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,

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


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.