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!