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Suppose we have a spin state for a spin 1 particle in $S_z$ basis defined by a column matrix $|Ψ\rangle=(a,b,c)$.What is the probability of getting $\hbar$ if we measure $S_z$ for a state $S_x|Ψ\rangle$ ?

I don't know how to solve this kind of problems and so it would be of great help if someone tells me. Thanks in advance.

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  • $\begingroup$ you can start by finding out what is the matrix representation of $S_x$ $\endgroup$ – glS Oct 17 '16 at 9:13
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The representation of $S_z$ in its own basis is trivial to write down: $$S_z=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1 \end{pmatrix}$$ in units $\hbar=1$. Now, using the standard arguments of angular momentum algebra, we have: $$S_+=S_x +iS_y$$ $$S_-=S_x -iS_y$$ Further,we know that $$S_+ \begin{pmatrix}0\\0\\1 \end{pmatrix}=\begin{pmatrix}0\\1\\0 \end{pmatrix}$$ $$S_+ \begin{pmatrix}0\\1\\0 \end{pmatrix}=\begin{pmatrix}1\\0\\0 \end{pmatrix}$$ $$S_+ \begin{pmatrix}1\\0\\0 \end{pmatrix}=0$$ Knowing the action of an operator on all basis elements, uniquely determines the operator. Simply reading off the operator from the above relations: $$S_+=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0 \end{pmatrix}$$ Similarly we have, $$S_-=\begin{pmatrix}0&0&0\\1&0&0\\0&1&0 \end{pmatrix}$$ Hence, we finally have the expression: $$S_x=\frac{1}{2}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0 \end{pmatrix}$$ Now, its plain as day to see that the probability of measuring spin up is simply: $\frac{|b|^2}{4}$

Hope this helps.

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  • $\begingroup$ How did you get the expression for $S_z$? It is not trivial for me to see why it should be the expression that you gave. $\endgroup$ – flippiefanus Oct 18 '16 at 5:02
  • $\begingroup$ well, $S_z$ better be diagonal in its own basis(this is the meaning of choosing the basis formed out of the eigenvectors of $S_z$). It is known that the eigenvalues are -1,0,1. These must occupy the diagonal positions in $S_z$. And thats what has been written down. $\endgroup$ – Anonjohn Oct 18 '16 at 12:25
  • $\begingroup$ Is there any reason for the particular ordering of the eigenvalues? The resulting matrices should obey particular commutation relations. Can you show that they do, or is there a quick way to see it? $\endgroup$ – flippiefanus Oct 18 '16 at 12:30
  • $\begingroup$ To answer your first question. No there is no strict ordering to the eigenvalues. However, It is convention to assign the canonical basis vector $e_1=\begin{pmatrix} 1&0&0\end{pmatrix}^T$, the highest eigenvalue. To answer your second question, recall that action of $S_+$ and $S_-$ on the basis vectors are found using the commutation relationships. So, checking if the final answers satisfy them is a mere consistency check. A good way to see it immediately is to calculate $S_+ S_z |j,m\rangle$using the commutation relationships. $\endgroup$ – Anonjohn Oct 18 '16 at 12:46

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