Griffiths Quantum Mechanics: Solving Angular Momentum with Commutators

  • Thread starter Thread starter Bobbo Snap
  • Start date Start date
  • Tags Tags
    Commutators
Bobbo Snap
Messages
29
Reaction score
0
I'm having a lot of trouble following Griffith's quantum mechanics text. I'm in section 4.3 which discusses angular momentum using commutators. The text proceeds as follows:
[L_x, L_y] = [yp_z - zp_y, zp_x - xp_z]\\<br /> =[yp_z, zp_x] - [yp_z, xp_z] - [zp_y, zp_x] + [zp_y, xp_z]\\<br /> =[yp_z, zp_x] + [zp_y, xp_z] \qquad (1)
Ok, I follow the previous, the last step drops the two middle terms because they do commute. It's the next step I don't get:
[L_x, L_y] = yp_x[p_z, z] + xp_y[z, p_z]
How is Griffiths able to factor out the yp_x \text{ and } xp_y? When I expand eqn (1), I get:
[L_x, L_y] = yp_zzp_x-zp_xyp_z + zp_yxp_z - xp_zzp_y
I don't see how he can factor those out without commuting position operators with momentum operators which you can't do, right?
 
Physics news on Phys.org
Oh, never mind. I think I can commute momentum and position operators acting on different variables. So xp_z = p_zx.
 
That's right.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top