0
$\begingroup$

On the 5th slide of http://web.ift.uib.no/~lipniack/phys201_v08/angularmomentum.pdf, they say that: $$L_y=\frac{i}{2}(L_+-L-)$$ And this is not the only place I have seen this (http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html for instance use this implicitly). However, if I start with $L_{\pm}=L_x\pm i L_y$ and then do $L_+-L_-$ I get: $$L_+-L_-=2iL_y$$ $$L_y=\frac{1}{2i}(L_+-L-)$$ So where do they get the formula:

$$L_y=\frac{i}{2}(L_+-L-)$$ from?

$\endgroup$

closed as off-topic by ACuriousMind, HDE 226868, Kyle Kanos, Qmechanic Aug 16 '15 at 15:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, HDE 226868, Kyle Kanos, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

You're right, they are mistaken.
You can verify this by noticing that (Using their $L_y$): $$L_+=L_x+iL_y=\frac{1}{2}(L_++L_-)+i\frac{i}{2}(L_+-L_-)=L_-$$ Which is obviously wrong. (Do note however that their matrices are correct)

$\endgroup$
  • $\begingroup$ How can their matrices be correct since if you do: $$L_y|l=1,m_l=1>=\frac{1}{2i}(L_+-L_-)|1,1>$$ $$=\frac{-\hbar}{\sqrt{2}i}|1,0>$$ and not $$=\frac{\hbar}{\sqrt{2}i}|1,0>$$ as they indicate. $\endgroup$ – Quantum spaghettification Aug 16 '15 at 9:23
  • $\begingroup$ See easyspin.org/documentation/spinoperators.html for the matrices $\endgroup$ – Omry Aug 16 '15 at 9:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.