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Constructing the input state given the probabilities

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    The spin components of a beam of atoms prepared in the state ##|\psi_{in}\rangle## are measured and the following experimental probabilities are obtained,
    $$P_{+}=\frac{1}{2}, \quad P_{+x}=\frac{3}{4}, \quad P_{+y}=0.067$$
    $$P_{-}=\frac{1}{2}, \quad P_{-x}=\frac{1}{4}, \quad P_{-y}=0.933$$

    2. Relevant equations
    ##P_+=|\langle+|\psi_{in}\rangle|^2 \quad## Probability of the input state to be in the up state (z-direction)

    3. The attempt at a solution
    I can write ##|\psi_{in}\rangle## as a linear combination of the up and down state,

    $$|\psi_{in}\rangle = a|+\rangle + b|-\rangle$$

    by projecting the input state to the up state, we get ##a=\frac{1}{2}\,## and similarly for the down state, ##\, b=\frac{1}{2}##.
    Note that the coefficients are complex numbers so there is a phase term, but the overall phase is not physically interesting and we only consider the phase difference of the coefficients. So,

    $$|\psi_{in}\rangle = \sqrt{\frac{1}{2}}(|+\rangle + e^{i\theta}|-\rangle)$$

    to determine ##\theta##, we project the previous equation to the ##|+\rangle_x = \sqrt\frac{1}{2}(|+\rangle + |-\rangle)## and ##|-\rangle_x = \sqrt\frac{1}{2}(|+\rangle - |-\rangle)##

    $$|_x\langle+|\psi_{in}\rangle|^2 = \frac{3}{4} = \frac{1}{4}(1+e^{i\theta})(1+e^{-i\theta}) \rightarrow 1 = e^{i\theta} + e^{-i\theta} = 2cos(\theta)$$

    This shows that ##\theta = \frac{\pi}{3}##, by similar projection for ##|-\rangle_x##, we get ##\theta = \frac{2\pi}{3}##
    In this case, I have to know which ##\theta## I should choose so I think I should project it to ##|+\rangle_y## and ##|-\rangle_y##. But should I just do trial and error and see where the results make sense?
     
    Last edited: Feb 12, 2017
  2. jcsd
  3. Feb 12, 2017 #2

    TSny

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    OK
    I get something else here.

    Try the same way you did for the projection onto ##|+\rangle_x## and ##|-\rangle_x##
     
  4. Feb 12, 2017 #3
    I'm sorry, it should be ##\theta = \frac{2\pi}{3}##, I edited my post.
     
  5. Feb 12, 2017 #4
    What I'm thinking is to project the input state to ##|+\rangle_y## and ##|-\rangle_y## then use the angles I've got and see which one would correspond to the probability given in the y measurement. Do you think this is a valid way, so far ##\theta = \frac{2\pi}{3}## matches the results.
     
  6. Feb 12, 2017 #5

    TSny

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    The value of ##\theta## for the projection onto ##|-\rangle_x## must be consistent with ##\theta## for projection onto ##|+\rangle_x##.
     
  7. Feb 12, 2017 #6
    So you mean ##\theta = \frac{2\pi}{3}## is not the angle I want?
     
  8. Feb 12, 2017 #7

    TSny

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    I don't think so. For the projection onto ##|+\rangle_x## there is another possibility. That is, ##\cos \theta = 1/2## has more than one solution. Similarly, for ##|-\rangle_x## there will be more than one solution for ##\theta##. Hopefully, one of the solutions for ##|+\rangle_x## matches one of the solutions for ##|-\rangle_x##.
     
  9. Feb 12, 2017 #8
    I think to identify the correct angle is to test the angles on the projection of ##|\psi_{in}\rangle## onto ##|+\rangle_y## and ##|-\rangle_y##
    I've found that if I use ##\theta = \frac{\pi}{3}##. The probability turns out to be what the experimental results say.
     
    Last edited: Feb 12, 2017
  10. Feb 12, 2017 #9

    TSny

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    Yes, I think that's the correct answer. It should give you the right probability for ##|-\rangle_x## also.
     
  11. Feb 13, 2017 #10
    Yes, I just need to correct myself in my initial post, the angles that I should get for both ##|+\rangle_x## and ##|-\rangle_x## are ##\theta = ±\frac{\pi}{3}##, I made a lousy mistake when projecting ##|\psi_{in}\rangle## onto ##|-\rangle_x##. It should be the same for both states, and I should verify which angle is the right one by projecting ##|\psi_{in}\rangle## onto ##|+\rangle_y## and ##|-\rangle_y##.
     
  12. Feb 13, 2017 #11

    TSny

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    I'm not quite getting that. I'm geting that it's ##\theta = ±\frac{\pi}{3}## for ##|+\rangle_x##, but it's ##\theta = \frac{\pi}{3}## and ##\theta = ## something else for ##|-\rangle_x##.
     
  13. Feb 13, 2017 #12
    ##|-\rangle_x = \sqrt\frac{1}{2}(|+\rangle-|-\rangle), \quad |\psi_{in}\rangle = \sqrt\frac{1}{2}(|+\rangle + e^{i\theta}|-\rangle)##

    ##\langle-|\psi_{in}\rangle = \frac{1}{2}(1-e^{i\theta}) \quad \rightarrow \quad |\langle-|\psi_{in}\rangle|^2 = \frac{1}{4} = \frac{1}{4}(1-e^{i\theta})(1-e^{-i\theta}) = \frac{1}{4}(2-e^{i\theta}-e^{-i\theta}) = \frac{1}{4}(2-2cos(\theta)) = \frac{1}{2}(1-cos(\theta))##
    ##\rightarrow \frac{1}{4} = \frac{1}{2}(1-cos(\theta)) \quad \rightarrow \quad cos(\theta) = \frac{1}{2}##
    Thus ##\theta = ±\frac{\pi}{3}##
     
  14. Feb 13, 2017 #13

    TSny

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    Yes, you are right. I didn't bother to look back at my notes and I was confusing the case of ##|-\rangle_x## with ##|+\rangle_y## and ##|-\rangle_y## (where I think you get ##\theta = \pi/3## and ##\theta = 2\pi/3##). Sorry, for causing you to have to take the time to correct me.

    Good work.
     
  15. Feb 13, 2017 #14
    At least we got the chance to discuss this. Thanks for the discussion!
     
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