Results

To fix the notation, let us begin by recalling the basic standard definitions and properties pertaining to the translation symmetry. Let $\vert\Phi\rangle$ denote a normalized Slater determinant built of single-particle orbitals that are localized in space. Since the total momentum operator $\hat{\bm{P}}=\sum_{i=1}^A \hat{\bm{p}}_i$ is the generator of translation, states $\vert\Phi\rangle$ can be shifted by $\bm{R}$ to an arbitrary location in space as

$$\vert\Phi(\bm{R})\rangle = \exp({\textstyle{\frac{i}{\hbar}}}\bm{R}\cdot\hat{\bm{P}})\vert\Phi\rangle.$$ (1)

Then, the eigenstates of $\hat{\bm{P}}$, the so-called projected states $\vert\bm{P}\rangle$, can be built as linear combinations of $\vert\Phi(\bm{R})\rangle$, that is,

$$\vert\bm{P}\rangle = \frac{1}{(2\pi\hbar)^{3}}\int {\rm {d}}... ...{\frac{i}{\hbar}}}\bm{R}\cdot\bm{P})\vert\Phi(\bm{R})\rangle ,$$ (2)

and $\hat{\bm{P}}\vert\bm{P}\rangle=\bm{P}\vert\bm{P}\rangle$. The normalization condition of states $\vert\bm{P}\rangle$ is chosen in such a way that the original Slater determinant is a simple integral thereof,

$$\vert\Phi\rangle \equiv \vert\Phi(\bm{0})\rangle= \int {\rm {d}}^3 \bm{P} \vert\bm{P}\rangle .$$ (3)

The Slater determinant $\vert\Phi\rangle$ is, therefore, a normalized wave packet built of non-normalizable center-of-mass plane waves $\vert\bm{P}\rangle$. For a system described by a translationally invariant Hamiltonian $\hat{H}$, $[\hat{H},\hat{\bm{P}}]=0$, one can determine the average energy $E_\Phi(\bm{P})$ of each plane wave, which is called the projected energy, as

$$E_\Phi(\bm{P}) = {\langle\Phi\vert\hat{H}\vert\bm{P}\rangle}/ {\langle\Phi \vert\bm{P}\rangle} .$$ (4)