# How Creation and Annihilation operator transform under an unitary transformation?

\begin{align} \hat{\mathcal H}= \sum_{i,j} \hat{\psi}^{\dagger}_i H_{i,j}\hat{\psi}_j \end{align} The $$\mathcal H$$ is the full second quantized Hamiltonian for a system and $$H$$ is the single particle Hamiltonian in basis $$\left|i\right>= \hat{\psi}^{\dagger}_i \left|0\right>$$, where $${i}={1,2,\ldots, N}$$. And $$H$$ commutes with $$U$$ $$U H U^{\dagger}= H$$ Under a unitary transformation an annihilation and a creation operator transform as \begin{align} \hat{\mathcal U} \hat{\psi}_i \hat{\mathcal U}^{-1} &= \sum_{j} U_{i,j}^{\dagger} \hat{\psi_j} \\ \hat{\mathcal U} \hat{\psi}_i^{\dagger} \hat{\mathcal U}^{-1} &= \sum_{j}\hat{\psi_j}^{\dagger} U_{j,i} \end{align} Here $$\hat{\mathcal U}$$ is the unitary operator acting on Fermion Fock space. and $$U$$ is the unitary operator acting on single particle Hilbert space.

I want to proof the last two equations. I have no idea where to start form. Any help is highly appreciated.
Source of doubt: Topological phases: Classification of topological insulators and superconductors of non-interacting fermions, and beyond (section 3.1.1)

• The equations could be considered a definition of $U$. Do you have another definition and want to show equivalence? – fqq Aug 29 '19 at 12:18
• can you provide any source in this regard? @fqq – Galilean Aug 29 '19 at 13:35
• You don't need a source to read about it. It's just plug and chug from the definition, Every book I have ever read says neither more nor less that what you have here, – mike stone Aug 29 '19 at 13:37

I would say that your two equations are the definition of how a unitary transformation $$U$$ acting on the single-particle Hilbert space induces a unitary transformation $$\mathcal U$$ on the many-particle Fock space. They need to be suplimented with the additional equation $${\mathcal U}|0\rangle = |0\rangle$$ where $$|0\rangle$$ is the no-particle Fock-space state. Once this is done the action of $${\mathcal U}$$ on any Fock-space state is determined.
It is important that $$U$$ be unitary, so that the (anti)-commutation relations $$[\hat \psi_i,\hat \psi^\dagger_j]_\pm = \delta_{ij}$$ be preserved.
• In this perspective (with which I agree) one could want to prove that $\mathcal{U}$ is a linear operator, and that it is unitary. – fqq Sep 3 '19 at 10:49
• $\mathcal U$ is defined to be linear, but one does need to prove unitarity, but this is easy --- at least at the formal level. Of course Fock spaces are tricky analytically. – mike stone Sep 3 '19 at 12:21