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Eigenvalues of 4x4 Hermitian Matrix (Observable)

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the allowed energies for a spin-3/2 particle with the given Hamiltonian:
    [tex]\hat{H}=\frac{\epsilon_0}{\hbar}(\hat{S_x^2}-\hat{S_y^2})-\frac{\epsilon_0}{\hbar}\hat{S_z}[/tex]

    3. The attempt at a solution
    The final matrix I get is:

    \begin{pmatrix}
    \frac{3}{2} & 0 & \hbar\sqrt{3} & 0\\
    0& \frac{\hbar}{2}-\frac{1}{2} & 0 &\hbar\sqrt{3} \\
    \hbar\sqrt{3}& 0 & \frac{\hbar}{2}+\frac{1}{2} & 0\\
    0& \hbar\sqrt{3} & 0 & \frac{3}{2}
    \end{pmatrix}

    My question is: Is there a more quick way to find the eigenvalues of a 4x4 hermitian matrix than going trough the tedious calculation of [itex]det(\hat{H}-\lambda I)=0[/itex]?
     
  2. jcsd
  3. Mar 18, 2012 #2

    tiny-tim

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    Hi Gunthi! :smile:
    If you swap the second and third rows, it becomes two 2x2 matrices. :wink:
     
  4. Mar 19, 2012 #3
    Hi tiny-tim! :smile:

    I would also have to switch the 2nd and 3rd columns right? Then I would just calculate the eigenvalues of the 2x2 matrices separately?

    I've been searching for properties of block matrices that could justify this, but to no avail. Is there a theorem that demonstrates this property? Or could you explain how this works?

    I guess I'm rustier than I thought at my algebra :redface:
     
  5. Mar 19, 2012 #4

    tiny-tim

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    Hi Gunthi! :smile:
    We're only re-arranging

    instead of the basis x y z t (or whatever), we're using x z y t :wink:

    To put it another way, can't you immediately see, just by looking at it, that the matrix is in two parts that operate completely separately? :smile:
     
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