I have a large hamiltonian but I know that it posseses some symmetries. How do you reduce the hamiltonian in order to find the eigenenergies?

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    $\begingroup$ Well, you can just diagonalize it directly without regard to the symmetry and it will work just fine. Are you looking for a method to find the joint eigenbasis? Or for a method which explicitly uses the symmetry to cut down on computational cost? $\endgroup$ – Emilio Pisanty Oct 2 '18 at 21:11
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    $\begingroup$ If it has symmetries and you diagonalize the hamiltonian, they should automatically appear in the corresponding matrix in a way that helps you calculate the eigenvalues (or you can apply them at your convenience after building the initial matrix). I'm not sure if it's your case, but if you manage to reduce it to a matrix that is diagonal-by-blocks you should be able to easily compute the eigenvalues. $\endgroup$ – Charlie Oct 2 '18 at 22:12

If $g$ is a symmetry operation, then by definition your Hamiltonian will satisfy \begin{align} gHg^{-1}=H \qquad \Leftrightarrow \qquad gH=Hg\, , \end{align} or if $g=\exp{(-i \theta K)}$ then $[H,K]=0$. The key point is that either eigenstates of $g$ or $K$ will also be eigenstates of the Hamiltonian, which will help you block diagonalize $H$ if some eigenvalues $\lambda_g$ of $g$ are repeated, or will completely diagonalize $H$ is none of the eigenvalues of $g$ are repeated.

Normally of course your Hamiltonian might act in an infinite-dimensional Hilbert space, so if the representations of $g$ are finite-dimensional, then you can work inside each finite dimensional block or at least identify which subspaces are the most important and then truncate.

The classic example is a system of particles on a ring interacting with nearest neighbours $$ H=\sum_{n=1}^N E_0\vert n\rangle \langle n\vert +\sum_{n=1}^N \left( W\vert n\rangle \langle n+1\vert +W\vert n+1\rangle\langle n\vert\right) $$ where the states $\{\vert n\rangle, n=1,\ldots N\}$ at each site are assumed orthogonal, and where the periodic condition $\vert N+j\rangle =\vert j\rangle$ holds.

Clearly this Hamiltonian commutes with the operators $\hat C_+=\sum_{n=1}^N \vert n\rangle \langle n+1\vert$, which "rotates" state $\vert n+1\rangle$ to $\vert n\rangle$, i.e. "rotates" the ring by one position to the left. Since $(\hat C_+)^N=\hat 1$, the $k$'the eigenvalue of $\hat C_+$ must satisfy $\lambda_k^N=1$ so that these are easy to find. In turn, the eigenvectors $\vert k\rangle$ of $\hat C_+$ are easy to find: the $p$'th component has amplitude $e^{2\pi i k p/N}$ and, as one can verify that $\lambda_k$ occurs at most once in the spectrum of $\hat C_+$, the eigenvectors $\vert k\rangle$ must also be eigenstates of $H$, so that we an get the eigenvectors of $H$ without finding the eigenvectors of a triangular matrix. Once we have the eigenvectors, it's child's play to get the eigenvalues of $H$.

In more sophisticated examples one may get only a block diagonal form. For instance, for a central potential one then has $[H,\vec L\cdot \vec L]=0$ so that, by finding the states of "good" angular momentum $\ell$, one automatically block diagonalizes the Hamiltonian.

If some eigenvalues of $g$ occur more than once then the job is more complicated. There is no guarantee that $H$ will be diagonal but what is known is that $H$ cannot connect states with different eigenvalues of $g$, so one can proceed by first finding the eigenstates of $g$ and then diagonalizing $H$ within the subspace of eigenstates with the same eigenvalue.

In practice, it is not always easier to find the eigenvectors of $g$, at least analytically. Even when this can be done, changing the basis to one where $H$ is block diagonal is not necessarily easy to do explicitly by hand. Possibly the simplest example of this situation is the coupling of two sets of states having angular momenta $\ell_1$ and $\ell_2$: to obtain states of good total angular momentum $L$ requires Clebsch-Gordan technology, which is well known but still requires some effort.

Also it may well be that some part in $H=H_0+H_1$ might "break the symmetry", i.e. $[H_0,g]=0$ but $[H_1,g]\ne 0$, in which case eigenstates of $g$ will be eigenstates of $H_0$ (barring repeated eigenvalues) and one can hope to treat $H_1$ as a perturbation.


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