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[PhD Qualifier] Spin singlet state

  1. Jul 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider a system made up of two spin 1/2 particles. Initially the system is prepared in a spin singlet state with total spin [tex]S_{total}=0[/tex]. We'll define the spin components of one particle as [tex](s_{1x},s_{1y},s_{1z})[/tex] and of the other as [tex](s_{2x},s_{2y},s_{2z})[/tex]

    a. If no measurements are made on particle 2, what is the probability that a measurement of the:
    i. [tex]s_{1z}[/tex] component of particle 1 will yield [tex]+\frac{\hbar}{2}[/tex]?
    ii. [tex]s_{1x}[/tex] component of particle 1 will yield [tex]+\frac{\hbar}{2}[/tex]?

    b. Now assume that a measurement of [tex]s_{2z}[/tex] component of particle 2 yielded [tex]+\frac{\hbar}{2}[/tex]. What is the probability that a subsequent measurement of the:
    i. [tex]s_{1z}[/tex] component of particle 1 will yield [tex]+\frac{\hbar}{2}[/tex]?
    ii. [tex]s_{1x}[/tex] component of particle 1 will yield [tex]+\frac{\hbar}{2}[/tex]?

    2. Relevant equations

    Spin singlet state: [tex]|\psi>=\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow>-|\downarrow\uparrow>\right)[/tex]

    3. The attempt at a solution

    a. i. The state is just random, right? p=1/2
    ii. All we're doing is changing the axis we're measuring - nothing physical has set it. p=1/2

    b. i. The total spin of the system is 0. Since particle 2 was spin up, particle 1 must be spin down. p=0
    ii. The measurements of the x and z components of the spin don't commute, so measuring [tex]s_{1x}[/tex] destroys the information we knew about [tex]s_{2z}[/tex], and we're back to case a ii. p=1/2

    Those are my answers and my reasoning - am I correct in my thinking?
     
  2. jcsd
  3. Jul 27, 2008 #2
    I agree with you and all your answers check out for me.
     
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