Finding Eigenvectors of 2-state system

In summary, the method outlined in Cohen-Tannoudji page 423 allows for the eigenvectors of a 2-state system to be found in a less cumbersome way. However, this step is confusing and requires help from someone more experienced.
  • #1
Kekeedme
5
3
TL;DR Summary
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?
20230412_223040.jpg
 
  • Like
Likes vanhees71
Physics news on Phys.org
  • #2
Kekeedme said:
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.
 
  • Like
Likes vanhees71 and topsquark
  • #3
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?
 
Last edited:
  • #4
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.
 
  • #5
Kekeedme said:
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.
 
  • #6
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
 
  • #7
Kekeedme said:
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.
 
  • Like
Likes Kekeedme
  • #8
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
 
  • Like
Likes PeterDonis and DrClaude

FAQ: Finding Eigenvectors of 2-state system

What is an eigenvector in the context of a 2-state system?

An eigenvector in a 2-state system is a non-zero vector that, when multiplied by a given 2x2 matrix (representing the system), results in a scalar multiple of itself. This scalar is known as the eigenvalue. In essence, the direction of the eigenvector remains unchanged by the transformation represented by the matrix.

How do you find the eigenvalues of a 2x2 matrix?

To find the eigenvalues of a 2x2 matrix, you solve the characteristic equation, which is obtained by setting the determinant of the matrix minus lambda times the identity matrix to zero. Mathematically, for a matrix A, this is written as det(A - λI) = 0. Solving this equation for λ gives the eigenvalues.

What is the significance of eigenvectors and eigenvalues in a 2-state system?

Eigenvectors and eigenvalues are significant in a 2-state system because they provide insight into the system's behavior. Eigenvalues can indicate stability, oscillatory behavior, or exponential growth/decay, while eigenvectors indicate the directions along which these behaviors occur. They are fundamental in various applications including quantum mechanics, vibration analysis, and stability analysis.

Can a 2x2 matrix have complex eigenvalues and eigenvectors?

Yes, a 2x2 matrix can have complex eigenvalues and eigenvectors, especially if the matrix has no real roots for its characteristic equation. Complex eigenvalues often occur in systems with oscillatory behavior, and the corresponding eigenvectors will also have complex components.

How do you normalize an eigenvector in a 2-state system?

To normalize an eigenvector in a 2-state system, you divide the eigenvector by its magnitude. The magnitude is calculated as the square root of the sum of the squares of its components. For an eigenvector v = [v1, v2], the normalized eigenvector v' is given by v' = v / ||v||, where ||v|| = sqrt(v1^2 + v2^2).

Similar threads

Replies
2
Views
1K
Replies
24
Views
2K
Replies
5
Views
2K
Replies
14
Views
2K
Replies
9
Views
1K
Replies
29
Views
3K
Back
Top