# Hubbard Model Hamiltonian in matrix form using basis

I am reading material on Hubbard Model (please see this link) "The limits of Hubbard model" by Grabovski, and I have difficulty deriving/calculating hamiltonian in chapter 8. eq.8.2. How did he derive Hamiltonian matrix $$H_1,H_2$$? Below is my attempt and ignoring other terms U and $$\mu$$,

$$H=-t\sum_{,s}(c_{i,s}^\dagger c_{j,s}+h.c.)$$

$$H=-t(c_{1,\uparrow}^\dagger c_{2,\uparrow}+c_{1,\downarrow}^\dagger c_{2,\downarrow}+c_{2,\uparrow}^\dagger c_{1,\uparrow}+c_{2,\uparrow}^\dagger c_{1,\uparrow} )$$

$$H|\downarrow_1\rangle=-t(c_{1,\uparrow}^\dagger c_{2,\uparrow}+c_{1,\downarrow}^\dagger c_{2,\downarrow}+c_{2,\uparrow}^\dagger c_{1,\uparrow}+c_{2,\uparrow}^\dagger c_{1,\uparrow} )|\downarrow_1\rangle$$

$$H|\downarrow_1\rangle=-t(c_{1,\uparrow}^\dagger c_{2,\uparrow}|\downarrow_1\rangle+c_{1,\downarrow}^\dagger c_{2,\downarrow}|\downarrow_1\rangle+c_{2,\uparrow}^\dagger c_{1,\uparrow}|\downarrow_1\rangle+c_{2,\uparrow}^\dagger c_{1,\uparrow}|\downarrow_1\rangle )$$

Note that others have different notation for $$|\downarrow_1\rangle$$, instead they use $$|1 \downarrow\rangle$$. At this point I do not know how to proceed from here. I know that $$\beta$$ are the basis vectors for Hamiltonian. The answer seems to be $$H|\downarrow_1\rangle=-t|\downarrow_2\rangle$$. After that I do not know how he gets matrix from these? Some notes/examples for this will be appreciated.

added: This is my definition of the following (if I am correct),

$$c^{\dagger}_j= \begin{bmatrix}0&0\\1&0\end{bmatrix}e^{+i\zeta_jt/h}$$

$$c_j= \begin{bmatrix}0&1\\0&0\end{bmatrix}e^{-i\zeta_jt/h}$$

$$|\uparrow\rangle= \begin{bmatrix}1\\0\end{bmatrix}$$ and $$|\downarrow\rangle= \begin{bmatrix}0\\1\end{bmatrix}$$

How the following equation was computed,

$$c_{1,\uparrow}^\dagger c_{2,\uparrow}|\downarrow_1\rangle=c_{1,\uparrow}^\dagger c_{2,\uparrow}|1\downarrow\rangle=?$$

By definition, $$H_1$$ is the matrix in the basis $$\beta$$ of $$H$$ restricted to the eigenspace $$N=1$$ (this is well defined since $$H$$ commutes with $$N$$ therefore they admit a simultaneously diagonal eigenspace decomposition). Since $$\beta$$ is orthonormal, you simply have: $$[H_1]_{\sigma_i,\sigma'_j} = \langle \sigma_i | H |\sigma'_j\rangle$$ Using the same order as in the paper, from $$H|\downarrow_1\rangle=-t|\downarrow_2\rangle$$, you get the first column of $$H_1$$: $$(0,0,-t,0)$$, and more generally using: $$H|\downarrow_1\rangle = 0|\downarrow_1\rangle+0|\uparrow_1\rangle-t|\downarrow_2\rangle+0|\uparrow_2\rangle \\ H|\uparrow_1\rangle = 0|\downarrow_1\rangle+0|\uparrow_1\rangle+0|\downarrow_2\rangle-t|\uparrow_2\rangle \\ H|\downarrow_2\rangle = -t|\downarrow_1\rangle+0|\uparrow_1\rangle+0|\downarrow_2\rangle+0|\uparrow_2\rangle \\ H|\uparrow_2\rangle = 0|\downarrow_1\rangle-t|\uparrow_1\rangle+0|\downarrow_2\rangle+0|\uparrow_2\rangle$$ you read off directly the transpose of $$H_1$$.

The same method applies to $$H_2$$.

Hope this helps and tell me if you find some mistakes.

It is usually best to use the anti-commutation rules of the creation/annihilation operators rather than a matrix representation. I'll remind that: $$\{c_i,c_j^\dagger\} = \delta_{ij}$$ the other combinations anti-commute. Writing $$|0\rangle$$ the zero particle state which is annihilated by the $$c_i$$ by definition, you also have by definition: $$|\downarrow_1\rangle = c_{\downarrow_1}^\dagger|0\rangle$$. Using the two previous facts, you can easily compute your expression, the detailed calculation gives: $$c_{\uparrow_1}^\dagger c_{\uparrow_2}|\downarrow_1\rangle = c_{\uparrow_1}^\dagger c_{\uparrow_2}c_{\downarrow_1}^\dagger|0\rangle \\ =-c_{\uparrow_1}^\dagger c_{\downarrow_1}^\dagger c_{\uparrow_2}|0\rangle \\ =0$$
This is to be expected, the operator corresponds to the hopping of $$\uparrow$$ states so does not affect $$\downarrow$$ states. A calculation that leads to a nontrivial result would be:
$$c_{\uparrow_1}^\dagger c_{\uparrow_2}|\uparrow_2\rangle = c_{\uparrow_1}^\dagger c_{\uparrow_2}c_{\uparrow_2}^\dagger|0\rangle \\ =c_{\uparrow_1}^\dagger (1-c_{\uparrow_2}^\dagger c_{\uparrow_2})|0\rangle \\ =c_{\uparrow_1}^\dagger|0\rangle \\ =|\uparrow_1\rangle \\$$
I don't know where you found your matrix definitions, but know that such formulas cannot exist in our case, since the $$c_i,c_j^\dagger$$ do not commute in with the total number operator, hence cannot be expressed in the $$\beta$$ basis (plus, I don't understand what the $$\zeta_i$$ represent).
• Still how do you calculate $t(c_{1,\uparrow}^\dagger c_{2,\uparrow}|\downarrow_1\rangle$? What is the matrix representation of creation/anihilation and how to enumerate spin up and down? Commented May 27, 2022 at 19:53