Finding Eigenvectors of 2-state system

[Kekeedme]

TL;DR

I am trying to understand a method for determining the eigenvectors of 2-state system as explained in Cohen-Tannoudji. I am having trouble with a step he seems to have skipped

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 enlighten me about this step please?

 

[PeterDonis]

Can someone please enlighten me about this step please?

Try writing $\tan \theta$ as $\sin \theta / \cos \theta$ so that you can factor out $1 / \cos \theta$ from both terms on the LHS of (20) (and thus you can eliminate that factor since the RHS of (20) is zero) and then look at the standard double angle and half angle formulas for trig functions.

 

[Kekeedme]

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 playing with the trig identities but I can't seem to see what I am missing. Do you perhaps see what I am missing, please?

 

[Kekeedme]

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.

 

[DrClaude]

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.

Notice that the equation you have involves $\cos \theta$ and $\sin \theta$, while CCT gives it in terms of $\cos \theta/2$ and $\sin \theta/2$, so you should think of using half-angle formulas.

 

[Kekeedme]

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

 

[DrClaude]

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

In the direction you want, they will bring about squares, not square roots. There is a way to remove the square afterwards.

 

[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