# Two particles with spin - measurements

1. Aug 28, 2016

### gasar8

1. The problem statement, all variables and given/known data

We have got two particles with $S_1=1$ and $S_2=1$. We know that $S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle$ and $S_{2x}|\psi_2\rangle = \hbar |\psi_2\rangle.$

a) Find wave function $|\psi_1\rangle$ in $S_{1z}$ basis and $|\psi_2\rangle$ in $S_{2z}$ basis.
b) We measure $S^2$ of total spin. What are possible outcomes and what are their probabilities?
c) Find expectation value and uncertainty of $S^2$.
d) We measure x component of total spin. What are possible outcomes and what are their probabilities?

3. The attempt at a solution

a) $|\psi_1\rangle = |11\rangle \\ |\psi_2\rangle = {1 \over 2} |1-1\rangle + {1 \over \sqrt{2}} |10\rangle+ {1 \over 2} |11\rangle.$ Can someone just check this?
b)\begin{align*} |\psi_{12}\rangle&={1 \over 2}|1\rangle|-1\rangle+{1 \over \sqrt{2}} |1\rangle|0\rangle+{1 \over 2}|1\rangle|1\rangle=\\ &={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle \end{align*}
For $S^2|\psi_{12}\rangle=\hbar^2 s(s+1)|\psi_{12}\rangle,$ we get:
\begin{align*} &Results \ \ \ \ &Probability\\ &6\hbar^2 &{13\over24}\\ &2\hbar^2 &{3 \over 8}\\ &0 &{1 \over 12} \end{align*}

c) Expectation value is $\langle S^2 \rangle = \langle \psi|S^2|\psi\rangle=4\hbar^2,$ but I can't find uncertainty? I am thinking in this way:
$$\delta_{S^2}=\sqrt{\langle S^2\rangle- \langle S \rangle ^2} or \\ \delta_{S^2}=\sqrt{\langle S^4\rangle- \langle S^2 \rangle ^2}?$$

d) How do I find outcomes and probabilities? I tried with $S_x=\frac{S_++S_-}{2}$, but got some weird wavefunction (which was not normalized), from which I can't find anything. Then I was thinking about Pauli matrices, so that possible outcomes would only be their eigenvalues, so $\pm {\hbar \over 2}$, but how can I apply this matrix to my wavefunction of 1x1 spins. I found something on wiki - Pauli matrices for such spins - and tried but got nothing...

2. Aug 28, 2016

I agree with your answer to "a". (I computed the (+1)eigenstate of the $L_x=(L_+ + L_-)/2$ operator using the three z- angular momentum states as a basis. Perhaps rotating the $m_z=+1$ eigenstate 90 degrees would be quicker, but my quantum mechanics is a little rusty.) I think "b" uses the Clebsch-Gordon coefficients, but I would need to do a review of the topic to check your part "b".

3. Aug 28, 2016

### blue_leaf77

Yes that's right.
It's been not yet specified on which state this measurement is applied on?
It seems like you are trying to form $|\psi_1\rangle |\psi_2\rangle$ and then express it in the basis $\{|S;m\rangle\}$ however the last vector is not normalized, I wonder if you put in the CG coefficients correctly.

Last edited: Aug 28, 2016
4. Aug 29, 2016

### gasar8

We measure the $S^2$ of both particles together, so because of that, I tried to write my wavefunction in $|S,m\rangle$ form. Is this right?

Yes, sorry, I made a mistake when I was copying it from my handwriting. I forgot a $|10\rangle$ state, so the whole function is:
$$|\psi_{12}\rangle={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{8}}|10\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle,$$
but this element is already included in the rest of my calculations in this exercise.

5. Aug 29, 2016

### blue_leaf77

Alright.
The second one.
No, you cannot use Pauli matrices because they are only for spin -1/2 particles. For this problem I think it will be easier if you express $|\psi_{12}\rangle$ in terms of $|S_1m_{1x}\rangle |S_2m_{2x}\rangle$ basis because they are eigenvectors of $S_x = S_{1x}+S_{2x}$.

