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hosein
 3 years ago
Best ResponseYou've already chosen the best response.0\[S _{y}\] is y component of s & s is spin.solve it in two dimension

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0i know what do that. but final answer has not matched with text

opiesche
 3 years ago
Best ResponseYou've already chosen the best response.0I was going to just take a wild guess, but really, I have no idea :P

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1First, can you write out \( S_y \) explicitly?

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0your goal is operation of Sy ? in matrix definition Sy is two dimensional matrix do you know or write it?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Whatever it is, write it out explicitly.

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0Sy= \[\left[\begin{matrix}h'/2 & ih'/2 \\ih'/2 & h'/2\end{matrix}\right]\] h'=h/2pi

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Ok, so you need to find first the eigenvalues of the matrix. I.e., solve the equation, \[ \det(S_y  \lambda I_2) = 0 \] for \( \lambda \).

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0yes know this but final answer don't match with text

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1What did you find for eigenvalues?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1I have from that definition of Sy, that the eigenvalues are 0 and h'

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0eigen value is match eigenvector... i didn't find

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1I'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 \[ S_y  \lambda I_2 \]

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1For example, for \( \lambda = 0 \), the null space is one dimensional and has as a basis vector v = (1,i)^t

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1that's an eigenvector corresponding to eigenvalue of zero. Now find the other null space and the eigenvector corresponding to the eigenvalue of \( \lambda = h' \).

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0your eigenvalues is zero?

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0eigenvalue is +h'/2&h'/2

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1One of the eigenvalues is zero, because the characteristic equation is \[ \lambda^2  h' \lambda = 0 \]

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0no eigenvalue is +h'/2&h'/2. see this: 4.29 part a

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0this book is solution of Introduction to quantum mechanics by david.j griffiths

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1You gave the wrong \( S_y \). Notice the matrix in the book has zeros along the diagonal, unlike the matrix you typed above.

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0compare your Sy with mine

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Hence, 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.

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0eigenvalue is ok thanx,now eigenvectors

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1The 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.

hosein
 3 years ago
Best ResponseYou've already chosen the best response.0i knew method but final answer don't matched can you give me details of solution ?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1For 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.
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