# Is the Lowdin Orthogonalization used in diagonalizing the atomic orbitals really a basis change?

This is kind of more on the mathematics side of quantum chemistry, but I can't quite figure out why the Lowdin Orthogonalization is called a basis set change. I get how it works from the perspective of matrices, and that it creates an orthonormal set as long as you have linearly independent input, but I don't get why it works as a change of basis.

Typically to change the basis set of a matrix $$M$$ you use a non-singular linear transformation $$A$$, which need not be unitary, as such: $$M'=A^{-1}MA.$$ Quantum mechanics, we are dealing with Hilbert Spaces, which produce an isometry between the Hilbert space and the dual (I think?), and then for post HF methods most methods often use unitary transformations, so that $$A^\dagger=A^{-1}$$, and so it becomes trivial.

The Lowdin transformation between nonorthogonal AO $$|\phi\rangle$$ and orthogonal AO $$|\phi\rangle_\perp$$ involves the overlap matrix $$S$$ as: $$|\phi \rangle_\perp = S^{-\frac{1}{2}}|\phi\rangle$$ Then, the corresponding bra just has the adjoint, which is also $$S^{\frac{-1}{2}}$$ because the matrix is self-adjoint: $$\langle \phi|_\perp = \langle \phi|(S^{-\frac{1}{2}})^\dagger = \langle \phi|S^{-\frac{1}{2}}$$

But, considering the matrix analog, why don't we use the inverse?

For instance, we have the identity: $$D^{AO} = C D^{MO}C^T$$ where $$D$$ is the 1-electron reduced density matrix. In my thinking this has to be a transformation from the MO to the AO on the right of $$D^{MO}$$, and from the AO to the MO on the left, which gives you $$D^{AO}$$. However, $$C$$ is not unitary obviously, because it comes from the Lowdin procedure, and so while $$C$$ on the left is from the AO to the MO, $$C^{-1}$$ is not equivalent to $$C^T$$.

Furthermore to get $$D^{MO}$$ matrix in terms of the $$D^{AO}$$ you DO have to apply the inverse relationship with $$C^{-1}$$. $$D^{MO} = C^{-1}D^{AO}(C^T)^{-1},$$ because $$C^{-1}$$ and $$C$$ and related non-trivially through $$S$$.

Hopefully this is appropriate to ask here. I think the answer involves Hilbert spaces but I can't find a lot on non-orthogonal transformations. Thanks!

• I might be misunderstanding your question, but isn't matrix inversion more computationally intensive than taking the adjoint?
– Zhe
Sep 25 '18 at 17:17
• It is, but unless the matrix is unitary, then the adjoint and the inverse are not the same, so applying will not yield the same result. Here, the C matrix for instance is not unitary. Sep 25 '18 at 17:56

## 1 Answer

Consider a nonsingular, linear transformation $$\mathbf A$$ of a set of vectors arranged into a matrix $$\mathbf V$$ $$\mathbf V'=\mathbf {VA}$$

We can say the matrix $$\mathbf V'$$ is orthogonal (and thus that we have linearly transformed the vectors to form an orthogonal basis) if $$\mathbf {V'^{\dagger}V'}=\mathbf {A^{\dagger}V^{\dagger}VA}=\mathbf {A^{\dagger}SA}=\mathbf{1}$$

In general, this occurs when $$\mathbf{A}=\mathbf{S}^{-1/2}\mathbf{B}$$, where $$\mathbf{B}$$ can be any unitary matrix. The Lowdin orthogonalization commonly used is just the case where $$\mathbf{B=1}$$, so we can see that we are properly changing the basis.

But your issue is specifically with transformations of the density matrix, where the transformation matrix is not unitary in general. The standard change of basis formalism would suggest that to go to AO basis from MO, we would do a transformation something like this: $$\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{-1}=\mathbf{D}^{AO}\mathbf{S}$$ Ensuring orthogonality of the MOs requires that $$\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}=\mathbf{1}$$, which we can sub in to the equation above to give: $$\mathbf{C}\mathbf{D}^{MO}\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}\mathbf{C}^{-1}=\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}\mathbf{S} =\mathbf{D}^{AO}\mathbf{S} \to \mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}=\mathbf{D}^{AO}$$

So the form in terms $$\mathbf C$$ and its adjoint is a direct result of the conventional similarity transform combined with the orthonormality condition on the MOs.

This latter derivation was adapted from$$^1$$ which can be found here.

As to why we avoid using transformations involving the inverse of the transformation matrix, I suspect Zhe is correct that it is a matter of cost, since obtaining the transpose/adjoint of a matrix requires almost no work at all, while an inverse is relatively difficult to obtain.

1. T. Helgaker et. al Chemical Physics Letters 327 (2000). 397–403
• Well, I have no problem with this. It's just the inverse and the transpose are not 1-1 for a non-unitary matrix. So the effect of $CDC^{-1}$ should not be the same as $CDC^{T}$, and it isn't. And generally, the first of those is presented as the 'change of basis' sort of formalism. Sep 25 '18 at 17:50
• @ScottS. I've edited my response; let me know if this better addresses your concern. Sep 25 '18 at 18:51
• Ah, okay this is better, thank you @Tyberius. The paper helps as well. The proper similarity transformation here is actually between $D^{AO}S$ and $D^{MO}$, not $D^{AO}$ and $D^{MO}$. This makes sense, as now the eigenvalues are preserved.....I guess then, my only question is, what is the interpretation of $D^{AO}S$? And so, it seems going from $D^{AO}$ to $D^{MO}$ might not actually be a simple change of basis because you have to change the matrix eigenvalues, i.e. not a similarity transformation? Sep 25 '18 at 19:35
• @ScottS. I'm not sure what it means. If you orthogonalized the AOs, then it becomes a conventional similarity transform, but otherwise I'm not sure how the basis are related. My best guess is that they are in some sense "similar with respect to a metric" for lack of better terminology. I posted a question on Math SE about if there is a clear meaning behind this pseudo-similarity transform. Sep 25 '18 at 21:02
• Thank you @Tyberius, that makes sense, and I think it helped me move in the right direction. I started looking into metric vector spaces and bilinear forms....a particular book$^\dagger$ was helpful. It seems that there is a congruence relation between the two density matrices $D^{AO}=CD^{MO}C^T$. Then, these would share the same bilinear (or sesquilinear) form just in different bases. Also it is not an isometry, meaning it does not preserve length. If it did, then it would be unitary. I also am not positive about $D^{AO}S$, but it might just be an artifact of the congruence relation? Sep 26 '18 at 7:29