6. Aug 29, 2016

### gasar8

So if I understand this correctly, I must use first $S^2$ on my wavefunction to get $\langle S^2\rangle$ and just two times $S^2$ on wavefunction to get $\langle S^4 \rangle$. If I do this, I get:
\begin{align*} \langle S^2 \rangle &= \langle \psi|S^2|\psi\rangle=4\hbar^2,\\ \langle S^4 \rangle &= \langle \psi|(S^2)^2|\psi\rangle=21\hbar^4\\ \Longrightarrow \delta_{S^2} &=\sqrt{5} \ \hbar^2 \end{align*}

Hmm, how do I do that, how can I get $m_x$? Do $m_x$ and $m_z$ have any connection?

7. Aug 29, 2016

### blue_leaf77

Yes.
For example for the first particle, you can find the matrix form of $S_{1x}$ for $S_1=1$ in $|S_1 m_{1z}\rangle$ basis and then find its eigenvectors. There will be three eigenvectors of $S_{1x}$ for $S_1=1$. Then invert them such that each of $|S_1 m_{1z}\rangle$ is expressed in terms of $|S_1 m_{1x}\rangle$ basis.

8. Aug 29, 2016

### gasar8

I think I wil need more help and hints here. So I am thinking this way:
\begin{align*} |S,S_z\rangle&=\alpha |S,S_x\rangle\\ |1,1\rangle &= |10\rangle \ \ \textrm{because the z component is 1, x and y must be 0 - right thinking???}\\ |1,0\rangle &= {1 \over \sqrt{2}} \big(|11\rangle +|1-1\rangle \big)\\ |1,-1\rangle &= |10\rangle\\ \end{align*}

So the matrix would be: $\left( \begin{array}{cc} 0 & 1 & 0\\ {1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}}\\ 0 & 1 & 0 \end{array} \right).$

I found something similar here: https://en.wikipedia.org/wiki/Spin_(physics)#Higher_spins

9. Aug 29, 2016

### blue_leaf77

Nope.
The matrix of $S_x$ for $S=1$ spin in $|S,m_z\rangle$ basis is
$\frac{\hbar}{\sqrt{2}} \left( \begin{array}{cc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array} \right).$
Now find the eigenvectors of this matrix, then write them in vector form using $|S,m_z\rangle$ basis.

10. Aug 29, 2016

### gasar8

Ahaaa, so I was thinking wrong. I have to apply an operator $S_x$ on every $|S,m_z\rangle$ state and then write a matrix, so:
\begin{align*} S_x|1,1\rangle &= {\hbar \over \sqrt{2}} |10\rangle \\ S_x|1,0\rangle &= {\hbar \over \sqrt{2}} \big(|11\rangle +|1-1\rangle \big)\\ S_x|1,-1\rangle &= {\hbar \over \sqrt{2}} |10\rangle, \end{align*}

from where I get your matrix. :)

Ok, the eigenvectors are:

\begin{align*} \vec{x_1} &= \bigg({1 \over 2},-{1 \over \sqrt{2}},{1 \over 2}\bigg) \\ \vec{x_2} &= \bigg({1 \over 2},{1 \over \sqrt{2}},{1 \over 2}\bigg) \\ \vec{x_3} &= \bigg(-{1 \over \sqrt{2}},0,{1 \over \sqrt{2}}\bigg) \\ \end{align*}

I'm not sure how to do it.

11. Aug 29, 2016

### blue_leaf77

I will give one example. Let's take $\vec x_1 = |S_x;-1\rangle$,
\begin{aligned} \vec x_1 &= |S_x;-1\rangle = ({1\over 2}, {-1\over \sqrt{2}},{1\over 2})^T = {1\over 2}(1, 0,0)^T - {1\over \sqrt{2}}(0, 1,0)^T + {1\over 2}(0, 0,1)^T \\ &= {1\over 2}|S_z;1\rangle - {1\over \sqrt{2}}|S_z;0\rangle + {1\over 2}|S_z;-1\rangle \end{aligned}

12. Aug 29, 2016

### gasar8

Aha, I just didn't know how to write $|S,m_z\rangle$ as a basis, so I only choose (1,0,0),(0,1,0),(0,0,1).

