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A naive question about the Second Quantization?

+ 3 like - 0 dislike
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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.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
asked Nov 30, 2013 in Theoretical Physics by Kai Li (975 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Emilio Pisanty
answered Nov 30, 2013 by Emilio Pisanty (330 points) [ no revision ]
@ Emilio Pisanty Thanks for your answer.

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

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