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At first consider the Hamiltonian of the $k$th$\m$series term
\begin{equation}
\mc H_k\e
\begin{bmatrix}
\hp{\m i}\epsilon_k & \alpha & ia & \!\!\!\!\m ia & \!\!\!\!\m a & 0 \vp\\
\hp{\m i}\alpha^* & \epsilon_k & 0 & 0 & 0 & a \vp\\
\m ia & 0 & \epsilon_{k+Q_1} & 0 & 0 & 0 \vp\\
\hp\m ia & 0 & 0 & \epsilon_{k+Q_2} & 0 &0 \vp\\
\m a & 0 & 0 & 0 & \epsilon_{k-Q_3} & \alpha \vp\\
\hp{\m i}0 & a & 0 & 0 & \alpha^* & \epsilon_{k-Q_3}\vp
\end{bmatrix}
\tl{01}
\end{equation}
To simplify the notation we change to the expression
\begin{equation}
\mc K\e
\begin{bmatrix}
\begin{array}{c|c|c|c|c|c}
r & g\p ih & ib & \m ib & \m b & 0 \vp\\
\hline
g\m ih & r & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} b \hp{hh} \vp\\
\hline
\m ib & 0 & s & 0 & 0 & 0 \vp\\
\hline
ib & 0 & 0 & t & 0 & 0 \vp\\
\hline
\m b & 0 & 0 & 0 & u & g\p ih \vp\\
\hline
0 & b & 0 & 0 & g\m ih & u\vp
\end{array}
\end{bmatrix}
\tl{02}
\end{equation}
where $\;b,g,h,r,s,t,u\;$ are real numbers.
The matrix $\;\mc K\;$ is a $\;6\times 6\;$ hermitian matrix so it has 6 real eigenvalues $\;\lambda_\rho\:(\rho\e 1,\cdots, 6)$, roots of its characteristic polynomial
\begin{equation}
\mr P\plr{\lambda}\e a_6\lambda^6 \p a_5\lambda^5 \p a_4\lambda^4 \p a_3\lambda^3 \p a_2\lambda^2 \p a_1\lambda\p a_0
\tl{03}
\end{equation}
where $\;a_\sigma\:(\sigma\e 0,\cdots, 6)\;$ real coefficients.
The problem to obtain analytical expressions for the eigenvalues is equivalent to the problem of finding analytical solutions of a 6th degree polynomial equation. We know that the latter is impossible in general, except of special cases.
The least we could help here is to give below the expressions of the coefficients $\;a_\sigma\;$ in terms of the variables $\;b,g,h,r,s,t,u\;$ so that to obtain numerically the eigenvalues for given values of these variables.
\begin{equation}
\begin{split}
a_6 & \e 1\\
a_5 & \e -2r-s-t-2u\\
a_4 & \e -4b^2-2g^2-2h^2+r^2+2rs+2rt+st+4ru+2su+2tu+u^2\\
a_3 & \e 4b^2r+2g^2r+2h^2r+3b^2s+2g^2s+2h^2s-r^2s+3b^2t+2g^2t+2h^2t-r^2t-2rst\\
& \hp\e +6b^2u+2g^2u+2h^2u-2r^2u-4rsu-4rtu-2stu-2ru^2-su^2-tu^2\\
a_2 & \e 3b^4+4b^2g^2+g^4+4b^2h^2+2g^2h^2+h^4-g^2r^2-h^2r^2-3b^2rs-2g^2rs-2h^2rs-3b^2rt\\
& \hp\e -2g^2rt-2h^2rt-2b^2st-2g^2st-2h^2st+r^2st-6b^2ru-4b^2su-2g^2su-2h^2su+2r^2su\\
& \hp\e -4b^2tu-2g^2tu-2h^2tu+2r^2tu+4rstu-2b^2u^2-g^2u^2-h^2u^2+r^2 u^2+2rsu^2+2rtu^2+stu^2\\
a_1 & \e -2b^2g^2r-2b^2h^2r-2b^4s-3b^2g^2s-g^4s-3b^2h^2s-2g^2h^2s-h^4s+g^2r^2s+h^2r^2s-2b^4t\\
