Probability of $\frac{-1}{\sqrt2} S_x + S_z $

Question

I have a State $\left|\Psi\right>=\frac{\left|1\right>+\left|0\right>}{\sqrt{2}},$ in the $z$-Spin basis and want to calculate the probability of this state for the eigenvectors of the operator $\frac{-1}{\sqrt2} S_x + S_z $ which are $\begin{pmatrix} 1-\sqrt2\\ 1 \end{pmatrix} and \begin{pmatrix} 1+\sqrt2\\ 1 \end{pmatrix}$(In the $z$-basis). So I take the norm squared of$ \langle\begin{pmatrix} 1\pm\sqrt2\\ 1 \end{pmatrix}|\Psi\rangle.$ Which gives me 1 in both cases which is no good for a probability.Where am I wrong?

Answer 1

Your operator is $\frac{1}{\sqrt{2}}S_x +S_z$. The eigenvectors of this operator are not what you have written down. They are

$v_1 = \frac{1}{\sqrt{5-2\sqrt{6}}}\begin{pmatrix} 1\\ \sqrt{3} - \sqrt{2} \end{pmatrix}$

$v_2 = \frac{1}{\sqrt{5+2\sqrt{6}}}\begin{pmatrix} 1\\ -\sqrt{3} - \sqrt{2} \end{pmatrix}$

Thus, the probability of finding the state $\left|\Psi\right>=\frac{\left|1\right>+\left|0\right>}{\sqrt{2}} = \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{pmatrix}$ is

$\langle v_1\vert\psi\rangle^2 = \frac{1}{6}(3+\sqrt{3})$

$\langle v_2\vert\psi\rangle^2 = \frac{1}{6}(3-\sqrt{3})$