Inversion on fermionic operators

Question

Inversion is the transformation defined by $$\hat{I}\hat{x}\hat{I}^{-1}=-\hat{x},$$ $$\hat{I}\hat{k}\hat{I}^{-1}=-\hat{k},$$ $$\hat{I}i\hat{I}^{-1}=i.$$

It is therefore a unitary operator. Let $\hat{H}$ be a Hamiltonian of free fermions on a lattice with sites labeled by a subindex in the fermionic operators. Unitary operators act on fermionic field as linear applications$$\hat{I}\hat\psi_l\hat{I}^{-1}=\sum_l U_{lm}\hat{\psi}_m.$$

Now, from this fact, one has that $$\hat{I}\hat{\psi}_k\hat{I}^{-1}=\frac{1}{\sqrt{L}}\sum_l\hat{I}\hat\psi_l e^{-ikx_l}\hat{I}^{-1}=\frac{1}{\sqrt{L}}\sum_l\hat{I}\hat\psi_l\hat{I}^{-1}e^{-ikx_l}\\=\frac{1}{\sqrt{L}}\sum_{lm}U_{lm}\hat\psi_me^{ikx_l}=\frac{1}{\sqrt L}\sum_{l\Delta}U_\Delta\hat{\psi}_{l+\Delta} e^{ikx_l}=\frac{1}{L}\sum_{l\Delta q}U_\Delta \hat\psi_q e^{i(k-q)x_l}e^{-iq\Delta}\\=\sum_\Delta U_\Delta e^{-ik\Delta}\hat{\psi_k}=U_k\hat{\psi}_k.$$ where I used translational invariance and defined $\Delta=l-m$. However, I understand that parity must act inverting the momentum as $\hat{I}\hat\psi_k\hat{I}^{-1}=U\psi_{-k}$. What is the origin of this difference between the conceptual action of this operator and the result of the above calculation?

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EDIT: I am not asking to solve any exercise, I just have noticed a conceptual contradiction between the result that I obtain and the underlying logic of the transformation.

Answer 1

The matrix $𝑈_{𝑙𝑚}$ is not translationally invariant, i.e. it is not simply a function of $Δ=𝑙−𝑚$. This is because you pick an origin when you think of inversions.

This is also clear from writing down the form of $𝑈_{𝑙𝑚}$. If you imagine you have a one-dimensional chain of rotationally symmetric ($s$) orbitals, we can write $U_{lm} = \delta_{l,-m}$. Then your expression becomes \begin{align} \hat{I}\hat{\psi}_k \hat{I}^{-1} &= \frac{1}{\sqrt{L}}\sum_{lm} U_{lm} \hat{\psi}_m e^{i k x_l} \\ &= \frac{1}{\sqrt{L}}\sum_{l} \hat{\psi}_{-l} e^{i k x_l} \\ &= \frac{1}{\sqrt{L}}\sum_{l'} \hat{\psi}_{l'} e^{-i k x_{l'}} \\ &= \hat{\psi}_{-k} \end{align} consistent with the usual expectation.