& \hp\e -3b^2g^2t-g^4t-3b^2h^2t-2g^2h^2t-h^4t+g^2r^2t+h^2r^2t+2b^2rst+2g^2rst+2h^2rst-2b^4u\\
& \hp\e +4b^2rsu+4b^2rtu+2b^2stu+2g^2stu+2h^2stu-2r^2stu+2b^2ru^2+b^2su^2+g^2su^2+h^2su^2\\
& \hp\e -r^2su^2+b^2tu^2+g^2tu^2+h^2tu^2-r^2tu^2-2rstu^2\\
a_0 & \e b^2g^2rs+b^2h^2rs+b^2g^2rt+b^2h^2rt+b^4st+2b^2g^2st+g^4st+2b^2h^2st+2g^2h^2st+h^4st-g^2r^2st\\
& \hp\e -h^2r^2st+b^4su+b^4tu-2b^2rstu-b^2rsu^2-b^2rtu^2-g^2stu^2-h^2stu^2+r^2stu^2\\
\end{split}
\tl{04}
\end{equation}
Note that
\begin{align}
a_5 & \e -2r-s-t-2u \e \m \texttt{trace}\plr{\mc K}\e \m \sum\limits_{\rho\e 1}^{\rho\e 6}\lambda_\rho
\tl{05.1}\\
a_0 & \e b^2g^2rs\p\cdots\cdots\p r^2stu^2 \e \texttt{det}\plr{\mc K}\e \prod\limits_{\rho\e 1}^{\rho\e 6}\lambda_\rho
\tl{05.2}
\end{align}
$\hebl$
EXAMPLE A
Consider that
\begin{equation}
h\e 0\,, \qquad b\e g\e r\e s\e t\e u \e 1
\tl{A-01}
\end{equation}
that is
\begin{equation}
\mc K\e
\begin{bmatrix}
\begin{array}{c|c|c|c|c|c}
1 & 1 & i & \m i & \m 1 & 0 \vp\\
\hline
\hp{gg} 1 \hp{hh} & \hp{gg} 1 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} b \hp{hh} \vp\\
\hline
\m i & 0 & 1 & 0 & 0 & 0 \vp\\
\hline
i & 0 & 0 & 1 & 0 & 0 \vp\\
\hline
\m 1 & 0 & 0 & 0 & 1 & 1\vp\\
\hline
0 & 1 & 0 & 0 & 1 & 1\vp
\end{array}
\end{bmatrix}
\tl{A-02}
\end{equation}
The characteristic polynomial is
\begin{equation}
\mr P\plr{\lambda}\e \lambda^6 \m 6\lambda^5 \p 9\lambda^4 \p 4\lambda^3 \m 13\lambda^2 \p 2\lambda\p 3
\tl{A-03}
\end{equation}
As shown in Figure-01 a graphical solution yields the following eigenvalues
\begin{equation}
\begin{bmatrix}
\hp g \lambda_1 \hp h \vp\\
\hp g \lambda_2 \hp h \vp\\
\hp g \lambda_3 \hp h \vp\\
\hp g \lambda_4 \hp h \vp\\
\hp g \lambda_5 \hp h \vp\\
\hp g \lambda_6 \hp h \vp
\end{bmatrix}\e
\begin{bmatrix}
\hp g \m 1.00000000 \hp h \vp\\
\hp g \m 0.41421356 \hp h \vp\\
\hp g \hp\m\: 1.00000000 \hp h \vp\\
\hp g \hp\m\: 1.00000000 \hp h \vp\\
\hp g \hp\m\: 2.41421356 \hp h \vp\\
\hp g \hp\m\: 3.00000000 \hp h \vp
\end{bmatrix}
\tl{A-04}
\end{equation}
while the result with MATHEMATICA is
\begin{equation}
\begin{bmatrix}
\hp g \lambda_1 \hp h \vp\\
\hp g \lambda_2 \hp h \vp\\
\hp g \lambda_3 \hp h \vp\\
\hp g \lambda_4 \hp h \vp\\
\hp g \lambda_5 \hp h \vp\\
\hp g \lambda_6 \hp h \vp
\end{bmatrix}\e
\begin{bmatrix}
\hp g \m 1 \hp h \vp\\
\hp g \hp\m 1\m\sqrt{2} \hp h \vp\\
\hp g \hp\m\: 1 \hp h \vp\\
\hp g \hp\m\: 1 \hp h \vp\\
\hp g \hp\m\: 1\p\sqrt{2} \hp h \vp\\
\hp g \hp\m\: 3 \hp h \vp
\end{bmatrix}
\tl{A-05}
\end{equation}
in full agreement with \eqref{A-04}.
