Griffiths Quantum Mechanics: Solving Angular Momentum with Commutators

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SUMMARY

The discussion focuses on Griffiths' treatment of angular momentum in quantum mechanics, specifically in section 4.3, which utilizes commutators. The key equation discussed is [L_x, L_y] = [yp_z, zp_x] + [zp_y, xp_z], where the middle terms are eliminated due to their commutation. The user expresses confusion over Griffiths' ability to factor out yp_x and xp_y in the subsequent step, but later realizes that commuting momentum and position operators acting on different variables is permissible, clarifying their understanding of the commutation relations.

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  • Understanding of quantum mechanics fundamentals
  • Familiarity with commutator notation in quantum mechanics
  • Knowledge of angular momentum operators in quantum mechanics
  • Basic grasp of operator algebra and commutation relations
NEXT STEPS
  • Study Griffiths' Quantum Mechanics, specifically section 4.3 on angular momentum
  • Learn about commutation relations between position and momentum operators
  • Explore the implications of angular momentum in quantum systems
  • Review operator algebra techniques in quantum mechanics
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Students of quantum mechanics, physicists working with angular momentum, and anyone seeking to deepen their understanding of commutators in quantum theory.

Bobbo Snap
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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?
 
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Oh, never mind. I think I can commute momentum and position operators acting on different variables. So xp_z = p_zx.
 
That's right.
 

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