What will happen to Hamiltonian matrix, eigenvalues and eigenvector on a non-eigenbasis?

In a simple example, Most of the Hamiltonians are talked about on its eigenbasis or a basis that can be transformed from the eigenbasis. With this, the eigenvalues do not change even if on a different basis, but the Eigenvector will change depending on which basis. (See example on 1.6 in https://www2.ph.ed.ac.uk/~gja/qp/qp1.pdf)

Question:
But in practice, most of the Hamiltonians cannot be solved analytically. If I use a basis which is not an eigenbasis $$\{|\psi^{non-e}_{n}\rangle\}$$ of the Hamiltonian $$H$$, what will happen?

For example, On a two-body system, what if I use the Gaussian form wave function ($$\psi_{nlm}^{Gaussian}\propto e^{-\alpha r^2}$$) on a Hamiltonian with potential in Coulomb form $$V_{ij}=\frac{A}{r}$$ (which the exact solution of Coulomb potential is proportional to $$\psi_{nlm}^{Coulomb}\propto e^{-\beta r}$$). (Or more practically, a Cornell form $$V_{ij}=\frac{A}{r}+Br$$)

Can I get the matrix element of $$H$$, Eigenvalues of $$H$$ on this basis $$\{|\psi^{non-e}\rangle\}$$?, and will the Eigenvalues be useful?

Edit 1: A similar thing about this question should be the variational method, $$\bar{H}=\frac{\langle \psi_{trial}|H|\psi_{trial}\rangle}{\langle\psi_{trail}|\psi_{trail}\rangle}.$$ But the variational method is valid only for the ground state. Well, in this case, it is a set of basis function which is similar to the trail function. $$\bar{H}=\begin{pmatrix}\bar{H}_{ij}&\cdots\\\vdots&\ddots\end{pmatrix},~where~\bar{H}_{ij}=\langle \psi^{trial}_i|H|\psi^{trial}_j\rangle.$$ Is it possible?

In an arbitrary basis the Hamiltonian is simply a non-diagonal matrix, and the eigenvalues represent the energy levels. Let me give you a much much simpler example: consider the Hamiltonian describing a $$1/2$$ fermion in a magnetic field along $$x$$ axis $$H = - \gamma S_x$$ where $$S_x$$ is the spin x operator. Now the Hilbert space is two-dimensional and if you take as a basis the eigenstates of the spin z operator $$S_z$$, and you call them $$\left| \uparrow \right\rangle$$ and $$\left| \downarrow \right\rangle$$, then the spin x operator, and hence the Hamiltonian are not diagonal $$S_x = \frac{\hbar}{2} \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right).$$ The eigenvalues and eigenvectors of $$H$$ then are $$E_0 = - \gamma \hbar/2 \;\;\;\; , \;\;\;\; \left| \rightarrow \right\rangle = \frac{1}{\sqrt{2}} \left( \left| \uparrow \right\rangle + \left| \downarrow \right\rangle \right)$$ $$E_1 = \gamma \hbar/2 \;\;\;\; , \;\;\;\; \left| \leftarrow \right\rangle = \frac{1}{\sqrt{2}} \left( \left| \uparrow \right\rangle - \left| \downarrow \right\rangle \right)$$.
If instead you decide from the beginning to use the eigenstates of $$S_x$$ (i.e. $$\left| \rightarrow \right\rangle$$ and $$\left| \leftarrow \right\rangle$$ ) as a basis for the Hilbert space, then the spin x operator and hence the Hamiltonian are diagonal:
$$S_x = \frac{\hbar}{2} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right).$$
• Thank you, but well, that is a simple case. As I mentioned the basis of $\{|\leftarrow\rangle, |\rightarrow\rangle\}$ can be completely transformed into basis $\{|\uparrow\rangle, |\downarrow\rangle\}$. But for this case, two bases cannot be completely transformed into each other. Feb 13, 2021 at 15:44
• This is very general: all the accessible states of the system form an Hilbert space, which is a vector space. Chosing a basis means by definition that any element of the space can be written as a linear combination of the basis, but there are two techincal complications in your problem: 1) the Hilbert space is infinite dimensional 2) all the states depend on a continuum label, which is the coordinate $r$. So if you want to use numerics you have to fix this problems, but in principle the eigenvalues of $H$ are the energy levels and the eigenstates are linear combinations of the basis Feb 13, 2021 at 16:02