ihatelolcats said:
i'm not sure what you mean by expand in terms of Sy
i have \chi = 1/sqrt(17) (\stackrel{4}{0}) + 1/sqrt(17) (\stackrel{0}{i})
It's actually
\chi = \frac{4}{\sqrt{17}} \begin{pmatrix}1 \\ 0\end{pmatrix} + \frac{1}{\sqrt{17}} \begin{pmatrix} 0 \\ 1 \end{pmatrix}
where, presumably,
\begin{pmatrix} 1 \\ 0 \end{pmatrix}
is the S
z spin-up eigenstate and
\begin{pmatrix} 0 \\ 1 \end{pmatrix}
is the S
z spin-down eigenstate. You can determine the probability of finding the particle in a spin state by squaring the modulus of the corresponding coefficient. For example, in this case, the probability of finding it in the spin-up state would be 16/17, and in the spin-down state, 1/17.
But this problem is asking you to find the probabilities using the y-axis instead of the z-axis, so what you want to do is find the eigenstates of S
y by diagonalizing the matrix
\hat{S}_y = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}
and express \chi as a linear combination of them to determine the coefficients you need. In linear-algebra-speak, you want to change basis from the S
z basis to the S
y basis.
This is effectively what you always do to find probabilities of a measurement. You find the eigenstates of the observable, express the state in terms of those eigenstates, and calculate the probabilities by squaring the modulus of the coefficients.
i'm still going to have to take an expectation of something to get a probability...
do you know of a reference or something online i can use to learn how to do this? i have griffiths intro to QM textbook, but like i said it isn't helpful on this. due friday heh.
An expectation value just gives you an average. It doesn't give you a probability.
