I'm trying to work out the expression for the matrix elements in Section VI of this paper (PRA 72, 053604).

There is a point I need to expand the contact interaction term

$\hat{V} = U \int dr\,\, \hat{\psi}^{\dagger}(r)\hat{\psi}^{\dagger}(r)\hat{\psi}(r)\hat{\psi}(r) $

in the Bloch function basis ($\hat{\psi}(r) = \sum_{k} u_{k}(r) \hat{a}_k$). I know that in the free-particle case, the amplitude $u_k (r) = e^{ikr}$ (plane-wave), and this essentially provides the constraint of momentum conservation when you write $\hat{V}$ in momentum space. That is, if each $\hat{\psi}$ provides amplitude of $e^{ik_ir}$, then you get $\delta(k_1+k_2-k_3-k_4)$ from integrating $e^{-i(k_1+k_2-k_3-k_4)r}$ over the real space. This allows us to say that two colliding momentum states exchange momentum when $\hat{V}$ acts on them.

I'm not sure how, in the case of particle in periodic potential, you mathematically obtain the conservation of quasimomentum (modulo $k_{lattice}$) when you have $u_{k}(r)$ equals Bloch function.


1 Answer 1


I think I know the answer now. You have to use the fact that Bloch functions behave under discrete translations as $u_{k}(r+R) = e^{ikR}u_{k}(r)$, where $R$ is a Bravais lattice vector.

I will be sloppy with $\sum_{k}$ vs $\int\,dk$ below. Also, for simplicity and without loss of generality, I restrict myself to a single band in a 1D lattice.

If we expand the field operators in the Bloch function basis:

$\hat{\psi}(r) = \int dk\, \hat{c}_{k} u_{k}(r) $

Then we can write the interaction term $\hat{V}$ as

$\hat{V} = \int \prod_{i=1,2,3,4} dk_{i}\,\hat{c}_{k_1}^{\dagger}\hat{c}_{k_2}^{\dagger}\hat{c}_{k_3}\hat{c}_{k_4} \left(\int dx\, \psi_{k_1}^{*}(r)\psi_{k_2}^{*}(r)\psi_{k_3}(r)\psi_{k_1}^{*}(r)\right)$

Now if we focus on the integral within the parentheses, we can divide the integration region into blocks of unit cells. Using the Bloch function property above, we get

$\sum_{R = \text{lattice vectors}}e^{-i(k_1 + k_2 - k_3 -k_4)R}\int_{\text{unit cell}} dx\, \psi_{k_1}^{*}(r)\psi_{k_2}^{*}(r)\psi_{k_3}(r)\psi_{k_1}^{*}(r)$

Then I presumably get the conservation of quasimomentum modulo $\frac{2\pi}{R}$ from $\sum_{R = \text{lattice vectors}}e^{-i(k_1 + k_2 - k_3 -k_4)R}$.


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