What is an operator times one of its eigenstates? I am trying to get a hold of caluclating with matrix elements. I have a Hamiltionian $\hat{H}$ in a two-dimensional Hilbert space, having eigenstates $\psi_1$ and $\psi_2$. My professor wrote down these equations:
$$\hat{H} \psi_1 = H_{11} \psi_1 + H_{12}\psi_2 \\
\hat{H} \psi_2 = H_{21} \psi_1 + H_{22}\psi_2 $$
He said these are alternative way to write the matrix elements of $\hat{H}$. However, I fail to see why is this. I tries to look it up in a linear algebra textbook, maybe this is a special property of matrix multiplications.
I have tried the following:
$$H_{11} = \psi_1^* \hat{H} \psi_1 \\
H_{11} \psi_1  = \psi_1^* \hat{H} \psi_1 \psi_1 $$
I did the same thing with $H_{12}$, added the equations together, but still can not see anything that resembles the original system.
Where do those two equations come from?
 A: The idea is that the two states $\{\psi_1,\psi_2 \}$ form a basis for the Hilbert space, this means that any other vector can be written as a linear combination of these two states, in particular the result of operating $H$ on, say, $\psi_{1}$
$$
H\psi_1 = H_{11}\psi_1 + H_{12}\psi_2 \\
H\psi_2 = H_{21}\psi_1 + H_{22}\psi_2
$$
Note that up to this point, the coefficients $H_{ij}$ are just complex number in the expansion. Now, if $\{\psi_1,\psi_2\}$ form an orthonormal basis we can write
$$
\langle\psi_1 | H \psi_1\rangle = \langle\psi_1 |H_{11}\psi_1 + H_{12}\psi_2 \rangle = H_{11}\underbrace{\langle \psi_1|\psi_1\rangle}_{=1} + H_{12}\underbrace{\langle \psi_1|\psi_2\rangle}_{=0} = H_{11}
$$
You can test the rest, but in general
$$
H_{ij} = \langle \psi_i|H|\psi_j\rangle
$$
A: I highly doubt that there being "alternative ways to write the matrix elements" mentioned by the professor has any conceptual significance. After choosing a basis of the two-dimensional Hilbert space, operators can be expressed as matrices,
$$H=\begin{pmatrix}H_{11} & H_{12}\\ H_{21} & H_{22}\end{pmatrix}$$ where $H_{11}, H_{22}$ are real and $H_{12}^*=H_{21}$.
The equations you cite simply mean that we have chosen the basis to be $\{\psi_1, \psi_2\}$. This is an orthonormal basis. In particular for a self-adjoint (as observables must be) matrix, there is always an orthonormal basis of eigenvectors. So the two eigenvectors in this basis are represented as $\psi_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $\psi_2=\begin{pmatrix}0\\1\end{pmatrix}$. Note that given an orthonormal Basis of an n-dim. Hilbert space $\{\psi_1, \dots, \psi_n\}$ we may express the Matrix elements of any operator as scalar products:
$H_{ij}:=\langle \psi_i, H\psi_j\rangle=\langle\psi_i|H|\psi_j\rangle$
A: Because $\psi_1$ and $\psi_2$ are eigenstates of the Hamiltonian, we know:
$$\hat{H}\psi_1 = E_1 \psi_1, \\ \hat{H}\psi_2 = E_2 \psi_2.$$
Let's consider the matrix element: $H_{ij} = \langle \psi_i|\hat{H}|\psi_j\rangle$.
If $\psi_n$ is an eigenstate of the Hamiltonian, then: $H_{ij} =\langle \psi_i|\hat{H}\psi_j\rangle = \langle \psi_i|E_j\psi_j\rangle$, which as $E_n$ is just a scalar, becomes:
$$H_{ij} =E_n\langle \psi_i|\psi_j\rangle = \delta_{ij}E_n,$$
where I've used that the eigenstates of a Hermitian operator are orthogonal.
That means in your first equations, the off-diagonal $H_{12}$, and $H_{21}$ terms are both zero.  So they are valid, but a bit of a weird thing to write out.
You can get far by thinking in terms of vectors.  Here is what your equation would say:
$$
\left( \begin{array}{cc}
E_1 & 0 \\
0 & E_2
\end{array} \right)
%
\left( \begin{array}{cc}
1 \\
0 
\end{array} \right)  = E_1\times \left( \begin{array}{cc}
1 \\
0 
\end{array} \right) + 0\times \left( \begin{array}{cc}
0 \\
1 
\end{array} \right).
$$
So while it is valid, it seems a bit odd to include the zero. 
