Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

$$S_x= \frac{\hbar}{2}\quad\begin{pmatrix}0&1\\1&0\end{pmatrix}\quad$$ $$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$ 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?

share|improve this question
1  
What do you know about linear algebra? –  Kyle Kanos Dec 4 '13 at 4:12
add comment

2 Answers

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

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

share|improve this answer
add comment

$\lambda$ 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}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.