Probability of measuring a specific value in a spin-1/2 system

[mtd]

Homework Statement

A spin-1/2 system in the state $\left|ψ\right\rangle = \left|0.5, z\right\rangle$ of the $S_{z}$ spin operator has eigenvalue $s = +\hbar/2$. Find the expectation values of the $S_{z}$ and $S_{x}$ operators.

Homework Equations

$\left\langle S_{x,z}\right\rangle = \left\langle ψ \right| S_{x,z}\left|ψ\right\rangle$

The Attempt at a Solution

Multiplied out above equations to find $\hbar z/2$ and $\hbar (0.25 - z^{2})/2$ for the x and z directions, respectively. I assume $z$ is just "some variable" - is it safe to normalize the eigenstate and set z equal to root 0.75?

Homework Statement

Find the probability of measuring $\hbar /2$ in a measurement of $S_{x}$ in the same system.

Homework Equations

The probability of measuring the eigenvalue $a_{n}$ in a measurement of the observable $A$ is $P \left( a_{n} \right) = \left| \left\langle b_{n} |ψ \right\rangle \right| ^{2}$ where $\left|b_{n}\right\rangle$ is the normalised eigenvector of $A$ corresponding the the eigenvalue

The Attempt at a Solution

I believe this should just be the eigenvalue squared i.e. $\hbar ^{2}/4$, but I'm not sure if or why this is the case.

 

[vanhees71]

$S_x$ and $S_z$ are the operators, representing the spin-x and -z components. Usually you write them with help of the Pauli spin matrices as
$$S_j=\frac{\hbar}{2} \sigma_j, \quad j \in \{x,y,z\}.$$

In the 2nd problem just calculate what you've written down and not guess some eigenvalue. Think also about the question, how can a probability be a dimensionful quantity as your result suggests?

 

[mtd]

I used the Pauli matrices to get my solution for the first question. For the second question I found the eigenvector for $S_{x}$ to be $(1,1)$ and the probability of measuring $\hbar /2$ to be $(z+0.5)^{2}$. However, if the eigenstate of the system is normalised and $z = \sqrt{3/4}$ then I must be incorrect.

Thank you

edit: Using the other eigenvector $(1,−1)$ yielded $(z−0.5)^{2} = 0.134$ - my first choice of eigenvector corresponded to $-\hbar /2$ rather than $\hbar /2$. I think I've solved the question, but why is the probability incorrect when I use the negative eigenvalue? I believe it is a valid value of $s_{x}$

 

[vanhees71]

I don't understand what $z$ should mean. In the formula you quote, there is no variable $z$.

You have $|1/2,x \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (I've normalized the vector properly for you). Your system is prepared in the spin state $|\psi \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Now you should be able to calculate the probability to measure $\hbar/2$ for $\sigma_x$!

 

[mtd]

Strange, the system state should be displayed like this:

edit: Okay, I think I understand why the system state should be $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, i.e. $z = 0$ - by using the $S_{z}$ operator on the given eigenstate you can see the z has to be zero.

Using this information I find expectation values of 0 and $\hbar /2$ for x and z respectively. However with $z = 0$ I find that probability of measuring $\hbar /2$ for $S_{x}$ is 1 which is inconsistent with the expectation value.

 

[vanhees71]

Argh! Sorry, I've overlooked this. I understood the question such that $|\psi \rangle$ should be the eigenstate of $\hat{s}_z$ with eigenvalue $\hbar/2$. Then, in the usual convention to choose the Pauli matrices with respect to the $\hat{s}_z$ eigenbasis, $z=0$.

For the probability you have to normalize the eigenvector of $\hat{\sigma}_x$ to 1!

 

[mtd]

Aha, got it (P = 0.5). Must remember to normalise eigenkets. Thanks!

 

[vanhees71]

good!