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hosein Group Title

please solve this : find eigen value & eigen vector of opration Sy

  • 2 years ago
  • 2 years ago

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  1. hosein Group Title
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    \[S _{y}\] is y component of s & s is spin.solve it in two dimension

    • 2 years ago
  2. hosein Group Title
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    i know what do that. but final answer has not matched with text

    • 2 years ago
  3. opiesche Group Title
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    I was going to just take a wild guess, but really, I have no idea :P

    • 2 years ago
  4. JamesJ Group Title
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    First, can you write out \( S_y \) explicitly?

    • 2 years ago
  5. hosein Group Title
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    your goal is operation of Sy ? in matrix definition Sy is two dimensional matrix do you know or write it?

    • 2 years ago
  6. JamesJ Group Title
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    Whatever it is, write it out explicitly.

    • 2 years ago
  7. hosein Group Title
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    Sy= \[\left[\begin{matrix}h'/2 & -ih'/2 \\ih'/2 & h'/2\end{matrix}\right]\] h'=h/2pi

    • 2 years ago
  8. hosein Group Title
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    is operation

    • 2 years ago
  9. JamesJ Group Title
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    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
  10. hosein Group Title
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    yes know this but final answer don't match with text

    • 2 years ago
  11. JamesJ Group Title
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    What did you find for eigenvalues?

    • 2 years ago
  12. JamesJ Group Title
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    I have from that definition of Sy, that the eigenvalues are 0 and h'

    • 2 years ago
  13. hosein Group Title
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    eigen value is match eigenvector... i didn't find

    • 2 years ago
  14. JamesJ Group Title
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    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
  15. JamesJ Group Title
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    For example, for \( \lambda = 0 \), the null space is one dimensional and has as a basis vector v = (1,i)^t

    • 2 years ago
  16. JamesJ Group Title
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    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
  17. hosein Group Title
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    your eigenvalues is zero?

    • 2 years ago
  18. hosein Group Title
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    eigenvalue is +h'/2&-h'/2

    • 2 years ago
  19. JamesJ Group Title
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    One of the eigenvalues is zero, because the characteristic equation is \[ \lambda^2 - h' \lambda = 0 \]

    • 2 years ago
  20. hosein Group Title
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    no eigenvalue is +h'/2&-h'/2. see this: 4.29 part a

    • 2 years ago
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  21. hosein Group Title
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    this book is solution of Introduction to quantum mechanics by david.-j griffiths

    • 2 years ago
  22. JamesJ Group Title
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    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
  23. hosein Group Title
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    compare your Sy with mine

    • 2 years ago
  24. JamesJ Group Title
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    • 2 years ago
  25. hosein Group Title
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    so...

    • 2 years ago
  26. JamesJ Group Title
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    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
  27. hosein Group Title
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    eigenvalue is ok thanx,now eigenvectors

    • 2 years ago
  28. JamesJ Group Title
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    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
  29. hosein Group Title
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    i knew method but final answer don't matched can you give me details of solution ?

    • 2 years ago
  30. JamesJ Group Title
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    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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