# A naive question about the Second Quantization?

+ 3 like - 0 dislike
393 views

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

+ 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 (520 points)
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.