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Determine the state of a particle

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data

    It is known that there is a 36% probability of obtaining [itex]S_z = \hbar/2[/itex] and therefore a 64% chance of obtaining [itex]S_z = -\hbar/2[/itex] if a measurement of [itex]S_z[/itex] is carried out on a spin 1/2 particle. In addition, it is known that the probability of finding the particle with [itex]S_x = \hbar/2[/itex], that is, in the state |+x>, is 50%. Determine the state of the particle as completely as possible from this information.


    2. Relevant equations



    3. The attempt at a solution

    So given the above information, I know the following:

    |<+x |[itex]\Psi>|^{2}[/itex] = .5

    |<+z |[itex]\Psi>|^{2}[/itex] = .36

    |<-z |[itex]\Psi>|^{2}[/itex] = .64

    Therefore, [itex]\Psi[/itex] can be written as:

    [itex]| \Psi> = (1/\sqrt{2})[/itex] |+x> + [itex](1/\sqrt{2})[/itex] |-x>

    or

    [itex]| \Psi>[/itex] = .36 |+z> + .64 |-z>


    This is what I'm thinking at least, but I feel like I'm missing something crucial. For instance, do I need to consider a phase? I feel like I shouldn't have to since the probability amplitude will not change even if there is some phase change. I'd definitely appreciate any thoughts on whether or not this is what the question's looking for. It seems a bit too straight forward, so I'm thinking I'm missing something. Thanks so much.
     
  2. jcsd
  3. Sep 4, 2014 #2

    Matterwave

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    First of all, your z-basis ##\Psi## is not right since the square modulus of the coefficients don't add up to 1 (you're missing square-roots). Second, have you tried to see if the first and second ##\Psi## are compatible with each other? In other words, are they really the same ##\Psi##?
     
  4. Sep 4, 2014 #3
    Thanks for the reply.

    You're right, I missed the square roots on z-basis expression. Thanks!

    I haven't checked whether [itex]\Psi[/itex] is really the same. I know that the inner product of [itex]\Psi[/itex] with itself should give me 1 (if indeed it's the same). But wouldn't I need to somehow go between x and z to do that?
     
  5. Sep 4, 2014 #4

    Matterwave

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    Do you know how |x+> is related to |z+> and |z->?
     
  6. Sep 5, 2014 #5
    You have to find a relation between ##|+x>##, ##|-x>## and ##|+z>##,##|-z>##.
    You can do it with Pauli's Operator:
    $$|\pm x>=e^{i \frac{\theta}{2} \vec{k} \cdot \hat{\sigma}}|\pm z>$$
    where ##\vec{k}=\vec{e_1}\times \vec{e_3}/|\vec{e_1}\times \vec{e_3}|## with ##\vec{e_1}## and ##\vec{e_3}## euclidean basis vector relative to the x axis and the z axis. Furthermore ##\cos{\theta}=\vec{e_1}\cdot \vec{e_3}##.
    Explicitly, the 2 ket are given by:
    $$|+x>=|+z>\cos{\frac{\theta}{2}}+|-z>\sin{\frac{\theta}{2}}$$
    $$|-x>=-|+z>\sin{\frac{\theta}{2}}+|-z>\cos{\frac{\theta}{2}}$$

    i hope that i haven't done mistakes.
     
  7. Sep 5, 2014 #6
    I thought it was realated via:

    |+x> =1/[itex]\sqrt{2}[/itex] (|+z> + 1/[itex]\sqrt{2}[/itex] (|-z>

    but I'm not sure how this changes for |-x>.
     
  8. Sep 5, 2014 #7

    Matterwave

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    Luca gave you the answer in his post, there's a sign change basically.

    What I wanted you to realize though is that the first ##\Psi## you wrote down in your first post is actually ##\left|+z\right>## so it is certainly not equal to the second ##\Psi## (even with the square root corrections) you wrote down!
     
  9. Jan 12, 2015 #8
    I'm trying to solve the same problem. Why does the first Ψ that he wrote down represent |+z>? Also is the second Ψ not sufficient enough to describe the state of the particle?
     
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