Thouless theorem

The Thouless theorem states that two non-orthogonal determinants are related by an exponential operator of single excitations: \[\begin{equation} \require{physics} \ket{^w\Phi} = \exp(\mathcal{Z})\ket{^x\Phi}\bra{^w\Phi}\ket{^x\Phi} \end{equation}\]

To derive Thouless theorem, we start by stating that we can re-write one set of creation (or annihilation) operators with respect to another, if both form a complete set \[\begin{equation} \require{physics} ^\dagger b = \sum_p ^\infty u_{\alpha p}a_p^\dagger \thinspace , \end{equation}\] which is essentially a change of basis operation.

The summation can be split over the occupied and virtual indices respectively \[\begin{equation} \require{physics} ^\dagger b = \sum_i ^{occ} u_{\alpha i}a_i^\dagger + \sum_a ^\infty u_{\alpha a}a_a^\dagger \thinspace , \end{equation}\]

Since our two determinants are non-orthogonal, we can choose n intermediate normalization setting as such \[\begin{equation} \require{physics} \bra{^w\Phi}\ket{^x\Phi} = 1 \thinspace , \end{equation}\] which implies that \(u_{\alpha i}\) is unitary where it runs from \(1\) to \(N_{occ}\). The \(N_{occ} \times N_{occ}\) block is thus invertible. \[\begin{equation} \require{physics} u_{\alpha i}^{-1} = U_{i \alpha} \thinspace . \end{equation}\] This leads to the following relations: \[\begin{align} \require{physics} \vb{U}\vb{u}_{occ, occ} &= \vb{I} \thinspace \text{or} \thinspace \sum_{\alpha = 1} ^{occ} U_{i \alpha}u_{\alpha j} = \delta_{ij} \thinspace , \\ \vb{u}_{occ, occ}\vb{U} &= \vb{I} \thinspace \text{or} \thinspace \sum_{i = 1} ^{occ} u_{\alpha i}U_{i, \beta} = \delta_{\alpha \beta} \thinspace , \\ \vb{U}\vb{u}_{occ, occ+1 : \infty} &= \vb{T} \thinspace \text{or} \thinspace \sum_{\alpha = 1} ^{occ} U_{i \alpha}u_{\alpha a} = t_{ai} \thinspace . \end{align}\]

Because \(\vb{U}\) is unitary, we can write \(N_{occ}\) linear combinations of the creation (or annihilation) operators \(d^\dagger_i\) \[\begin{align} d^\dagger_i &= \sum_{\alpha = 1} ^{occ} U_{i \alpha} b^\dagger_\alpha \\ &= \sum_{\alpha = 1} ^{occ} U_{i \alpha} (\sum_{j = 1} ^{occ} u_{j\alpha}a^\dagger_j + \sum_a ^\infty u_{\alpha a}a_a^\dagger) \\ &= \sum_{\alpha = 1} ^{occ}\sum_{j = 1} ^{occ} U_{i \alpha} u_{j\alpha} a^\dagger_j + \sum_{\alpha = 1} ^{occ}\sum_a ^\infty U_{i \alpha} u_{\alpha a} a_a^\dagger \\ &= \sum_{j = 1} ^{occ} \delta_{ij} a^\dagger_j + \sum_a ^\infty t_{ai} a^\dagger_a \\ &= a_i^\dagger + \sum_a ^\infty t_{ai} a^\dagger_a \thinspace , \end{align}\] where we replaced \(b_\alpha^\dagger\) by its expansion in \(a^\dagger\), split in its occupied and virtual components and subsequently used the relations between \(\vb{u}\) and \(\vb{U}\) to simplify the expression.

To verify Thouless theorem and define its operator \(\mathcal{Z}\) we use the new \(d^\dagger_i\) operators to construct a new single slater determinant \[\begin{align} \require{physics} \ket{\psi} &= \prod^N_i d^\dagger_i \ket{0} \\ &= \prod^N_i (a_i^\dagger + \sum_{a=N+1} ^\infty t_{ai} a^\dagger_a) \ket{0} \\ &= \prod^N_i (1 + \sum_{a=N+1} ^\infty t_{ai} a^\dagger_a a_i) a_i^\dagger\ket{0} \\ &= \prod^N_i (1 + \sum_{a=N+1} ^\infty t_{ai} a^\dagger_a a_i) \ket{\psi_0} \\ &= \prod^N_i \prod^\infty_{a=N+1}(1 + t_{ai} a^\dagger_a a_i) \ket{\psi_0} \\ &= \exp(\mathcal{Z})\ket{\psi_0} \thinspace , \end{align}\] with \(\mathcal{Z}\) defined as \[\begin{equation} \mathcal{Z} = 1 + t_{ai} a^\dagger_a a_i \thinspace . \end{equation}\]