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Eigenspinor Sx for Spin 1 Particle

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data



    I use Griffiths and see the example for spin 1/2 to derive the eigenspinor of Sx. I just can't seem to follow how he get from there or how he is measuring the probability for a given state.

    2. Relevant equations

    I have correctly derived the spin operator for Sx for a spin 1 particle:
    Sx = [itex]ħ/\sqrt{2}\left| \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right| [/itex]


    3. The attempt at a solution


    Trying to solve the equation

    [itex]ħ/\sqrt{2}\left| \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right| \left| \begin{matrix} a \\ b \\ c \end{matrix} \right|

    =

    λ \left| \begin{matrix} a \\ b \\ c \end{matrix} \right|



    [/itex]

    where a, b, c are the three spin states: up, zero, down for a spin 1 particle. And of course λ are the eigenvalues for this operator, which I found to be λ = 0, +/- ħ.


    I'm stuck on how to proceed further. How do I find the eigenspinsors and how do I move on to make probability measurements for each state, given some initial state?

    I'd appreciate any help at all since I've made myself quite confused. Thank you!
     
  2. jcsd
  3. May 6, 2012 #2
    Write down that matrix equation for each eigenvalue. In each case, then, you will have a system of three equations, which you can solve for [itex]a, b, c[/itex] any way you want (for example, finding the reduced row echelon form of the matrix, or solving the equations one at a time using substitution).
     
  4. May 6, 2012 #3

    Ok, thanks. I did that and.. do I assume that for λ = 0, the solution is trivial?

    Using + ħ I get these relations
    [itex]\frac{1}{\sqrt{2}}[/itex]b = a
    [itex]\frac{1}{\sqrt{2}}[/itex](a+c) = b
    [itex]\frac{1}{\sqrt{2}}[/itex]b = c

    And using - ħ I get these relations
    [itex]\frac{1}{\sqrt{2}}[/itex]b = -a
    [itex]\frac{1}{\sqrt{2}}[/itex](a+c) = -b
    [itex]\frac{1}{\sqrt{2}}[/itex]b = -c

    From these I get

    b = +/- [itex]\sqrt{2}[/itex]a
    b = +/- [itex]\sqrt{2}[/itex]c
    a + c = +/- [itex]\sqrt{2}[/itex]b

    So do I use either value for b? And do I assume from these that both a and c are

    a = +/-[itex]\frac{1}{\sqrt{2}}[/itex]b
    c = +/-[itex]\frac{1}{\sqrt{2}}[/itex]b

    I'm sorry if I'm not seeing far enough, but how would I proceed from here?
     
  5. May 6, 2012 #4
    The trivial solution is one solution. Work it out to see if there are others.

    You have to solve each case for separate eigenvalues as a coherent set [itex](a,b,c)[/itex]. So, for example, for the positive eigenvalue case, you have a set of three equations in three unknowns, which must be solved independently of what happens for either eigenvalues. The resulting set [itex](a,b,c)[/itex] is only associated with the eigenvalue it was derived from.
     
  6. May 6, 2012 #5
    Thank you... I still don't seem to understand. Work out the trivial solution? And taking the positive eigenvalue case, I seem to get tautological answers, like b = b, and a = a. Do you mean I need to come up with any column vector that satisfies this relations? Like based on this same positive eigenvalue, would I choose

    [itex](a,b,c) = (\frac{1}{\sqrt{2}}, 1, \frac{1}{\sqrt{2}})[/itex]

    and for the negative

    [itex](a,b,c)= (\frac{-1}{\sqrt{2}}, -1, \frac{-1}{\sqrt{2}})[/itex]


    For the zero eigenvalue, would I just get (0,0,0)? But that not be included in the basis since it's linearly dependent? Or would they all be equal and I could choose (1,1,1)?

    Then would the generic spinor for S_x be

    [itex] X = (\frac{a+c}{\sqrt{2}} + b) X_+ + (\frac{-a-c}{\sqrt{2}} - b)X_- [/itex]

    with the zero state not included?

    Sorry, I'm still not sure how to proceed..
     
  7. May 6, 2012 #6

    vela

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    Note that to find the eigenvalues and eigenvectors satisfy the matrix equation (A-λI)x=0, where A is your matrix, I is the identity matrix, and x is the eigenvector. The only way this equation can have a non-trivial solution is when det(A-λI)=0, so when you plug your eigenvalues back into the equations to solve for x, you're going to necessarily end up with a dependent system. You will only be able to solve for x up to a multiplicative factor.

    You need to find non-trivial solutions for each eigenvalue. The trivial solution x=0 will obviously always satisfy the equation, but eigenvectors are, by definition, non-trivial solutions to the equation (A-λI)x=0.
     
  8. May 6, 2012 #7

    vela

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    Note that if you add the first and third equations together, you get the second equation, so you can get rid of the second equation as it's not independent. Solving for everything in terms of b, you get
    \begin{align*}
    a &= \frac{1}{\sqrt{2}} b \\
    b &= b \\
    c &= \frac{1}{\sqrt{2}} b
    \end{align*} Written in vector notation, you'd have
    $$\begin{pmatrix} a \\ b \\ c \end{pmatrix} = b\begin{pmatrix} \frac{1}{\sqrt{2}} \\ 1 \\ \frac{1}{\sqrt{2}} \end{pmatrix}$$ So your eigenvector is a multiple of (1/√2, 1, 1/√2). Typically, you choose the multiple so that the vector is normalized.
     
  9. May 6, 2012 #8

    vela

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    Both of these can't be correct. Eigenvectors for different eigenvalues are linearly independent. Your vectors, in contrast, are multiples of each other.
     
  10. May 7, 2012 #9
    Oh... I can't believe I made that error! I found the spinor for X+/- by varying b only. Same for the X0 part. I guess I was frazzled by the computation. Thanks vela; I'll follow your suggestions and post back here if I need more help. Thanks again :)
     
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