To give some context, this question stems from the physical scenario of 6 ions forming a molecule, where each ion is a "site" in which a single electron is able to hop between ions. Consider a crystal lattice with periodic boundary conditions as an analogue. Periodic boundary conditions are relevant here where the basis vector $|e_j\rangle$ represents the electron being present on the $j^{th}$ ion, where $j = 1,2,3,4,5,6$. Then to enforce periodicity, $|e_j\rangle = |e_{j\pm 6}\rangle$.
The state of the system is given as $$|\Psi\rangle = \sum_{j=1}^{6}a_j|e_j\rangle$$ where $$\sum_{j=1}^{6}|a_j|^2 = 1$$
Now for the math:
I have kinetic and potential operators for the single electron as:
$$\hat{\mathcal{K}} = (-f) \sum_{j=1}^{6}(|e_j\rangle\langle e_{j+1}| + |e_{j+1}\rangle\langle e_j|) + (-f')\sum_{j=1}^{3}(|e_j\rangle\langle e_{j+3}| + |e_{j+3}\rangle\langle e_j|)$$
$$\hat{\mathcal{V}} = \sum_{j=1}^{6}\varepsilon_j |e_j\rangle\langle e_j|$$
respectively. Then, the Hamiltonian, $\hat{\mathcal{H}}$ should be a sum of these two operators, $\hat{\mathcal{H}} = \hat{\mathcal{T}}+\hat{\mathcal{V}}$.
I don't understand how to compute the sums within these operators in order to form a matrix.
More specifically, how do I approach the sums:
$$\sum_{j=1}^{6}(|e_j\rangle\langle e_{j+1}| + |e_{j+1}\rangle\langle e_j|) $$$$\sum_{j=1}^{3}(|e_j\rangle\langle e_{j+3}| + |e_{j+3}\rangle\langle e_j|)$$
For a two ion molecule, we have that $|e_j\rangle\langle e_{j+1}| = |e_1\rangle\langle e_{2}|$ for $j=1$. This yields:
$$|e_1\rangle\langle e_2| = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} $$
and similarly, $$|e_2\rangle\langle e_1| = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ \end{bmatrix}$$
Do I need to compute the same steps above for each $j$ in the 6 ion case, and add the resulting 6 x 6 matrices which arise? Or is there an easier way to obtain the result of these sums?
Then finally, once I reduce $\hat{\mathcal{K}}$ and $\hat{\mathcal{V}}$ down to three matrices, do I just add them up? I feel that the second sum in $\hat{\mathcal{K}}$, which only goes from $j = 1,2,3$ will not allow me to add them easily..