Two particles with spin - measurements

[gasar8]

Homework Statement

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?

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...

 

[Charles Link]

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".

 

[blue_leaf77]

a) Find wave function |ψ1⟩|ψ1⟩|\psi_1\rangle in S1zS1zS_{1z} basis and |ψ2⟩|ψ2⟩|\psi_2\rangle in S2zS2zS_{2z} basis.

Yes that's right.

b) We measure S2S2S^2 of total spin. What are possible outcomes and what are their probabilities?

It's been not yet specified on which state this measurement is applied on?

b)$$\begin{align*} |\psi_{12}\rangle&={1 \over 2}|1\rangle|-1\rangle+{1 \over \sqrt{2}} |1\rangle|0\rangle+{1 \over 2}|1\rangle|1\rangle=\\ &={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle \end{align*}$$

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.

 

[gasar8]

It's been not yet specified on which state this measurement is applied on?

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?

I wonder if you put in the CG coefficients correctly

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.

 

[blue_leaf77]

$$|\psi_{12}\rangle={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{8}}|10\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle,$$
but this element is already included in the rest of my calculations in this exercise.

Alright.

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:
$$\delta_{S^2}=\sqrt{\langle S^2\rangle- \langle S \rangle ^2} or \\ \delta_{S^2}=\sqrt{\langle S^4\rangle- \langle S^2 \rangle ^2}?$$

The second one.

d) How do I find outcomes and probabilities? I tried with Sx=S++S−2Sx=S++S−2S_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 ±ℏ2±ℏ2\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...

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 $S_x = S_{1x}+S_{2x}$.

 

[gasar8]

The second one.

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:
$$\begin{align*} \langle S^2 \rangle &= \langle \psi|S^2|\psi\rangle=4\hbar^2,\\ \langle S^4 \rangle &= \langle \psi|(S^2)^2|\psi\rangle=21\hbar^4\\ \Longrightarrow \delta_{S^2} &=\sqrt{5} \ \hbar^2 \end{align*}$$

express $|\psi_{12}\rangle$ in terms of $|S_1m_{1x}\rangle |S_2m_{2x}\rangle$basis

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

 

[blue_leaf77]

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:
$$\begin{align*} \langle S^2 \rangle &= \langle \psi|S^2|\psi\rangle=4\hbar^2,\\ \langle S^4 \rangle &= \langle \psi|(S^2)^2|\psi\rangle=21\hbar^4\\ \Longrightarrow \delta_{S^2} &=\sqrt{5} \ \hbar^2 \end{align*}$$

Yes.

Hmm, how do I do that, how can I get mxmxm_x? Do mxmxm_x and mzmzm_z have any connection?

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.

 

[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

 

[blue_leaf77]

$$\begin{align*} |S,S_z\rangle&=\alpha |S,S_x\rangle\\ |1,1\rangle &= |10\rangle \ \ \textrm{because the z component is 1, x and y must be 0 - right thinking?}\\ |1,0\rangle &= {1 \over \sqrt{2}} \big(|11\rangle +|1-1\rangle \big)\\ |1,-1\rangle &= |10\rangle\\ \end{align*}$$

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.

 

[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:

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

write them in vector form using |S,mz⟩|S,mz⟩|S,m_z\rangle basis.

I'm not sure how to do it.

 

[blue_leaf77]

I'm not sure how to do it.

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

 

[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:
$$\begin{align*} |1-1\rangle &= {1 \over 2} |11\rangle - {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \\ |10\rangle &= - {1 \over \sqrt{2}} |11\rangle + {1 \over \sqrt{2}} |1-1\rangle \\ |11\rangle &= {1 \over 2} |11\rangle + {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \end{align*}$$

 

[blue_leaf77]

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:
$$\begin{align*} |1-1\rangle &= {1 \over 2} |11\rangle - {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \\ |10\rangle &= - {1 \over \sqrt{2}} |11\rangle + {1 \over \sqrt{2}} |1-1\rangle \\ |11\rangle &= {1 \over 2} |11\rangle + {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \end{align*}$$

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.

 

[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.

 

[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.

 

[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?

 

[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.

 

[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?

 

[blue_leaf77]

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?

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

 

[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!

 

[blue_leaf77]

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!

You are welcome.

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→xz→xz \rightarrow x and x→−zx→−zx \rightarrow -z and only the inital condition changes, so S1z|ψ1⟩=ℏ|ψ1⟩S1z|ψ1⟩=ℏ|ψ1⟩S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle and S2x|ψ2⟩=−ℏ|ψ2⟩S2x|ψ2⟩=−ℏ|ψ2⟩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.

Actually, since the problem asks you to calculate the probability for the x components given the system in the state as described by the initial condition, you still need to know the connection between the z eigenvectors and x eigenvectors. But maybe your prof still has some more steps with his approach to reach the same final answer.