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

In summary: However, if you want to find the eigenspinor for spin in the z-direction, you will need to use the projection operator to project the state onto the z-axis. So, in summary, we discussed how to construct the spin matrix for spin = 1 case using raising and lowering operators. We also looked at how to find the eigenspinor for a given eigenvalue of the spin matrix. Finally, we discussed how to find the probabilities for measuring a certain spin value in the z-direction and how to find the eigenspinor for spin in the z-direction using the projection operator.
  • #1
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.
 

1. What is the formula for calculating the probability of measuring ##S_y## for a spin 1 particle?

The formula for calculating the probability of measuring ##S_y## for a spin 1 particle is P(##S_y##) = (1 + ##S_y##) / 2, where ##S_y## is the spin operator in the y-direction.

2. How is the spin operator represented in the calculation?

The spin operator, ##S_y##, is represented by a matrix in the calculation. For a spin 1 particle, the matrix has a size of 3x3 and contains the values 0, 1, and -1 along the diagonal and 0s in all other positions.

3. Can the probability of measuring ##S_y## be greater than 1?

No, the probability of measuring ##S_y## cannot be greater than 1. The maximum value for the probability is 1, which occurs when the spin of the particle is perfectly aligned with the y-direction.

4. How does the probability of measuring ##S_y## change with different spin states?

The probability of measuring ##S_y## depends on the spin state of the particle. If the spin state is aligned with the y-direction, the probability will be 1. If the spin state is aligned with the x-direction, the probability will be 0. If the spin state is a superposition of both directions, the probability will be between 0 and 1.

5. Can the probability of measuring ##S_y## be negative?

No, the probability of measuring ##S_y## cannot be negative. The probability is always a positive value between 0 and 1, representing the likelihood of measuring a specific spin state for the particle.

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