8
$\begingroup$

Consider a many body problem of a small cluster, e.g. the 'Hubbard-Cluster' (albeit the question may be of relevance for other Hamiltonians as well):

$$\mathcal{H}=\sum_{<ij>\sigma} t_{ij} (c^\dagger_{i\sigma}c^{}_{j\sigma}+ c.c.) + U\sum_i n_{i\uparrow}n_{i\downarrow} -\mu N$$

It is well understood that when such an operator commutes with an oberservable like the density $N=\sum_i n_i,\; n_i=n_{i\uparrow}+n_{i\downarrow}$ and/or the magnetic moment $M=\sum_i m_i,\; m_i=n_{i\uparrow}-n_{i\downarrow}$ these are good quantum numbers and under an appropriate sorting of fock states $⟨\psi_\alpha|$ the Hamiltonian matrix

$$\mathcal{H}_{\alpha\beta} = ⟨\psi_\alpha|\mathcal{H}|\psi_\beta⟩$$

decomposes into blocks of constant particle number and magnetic moment. However in most of the literature it is also mentioned that if the cluster is invariant under a symmetry operation $S$ the problem may be further simplified, i.e. the Hamiltonian be decomposed into yet smaller blocks by a unitary transformation. Now here are my questions:

  1. Is there any systematic understanding of this simplification? Given a symmetry $S$, what is the unitary transformation simplifying the Hamiltonian?

  2. Once a symmetry operation has been found, how small are the resulting blocks? Can their size be predicted?

  3. If there are several symmetry operations at hand, which one results in the greatest simplification of the problem?

  4. How can exact solutions be found, using symmetry related unitary transformations?

$\endgroup$

1 Answer 1

7
+50
$\begingroup$

You may find the following paper useful: A Symbolic Solution of the Hubbard Model for Small Clusters, by J. Yepez.

You may also want to review group theory for condensed matter physics, because your questions essentially span the basics of group and representation theory. Many texts give good overviews of the fundamentals of group theory as applied to solid state crystals, and some are available on-line, for instance "Symmetry in Condensed Matter Physics" by P.G. Radaelli (U. of Oxford link) or "Applications of Group Theory to the Physics of Solids" by M.S.Dresselhaus (MIT link).

Anyway, the basic ideas are as follows:

  1. Is there any systematic understanding of this simplification? Given a symmetry S, what is the unitary transformation simplifying the Hamiltonian?

Yes, there is a sistematic way to do this, and it has to do with the group of symmetry transformations of the Hamiltonian $H$. A simplifying unitary transformation can be found starting from any arbitrary basis of states as follows:

i) In the given basis generate the matrix representations for the Hamiltonian and the symmetry generators. These matrices produce in general a reducible representation of the symmetry group. ii) Use the generator matrices and group theoretical techniques to construct projectors on states that are invariant under the symmetry transformations. These new states are not energy eigenstates, just symmetry invariant states. iii) Construct the new basis states and the unitary operation that transforms the original basis into the new one. This is the unitary transformation you are looking for. See below for further details.

  1. Once a symmetry operation has been found, how small are the resulting blocks? Can their size be predicted?

Yes, the size of the blocks and the degeneracy of the energy eigenstates is determined by the nature of the symmetry group. The new symmetry invariant basis contains groups of states that transform into each other under the symmetry transformations, but cannot be decomposed into any smaller groups with the same property. Each of these groups generates what is called an irreducible representation of the symmetry group. The unitary transformation described above decomposes the original reducible representation into some number of such irreducible representations. The states corresponding to any given irreducible representation mix only among themselves in the Hamiltonian matrix. Therefore the transformation to symmetry invariant states and the decomposition into irreducible representations resolves the matrix of the Hamiltonian into a simpler block diagonal form. The possible dimensions of the blocks are known and are determined by the structure of the symmetry group, independently of the particular form of the Hamiltonian. That is, Hamiltonians of completely different systems that share the same symmetry group will have block-diagonal forms with blocks of the same pre-determined dimensions. These are the dimensions of the irreducible representations of the symmetry group. They differ from group to group, but have been calculated and tabulated for all important symmetry groups.

  1. If there are several symmetry operations at hand, which one results in the greatest simplification of the problem?

The general rule is that operations of higher symmetry generate reducible representations with respect to operations of lower symmetry. Therefore the operation that actually determines the resolution into the smallest blocks is the one of lowest symmetry among the operations of the symmetry group.

  1. How can exact solutions be found, using symmetry related unitary transformations?

Basically symmetry simplifications of the Hamiltonian matrix produce decompositions into much smaller diagonal blocks (irreducible representations) that can then be diagonalized independently. So the procedure is to diagonalize a block, generate the associated unitary transformation, and then use the latter to find the exact eigenstates. Sometimes the diagonalization can be done analytically, but even if it needs to be done numerically, it is still a much simpler problem than the original one.

$\endgroup$
2
  • $\begingroup$ I didn't have time to check all of this thoroughly but that's is enough material for the bounty. I wasn't expecting an answer anymore after this initial responsiveness. Thanks anyway! $\endgroup$ Commented Sep 20, 2015 at 14:43
  • $\begingroup$ Welcome. I was actually surprised nobody took it up or even mentioned some pointers in a comment. $\endgroup$
    – udrv
    Commented Sep 20, 2015 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.