# Self-consistent non-linear eigenproblem

Given two Hermitian matrices $$\mathbf{A}$$ and $$\mathbf{B}$$ of dimension $$M \times M$$, search for a set of $$N \leq M$$ orthonormal vectors $$\mathbf{v}_n$$ for which

$$\left( a_n \mathbf{A}+\mathbf{B} \right) \mathbf{v}_n=\lambda_n \mathbf{v}_n\;$$

where $$\;a_n=\mathbf{v}_n^\dagger \mathbf{A} \mathbf{v}_n$$, holds such that $$\sum_n \lambda_n \rightarrow \max$$.

Here, $$\dagger$$ denotes the complex conjugate transpose of a matrix/vector.

Is this problem solvable and if, how to solve it?

My thoughts so far:

First, the problem makes sense. Since $$\mathbf{A}$$ and $$\mathbf{B}$$ are hermitian both the $$a_n$$ and $$\lambda_n$$ are real and it makes sense to ask for the maximum of the sum.

My first idea was to solve the problem iteratively:
For each $$n=1,\dots,N$$ guess an initial $$a_n$$ (e.g. $$a_n=1$$), solve the eigenvalue problem and choose as $$\mathbf{v}_n$$ the eigenvector corresponding to the $$n$$-th largest eigenvalue $$\lambda_n$$, with this update $$a_n$$ and reiterate until self-consistency is reached. The problem with this method (besides possible convergence issues) is that each $$\mathbf{v}_n$$ comes from an individual diagonalization and thus the resulting $$\mathbf{v}_n$$ are not necessarily orthogonal.

The second idea/question that came to my mind was if the problem can be cast into a non-linear (quadratic?) eigenproblem that then can be solved with standard techniques?

This problem comes from an optimization problem in quantum physics. In the following, I will use the Bra-Ket notation commonly used in physics ($$\langle f|g \rangle = \int \mathrm{d}\mathbf{r}\; f^\ast(\mathbf{r}) g(\mathbf{r})$$ with $$\ast$$ denoting the complex conjugate). The goal is to minimize the functional $$\Omega[\lbrace w_n\rbrace ] = \sum_n [\langle w_n|\hat{\mathbf{R}}^2|w_n \rangle - \langle w_n|\hat{\mathbf{R}}|w_n \rangle^2]\rightarrow \mathrm{min}\;,$$ where $$\hat{\mathbf{R}}$$ is an operator and $$\lbrace|w_n\rangle \rbrace$$ a set of $$N$$ orthonormal states/functions ($$\langle w_m|w_n \rangle = \delta_{mn}$$). Taking the derivative $$\frac{\delta \Omega}{\delta w_n^\ast} = \hat{\mathbf{R}}^2 |w_n\rangle - 2\langle w_n|\hat{\mathbf{R}}|w_n \rangle \hat{\mathbf{R}} |w_n\rangle$$ and adding orthonormality as a constraint with Lagrangian multipliers $$\Lambda_{mn}$$, we need to solve $$\frac{\delta \Omega}{\delta w_n^\ast} + \sum_m \Lambda_{mn}\frac{\delta}{\delta w_n^\ast}[\langle w_n|w_m\rangle - \delta_{mn}] = 0\;.$$ Using a unitary transformation that diagonalizes $$\Lambda$$, this can be written as $$\frac{\delta \Omega}{\delta w_n^\ast} + \lambda_n |w_n\rangle = 0\;.$$ Now, expressing the states in an orthonormal basis $$\lbrace|g_i\rangle\rbrace$$ ($$|w_n\rangle = \sum_{i=1}^M v_{in}|g_i\rangle$$), this leads me to the above stated problem with the matrices $$\mathbf{A}$$ and $$\mathbf{B}$$ being the follwing matrix elements of the operators $$A_{ij} = \sqrt{2}\langle g_i|\hat{\mathbf{R}}|g_j\rangle$$ and $$B_{ij} = -\langle g_i|\hat{\mathbf{R}}^2|g_j\rangle$$.
• Don't $A$ and $B$ commute in your motivating problem? It seems to me that we should have $B=-A^2/2$. – user856 Oct 24 '19 at 13:15
• @Rahul I think, you are right under the following assumptions: first, with $A^2$ you mean $\mathbf{A}^2=\mathbf{A}\mathbf{A}$ (and not the squares of the elements) and second, $\hat{\mathbf{R}} |g_i \rangle \in \mathrm{span}\lbrace |g_i \rangle \rbrace$. Let's assume that these assumptions are fulfilled. What does this additional piece of information mean for the solution? – Basti Oct 24 '19 at 13:49
• Ah, I guess you're right about the second assumption. Anyway, then $A$ and $B$ are simultaneously diagonalizable, and their unit eigenvectors form a set of orthonormal $v_n$'s satisfying your condition. I'm not sure how to prove whether there are other feasible vectors. – user856 Oct 24 '19 at 16:01