Schrodinger half spin states expectation values

1. Nov 6, 2013

crowlma

1. The problem statement, all variables and given/known data

What is the expectation value of $\hat{S}_{x}$ with respect to the state $\chi = \begin{pmatrix} 1\\ 0 \end{pmatrix}$?
$\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$

2. Relevant equations

$<\hat{S}_{x}> = ∫^{\infty}_{-\infty}(\chi^{T})^{*}\hat{S}_{x}\chi$

3. The attempt at a solution

So I have $(\chi^{T})^{*}$ as equalling (1 0), giving me: $\frac{\hbar}{2} ∫^{\infty}_{-\infty}\begin{pmatrix} 1&0 \end{pmatrix}\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix}$ which simplifies to $\frac{\hbar}{2} ∫^{\infty}_{-\infty}\begin{pmatrix} 1&0 \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix} = 0$. Is this right?

2. Nov 7, 2013

MisterX

This is a discrete system and it doesn't make sense to have an integral there. Also we wouldn't need a sum or integral since that is already implied by $(\chi^{T})^{*}\hat{S}_{x}\chi$. There is a mistake in the column vector in the last line as well, although you somehow came up with the correct answer.

$\langle \hat{S}_x \rangle = (\chi^{T})^{*}\hat{S}_{x}\chi$