Recent content by Kekeedme

This seems amazing! Thank you so much for this resource

Thank you so much for the reply. So at first, does it mean that we should indeed get this expression by ignoring zero-point energy and equation the classical field energy to that of the single-photon? For the V portion, that is very helpful, and also very unfortunate that they have this...

In the book, they present the solution to the maxwell equation for a wave propagating in 1-D. My only issue is determining where the specific expression for the amplitude of the electric field comes from. Their solution is: Even if we use the classical energy in the field and equate it to the...

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...