Eigenvalues of 4x4 Hermitian Matrix (Observable)

AI Thread Summary
The discussion revolves around finding the eigenvalues of a 4x4 Hermitian matrix representing the Hamiltonian of a spin-3/2 particle. A participant suggests that by swapping the second and third rows, the matrix can be simplified into two separate 2x2 matrices, potentially easing the calculation of eigenvalues. Another participant confirms that this row swap would also require a corresponding column swap. The conversation touches on the properties of block matrices and whether there is a theorem supporting this simplification method. Ultimately, the participants emphasize the separability of the matrix components for easier analysis.
Gunthi
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Homework Statement


Find the allowed energies for a spin-3/2 particle with the given Hamiltonian:
\hat{H}=\frac{\epsilon_0}{\hbar}(\hat{S_x^2}-\hat{S_y^2})-\frac{\epsilon_0}{\hbar}\hat{S_z}

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 det(\hat{H}-\lambda I)=0?
 
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Hi Gunthi! :smile:
Gunthi said:
Is there a more quick way to find the eigenvalues of a 4x4 hermitian matrix than going trough the tedious calculation of det(\hat{H}-\lambda I)=0?

If you swap the second and third rows, it becomes two 2x2 matrices. :wink:
 
tiny-tim said:
Hi Gunthi! :smile:


If you swap the second and third rows, it becomes two 2x2 matrices. :wink:

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:
 
Hi Gunthi! :smile:
Gunthi said:
… 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?

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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