The Hamiltonian elements in Anderson dimer

[hokhani]

TL;DR Summary

I don't know the occur of - sign in some elements of the Hamiltonian

In a system with two orbitals $c$ and $d$ (each with two spin degrees of freedom), consider the Hamiltonian $H=V(d^{\dagger}_{\uparrow} c_{\uparrow} + c^{\dagger}_{\uparrow}d_{\uparrow}+d^{\dagger}_{\downarrow} c_{\downarrow} + c^{\dagger}_{\downarrow}d_{\downarrow})$. Also suppose that the many body states are indexed as $|n_{d_{\uparrow}}, n_{d_{\downarrow}}, n_{c_{\uparrow}} , n_{c_{\downarrow}}\rangle$ where $n$ shows the occupation number of each spin-orbital.
As mentioned in the literature, $\langle1001|H|1100\rangle=V$ and $\langle0110|H|1100\rangle=-V$. However, according to my calculations, $H|1100\rangle=V(|0110\rangle+|1001\rangle)$ which gives $V$ for the two matrix elements $\langle1001|H|1100\rangle$ and $\langle 0110|H|1100\rangle$. I would like to know where my calculations goes wrong! Any help is appreciated.

 

[Haborix]

Can you provide a citation to an example of the literature you refer to?

 

[hokhani]

Can you provide a citation to an example of the literature you refer to?

Of course. In the following virtual lecture on youtube at the minute 52 (the blue text):

The Hamiltonian is written in the subspace $Q=2, S_z=0$ and I am only interested in the difference between the arrays $H_{1,3}$ and $H_{2,3}$ so I neglected the other terms of the Hamiltonian in my main post.

 

[Haborix]

I think you've probably just gotten the order of the basis vectors mixed up. I get $H|1100\rangle=V(-|0110\rangle+|1001\rangle)+(2\epsilon+U)|1100\rangle$. In the notation of the video I can represent $|1100\rangle$ as $$\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix}.$$

EDIT: I think I may have misunderstood your question. Are you essentially asking how to get the matrix representation of the Hamiltonian?

 

[hokhani]

I think you've probably just gotten the order of the basis vectors mixed up. I get $H|1100\rangle=V(-|0110\rangle+|1001\rangle)+(2\epsilon+U)|1100\rangle$. In the notation of the video I can represent $|1100\rangle$ as $$\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix}.$$

EDIT: I think I may have misunderstood your question. Are you essentially asking how to get the matrix representation of the Hamiltonian?

Thank you so much for your attention. Could you please explain how did you obtain $c^{\dagger}_{\uparrow} d_{\uparrow} |1100 \rangle =- |0110 \rangle$? In other words, my question is about the appearance of the minus sign.
EDIT: Your answer is quite to the point and I am lookig for the reason for the minus sign.