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This is about a matrix element of a second quantized operator.

Consider the operator $$ U=\sum_{\alpha\beta}U_{\alpha\beta}c^{+}_{\alpha}c_{\beta} $$

Something strange emerges if we calculate again the matrix element of U,

$$ <\gamma|U|\delta>=\sum_{\alpha\beta}<\Omega|c_{\gamma}U_{\alpha\beta}c^{+}_{\alpha}c_{\beta}c^{+}_{\delta}|\Omega> $$

Where $\Omega$ is the ground state with all electrons below the Fermi sea. And then use Wick's theorem, we will not get $U_{\gamma\delta}$, but $U_{\gamma\delta}(1-f_{\gamma})(1-f_{\delta})+U_{\alpha\alpha}f_{\alpha}(1-f_{\gamma})\delta_{\gamma,\delta}$. Where f is the Fermi distribution function. It is a heaviside step function at zero temperature. What is the reason?

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Can this question possibily origins from fermi sea as ground state? Since if we take vacuum as ground state, $<\gamma|U|\delta>=U_{\gamma\delta}$ –  a0087946gy Sep 21 '12 at 9:56
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Could you please add tot he question what is $f_\alpha$ and $\delta(\gamma,\delta)$? –  Luboš Motl Sep 21 '12 at 11:42

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Well, the first $U_{\gamma\delta}$ term is multiplied by $(1-f_\gamma)$ as well as by $(1-f_\delta)$ because the full matrix element has to vanish for states $\gamma,\delta$ with $f_\gamma=1$ as well as for those with $f_\delta=1$. That has a simple reason. These states are already occupied in $|\Omega\rangle$ so the whole state $$c^\dagger_\delta|\Omega\rangle$$ that appears on the right end of your matrix element vanishes for $f_\delta=1$ because you're trying to create another fermion into the state $\delta$ but one is already there. (Here I am assuming that we are using the pre-Dirac-sea convention for the creation/annihilation operators and we haven't switched them for the Dirac sea states, so all the creation/annihilation operators create/annihilate particles of the same charge, not holes.) Two fermions in the same state are prohibited by Pauli's exclusion principle so the state – and the matrix element of any operator involving this state – has to vanish. The same comment applies to the $\langle\Omega|$ factor on the bra-side of your matrix element.

If you wanted to address the matrix elements involving the occupied states, the right way to create excitations is to create holes in these states, i.e. consider $c_\delta|\Omega\rangle$ without the dagger for states with $f_\delta=1$ and similarly (with dagger, instead of without) for the bra-state and $\gamma$. Quantum field theory often naturally works with fields that combine creation as well as annihilation operators so their matrix elements usually don't vanish.

The second term starting with $U_{\alpha\alpha}$ is proportional to the identity matrix – I suppose this is what your notation $\delta(\gamma,\delta)$ means: it should be $\delta_{\gamma\delta}$ – and in physical applications, it is ultimately ignored because it has something to do with the normal ordering (of infinitely many states, in typical cases). When one talks about the total charge or total energy etc., one must be ready that the right definition may have an additive shift. And indeed, the right additive shift in your case would be one that cancels the unwanted normal-ordering constant, your second term. In the condensed matter system, the additive shift to the charge has contributions from the nuclei and there are easy ways to see what the charge of individual states should be. Almost all accessible low-energy states of matter are nearly neutral so the infinite terms in the charge have to cancel automatically or by hand.

As long as gravity is turned off, a nonzero shift of the energy wouldn't be a problem because only energy differences affect dynamics. But one still wants the vacuum to have $E=0$, especially in Lorentz-invariant theories, so an additive shift that cancels the divergent constant may be needed.

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Thanks for your reply. Your suggestion is correct, I indeed left the excition of hole. –  a0087946gy Sep 21 '12 at 13:43
    
Let redefine the notion: the creation operator $\phi_{k}=c_k$ if $k<k_F$; $\phi_{k}=c^+_k$ if $k>k_F$. And $\bar{\phi_k}$ to be the hermit conjugate of $\phi_k$. Then, $U=\sum_{\alpha\beta}U_{\alpha\beta}\phi_{\alpha}\bar{\phi_{\beta}}$, we can get $<\gamma|U|\delta>=U_{\gamma\delta}$ if we do in zero temperature and throw away the second term. That is correct. But what if the finite temperature case? For example, when $\gamma,\delta>k_F$, we can get $<\gamma|U|\delta>=U_{\gamma\delta}(1-f_{\gamma})(1-f_{\delta})$. Here $f$ is fermi distribution rather than $\theta$ function. –  a0087946gy Sep 21 '12 at 13:45

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