On Friday we had a great visit from Weitao Yang. One thing (among many others) I learnt from him was his definition of static versus dynamic correlations in quantum chemistry. This is something that people talk about but I have trouble getting a clear definition that I can understand.

Weitao illustrated this using the two-site Hubbard model and the hydrogen molecule.

Near the equilibrium geometry of the molecule, U/t is not large and a Hartree-Fock wavefunction is qualitatively but not quantitatively correct. The corrections to this are dynamical correlations.

At large separations, U/t is large, and the wavefunction approaches a Heitler-London state, and a Hartree-Fock wavefunction is qualitatively incorrect, e.g., it claims the H atoms will not bond to each other! This is static correlations.

Somehow I wonder if there is a way to quantify this distinction in terms of entanglement measures.

Subscribe to:
Post Comments (Atom)

There can be no clear distinction, ultimately. There is only one type of off-diagonal density matrix element.

ReplyDeleteThe best distinction I've heard is that static correlation is correlation mandated by permutation symmetry, while dynamic correlation is "real correlation" - entanglement of the 1-body states by the 2-body Coulomb term.

Spin and space variables can only be separated for systems with low "correlation rank". Two electrons is about it.

This is due to a problem not just for physics, but also for all applied mathematics. There is no analogue of the singular value decomposition (SVD) for tensors with more than two indices. The SVD is known by a different name in quantum theory: it is called the Schmidt decomposition.

This is the problem: once the spin and space variables have been separated using the SVD, further decomposition into orthogonal space states won't work. If one decomposes the orbitals into "corresponding" spin orbitals, then these sets are generally non-orthogonal.

The problem of >2-index SVD analogues is one of the most important problems of modern tensor algebra, as I understand it.

This is sufficiently important as to merit a rant (i.e. two consecutive posts without response).

ReplyDeleteThe SVD is the pinnacle of linear algebra. This means that the real problem here is that quantum mechanics is a linear theory. There can be no definite separation of 'static' vs. 'dynamic' correlations in such a theory.

When a student is told (as I was) that the SVD (or Schmidt decomposition) can only be invoked ONCE in general, the immediate retort is 'why not?'. This is a good question. After all, if the SVD for a matrix A is written:

A=U.S.Vt

where U and V are orthogonal and S is diagonal, why can't you just apply it again to either U or V? Of course you can. The point is that it doesn't get you anywhere. One sees this immediately if one knows that the singular values of A are also the eigenvalues of A.At and the singular vectors (columns of U and V) are the eigenvectors of A.At and At.A. If U and V are orthogonal, then U.Ut=I, Vt.V=I, I being the identity. Note that the spaces need not be of the same rank here, so the two identities can be different (they are identities on different spaces).

This means that there is no point in performing SVDs on U and V, because the singular values of either of these matrices are all degenerate and equal to 1. Also, the singluar vectors of U are already the columns of U.

The significance of the Schmidt Decomposition, like the significance of the SVD, is that it tells you how 'compressible' a representation is. This compressibility, in quantum chemistry, is a measure of how 'multireference' a problem is. You don't get more compression by performing the SVD multiple times in succession.

Coming from quantum information theory, I was quite intrigued by this post. But I'm a little confused by the first comment because it would seem that the definition given does clearly differentiate static and dynamic correlations.

ReplyDeleteStatic correlations come from the permutation antisymmetry alone, so if we start in a product state with distinguishable particles we have to antisymmetrize and use the Slater determinant. Such states are usually accepted to be unentangled (if we want to think of them as really entangled, we face the problem of trying to use the entanglement in a way which doesn't require us to distinguish between the constituent fermions).

Dynamic correlations are correlations beyond this (I'm assuming), which can only come from having a superposition of Slater determinants. And this matches up with the notion of entanglement, so it would seem that dynamic correlations could be quantified by the measure of entanglement.

I am delighted some discussion is starting!

ReplyDeleteI found reading the wikipedia entry on singular value decomposition necessary and helpful to understand Seth's post. So I recommend that.

Responding to "complementary slackness", it is not quite so straightforward.

static correlations require more than one slater determinant. e.g., the Heitler-London state has maximal entanglement of spin

an earlier post I wrote may help....

I will try and get back to this..