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Spin-1 particles' spin operator

  1. Oct 13, 2013 #1
    The S[itex]_{z}[/itex] operator for a spin-1 particle is

    S[itex]_{z}[/itex]=[itex]\frac{h}{2\pi}[/itex][1 0 0//0 0 0//0 0 -1]

    I'm given the particle state

    |[itex]\phi[/itex]>=[1 // i // -2]

    What are the probabilities of getting each one of the possible results?

    Now... we can say the possible measure results will be 1,0,-1 and the autovectors associated to each result is

    v[itex]_{1}[/itex] = [1 // 0 // 0]

    v[itex]_{0}[/itex] = [0 // 0 // 0]

    v[itex]_{-1}[/itex] = [0 // 0 // -1]

    I 've normalized the given state. It gives me

    |[itex]\phi[/itex]>=[itex]\frac{1}{\sqrt{6}}[/itex][1 // i // -2]

    which can be written as

    |[itex]\phi[/itex]>=[itex]\frac{1}{\sqrt{6}}[/itex](v[itex]_{1}[/itex] + [ 0 // i // 0] + 2v[itex]_{-1}[/itex])

    I would expect to write this state to be a combination of v[itex]_{1}[/itex], v[itex]_{0}[/itex] and v[itex]_{-1}[/itex] but this vector [ 0 // i // 0 ] is tricking me. What measure is associated to it and what is the probability to measure zero?

    Thank you.
    Last edited: Oct 13, 2013
  2. jcsd
  3. Oct 13, 2013 #2


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    Your v[itex]_{0}[/itex] vector is written incorrectly.
  4. Oct 13, 2013 #3
    I'm sorry... what's wrong with it?
  5. Oct 13, 2013 #4


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    An eigenvector (autovector) is never the zero vector .
  6. Oct 13, 2013 #5
    So how can I find the eigenvector associated to the measure of the eigenvalue zero?

    What I did was to find the eigenvalues with the condition that


    where [itex]\omega[/itex] is the eigenvalue and I is the identity matrix. This procedure gave me


    I then used these results (one at a time) in


    and for [itex]\omega[/itex]=0 this equation gave me v=[0 // 0 // 0]

    I can see why the zero vector cannot be an eigenvector but how do I find the vector associated to [itex]\omega[/itex]=0 then?
  7. Oct 13, 2013 #6


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    Recheck your work for the ##\omega = 0## case. If you still get the zero vector then show your work for this case.
  8. Oct 13, 2013 #7
    Beginning from the eigenvalue equation


    we have

    [[itex]\hbar[/itex] 0 0 // 0 0 0 // 0 0 -[itex]\hbar[/itex]] * [v[itex]_{a}[/itex] // v[itex]_{b}[/itex] // v[itex]_{c}[/itex]]= [itex]\omega[/itex][v[itex]_{a}[/itex] // v[itex]_{b}[/itex] // v[itex]_{c}[/itex]]

    where v[itex]_{a}[/itex], v[itex]_{b}[/itex] e v[itex]_{c}[/itex] are the components of the vector [itex]\stackrel{\rightarrow}{v}[/itex].

    Applying the matrix operator S[itex]_{z}[/itex] we have the equations




    if [itex]\omega[/itex]=0 I see v[itex]_{a}[/itex] and v[itex]_{c}[/itex] are zero too but I'm not sure about v[itex]_{b}[/itex] because


    It could be anything. How can I find it?
  9. Oct 13, 2013 #8


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    Right, v[itex]_{b}[/itex] could be any number. But you want your eigenstates to be normalized.
  10. Oct 13, 2013 #9
    Right right... in this case I have the condition v[itex]^{2}[/itex][itex]_{b}[/itex]=1. And then v[itex]_{b}[/itex]=[itex]{\pm}[/itex]1.

    I could define a normalization constant N which would be N=[itex]\frac{1}{\sqrt{v_{b}}}[/itex] but it is simpler to choose v[itex]^{2}[/itex][itex]_{b}[/itex]=[itex]{\pm}[/itex]1.

    Am I free to choose the sign? I think I am. I don't see any constraint to my choice.
  11. Oct 13, 2013 #10


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    Yes, you can choose either sign (or even choose a complex number of magnitude 1). But, there is one choice that will look the best!
  12. Oct 13, 2013 #11
    In this case I will choose v[itex]_{b}[/itex]=1.

    Thank you!
  13. Oct 13, 2013 #12


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    Good. Note, for the vector V-1 you could have chosen [0, 0, 1] rather than [0, 0, -1].

    Then your eigenvectors are [1, 0, 0], [0, 1, 0], and [0, 0, 1] which is the "standard" representation.
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