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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_{<i,j>,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,

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

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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.

Edit (added question):

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).

Hope this helps.

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  • $\begingroup$ 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? $\endgroup$
    – Aschoolar
    Commented May 27, 2022 at 19:53

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