I am stuck calculating a simple expectation value for an operator, which is expressed in second quantization. I know the result, but I fail to proof it.
Lets say I have one-particle wave function $|\phi_n\rangle$ given by $|\phi_n\rangle=\sum_{j=1}^K |\alpha_j\rangle A_{j,n}$, where $K$ is the number of orbitals/sites in the system and the $A_{j,n}$ are the probability amplitudes of the oribitals $|\alpha_j\rangle$. The index $n$ labels the particles in the system, of which we have $N$.
The orbitals are orthonormal, i.e. $$\sum_j A_{j,m}^* A_{j,n} = \delta_{m,n}.$$
Let's ignore any spin degrees of freedom. The many particle wave function is now given by $$|\Psi\rangle = \left(\prod_{n=1}^N \sum_{j=1}^K \hat{c}^\dagger_j A_{j,n}\right) |\text{vac}\rangle,$$ where the $\hat{c}^\dagger_j$ is the usual creation operator on site $j$.
What I now want to calculate, is the expectation value $$\langle \Psi|H_\text{hop}|\Psi\rangle$$ with $$H_\text{hop}=-t \sum_{j=1}^K \hat{c}^\dagger_{j+1}\hat{c}_j + h.c.$$ the usual hopping Hamiltonian. I have the strong feeling (and one example calculation supported this), that the result is just $$\langle \Psi|H_\text{hop}|\Psi\rangle = -t \sum_{n=1}^N \sum_{j=1}^K ( A_{j,n}^* A_{j+1,n} + A_{j+1,n}^* A_{j,n} )$$
I think this result is trivially related to the Slater-Condon rules, but I fail to see the connection. Additionally, I fail to explicitly calculate the expectation value, which contains the sum and products of the creation/annihilation operators.
What is a good way to prove my result?