Expectation value in second quantization 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?
 A: For future convenience denote
$$
{\hat \phi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_j}
$$
$$
{\hat \chi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_{j+1}}
$$
The average you want to calculate reads then 
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\langle 0 | \prod_n{{\hat \phi}_n} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\hat \phi}^\dagger_n} |0\rangle
$$
Starting with the regular CCR-s,
$$
\left[ {\hat c}_j, {\hat c}^\dagger_k \right]_\mp = \delta_{jk},\;\;\;\;\; \left[ {\hat c}_j, {\hat c}_k\right]_\mp = \left[ {\hat c}^\dagger_j, {\hat c}^\dagger_k\right]_\mp = 0
$$
obtain
$$
\left[ \sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j}, {\hat \phi}^\dagger_n\right] = {\hat \chi}^\dagger_n
$$
and 
$$
{\hat \phi}^\dagger_n {\hat \chi}^\dagger_m  = \pm {\hat \chi}^\dagger_m {\hat \phi}^\dagger_n
$$
$$
{\hat \phi}_n {\hat \chi}^\dagger_m  = \pm {\hat \chi}^\dagger_m {\hat \phi}_n + \sum_j{A^*_{j+1, n}A_{j,m}}
$$
Now, in the average you need to calculate, use the above to successively move $\left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)$ past the orbital operators to its right: 
$$
\left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\hat \phi}^\dagger_n} = {\hat \phi}^\dagger_1 \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>1}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} =
$$
$$
= \prod_{m=1}^{m=2} {{\hat \phi}^\dagger_m} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>2}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n}  + {\hat \phi}^\dagger_1 {\hat \chi}^\dagger_2 \prod_{n > 2} {{\hat \phi}^\dagger_n} = 
$$
$$
= \prod_{m=1}^{m=2} {\hat \phi}^\dagger_m \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>2}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n}  \pm  {\hat \chi}^\dagger_2 \prod_{n \neq 2} {{\hat \phi}^\dagger_n} = \dots =
$$
$$
= \prod_m {{\hat \phi}^\dagger_m} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) + \sum_m{ (\pm 1)^{m-1} {\hat \chi}^\dagger_m \prod_{n \neq m} {{\hat \phi}^\dagger_n}}
$$
The first term will annihilate the rhs vacuum, so only the sum remains. Now bring in the orbital operators on the left and flip each ${\hat \chi}^\dagger_m$ past them:
$$
\left(\prod_n{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m = \pm \left(\prod_{n>1}{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m {\hat \phi}_1 + \left(\prod_{n\neq 1}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, 1}A_{j,m}} = 
$$ 
$$
\left(\prod_{n>1}{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m  \prod_{l=1}^{l=2} {{\hat \phi}_l} + \left(\prod_{n\neq 1}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, 1}A_{j,m}} \pm \left(\prod_{n\neq 2}{{\hat \phi}_n}\right)\sum_j{A^*_{j+1, 2}A_{j,m}} = \dots = 
$$
$$
= {\hat \chi}^\dagger_m \left(\prod_n{{\hat \phi}_n}\right) + \sum_l{(\pm 1)^{l-1}\left(\prod_{n\neq l}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, l}A_{j,m}} }
$$
The first term now annihilates the lhs vacuum, and after substituting everything the desired average becomes
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, l,m}{ (\pm 1)^{m-1} (\pm 1)^{l-1}  A^*_{j+1, l}A_{j,m} \langle 0 |\left(\prod_{n\neq l}{{\hat \phi}_n}\right)\left( \prod_{n' \neq m} {{\hat \phi}^\dagger_{n'}}\right)  |0\rangle } = 
$$
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, l,m}{ (\pm 1)^{m-1} (\pm 1)^{l-1}  A^*_{j+1, l}A_{j,m}\delta_{l,m} }
$$
and finally
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, m}{  A^*_{j+1, m}A_{j,m} }
$$
There may be some bugs I missed, but this is the general idea.
