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Symmetry adapted basis in quantum chemistry

My friend and collaborator James Whitfield just wrote a paper that uses a construction which I like.

James is a quantum chemist, and for those of you who don't know, quantum chemists know their stuff when it comes to quantum theory, particularly as it is applied to atoms and molecules. Part of the name of the game there is to use symmetry as much as possible in hopes to reduce the problem size and to understand the physics better.

I wanted to mention this construction since it could maybe be used in other things related to the Azimuth project, in particular in the so called, Baez-Fong version of Noether's theorem. Probably some of you already know it, but I personally did not think of creating a group out of an algebra in this simple way:

The "symmetry group" in quantum chemistry is defined as

$ \{A | [A, H] = 0, A^\dagger = A, det A \neq 0 \} $

Here the commutator $ [A, H] := A H - H A $, and $ A $ is equal to its own conjugate transpose, meaning that it is guaranteed to be a valid quantum generator (see the network theory series, e.g. part 12).

Normally people consider the algebra formed by elements of $A$. The product structure here is given by the commutator (see Lie algebra ).

The point of the above construction is that we can always add the center of the algebra to make $A$ invertible. For instance, we could let $A := A' + \alpha 1$ to make non-invertible $A'$ invertible. For those familiar with quantum theory, adding multiples of the identify to a generator won't change the evolution given by the exponential of the map $A$ and just shifts the spectrum of $A'$ and so one could always make arguments about the spectrum of either of these and translate simply between them.

If we have two elements $g$ and $h$ that satisfy the definition above (that is, they are valid symmetry group generators) we have that

$[g h, H] = g[h, H] + [g, H]h = 0 $

which implies closure. Consider then the inverse of $g$, $g^{-1}$. We have that

$[g g^{-1}, H] = g[g^{-1}, H] + [g, H]g^{-1}$

The second term on the right above vanishes from the definition of $g$. Since $g g^{-1} = 1$ and $[1, H] = 0$ we know that

$g[g^{-1}, H] = 0 $

$g$ does not equal $0$ and so $[g^{-1}, H]$ must vanish, which happens iff $g^{-1}$ commutes with $H$.

There are a few more things to be said about this. Probably some of you, particularly John who is an expert on symmetry and Lie algebras etc. would have a lot more to say. The reason I like it, is that it gives a way to add a group structure onto a Lie algebra.

From a practical standpoint, given some symmetry generator $Q$, we can consider invertible $Q'$ and then the group generated by $Q'$ defines a symmetry group.

I want to think about this more as it relates to enveloping algebras and q-deformations in addition to non-Hermitian generators found in stochastic mechanics.

Comments

  • 1.

    That's a nice construction! I'm too sleepy right now to say anything about it. If you want some intelligent remarks, just ask and I'll deliver 'em when I'm more awake.

    Comment Source:That's a nice construction! I'm too sleepy right now to say anything about it. If you want some intelligent remarks, just ask and I'll deliver 'em when I'm more awake.
  • 2.

    James Whitfield himself said he's joining Azimuth and on this thread, over time, we wanted to go over some of the things from his paper. James, are you out there yet? The next thing he does in his paper is consider Young diagrams... I certainly want to consider the construction above again and will pester you about it sometime in the future John!

    Comment Source:James Whitfield himself said he's joining Azimuth and on this thread, over time, we wanted to go over some of the things from his paper. James, are you out there yet? The next thing he does in his paper is consider Young diagrams... I certainly want to consider the construction above again and will pester you about it sometime in the future John!
  • 3.

    I just added James as a member a little while ago.

    Comment Source:I just added James as a member a little while ago.
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