Let's consider a single-particle(boson or fermion) with $n$ states $\phi_1,\cdots,\phi_n$(normalized orthogonal basis of the single-particle Hilbert space), and let $h$ be the single-particle Hamiltonian. As we all know, the second quantization Hamiltonian $H=\sum\left \langle \phi_i \mid h \phi_j \right \rangle c_i^\dagger c_j$ of $h$ should *not* depend on the single-particle basis we choose(where $c_i,c_i^\dagger$ are the bosonic or fermionic operators.), and this can be easily proved as follows:

Choose a new basis, say $(\widetilde{\phi}_1,\cdots,\widetilde{\phi}_n)=(\phi_1,\cdots,\phi_n)U$, where $U$ is a $n\times n$ unitary matrix. Further, from the math viewpoint, an inner product can has two alternative definitions, say $\left \langle \lambda_1\psi_1\mid \lambda_2 \psi_2 \right \rangle=\lambda_1^*\lambda_2 \left \langle \psi_1\mid \psi_2 \right \rangle(1)$ or $\lambda_1\lambda_2^* \left \langle \psi_1\mid \psi_2 \right \rangle(2)$.

Now, if we think $(\widetilde{c}_1^\dagger,\cdots,\widetilde{c}_n^\dagger)=(c_1^\dagger,\cdots,c_n^\dagger)U$ combined with the definition (1) for inner product, then it's easy to show that $\sum\left \langle \widetilde{\phi}_i \mid h \widetilde{\phi}_j \right \rangle \widetilde{c}_i^\dagger \widetilde{c}_j=\sum\left\langle \phi_i \mid h \phi_j \right \rangle c_i^\dagger c_j$; On the other hand, if we think $(\widetilde{c}_1,\cdots,\widetilde{c}_n)=(c_1,\cdots,c_n)U$ combined with the definition (2) for inner product, one can also show that $\sum\left \langle \widetilde{\phi}_i \mid h \widetilde{\phi}_j \right \rangle \widetilde{c}_i^\dagger \widetilde{c}_j=\sum\left\langle \phi_i \mid h \phi_j \right \rangle c_i^\dagger c_j$.

Which **combination** of transformation for operators and definition for inner product is more reasonable? I myself prefer to the former one.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy