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$$S_x= \frac{\hbar}{2}\quad\begin{pmatrix}0&1\\1&0\end{pmatrix}\quad$$ $$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$ How can I find the eigenvalue of $S_x$? My book says $$ \left| \begin{array}{cc} -\lambda & \frac{\hbar}{2} \\ \frac{\hbar}{2} & -\lambda\\\end{array} \right|=0 $$ So $\lambda=\frac{\hbar}{2} or \frac{-\hbar}{2}$ and therefore

$$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$ and $$S_x{X_-}^{(x)}=\frac{-\hbar}{2}{X_-}^{(x)}$$

My question is why do we need determinant=0? And what does the ${-\lambda}$ in the diagonal stand for?

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What do you know about linear algebra? – Kyle Kanos Dec 4 '13 at 4:12
up vote 2 down vote accepted

Take case for an $n\times n$ matrix $A$. To find its eigenvalues, first you write the eigenvalue equation for it.

$$Au=\lambda u$$

where $u$ are its eigenvectors. This can be rewritten in the following way

$$Au-\lambda u=(A-\lambda I)u=0$$

with $I$ the identity matrix. Let $A-\lambda I=B$, and we know that the equation $Bu=0$ has a non zero solution $u$ if and only if $\mathrm{det}B=0$. From this we find that the eigenvalues of $A$ are the $\lambda 's$ which satisfy the following equation

$$\mathrm{det}(A-\lambda I)=0$$

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$\lambda$ stands for the eigenvalue.

Eigenvalue equation is:

$S_xX=\lambda X$

$S_xX-\lambda X=0$

$(S_x-\lambda I)X=0$

Since X is eigenfunction, we seek solutions for $det(S_x-\lambda I)=0$

\begin{align} (S_x-\lambda I)= \begin{bmatrix} 0 & \frac{\hbar}{2} \\ \frac{\hbar}{2} &0 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} - \lambda & \frac{\hbar}{2} \\ \frac{\hbar}{2} & - \lambda \end{bmatrix} \end{align}

So to find eigenvalues you should just solve this:

\begin{align} \begin{vmatrix} - \lambda & \frac{\hbar}{2} \\ \frac{\hbar}{2} & - \lambda \end{vmatrix}=0 \end{align}

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