Stuck calculating probability of measuring ##S_y## for spin 1 particle

[earthling75]

Homework Statement

Calculate the probability of finding a particle with spin = 1 in a given state to have an eigenvalue of $\hbar in S_y$ basis.

Relevant Equations

$$S_{\pm}|s,m\rangle = \hbar \sqrt{s(s+1)-m(m\pm 1)}|s,(m\pm 1)\rangle$$
Probability $=|\langle \chi |\chi_+^y \rangle|^2$

I know how to construct Sy for spin = 1 case from the raising and lowering operators.
I get
$$
S_y=\frac{i\hbar}{\sqrt{2}}\begin{pmatrix}
0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \\
\end{pmatrix}
$$
From what I have seen, the eigenspinor for $\hbar$ is found by solving

$$
\frac{i\hbar}{\sqrt{2}}\begin{pmatrix}
0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
\alpha \\ \beta \\ \gamma
\end{pmatrix} =
\hbar \begin{pmatrix}
\alpha \\ \beta \\ \gamma
\end{pmatrix}
$$
That leaves me with three equations
$$
-\frac{i}{\sqrt{2}} \beta = \alpha$$
$$
\frac{i}{\sqrt{2}} \alpha - \frac{i}{\sqrt{2}}\gamma = \beta$$
$$
\frac{i}{\sqrt{2}} \beta = \gamma$$

I think I know how to construct the eigenspinor from these values. Is it simply
$$
\chi_{+}^y=\frac{1}{\sqrt{2}}\begin{pmatrix}
-\frac{i}{\sqrt{2}} \\ 1 \\ \frac{i}{\sqrt{2}}
\end{pmatrix}
$$?

The actual particle I'm trying to measure is in the state
$$
\chi = \frac{1}{2}
\begin{pmatrix} 1\\ i\sqrt{2}\\ -1
\end{pmatrix}
$$
but when I do the calculation, I get
$$
|\langle \chi_{+}^y|\chi\rangle|=\frac{1}{\sqrt{2}}\begin{pmatrix}
\frac{i}{\sqrt{2}}& 1& -\frac{i}{\sqrt{2}}
\end{pmatrix}\cdot \frac{1}{2}\begin{pmatrix}1\\ i\sqrt{2}\\-1
\end{pmatrix} =1
$$
What am I doing wrong?

 

[vela]

but when I do the calculation, I get
$$
| \langle \chi_{+}^y | \chi\rangle| =
\frac{1}{\sqrt{2}} \begin{pmatrix} -\frac{i}{\sqrt{2}} & 1 & \frac{i}{\sqrt{2}} \end{pmatrix}
\cdot
\frac{1}{2} \begin{pmatrix} 1 \\ i \sqrt{2} \\ -1 \end{pmatrix} = 1
$$ What am I doing wrong? If I follow the same procedure for $-\hbar$ or $0$ I get 1.

You forgot to conjugate the first matrix. Nevertheless, you should find a probability of 1. Note that $\lvert \chi \rangle$ is a multiple of the eigenstate.

 

[earthling75]

You forgot to conjugate the first matrix. Nevertheless, you should find a probability of 1. Note that $\lvert \chi \rangle$ is a multiple of the eigenstate.

I fixed that typo. So,
$$
\frac{1}{\sqrt{2}}\begin{pmatrix}
-\frac{i}{\sqrt{2}}\\ 1\\ \frac{i}{\sqrt{2}}
\end{pmatrix} =\left(-i\right) \frac{1}{2}\begin{pmatrix}1\\ i\sqrt{2}\\-1
\end{pmatrix}
$$
If I next try to measure $-\hbar$ in $S_z$ basis, I get

$$
\begin{pmatrix}
0 & 0 & -1 \\
\end{pmatrix} \cdot \frac{1}{2}\begin{pmatrix}1\\ i\sqrt{2}\\-1
\end{pmatrix}=\frac{1}{2}
$$
Probability $= \left(\frac{1}{2}\right)^2=\frac{1}{4}$.

Do I have to find the new eisgenspinor in $z$ basis or is $\begin{pmatrix} 0 & 0 & -1 \\ \end{pmatrix}$ okay?

 

[PeroK]

Do I have to find the new eisgenspinor in $z$ basis or is $\begin{pmatrix} 0 & 0 & -1 \\ \end{pmatrix}$ okay?

If you want to find the probabilities of the various measurement outcomes for spin in the z-direction, then you can simply read off the state vector column entries. That's the advantage of having the state expressed in the z-basis in the first place.