Does there exist a mathematically rigorous formulation of curved Hilbert Space?

Question

The Hamiltonian is traditionally defined as \begin{align} H_{\text{flat}} = U^{\dagger}DU \end{align} where $D$ is a diagonal matrix with real eigenvalues and $U^{\dagger}U=I$ are the unitary transformations to generalize $H_{\text{flat}}$ in any traditional base. Thus the hermitian relation $(H_{\text{flat}})^{\dagger}=H_{\text{flat}}$ holds.

However I discovered that some operators can have real eigenvalues, but are not necessarily hermitian.

For example the following matrix \begin{align} H_{\text{curved}} = \begin{pmatrix} 3 && 4 && 1 && 2 \\ 2 && 4 && 2 && 4 \\ 1 && 3 && 2 && 3 \\ 2 && 4 && 4 && 1 \end{pmatrix} \end{align} has the real eigenvalues $(10.73814 , -2.34185 , 1.78222 , -0.17850)$, but is not hermitian.

I further explored this generalization and discovered a way to understand how $H_{\text{curved}}$ differs from $H_{\text{flat}}$ by showing how the hilbert space is different between them. The diagonalization of $H_{\text{curved}}$ requires a left and right eigenvector matrix such that \begin{align} H_{\text{curved}} = RDR^{-1} \end{align} where $L = R^{-1}$ and $R^{\dagger}\neq R^{-1}$. The orthogonality of the eigenvectors is achieved from the following relation $LR=R^{-1}R=I$, however $R^{\dagger}R\neq I$ illustrates how the orthogonal hilbert space no longer makes sense in this non-traditional context. To make sense of this is to form the inner product of the eigenvectors as if it were in a curved base. To do this we define a metric of the form $G^{-1} = R^{-\dagger} R^{-1} = (R R^{\dagger})^{-1}$, such that the orthogonality relation $R^{\dagger}G^{-1}R=I$ holds.

I dubbed $H_{\text{curved}}$ as the curved hamilotian, because its hilbert metric $G^{-1}$ mimics similar properties of the metric of curved space time $g^{ab}$ in general relativity.

I present the following question using my own vocabulary: does there exist a mathematically rigorous formulation of curved Hilbert Space similar to how general relativity is treated?

Answer 1

i think you should read about abstract vector spaces with inner products. An Hermitian linear transformation is one that satisfy $$ \langle u,H \, v\rangle = \langle H\,u,v\rangle $$ For arbitrary vectors $u,v$. It happens that when you express this linear transformation in a orthonormal basis (with respect to the given inner product) satisfy $$ H^\dagger = H $$ In the sense you understand right now. When you define $G_{curved}$ you are just defining a different inner product in your vector space, but in the differential geometry sense you keep working in a flat space (so you gained nothing new), which is a different thing of the metric $g_{ab}$ from general relativity (which deal with a inner product defined in a different vector space attached to each point of your manifold). Try to look for a math book of abstract linear algebra, and try to understand the difference between a linear map and the matrix it represent in a given basis. Also as you could verify, every matrix with real eigenvalues is hermitian with respect some inner product (just declare the basis of eigenvector orthonormal to define the inner product).