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Qubit Probabilities and Angles between the measurements

  1. Nov 5, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Suppose that two measurements are made on a qubit in rapid succession. The first is δz and the second is δn. Suppose the first results is always +1. Calculate the probabilities of obtaining the results +/- 1 for the second measurement in terms of the angle θ between z and n. Show that <σn> = cos(θ) for the second measurement.

    2. Relevant equations
    δz = (1, 0), (0,1) In this problem we only use (1,0)
    δn = 1/sqrt(2) ( sqrt(1+nz)), sqrt(1-nz) e^(iθz) ), 1/sqrt(2) ( sqrt(1-nz), -sqrt(1+nz)e^(iθz) )
    <σn> = average

    3. The attempt at a solution

    The diagram of the situation and calculations are pretty straight forward here.

    The problem is, how do I calculate the probabilities in terms of the angle between the vectors? I don't know how to interpret this.

    prob(+1) = (1+nz)/2
    prob(-1) = (1-nz)/2
    prob(+1)+prob(-1) = 1 which checks out.

    The average value <σn> is equal to nz as we take prob(+1)-prob(-1).

    So, I have all the answers, but how do I do this in terms of the angle between them? I thought it was a definition that the average aka <σn> is equal to cos(θ)...


    In my notes, I have that prob(+1) = cos^2(θ/2)
    prob(-1) = sin^2(θ/2)
    but I'm not sure how this would help.

    Any hints on how to do this?
     
  2. jcsd
  3. Nov 6, 2015 #2

    DrClaude

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    I will use the Dirac notation to represent the eigenstates:
    $$
    \sigma_z \left| + z \right\rangle = \left| + z \right\rangle \quad \sigma_z \left| - z \right\rangle = -\left| - z \right\rangle \\
    \sigma_n \left| + n \right\rangle = \left| + n \right\rangle \quad \sigma_n \left| - n \right\rangle = - \left| - n \right\rangle
    $$
    You are correct in saying that after the first measurement (of ##\sigma_z##), the qubit is in state ##\left| + z \right\rangle##. You then need to calculate the probability of measuring the system in either of the eigenstates of ##\sigma_n##, namely ## \left| \langle +n \right|\left. +z \rangle \right|^2## and ##\left| \langle -n \right|\left. +z \rangle \right|^2##.
     
  4. Nov 6, 2015 #3

    RJLiberator

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    I believe I did with the probabilities of (+1) and (-1) in the initial post.

    Those probabilities represent the probabilities that the qubit takes a (+,+) approach or a (+,-) approach.

    Overall, i think I have everything right about this problem (my calculations), but I just don't understand how to express it in terms of the angle between them.
     
  5. Nov 6, 2015 #4

    DrClaude

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    Yes, I see that now. Well, what is nz?
     
  6. Nov 6, 2015 #5

    RJLiberator

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    Nz is the length of z, yes?
    So this would just be 1?
    Hm...

    Would this mean that prob(+1)=1 prob(-1) = 0
    average probability = 1.

    But this doesn't make sense to me as cos(angle between them) must be equal to the average probability. And since n is a generic vector, it can take o ndifferent angles between it and z, so it should fluctuate the probabilities or at least the average probability, I would think.
     
  7. Nov 6, 2015 #6

    DrClaude

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    For this to make sense, you need the eigenvectors to be equal when n points along the z axis. Therefore, what can nz represent?

    (Additional note: what you have written there is not σz and σn, but their eigenvectors, so be careful with σz = ...)
     
  8. Nov 6, 2015 #7

    RJLiberator

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    Hm. We have not learned about eigenvectors in this course.

    That is a good point, so Nz must equal 1 as it works out then. This nz value is the length of z.

    But okay, so Nz = 1, but this implies that the <σn> = 1 = cos(x) where x is the angle between z and n.
    But this angle is 0. So n=z, but that does not make sense to me with reference to the question.

    How can this generic vector always be = z? Why can't I just make the generic vector something else, which would have a different angle and thus a different average probability.
    Or perhaps my definition of average probability = cos(x) is wrong?
     
  9. Nov 6, 2015 #8

    DrClaude

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    My guess is that nz is actually ##\mathbf{n} \cdot \mathbf{z}##.
     
  10. Nov 6, 2015 #9

    RJLiberator

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    No, it should be defined as n_z.
    where
    n = n_x*x+n_y*y+n_z*z where x,y,z are x-hat, y-hat, and z-hat.

    With this representation, however, I can see that nz must equal to 1 as we are taking the z measurement in the +1 direction in our first measurement.
    So this seems to confirm that Nz = 1 and therefore it may be independent of the angle between n and z.

    Why does this make sense? Because when we take the inner product of z = (1 0) and generic vector n, in either case we do not see dependence on the angle between them, but rather only on dependence of N_z.

    Since N_z = 1, we see that the prob(-1) = 0 and prob(+1) = 1
    therefore the average probability is 1 which makes sense since it is N_z.

    But if cos(x) is supposed to equal the average probability, that would mean the angle is 0 degrees, which is the part that doesn't make sense to me. Perhaps I have that interpretation wrong.
     
  11. Nov 6, 2015 #10

    RJLiberator

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    I've found the problem.

    The thing is, we take the dot product between z and the generic vector n. This results in "n_z" since z = 1.

    But since n=1 the dot product is the magnitude of n and z and the cos(x) where x is the angle between them.

    Therefore n_z = cos(x) and all of my calculations follow from that.
     
  12. Nov 6, 2015 #11

    DrClaude

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    Correct. "nz" is indeed ##n_z##, but it comes from ##\mathbf{n} \cdot \mathbf{z}##, hence the appearance of ##\cos \theta##.

    I've been helping you in a couple of threads now, and I must say that there is something that bothers me a bit with the approach you are using to solve these problems. You seem to be using a lot of "ready-made" formulas to calculate the results. This may be how you are being thought the subject, and I'm not blaming you for this approach. But I think that you would really benefit from understanding what is going on one step removed. I suggest that you pick up a good book on quantum mechanics and make sure that you master the basics. In particular, if you can get your hand on J. S. Townsend A Modern Approach to Quantum Mechanics, you will find the first few chapters quite useful.
     
  13. Nov 6, 2015 #12

    RJLiberator

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    Thank you for the suggestion. I will look into it. This assignment is the first time (ever) that I've done 'actual' quantum mechanics. It's very basic material, but it is all new to me.

    As always, Thank you.
     
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