3
$\begingroup$

In real space, the Fermi-Hubbard model can be written as:

$$ H = -t \sum_{\langle i,j \rangle\sigma} (c^\dagger_{i\sigma} c_{j\sigma} + c.c.) + U \sum_in_{i\uparrow}n_{i\downarrow}$$

The only difference between having periodic boundary conditions and not having them is that the nearest-neighbor pair, $\langle i, j \rangle$ will also include the pair $(N, 1)$ for periodic boundary conditions.

If we apply a transform to move into k-space, the real space summation really comes into effect when simplifying the exponential (in the form: $ e^{i(k-k')\cdot r_i}$). (I'm assuming that the sites are equally spaced.) I'm wondering if the k-space Hamiltonian is the same, whether or not we have periodic boundary conditions in real space. It seems that the only difference is that our sum over all real sites, $i$, is restricted from 1 to $N$ in the periodic boundary condition case (in order to include $(N, 1)$ at the end, but goes from 1 to $N-1$ in the non-periodic boundary condition case.

The delta function comes (for example in the kinetic term) from:

$$ -\frac{t}{N^2}\sum_{i \sigma}\sum_{kk'}e^{-i(k-k')r_i}e^{ik'\delta}a^\dagger_{k\sigma}a_{k'\sigma} + ...$$ where the $a$'s are my k-space creation/annihilation operators. I'm assuming $\sum_ie^{-i(k-k')r_i} = N\delta_{kk'}$ regardless of whether we sum $i$ to $N$, or $i$ to $N-1$.

Is this true? Is there also a way to physically think about this, to justify whether the k-space Hamiltonian should change?

$\endgroup$
1
  • $\begingroup$ Why should it be true? Try e.g. N=2. $\endgroup$ Jul 24, 2020 at 1:48

1 Answer 1

2
$\begingroup$

Short answer: yes and no.

Please keep in mind that $\sum_i e^{-i (\vec{k}-\vec{k}\prime) \cdot \vec{r}_i} = N \delta_{\vec{k} \vec{k}\prime}$ is not always valid in periodic boundary conditions, i.e., it implies that $\langle \vec{k} | i \rangle = e^{i \vec{k} \cdot \vec{r}_i}$, which is not true, since the electrons described by the Hubbard model are definitely not free electrons that can be described by plane waves!

Instead, you need to involve the Bloch theorem $\psi_i(\vec{r}+\vec{G})=e^{i \vec{k} \cdot \vec{G}} \psi_i(\vec{r})$ where $\vec{G}$ is an integer multiple of the lattice vectors, and $\psi_i(\vec{r}) = \langle \vec{r} | i \rangle$ is the electron wavefunction in real space for an electron occupying site $i$.

$\langle \vec{k} | i \rangle$ actually represents a basis transformation from $\vec{k}$-space to the local basis {i} space. The annihilation and creation operators are also transformed according to $\hat{a}_{\vec{k}\sigma}=\sum_i \langle \vec{k} | i \rangle \hat{c}_{i \sigma}$ and $\hat{a}^\dagger_{\vec{k}\sigma}=\sum_i \langle i | \vec{k} \rangle \hat{c}^\dagger_{i \sigma}$. As expected, these operators anticommute in both bases, $[\hat{c}_{i\sigma},\hat{c}^\dagger_{j\sigma}]_{+} = \delta_{ij}$ and $[\hat{a}_{\vec{k}\sigma},\hat{a}^\dagger_{\vec{k}\prime\sigma}]_{+} = \delta_{\vec{k}\vec{k}\prime}$

In fact, in $\vec{k}$-space, the kinetic term of the Hubbard model takes the form of $\sum_\sigma \epsilon(\vec{k}) \hat{a}^\dagger_{\vec{k}\sigma} \hat{a}_{\vec{k}\sigma}$, and $\epsilon(\vec{k})$ depends on the geometry of the system. For instance, with the 1-D Hubbard model (assume monoatomic chain), $\epsilon(\vec{k})=-2t \cos(|\vec{k}|a)$, where $a$ is the lattice constant. Thus, it indeed changes going from the local basis {i} to momentum basis $\vec{k}$.

(Side note: I wish I can send you my professor's lecture notes on this topic, where he lays out the algebra for the basis transformation.)

$\endgroup$
4
  • $\begingroup$ thanks for the info! I'm not entirely sure if I follow though. So when we take, say, a 3 site ring (3 site periodic BCs) I get exactly the kinetic term you mention. Should I expect it to be different for the 3 site chain? I guess I'm mathematically a bit lost on how to simplify the terms when we cannot reduce the exponential into some sort of delta function. $\endgroup$
    – Jlee523
    Jul 24, 2020 at 4:43
  • $\begingroup$ That's where the definition of the $t_{ij}$ comes in. It is usually defined as $t_{ij} = \langle i | j \rangle$ with the value being nonzero if site $i$ and site $j$ are nearest neighbors, i.e. $j = i \pm 1$. When they are all the same for all $i$ and $j$, we have the final expression $t_{ij} = t$. $\endgroup$
    – wyphan
    Jul 24, 2020 at 16:36
  • $\begingroup$ Also, Bloch theorem holds for a periodic boundary condition. Try to use that to get rid of the exponentials. $\endgroup$
    – wyphan
    Jul 24, 2020 at 17:04
  • $\begingroup$ Will give this a try. Thank you! $\endgroup$
    – Jlee523
    Jul 25, 2020 at 17:28

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.