Let $$u=\left( \begin{array}{cccc} c_1&c_2&c_3&c_4 \end{array} \right)^T$$ for $$\psi = c_1\psi_1 + c_2\psi_2 + c_3\psi_3+ c_4\psi_4$$

We assume that $\left<\psi_i|\psi_j\right> = \delta_{ij}$ so that $$\left<\psi|\psi\right> = \left<\phi|\phi\right>=|c_1|^2 + ... +|c_4|^2$$

Let $U$ be the representation of the symmetry group $C_{4v}$ on these 4-dimensional state-space of electrons.

I know for any $R \in C_{4v}$ , $T(R)u$ is also an eigenvector with the same eigenvalue: $$\tag{1}H(T(R)u) = \epsilon (T(R)u)$$ Here is my question:

  • Suppose our Hamiltonian is given as $$H= \left( \begin{array}{cccc} \epsilon_0 & t & t' &t \\ t & \epsilon_0 & t &t' \\ t' & t & \epsilon_0 &t \\ t & t' & t &\epsilon_0 \end{array} \right) $$

How can I find energy eigenvalues of $H$ by using the decomposition of $U$ and Eq$(1)$?

Note: I found the decomposition as $$U=A_1\oplus E$$


1 Answer 1


Well, you might not really want to overthink it. It all hinges on the Pauli matrix $\sigma_1$ which, yes, is the permutation group with two components, with eigenvalues $\pm 1$ for $v^T=(1,1)$, symmetric, and $w^T=(1,-1)$, antisymmetric, respectively.

Equivalently to your decomposition, you may write the hamiltonian as $$ H= \epsilon_0 I_4 + t~(\sigma_1+I_2)\otimes \sigma_1+t' ~\sigma_1\otimes I_2 , $$ where my convention inserts the right tensor product 2x2 factor into the entries of the left 2x2 factor. Of course the 3rd term times the 2nd term yields the second term, if the numerical coefficients t, t' are ignored.

It is then evident by inspection that $\epsilon_0$ is a common additive shift to all eigenvalues; and that the four eigenvectors are inevitably $v\otimes v$, $v\otimes w$, $w\otimes v$, $w\otimes w$, with manifest eigenvalues 2 t+t'; -2 t+t'; -t'; -t', respectively. So add $\epsilon_0$ to all of them to be done. By inserting the right vector into the components of the left one, you may recover the 4-vectors and check them, subverting the original algebraic implicitness intent of the exercise.

Note the 3rd term of the hamiltonian squares to the identity, the second one projects w out on the left tensor space, while the 3rd one acts indifferently (identically) on the right tensor space, whence the degeneracy between the 3rd and 4th eigenvectors, and the identical t' piece of the eigenvalue of the 1st and 2nd.


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