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- 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?

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?

Last edited: