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

  • #1
earthling75
3
0
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?
 
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  • #2
earthling75 said:
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.
 
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  • #3
vela said:
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?
 
  • #4
earthling75 said:
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.
 
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