Recent content by Kekeedme

Oh, I got it! I should divide through by ##\sin{\theta}## This will allow me to get ##-\tan{\frac{\theta}{2}}## as a factor of ##a## and then multiply through by ##\cos{\frac{\theta}{2}}## Thank you Peter and Dr Claude

Hello Dr Claude, I did try that. But I don't seem to see how to use them to go from (20) from CCT to (21) or even from what I wrote above. The double angle formulas involve ##sqrt##, which are not present in the expressions, or I can't make them appear

Then multiplying through by ##\exp{\frac{i\phi}{2}}## yields: $$\left(\cos{\theta}-1\right)\exp{\frac{i\phi}{2}} a - \left(\sin{\theta}\exp{\frac{-i\phi}{2}}\right)b=0$$ Which is a bit closer to the result, but not it.

Hello Peter Donis, thank you for your response. I did start trying to play with double and half angle trig identities, but you are right that I did not factor ##\frac{1}{\cos{\theta}}## first. When I do, I get: $$(\cos({\theta}) -1)a - (\sin({\theta})\exp{-i\phi})b = 0$$ From there, I have tried...

In Cohen-Tannoudji page 423, they try to teach a method that allows to find the eigenvectors of a 2-state system in a less cumbersome way. I understand the steps, up to the part where they go from equation (20) to (21). I understand that (20) it automatically leads to (21). Can someone please...