anonymous
  • anonymous
please solve this : find eigen value & eigen vector of opration Sy
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[S _{y}\] is y component of s & s is spin.solve it in two dimension
anonymous
  • anonymous
i know what do that. but final answer has not matched with text
anonymous
  • anonymous
I was going to just take a wild guess, but really, I have no idea :P

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JamesJ
  • JamesJ
First, can you write out \( S_y \) explicitly?
anonymous
  • anonymous
your goal is operation of Sy ? in matrix definition Sy is two dimensional matrix do you know or write it?
JamesJ
  • JamesJ
Whatever it is, write it out explicitly.
anonymous
  • anonymous
Sy= \[\left[\begin{matrix}h'/2 & -ih'/2 \\ih'/2 & h'/2\end{matrix}\right]\] h'=h/2pi
anonymous
  • anonymous
is operation
JamesJ
  • JamesJ
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 \( \lambda \).
anonymous
  • anonymous
yes know this but final answer don't match with text
JamesJ
  • JamesJ
What did you find for eigenvalues?
JamesJ
  • JamesJ
I have from that definition of Sy, that the eigenvalues are 0 and h'
anonymous
  • anonymous
eigen value is match eigenvector... i didn't find
JamesJ
  • JamesJ
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 \[ S_y - \lambda I_2 \]
JamesJ
  • JamesJ
For example, for \( \lambda = 0 \), the null space is one dimensional and has as a basis vector v = (1,i)^t
JamesJ
  • JamesJ
that's an eigenvector corresponding to eigenvalue of zero. Now find the other null space and the eigenvector corresponding to the eigenvalue of \( \lambda = h' \).
anonymous
  • anonymous
your eigenvalues is zero?
anonymous
  • anonymous
eigenvalue is +h'/2&-h'/2
JamesJ
  • JamesJ
One of the eigenvalues is zero, because the characteristic equation is \[ \lambda^2 - h' \lambda = 0 \]
anonymous
  • anonymous
no eigenvalue is +h'/2&-h'/2. see this: 4.29 part a
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anonymous
  • anonymous
this book is solution of Introduction to quantum mechanics by david.-j griffiths
JamesJ
  • JamesJ
You gave the wrong \( S_y \). Notice the matrix in the book has zeros along the diagonal, unlike the matrix you typed above.
anonymous
  • anonymous
compare your Sy with mine
JamesJ
  • JamesJ
anonymous
  • anonymous
so...
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
eigenvalue is ok thanx,now eigenvectors
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
i knew method but final answer don't matched can you give me details of solution ?
JamesJ
  • JamesJ
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.

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