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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?
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?
I'm really struggling with problem 32(a) on the homework. Anyone have any helpful suggestions to get me started?
You may want to find out the coefficient of kai first.
It appeared to me that this was just a plug and chug problem. On page 175, Griffiths gives the explicit probabilities of getting each value of spin in terms of the elements of the wavefunction.
Is A(t), as we discussed in class, the probability of finding the spin up or down concerning the measurement of a single particle rather than just a random particle out of a system of many? If so, what about repeated measurements (does the measurement process affect the spin)?
In response to your first question, I believe it's for the single particle. For a many-particle system, we would have a completely different wavefunction to describe the entire system. Unless you're referring to many independent single particle systems, in which case it should be the same as the single particle.
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.”
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.
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?
You need to deal with m=0 and <math>m = {\pm}1</math>. This will get you 3×3 matrix operators for spin 1 particles.
If you only use the m=0 case, you wont get the 3×3 matrix you're looking for. remember the formula
S^2|S m> = s(s+1)hbar^2 | s m>
By plugging in all three values of m we get three 3×1 (or 1×3?) matrices which collectively give us the 3×3 s^2 matrix. Obviously this isn't step by step, but I believe it's correct…
Yeah, I agree with you and we get three 3by1 matrices not 1by3.
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 3×3 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 3×3 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.
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?
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?
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.
Also, remember if the commutator of two operators is nonzero that their eigenstates do not share the same basis.
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.
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?
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.
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.
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.
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!