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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?
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.”
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…
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?