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Linear Combos of Y; eigenfunction of L_x

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    I have to find the linear combinations of Y_10, Y_11, and Y_1-1 that are eigenfunctions of L_y. There are three such combinations...

    2. Relevant equations

    3. The attempt at a solution

    Starting from (L_y)(psi_y)=(alpha)(psi_y),
    Using the relationshiP: psi_y= aY_11 + bY_10 + cY_1-1
    And: L_y=(i/2)(L_(-) - L_(+))

    I solved it to the point where I got to:
    (alpha)(psi_y) = (alpha)aY_11 + (alpha)bY_10 + (alpha)cY_1-1 = (i/sqrt(2))(bY_1-1 - bY_11 + (a-c)Y_10)

    What I have to solve is this system of equations for when alpha=0, 1, -1:
    (ib/sqrt(2)) = (alpha)c
    (-ib/sqrt(2) = (alpha)a
    i(a-c)/sqrt(2) = (alpha)b

    So for instance, when alpha=-1, the answers I know are: a=1/2, c=-1/2, b=-i/sqrt(2).

    Which gives one of the three linear combinations: psi_y=(1/2)Y_11 - (i/sqrt(2))Y_10 - (1/2)Y_1-1.

    However, I have no idea how to solve that system of equations, though I really feel like I should :( So I guess it really boils down to me not knowing the math..... help..?
  2. jcsd
  3. Nov 19, 2007 #2
    Do you know how to find eigenvalues and eigenvectors? You have three simultaneous relations between a,b and c. Write down the whole set as matrix multiplying the column vector (a,b,c)^t. Then use the fact that a matrix equation of the form [itex]A \mathbf{a} = \mathbf{0}[/itex] has nontrivial solutions iff det A = 0. The solution of the charcteristic polynomial gives you the possible values of alpha. Substituting those back into your matrix equation will give you the appropriate values of a,b and c.
  4. Nov 19, 2007 #3
    Can you lead me to a website that explains how to solve it that way? I think I've heard of that method but never done it myself...
  5. Nov 19, 2007 #4
    Actually, unless I misunderstood you, I guess I already know the possible values of alpha, 0, 1, and -1. I'm just having a brain freeze and cannot figure out how to solve this set of equations for values of a, b, and c after substituting in each value of alpha...
  6. Nov 19, 2007 #5
    You need to prove that alpha = -1,0,1. You do this by finding the eigenvalues of that matrix.

    Do you know how to write the system of equations

    (ib/sqrt(2)) = (alpha)c
    (-ib/sqrt(2) = (alpha)a
    i(a-c)/sqrt(2) = (alpha)b

    as a matrix?
  7. Nov 19, 2007 #6
    No :(
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