The Hamiltonian of tight binding model reads $H=-|t|\sum\limits_{<i,j>}c_i^{\dagger}c_j+h.c.$, why is there a negative sign in the hopping term?
[Physics] Sign of the hopping integral in tight binding model
quantum mechanicssolid-state-physics
Related Solutions
Starting with some background information from Wikipedia, we have that under time reversal the position is unchanged while the momentum changes sign.
In quantum mechanics we can express the action of time reversal on these operators as $\Theta\,\mathbf{x}\,\Theta^\dagger = \mathbf{x}$ and $\Theta\,\mathbf{p}\,\Theta^\dagger = -\mathbf{p}$. It is worth mentioning here that the time reversal operator, $\Theta$, is anti-unitary, which allows it to be expressed as $\Theta = UK$ where $U$ is unitary and $K$ is the complex conjugation operator.
As for the creation/annihilation operators used in second quantization the sign changes under $\Theta$ would suggest a transformation of $a_r \rightarrow a_r$ and $a_k \rightarrow a_{-k}$. If you are worried about the fact that $k$ represents a crystal momentum and not a true momentum you can just take the position transformation, which is perhaps more trustworthy, and use $a_k = \sum_r \, a_r \,\mathrm{exp}[ -\mathrm{i} k\cdot r]$ to verify $a_k \rightarrow a_{-k}$ directly.
Using these transformations you should be able to verify that the tight-binding Hamiltonian is invariant under time reversal in position and momentum space for a lattice with or without a basis. Keep in mind that you would generally take the complex conjugate of the coefficients in $H$, however in your case $t$ and $\epsilon_k$ are both real. Its important to remember though, mostly to make sure $H$ stays hermitian.
As far as your comment about $\sigma_y$, this is only necessary if you include spin. Spin changes sign under time reversal so $\Theta\,\mathbf{S}\,\Theta^\dagger = -\mathbf{S}$. In this case, we can formally write $\Theta = \mathrm{exp}[-i \pi J_y]\,K$, which is probably the relation you are alluding to.
According to J.J. Sakurai's Modern Quantum Mechanics one possible convention for the time-reversed angular momentum states is $\Theta | j,m\rangle = (-1)^m |j,-m\rangle$. This suggests that with spin indices the creation/annihilation operators transform like $a_{r,m} \rightarrow (-1)^m\,a_{r,-m}$ and $a_{k,m} \rightarrow (-1)^m \, a_{-k,-m}$ under time reversal. From what I understand, most spin Hamiltonians will be invariant under this transformation. An example when this is not the case would be in the presence of an external magnetic field which couples to the spins through a $\mathbf{S}\cdot \mathrm{B}-$like term.
It is interesting how even in the absence of an external field the groundstate of spin Hamiltonians can still sponanteously break the time reversal symmetry present in $H$, but rather than discuss this myself I will direct you to this very well written answer.
Under the substitution ℏk→ℏk−qA
$ \langle p{\mid}x\rangle =\langle 0{\mid} a_{p} a_{x}^{+}{\mid}0\rangle =exp(-ipx/ h) $
will become
$ \langle p{\mid}x\rangle =\langle 0{\mid} a_{p} a_{x}^{+}{\mid}0\rangle =exp(-i(p-qA)x/ h) $
effectively, the change in operator:
$ a_{p} a_{x}^{+} \rightarrow a_{p} a_{x}^{+} e^{iqAx/h} $
Then it looks as if:
$ a_{x}^{+} \rightarrow a_{x}^{+} e^{iqAx/h} $
$ a_{x} \rightarrow a_{x} e^{-iqAx/h} $
Actually, this is just $ \Phi \rightarrow e^{iqAx /h} \Phi $ for the substitution solution of Schrodinger equation.
Best Answer
The hopping term is given by
$$ t_{ij}=\int\limits_{\mathbf{r}}d\mathbf{r}\phi_i^*\left(\mathbf{r}\right)\left[-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf{r})\right]\phi_j\left(\mathbf{r}\right) $$
where $i$ and $j$ are the sites whose hopping you want to find, $\phi_{i,j}(\mathbf{r})=\phi(\mathbf{r}-\mathbf{r}_{i,j})$ ($\mathbf{r}_i$ being coordinate of $i$-th site) are the atomic orbitals, and $U(\mathbf{r})$ is the potential of the crystal lattice. So the sign depends on your choice for $U(\mathbf{r})$. If $U(\mathbf{r})$ is taken as negative (Coulomb potential) you are more likely to end up with a negative $t$ which can be a nuisance, so people just redefine it as $-|t|$.