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

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

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

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- JamesJ

First, can you write out \( S_y \) explicitly?

- anonymous

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

- JamesJ

Whatever it is, write it out explicitly.

- anonymous

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

- anonymous

is operation

- 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

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

- JamesJ

What did you find for eigenvalues?

- JamesJ

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

- anonymous

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

- 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

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

- 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

your eigenvalues is zero?

- anonymous

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

- JamesJ

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

- anonymous

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

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- anonymous

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

- JamesJ

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

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- anonymous

compare your Sy with mine

- JamesJ

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- anonymous

so...

- 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

eigenvalue is ok thanx,now eigenvectors

- 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

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

- 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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