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Consider Two Spin-1/2 Particles

  1. Dec 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider two spin 1/2 particles. Initially these two particles are in a spin singlet state. If a measurement shows that particle 1 is in the eigenstate of ##S_x = -\hbar/2##, what is the probability that particle 2 in this same measurement is in the eigenstate of ##S_z = \hbar/2##?
    2. Relevant equations
    ##\chi^{(x)}_{\pm} = \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{\pm1}{\sqrt{2}}\end{pmatrix}## and
    ##\chi^{(z)}_{+} = \begin{pmatrix}1\\0\end{pmatrix}##
    ##\chi^{(z)}_{-} = \begin{pmatrix}0\\1\end{pmatrix}##

    3. The attempt at a solution
    Okay so I am not entirely sure how to proceed, my instinct tells me to setup a linear combination of two measurements such that the linear combo is equal to one, i.e.:
    $$
    \begin{pmatrix}1\\1\end{pmatrix} = a
    \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{pmatrix} + b \begin{pmatrix}1\\0\end{pmatrix}
    $$ But that gives a or b =1. So that is not correct. What am i doing wrong?
     
  2. jcsd
  3. Dec 19, 2014 #2

    Orodruin

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    What is the state vector for the singlet state? Which state is the second spin going to be in if you measure the first to be in state Sx = - hbar/2?
     
  4. Dec 19, 2014 #3
    The singlet state is:
    $$|00\rangle = \frac{1}{\sqrt{2}}(\uparrow\downarrow-\downarrow\uparrow)$$
    As I understand it, the total spin that the two particles can carry is 0. So that would mean that if the first measurement is ##-\hbar/2## then the other must yield ##+\hbar/2## for the total spin to be zero.
     
  5. Dec 20, 2014 #4

    Orodruin

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    So what is the probability of a particle in the Sx = hbar/2 state to be in the Sz = hbar/2 state?
     
  6. Dec 20, 2014 #5
    So after some thought here is how I went about solving it: If particle one has ##S_x = -\hbar/2## then particle two must have ##S_x = \hbar/2## therefore the probability of measuring ##\hbar/2## for ##S_z## for particle two is: $$\left|\langle\chi^{(x)}_{+}|\chi^{(z)}_{+}\rangle\right|^2 = \begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \frac{1}{2}$$ I wanted to make sure that I could justify the answer mathematically.
     
  7. Dec 29, 2014 #6

    DEvens

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    Just for fun, consider what the result of the probability being 1/2 means. If you had no knowledge whatever of the state of the second particle, if it was completely random up or down, you would get half. But you have been told that it is part of a singlet state, and the other particle in that state is down. And the probability is still 1/2 of it being up. Does that seem correct?

    Consider exactly what it means to state that the system is in the singlet state, and to state that one of the particles is definitely in the down state. Can both of these statements be true at the same time? What does it mean to perform a measurement on the system? After the measurement, what will the system look like? (Hint: It won't be in the singlet state any more.) Think projection operators.
     
  8. Dec 29, 2014 #7

    Vanadium 50

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    But what is it given that you know it already for particle one?
     
  9. Dec 29, 2014 #8

    Orodruin

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    Yes, this can (and is) definitely true as the problem states that the spin of the particles are considered in different directions. As the original state is a singlet and the first particle is found in the spin down state in the x-direction, the second particle must now be in the spin up state in the x-direction. This spin state has probability 1/2 to be in the spin up state of the z-direction, as found by the OP.
     
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