## hosein Group Title please solve this : find eigen value & eigen vector of opration Sy 2 years ago 2 years ago

1. hosein Group Title

$S _{y}$ is y component of s & s is spin.solve it in two dimension

2. hosein Group Title

i know what do that. but final answer has not matched with text

3. opiesche Group Title

I was going to just take a wild guess, but really, I have no idea :P

4. JamesJ Group Title

First, can you write out $$S_y$$ explicitly?

5. hosein Group Title

your goal is operation of Sy ? in matrix definition Sy is two dimensional matrix do you know or write it?

6. JamesJ Group Title

Whatever it is, write it out explicitly.

7. hosein Group Title

Sy= $\left[\begin{matrix}h'/2 & -ih'/2 \\ih'/2 & h'/2\end{matrix}\right]$ h'=h/2pi

8. hosein Group Title

is operation

9. JamesJ Group Title

Ok, so you need to find first the eigenvalues of the matrix. I.e., solve the equation, $\det(S_y - \lambda I_2) = 0$ for $$\lambda$$.

10. hosein Group Title

yes know this but final answer don't match with text

11. JamesJ Group Title

What did you find for eigenvalues?

12. JamesJ Group Title

I have from that definition of Sy, that the eigenvalues are 0 and h'

13. hosein Group Title

eigen value is match eigenvector... i didn't find

14. JamesJ Group Title

I'm having a very hard time understanding what you're saying. But first do you agree with these eigenvalues. If so, the next step is to find the null space corresponding to each eigenvalue of $S_y - \lambda I_2$

15. JamesJ Group Title

For example, for $$\lambda = 0$$, the null space is one dimensional and has as a basis vector v = (1,i)^t

16. JamesJ Group Title

that's an eigenvector corresponding to eigenvalue of zero. Now find the other null space and the eigenvector corresponding to the eigenvalue of $$\lambda = h'$$.

17. hosein Group Title

18. hosein Group Title

eigenvalue is +h'/2&-h'/2

19. JamesJ Group Title

One of the eigenvalues is zero, because the characteristic equation is $\lambda^2 - h' \lambda = 0$

20. hosein Group Title

no eigenvalue is +h'/2&-h'/2. see this: 4.29 part a

21. hosein Group Title

this book is solution of Introduction to quantum mechanics by david.-j griffiths

22. JamesJ Group Title

You gave the wrong $$S_y$$. Notice the matrix in the book has zeros along the diagonal, unlike the matrix you typed above.

23. hosein Group Title

24. JamesJ Group Title

25. hosein Group Title

so...

26. JamesJ Group Title

Hence, yes, with the $$S_y$$ given in the solutions, the eigenvalues are exactly $$\lambda = \pm h'/2$$. Anyway, I've sure you've figured out the problem by now.

27. hosein Group Title

eigenvalue is ok thanx,now eigenvectors

28. JamesJ Group Title

The general procedure to find eigenvectors you follow this procedure: The eigenvectors corresponding to an eigenvalue $$\lambda_j$$ are vectors $$v$$ such that $( S_y - \lambda_jI_2 )v = 0$ That is, you are looking for vectors $$v$$ that lie in the null space of the matrix $S_y - \lambda_jI_2$ So for each eigenvector, look at that matrix and find the null space. That's where the eigenvectors come from in the solution.

29. hosein Group Title

i knew method but final answer don't matched can you give me details of solution ?

30. JamesJ Group Title

For example, for $$\lambda = h'/2$$, the matrix $S_y - \lambda I = \frac{h'}{2} \left( \begin{matrix} 1 & -i \\ i & -1 \end{matrix} \right)$ which has a null space spanned by the vector $$(i,1)^t$$. If you want to normalize that vector, then you have $\frac{1}{\sqrt{2}} { i \choose 1 }$ Now repeat that process for the other eigenvector. The procedure in the solutions isn't standard, but obviously it's not wrong.