So $|S,m_x\rangle = \alpha |S,m_z\rangle$ would be:
\begin{align*} |1-1\rangle &= {1 \over 2} |11\rangle - {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \\ |10\rangle &= - {1 \over \sqrt{2}} |11\rangle + {1 \over \sqrt{2}} |1-1\rangle \\ |11\rangle &= {1 \over 2} |11\rangle + {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \end{align*}

13. Aug 29, 2016

### blue_leaf77

Note that with the notation style you are using now, you are prone to being confused by the eigenvectors of $S_x$ and $S_z$. I suggest that you find a way to distinguish which kets belong to the eigenvectors of $S_x$ and which kets to the eigenvectors of $S_z$.
The next step would be to express the three kets appearing in the RHS in those three equations (which are eigenvectors of $S_z$) in terms of the three kets in the LHS (which are eigenvectors of $S_x$). If you are familiar with matrix inverse, you can employ coefficient matrix of the above system of linear equations to do the inversion.

14. Aug 29, 2016

### gasar8

Yes, I know, I was confused. :D

Aha, I have already tried with the $S_x$ matrix before, to write its inverse. So firstly, I have to write $S_x$ matrix, then find its eigenvectors, build another matrix from this eigenvectors and then find the inverse:
$$\left( \begin{array}{cc} {1 \over 2} & -{1 \over \sqrt{2}} & {1 \over 2} \\ -{1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}} \\ {1 \over 2} & {1 \over \sqrt{2}} & {1 \over 2} \end{array} \right)^{-1}= \left( \begin{array}{cc} {1 \over 2} & -{1 \over \sqrt{2}} & {1 \over 2} \\ -{1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}} \\ {1 \over 2} & {1 \over \sqrt {2}} & {1 \over 2} \end{array} \right)$$

EDIT: The inverse was wrong.

Last edited: Aug 29, 2016
15. Aug 29, 2016

### blue_leaf77

Having found $|S,m_z\rangle$ in $\{|S,m_x\rangle\}$ basis, you can now write
$$|\psi_{12}\rangle={1 \over 2}|S_{1z};1\rangle|S_{2z};-1\rangle+{1 \over \sqrt{2}} |S_{1z};1\rangle|S_{2z};0\rangle+{1 \over 2}|S_{1z};1\rangle|S_{2z};1\rangle$$
in $|S_{1x};m_{1x}\rangle|S_{2x};m_{2x}\rangle$ basis.

16. Aug 29, 2016

### gasar8

Do I have to normalize this matrix from my previous post first? Because if I write $|S,m_z\rangle =|1,1\rangle = {1 \over 2} |1,S_x=1\rangle + {1 \over 2} |1,S_x=0\rangle + {1 \over 2} |1,S_x=-1\rangle$ its not? Or am I wrong again?

17. Aug 29, 2016

### blue_leaf77

Actually I just checked the inverse of that matrix in post #14 using an online calculator and it returns a different answer from yours.

18. Aug 29, 2016

### gasar8

Uf, yes, I was wrong, I don't know what was I doing, I edited it now, so its correct.
But anyway, what now, when I have to write $|S_{1z};1\rangle|S_{2z};-1\rangle$ for example, there are 9 elements, at $|S_{1z};1\rangle|S_{2z};0\rangle$ there are 6 and at $|S_{1z};1\rangle|S_{2z};1\rangle$ nine again?

19. Aug 29, 2016

### blue_leaf77

Yes that's right. That's a lengthy expression but you are actually just one step from finished.

20. Aug 31, 2016

### gasar8

Ok, thank you very much blue_leaf77 for all your help and patience with me. Today I passed the Introductory quantum mechanic exam, pretty much because of your help at all my posted exercises. I would really like to thank you also for all your time you took for me, because I know how much time I spent on computer just for all my replies, so you had to take it a lot, too. Very big thanks!

BTW: After the exam I went to my professor with this threads exercise and he proposed a different way of solving it. He would just turn the coordinate system around y-axis so that $z \rightarrow x$ and $x \rightarrow -z$ and only the inital condition changes, so $S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle$ and $S_{2x}|\psi_2\rangle = - \hbar |\psi_2\rangle$ or something like that. And from there on, it just the whole exercise identical to a), b) and c) part.

But again, thank you for all your effort with me I really appreciate it!