Pauli spin matrix problem

Discussion in 'Physics & Math' started by neelakash, Feb 15, 2008.

  1. neelakash Registered Senior Member

    Messages:
    491
    If P^ is the momentum operator, and σ^ are the three Pauli spin matrices, the eigenvalues
    of (σ^.P^) are

    (a) (p_x) and (p_z) (b) (p_x)±i(p_y) (c) ± |p| (d) ± (p_x + p_y +p_z)

    Pauli matrices are related to rotation.So, (b) looks correct to me.

    [I am a Bachelor level student and this problem belongs to Masters level.I am trying to do this to see if any tricky method, known to me can be used to solve this.]
     
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  3. neelakash Registered Senior Member

    Messages:
    491
    HMmmm
    (b) is not correct as eigenvalues of Hermitian matrix are always real.

    (c) or (d) is correct

    I will choose (c) as we cannot possibly multiply pauli matrix (2x2) with a (3x1) momentum operator matrix [I say 3x1 matrix as three components are specified]
     
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  5. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    8,967
    I think you have to be careful. I think the equation you have is

    \(\vec{\sigma}\cdot \vec{p}\)

    where \(\vec{\sigma} = \left(\sigma_x,\sigma_y,\sigma_z\right)\) is a three vector.

    So the operator looks like

    \(\left(\begin{array}{cc}p_z&p_x+ip_y \\ p_x - ip_y&-p_z\end{array}\right)\)

    Check the minus signs because I always screw up the \(\sigma_y\). Now diagonalize this matrix and you'll find the eigenvalues.
     
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  7. thejusmaha Registered Member

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    1
    Thanks Ben , The eigen values are C rt,
    I think for the particular problem it is enough to find the eigen value by just looking at the trace and determinant
     
  8. Fednis48 Registered Senior Member

    Messages:
    725
    The "shortcut" to this problem is to recognize that the eigenvalues of a Pauli operator are always plus or minus 1, regardless of its direction. In other words, the Pauli vector operator dotted into a unit vector has eigenvalues plus or minus 1. Write momentum as its magnitude times a unit vector, and the answer becomes obvious.
     

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