?
 +
 +===The Doctor 11/19 12:02AM===
 +Equation 4.137 should solve the question.
 +
 +I looks like s is a fixed value for a particular particle while m can change depending on the state of that particle.  From what I can tell m is actually dependent on s which is independent of the other numbers.  It just has to do with the properties of the particle.
  
 +==David Hilbert's hat 11/20 11am==
 +Usually I think of s as being like a constant for each particle; all electrons have spins of 1/2, photons have spin of 1, and so on for every particle. Then when you measure it, spin can be up or down, so the <math> m_{s}</math> value can be plus or minus s, or any integer value in between (like <math> m_{l} </math> is for l). So the s value itself tells you something like the magnitude and the m value tells you which direction it's pointing in. 
  
  
classes/2009/fall/phys4101.001/q_a_1118.1258517894.txt.gz · Last modified: 2009/11/17 22:18 by gebrehiwet

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