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

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

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

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

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

JamesJBest ResponseYou've already chosen the best response.1
Whatever it is, write it out explicitly.
 2 years ago

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

JamesJBest ResponseYou've already chosen the best response.1
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 \( \lambda \).
 2 years ago

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

JamesJBest ResponseYou've already chosen the best response.1
What did you find for eigenvalues?
 2 years ago

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

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

JamesJBest ResponseYou've already chosen the best response.1
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 \[ S_y  \lambda I_2 \]
 2 years ago

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

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

hoseinBest ResponseYou've already chosen the best response.0
your eigenvalues is zero?
 2 years ago

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

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

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

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

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

hoseinBest ResponseYou've already chosen the best response.0
compare your Sy with mine
 2 years ago

JamesJBest ResponseYou've already chosen the best response.1
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.
 2 years ago

hoseinBest ResponseYou've already chosen the best response.0
eigenvalue is ok thanx,now eigenvectors
 2 years ago

JamesJBest ResponseYou've already chosen the best response.1
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.
 2 years ago

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

JamesJBest ResponseYou've already chosen the best response.1
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.
 2 years ago
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