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In the proof that [L^2,Lx] = 0 on page 161, how does Griffiths get from the first line to the second line?

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].

Where does the minus sign come from when you operate Sz on <math>\chi-</math> in eq. 4.144 (pg 174)?

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>

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.

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>.

I'm having a little trouble understanding Larmor precession (Example 4.3, pg 179). Basically a particle with spin 1/2 is at rest in a uniform magnetic field pointing in the k-hat direction. Griffiths shows that the energy of the particle is lowest when the dipole moment is parallel to the field lines, which makes intuitive sense. Where I don't follow is when he just chooses–seemingly arbitrarily–the values of a and b, in terms of an angle alpha, and assumes that the angle is constant in time… finally arriving at the conclusion that the dipole “rotates” about the z-axis with a frequency equal to the product of the gyromagnetic ratio and the magnitude of the field. -How can Griffiths assume that the angle alpha is constant in time? Wouldn't it collapse to zero, the lowest energy state? –or do we assume that the particle maintains a constant angular momentum with no losses to friction and alpha is the angle at which the rotational and magnetic torques balance? That sounds pretty reasonable…I may have just answered my own question.

I find myself answering my own question a lot, while I write on here to.

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?

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.

I knew there was something obvious that I was overlooking. Thanks!

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.

Also http://hyperphysics.phy-astr.gsu.edu/Hbase/spin.html seems to have a few extra details about the silver atoms.

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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