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--- classes:2009:fall:phys4101.001:q_a_1125 [2009/11/25 00:28] – x500_szutz003
+++ classes:2009:fall:phys4101.001:q_a_1125 [2009/11/30 08:56] (current) – x500_bast0052
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 **Main class wiki page: ** [[home]]
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 ====John Galt  11/23  5:35 PM====
 Does anyone see anything useful coming from taking the time derivative of A(t) and B(t) as described in class and setting it equal to zero? In a magnetic field, will there ever be a time period over which the probability of finding a particle spin up or spin down will not change at all?
+===Ralph 11/25 10:40 AM ===
+Well one case I can think of is when the wavefunction is observed repeatedly in immediate succession. Does this change if the particle is in a magnetic field?
 ==== Mercury 11/23/2009 10:49 pm ====
 I'm really struggling with problem 32(a) on the homework. Anyone have any helpful suggestions to get me started?
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 For the second question, I think you can think of it in the same manner as Griffiths discusses on pages 4-5 of the book with respect to repeated measurements.  Since the spin is part of the total wavefunction (which you may remember from 2601, otherwise I think we're discussing it next chapter), I think the idea is the same.  The wavefunction collapses upon measurement, so a repeated immediate measurement would provide the same value, since the wavefunction doesn't have time to "spread out."
+=== Yuichi ===
+Just to make sure we are not confused about a(t).  It's related to probability, and it's NOT.  |a(t)|^2 is the probability.  Since some of the time dependence of a(t) is in the complex phase, it will disappear when the probability is calculated.  //i.e.// even a(t) is time dependent, the probability may be constant.
 ====Pluto 4ever 11/24 4:13PM====
 I'm still a little confused about today's discussion problem. Do we only need to concern ourselves with the m = 0 state for when we are dealing with particles of spin 1, such as photons? Or does <math>m = {\pm}1</math> also come into effect when solving for the S matrices?
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 By plugging in all three values of m we get three 3x1 (or 1x3?) matrices which collectively give us the 3x3 s^2 matrix.
 Obviously this isn't step by step, but I believe it's correct...
+===Blackbox 11/25 10am ===
+Yeah, I agree with you and we get three 3by1 matrices not 1by3.
+===David Hilbert's Hat 11/25 11am===
+You have to solve for three different values of χ (they look like vectors in 3D space) that are linearly independent to give you a basis for each possible spin state, just like how Griffiths did a χ+ and a χ- case, you need to do a χ+, χ-, and a χ0 case. Once you've solved for all of them, since they're an orthonormal basis, you can put them all together for a 3x3 matrix that has the information about all of the states in it.
+The easiest way to do it is by taking what you know, the raising and lower operators acting on any spin state to raise or lower it, and having the 3x3 be entirely made up of variables - the a, b, c, d, e... stuff (you need 9 for this problem). Although this is a rather crude way of doing it because you end up taking one equation, using it to solve for 3 variables, and everything else is multiplied by zero. The easier way is to let any generic matrix A be sandwiched in between two states, like < s m' | A | s m >. Since you have the eigenvalues of A for any operator, you can pull those out and figure out each component by taking < s m' | s m > similar to what Griffiths did on p. 120, equation [3.81]. It seems more elegant to do it that way, but on homework or a test I'm just going to use the first method I think.
 ====Hydra  11/24 6pm====
-What does Griffiths mean when he says Lx Ly and Lz are incompatible observables?  And is he implying that L^2 is compatible with Lx Ly and Lz?
+What does Griffiths mean when he says Lx, Ly and Lz are incompatible observables?  And is he implying that L^2 is compatible with Lx Ly and Lz?
 Does it just mean that if you know Lx with more certainty, you know Ly and Lz with less certainty?
+=== Spherical Chicken stardate 313101.3 ===
+L^2 is commutable with Lx, Ly and Lz  because L^2 corresponds to the radius, where as Lx... corresponds to position of the actual vector.  So we're not going to change the radius, but when observing Lx we change  the position of Ly and Lz right?
+=== Andromeda 11/25 9:12Am ===
+being incompatible means that they do not commute and for every pair of operators that do not commute and so are incompatible there is an uncertainty relation.
+=== Ralph 11/25 10:35 am ===
+Also, remember if the commutator of two operators is nonzero that their eigenstates do not share the same basis.
+===John Galt 11/28 1:15 PM===
+What you said is mainly correct, but remember knowing Lx with more certainty means you know (Lz + Ly) with less certainty, not so much each one indiviually.
+====Andromeda 11/25 9:17AM====
+I have two general questions: first do we know if the date of the 4th test is going to change or not? and second I am still getting used to the new notation we are using. the concept isnt hard but the notation is still not like a second language to me. I wonder how man other students are still in this stage as well?
+===Zeno 11/25 10:20am===
+I agree with you, Andromeda, about the notation. It's going to take some getting used to. The concepts and the general ideas are all fairly simple and logical, it's just translating what the notation is saying or taking what you want to do and translating it into the notation to begin solving a problem that's difficult. I don't know about the exam. I'm guessing we'll be told today in lecture what it'll cover and when it'll be.
+===David Hilbert's Hat 11/25 12:30pm===
+The notation is a little tricky to learn but I've found that if anything looks simple and the notation is screwing you up then the easiest thing to do is refer back to chapter 3 and see if there is a simple one or two sentence that describes what you're looking for. Today in lecture Yuichi said the test will be on the 11th.
+===Pluto 4ever 11/25 6:23PM===
+I also had this problem as well, but,just as David Hilbert's Hat said, it is much easier to refer back to notation that you already know and apply it to the new material. Although, it does take some thinking and time to adjust to the new notation, and I still am, I find that after I read the sections a few times and start working out the problems it will eventually come to me.
+===Dark Helmet 11/28 10:12===
+I found that writing all the different notation out with what it means, kinda like making an equation sheet, helps cement it into my brain.  That and looking back at my calc 3 book for a refresher on linear algebra!
+===Devlin===
+I still find myself moving slowly because of the new notation as well.  I think it'll just take practice.
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