\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)

  • 1
    $\begingroup$ The equations could be considered a definition of $U$. Do you have another definition and want to show equivalence? $\endgroup$
    – fqq
    Aug 29, 2019 at 12:18
  • $\begingroup$ can you provide any source in this regard? @fqq $\endgroup$
    – Galilean
    Aug 29, 2019 at 13:35
  • 2
    $\begingroup$ 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, $\endgroup$
    – mike stone
    Aug 29, 2019 at 13:37

2 Answers 2


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.

As a definition, no "proof'' is to be expected or desired.

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.

  • $\begingroup$ In this perspective (with which I agree) one could want to prove that $\mathcal{U}$ is a linear operator, and that it is unitary. $\endgroup$
    – fqq
    Sep 3, 2019 at 10:49
  • $\begingroup$ $\mathcal U$ 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. $\endgroup$
    – mike stone
    Sep 3, 2019 at 12:21

On the one hand, the creation and annihilation operators transform just like any other operator (keeping in mind that here the operators and transformations are in the Fock space), i.e., $$ a,a^\dagger\rightarrow UaU^\dagger, Ua^\dagger U^\dagger. $$ On the other hand, a result of action of a creation/annihilation operator is a wave function (in the Fock space) that belongs to the same Hilbert space. Since wave function transforms as $$ |\phi\rangle \rightarrow U|\phi\rangle, $$ we should also have $$ a|\phi\rangle, a^\dagger|\phi\rangle \rightarrow Ua|\phi\rangle, Ua^\dagger|\phi\rangle. $$


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