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

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?

• need to normalize your eigenstates... – ZeroTheHero Oct 27 '18 at 14:13

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
$$\langle v_1\vert\psi\rangle^2 = \frac{1}{6}(3+\sqrt{3})$$
$$\langle v_2\vert\psi\rangle^2 = \frac{1}{6}(3-\sqrt{3})$$