Eigenvalues of Hamiltonian

by Krischi
Tags: eigenvalues, hamiltonian
 P: 3 Really, a 4$$\times$$4 matrix? If there are 4 base states, why can't I use a 2$$\times$$2 matrix? 4 states fit into a 2$$\times$$2 matrix, right? I tried this and calculated 2 eigenvalues, $$\lambda$$, but I am not sure, if the result is correct, since it "looks" to complicated (see my 1st reply)