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Two particles with spin - measurements

  1. Aug 28, 2016 #1
    1. The problem statement, all variables and given/known data

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

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

    3. The attempt at a solution

    a) [itex]|\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.[/itex] Can someone just check this?
    b)[tex]
    \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*}
    [/tex]
    For [itex]S^2|\psi_{12}\rangle=\hbar^2 s(s+1)|\psi_{12}\rangle,[/itex] we get:
    [tex]
    \begin{align*}
    &Results \ \ \ \ &Probability\\
    &6\hbar^2 &{13\over24}\\
    &2\hbar^2 &{3 \over 8}\\
    &0 &{1 \over 12}
    \end{align*}
    [/tex]

    c) Expectation value is [itex]\langle S^2 \rangle = \langle \psi|S^2|\psi\rangle=4\hbar^2,[/itex] but I can't find uncertainty? I am thinking in this way:
    [tex]\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}?[/tex]

    d) How do I find outcomes and probabilities? I tried with [itex]S_x=\frac{S_++S_-}{2}[/itex], 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 [itex]\pm {\hbar \over 2}[/itex], 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. jcsd
  3. Aug 28, 2016 #2

    Charles Link

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    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".
     
  4. Aug 28, 2016 #3

    blue_leaf77

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    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
  5. Aug 29, 2016 #4
    We measure the [itex]S^2[/itex] of both particles together, so because of that, I tried to write my wavefunction in [itex]|S,m\rangle[/itex] form. Is this right?


    Yes, sorry, I made a mistake when I was copying it from my handwriting. I forgot a [itex]|10\rangle[/itex] state, so the whole function is:
    [tex]
    |\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,
    [/tex]
    but this element is already included in the rest of my calculations in this exercise.
     
  6. Aug 29, 2016 #5

    blue_leaf77

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    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}##.
     
  7. Aug 29, 2016 #6
    So if I understand this correctly, I must use first [itex]S^2[/itex] on my wavefunction to get [itex]\langle S^2\rangle[/itex] and just two times [itex]S^2[/itex] on wavefunction to get [itex]\langle S^4 \rangle[/itex]. If I do this, I get:
    [tex]
    \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*}
    [/tex]

    Hmm, how do I do that, how can I get [itex]m_x[/itex]? Do [itex]m_x[/itex] and [itex]m_z[/itex] have any connection?
     
  8. Aug 29, 2016 #7

    blue_leaf77

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    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.
     
  9. Aug 29, 2016 #8
    I think I wil need more help and hints here. So I am thinking this way:
    [tex]
    \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*}
    [/tex]

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

    I found something similar here: https://en.wikipedia.org/wiki/Spin_(physics)#Higher_spins
     
  10. Aug 29, 2016 #9

    blue_leaf77

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    Nope.
    The matrix of ##S_x## for ##S=1## spin in ##|S,m_z\rangle## basis is
    [itex]
    \frac{\hbar}{\sqrt{2}}
    \left(
    \begin{array}{cc}
    0 & 1 & 0\\
    1 & 0 & 1\\
    0 & 1 & 0
    \end{array}
    \right).
    [/itex]
    Now find the eigenvectors of this matrix, then write them in vector form using ##|S,m_z\rangle## basis.
     
  11. Aug 29, 2016 #10
    Ahaaa, so I was thinking wrong. I have to apply an operator [itex]S_x[/itex] on every [itex]|S,m_z\rangle[/itex] state and then write a matrix, so:
    [tex]
    \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*}
    [/tex]

    from where I get your matrix. :)

    Ok, the eigenvectors are:

    [tex]
    \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*}
    [/tex]

    I'm not sure how to do it.
     
  12. Aug 29, 2016 #11

    blue_leaf77

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    I will give one example. Let's take ##\vec x_1 = |S_x;-1\rangle##,
    $$
    \begin{aligned}
    \vec x_1 &= |S_x;-1\rangle = ({1\over 2}, {-1\over \sqrt{2}},{1\over 2})^T = {1\over 2}(1, 0,0)^T - {1\over \sqrt{2}}(0, 1,0)^T + {1\over 2}(0, 0,1)^T \\
    &= {1\over 2}|S_z;1\rangle - {1\over \sqrt{2}}|S_z;0\rangle + {1\over 2}|S_z;-1\rangle
    \end{aligned}
    $$
     
  13. Aug 29, 2016 #12
    Aha, I just didn't know how to write [itex]|S,m_z\rangle [/itex] as a basis, so I only choose (1,0,0),(0,1,0),(0,0,1).

    So [itex] |S,m_x\rangle = \alpha |S,m_z\rangle[/itex] would be:
    [tex]
    \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*}
    [/tex]
     
  14. Aug 29, 2016 #13

    blue_leaf77

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    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.
     
  15. Aug 29, 2016 #14
    Yes, I know, I was confused. :D

    Aha, I have already tried with the [itex]S_x[/itex] matrix before, to write its inverse. So firstly, I have to write [itex]S_x[/itex] matrix, then find its eigenvectors, build another matrix from this eigenvectors and then find the inverse:
    [tex]
    \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)
    [/tex]

    EDIT: The inverse was wrong.
     
    Last edited: Aug 29, 2016
  16. Aug 29, 2016 #15

    blue_leaf77

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    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.
     
  17. Aug 29, 2016 #16
    Do I have to normalize this matrix from my previous post first? Because if I write [itex]|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 [/itex] its not? Or am I wrong again?
     
  18. Aug 29, 2016 #17

    blue_leaf77

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    Actually I just checked the inverse of that matrix in post #14 using an online calculator and it returns a different answer from yours.
     
  19. Aug 29, 2016 #18
    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 [itex] |S_{1z};1\rangle|S_{2z};-1\rangle [/itex] for example, there are 9 elements, at [itex] |S_{1z};1\rangle|S_{2z};0\rangle [/itex] there are 6 and at [itex] |S_{1z};1\rangle|S_{2z};1\rangle [/itex] nine again?
     
  20. Aug 29, 2016 #19

    blue_leaf77

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    Yes that's right. That's a lengthy expression but you are actually just one step from finished.
     
  21. Aug 31, 2016 #20
    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 [itex]z \rightarrow x[/itex] and [itex]x \rightarrow -z[/itex] and only the inital condition changes, so [itex]S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle[/itex] and [itex]S_{2x}|\psi_2\rangle = - \hbar |\psi_2\rangle[/itex] 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!
     
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