$\hebl$
EXAMPLE B
Consider that
\begin{equation}
b\e g\e s\e u\e 1\,,\: h\e -1\,,\: r\e 2\,,\: t\e 0.5
\tl{B-01}
\end{equation}
that is
\begin{equation}
\mc K\e
\begin{bmatrix}
\begin{array}{c|c|c|c|c|c}
2 & 1-i & i & \m i & \m 1 & 0 \vp\\
\hline
1+i & \hp{gg} 2 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} b \hp{hh} \vp\\
\hline
\m i & 0 & 1 & 0 & 0 & 0 \vp\\
\hline
i & 0 & 0 & 0.5 & 0 & 0 \vp\\
\hline
\m 1 & 0 & 0 & 0 & 1 & 1-i \vp\\
\hline
\hp{gg} 0 \hp{hh} & 1 & 0 & 0 & 1+i & 1\vp
\end{array}
\end{bmatrix}
\tl{B-02}
\end{equation}
The characteristic polynomial is
\begin{equation}
\mr P\plr{\lambda}\e \lambda^6 \m 7.5\lambda^5 \p 14.5\lambda^4 \p 2\lambda^3 \m 16.5\lambda^2 \p 1.5\lambda\p 4
\tl{B-03}
\end{equation}
As shown in Figure-02 a graphical solution yields the following eigenvalues
\begin{equation}
\begin{bmatrix}
\hp g \lambda_1 \hp h \vp\\
\hp g \lambda_2 \hp h \vp\\
\hp g \lambda_3 \hp h \vp\\
\hp g \lambda_4 \hp h \vp\\
\hp g \lambda_5 \hp h \vp\\
\hp g \lambda_6 \hp h \vp
\end{bmatrix}\e
\begin{bmatrix}
\hp g \m 0.807861\hp h \vp\\
\hp g \m 0.503229 \hp h \vp\\
\hp g \hp\m\:\, 0.698461 \hp h \vp\\
\hp g \hp\m\: 1.213848 \hp h \vp\\
\hp g \hp\m\: 2.907977 \hp h \vp\\
\hp g \hp\m\: 3.990804 \hp h \vp
\end{bmatrix}
\tl{B-04}
\end{equation}
while the result with MATHEMATICA is
\begin{equation}
\begin{bmatrix}
\hp g \lambda_1 \hp h \vp\\
\hp g \lambda_2 \hp h \vp\\
\hp g \lambda_3 \hp h \vp\\
\hp g \lambda_4 \hp h \vp\\
\hp g \lambda_5 \hp h \vp\\
\hp g \lambda_6 \hp h \vp
\end{bmatrix}\e
\begin{bmatrix}
\hp g \m 0.807861 \hp h \vp\\
\hp g \m 0.503229 \hp h \vp\\
\hp g \hp\m\:\, 0.698461 \hp h \vp\\
\hp g \hp\m\:\, 1.21385 \hp{hh} \vp\\
\hp g \hp\m\:\, 2.90798 \hp{hh} \vp\\
\hp g \hp\m\:\, 3.9908 \hp{hhh}
\end{bmatrix}
\tl{B-05}
\end{equation}
in full agreement with \eqref{B-04}.
Best Answer
You likely already know this, but for $N=1$ there are no on site interactions because you only have 1 particle, what can it interact with?
To answer your question, in $k$ space the second term in your Hamiltonian corresponds to a scattering event. To clearly see how this works, make a quick change of variables $k\to k+q$ and $k'\to k'-q$. Then you see that your new interaction term is $$\frac{U}{L}\sum_{k,k',q}c^\dagger_{k+q}c^\dagger_{k'-q}c_{k'}c_k$$ which destroys two particles with momenta (wavevectors) $k$ and $k'$ and creates two new particles with momenta (wavevectors) $k+q$ and $k'-q$. Which is exactly what you'd expect from a scattering event between two particles with conservation of momentum. Two initial particles with momenta $k$ and $k'$ interact and leave with momenta $k+q$ and $k'-q$. Like above, when you have only 1 particle this term is trivial - what can this particle scatter with?
The allowed values for $k$, $k'$ and $q$ are just what you'd expect them to be - they can take on any allowed value for momentum in your system $\to \frac{2\pi n}{M}$ with $n\in\{0,\pm 1,\pm2,...,\pm N/2\}$ (remembering that the two endpoints correspond to the same physical mode). Another way to say this, and likely more useful in practice, is that $k$, $k'$ and $q$ run over every site in $k$-space for your system. It's not longer telling you about "which particles are hopping from where to where", you're working in the momentum basis. Anything with a definite momentum is in a superposition of position basis states.
Hope that helps!