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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.

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The entirety of the modern quantum mechanics literature uses inner products that are linear in the second argument, and antilinear in the first one. Mathematicians often use the other convention, but I've never seen it used in physics. This is of course pure convention, but you will find grief, at least when you try to publish, if you go against the flock on this one.

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@ Emilio Pisanty Thanks for your answer. –  K-boy Nov 30 '13 at 12:55

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