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Pauli Matrices and orthogonal projections

  1. Jul 19, 2004 #1
    Ok, I'm working with the Pauli Matrices, and I've already gone through showing a few bits of information. I've got a good idea how to keep going, but I'm not exactly sure about this one--

    say M= 1/2(alphaI + a*sigma)

    where alpha E C, a=(ax, ay, az) a complex vector, a*sigma=ax sigmax+ay sigmay+ az sigmz, and I is the identity matrix.

    So, an orthogonal projection means that for a matrix P, P^2 and P dagger are both equal to P, right?

    Supposedly alpha and a can beonstrained so that M is an orthogonal projection.

    How would I go about doing that? :confused:

    Thanks muchly!
  2. jcsd
  3. Jul 19, 2004 #2
    Some hints

    Hey there !

    Here are some hints on a was to do it :

    You will only use the fact that [tex]M^2 = M[/tex]...

    1) Compute [tex] M^2 [/tex] :

    [tex] M^2 = \cfrac{1}{4}\left(\alpha I + \sum_i a_i \sigma_i \right)^2 = ...[/tex]

    You will find a term like [tex] \sum_i \sum_j a_i a_j \sigma_i \sigma [/tex]

    In order to reduce this term, use some common identities :

    [tex] \sigma_i \sigma_j = I \delta_{ij} + i \epsilon_{ijk} \sigma_k [/tex]


    [tex] \sigma_i \sigma_j + \sigma_j \sigma_i = 2 \delta_{ij} I[/tex]

    If you applied them correctly, you should get something quite simple. You then use the fact that [tex] M^2 = M [/tex] and you will find by identification :
    [tex] \alpha = 1 \text{ and } ||a|| = 1 [/tex], which is your final answer (Hopefully, I didn't mess up).

    If you're not confident with the use of the Levi-Civita symbol, another (more tedious) way is to write all in matrix notation (2x2), compute M^2 and put this equal to M... You will get 4 equations, which will reduce to the answer given above.

    Hope this helps,
  4. Jul 20, 2004 #3
    i must've screwed something up, for i'm getting a sqrt(2) for my alpha?

    maybe i screwed up something with the identities...
  5. Jul 20, 2004 #4
    oh, and i almost forgot, THANK YOU VERY MUCH!!!!
  6. Jul 20, 2004 #5
    can we solve the particle in a box problem using schrodingers equation?
  7. Jul 20, 2004 #6

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    Sure, it's easy; they do it in Halliday and Resnick.

    (But not with the Pauli spin matrices.)
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