Exact diagonalization (ED) refers to the procedure of diagonalizing the Hamiltonian matrix expressed in a complete basis that spans the entire Hilbert space of quantum system. The output of this procedure will be eigenvalues and eigenvectors of the Hamiltonian expressed as a linear combination of the chosen basis elements. There are three basic steps in carrying out an ED procedure:
- Decide on a reasonable complete set of basis elements
- Write the Hamiltonian operator as a matrix in the given basis
- Diagonalize the Hamiltonian matrix
Unfortunately, the “size” of the Hilbert space of a quantum system grows exponentially with system size, thus many tricks are needed to speed up the ED procedure. First reduce the problem by restricting with quantum numbers, then solve faster with iterative techniques.
ED is most easily understood in the context of 1D spin chain.
Consider a 1D chain of N spins interacting through the \(S=\frac{1}{2}\) Heisenberg Hamiltonian $$ \hat{H} = -\frac{1}{2}\sum_{<i,j>}\sum_{d=1}^3 J_{i,j}^d\sigma^d_i \sigma^d_j, $$ where \(i,j\) label lattice sites and the angle bracket denotes that \(i\) and \(j\) are neighboring spins. The \(J_{i,j}\) are coupling parameters. The spin operator for site \(i\) is the Pauli vector of Pauli matrices on site \(i\) $$\vec{\sigma}_i=\{\sigma_i^1,\sigma_i^2,\sigma_i^3\}. $$ In the simplest case where the coupling parameter is equal for all pairs of spins and in all 3 spatial dimensions, we have ourselves an XXX model. Consider periodic conditions where the virtual site \(N+1\) is identified with site \(1\), we finally arrive at the model Hamiltonian we wish to diagonalize $$\hat{H} = -\frac{1}{2}J\sum_{i=1}^{N} \sum_{l=1}^3 \sigma^l_i\sigma^l_{i+1}.$$ One more step of algebra will make the above Hamiltonian easy to represent in the \(S_z\) basis. That is, change basis/rotate/unitary transform to turn \(\sigma^x,\sigma^y\) into \(\sigma^+,\sigma^-\), i.e. \(\sigma^x\sigma^x+\sigma^y\sigma^y=\frac{1}{2}(\sigma^+\sigma^-+\sigma^-\sigma^+)\) $$\hat{H} = -J \sum_{i=1}^{N} \frac{1}{4}(\sigma^+\sigma^-+\sigma^-\sigma^+) + \frac{1}{2}\sigma^z\sigma^z.$$