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Based on Wick's theorem, the time-ordered product of operators can be written as a sum of normal-ordered product and products involving all types of contractions. Upon taking the ground state expectation value, people claim that the normal-ordered products will have zero expectation. I have no doubt regarding this if we are considering the bosonic system. However, when it comes to fermionic systems, the ground state is the filled fermi sea, and if in this case the normal-ordered product is something like $c^{\dagger}_kc_k$ with $k<k_F$, then its ground state expectation will not vanish.

If this is the case, then what would be its physical consequences?

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    $\begingroup$ The normal-ordered operator is defined as the difference between the operator and the ground state expectation value of that operator, e.g. $:\hat{O}:=\hat{O}-\langle \hat{O} \rangle$ $\endgroup$ – Chuan Chen Jul 24 '17 at 7:03
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/133426/2451 $\endgroup$ – Qmechanic Jul 24 '17 at 7:35
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    $\begingroup$ It is also worth noting that there exists a version of Wick's theorem for thermal states of free fermion systems. $\endgroup$ – Blazej Jul 24 '17 at 11:37
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The normal-ordered operator is defined as the difference between the operator and the ground state expectation value of that operator, e.g. $$[\hat{O}] \equiv \hat{O}-\langle \hat{O} \rangle$$ so the ground state expectation value of the normal-ordered operator should be zero. In your case of free fermi sea, if $k<k_F$, $$\begin{align} [c^{\dagger}_k c_k] & \equiv c^{\dagger}_k c_k-\langle c^{\dagger}_k c_k \rangle_0 \\ & = c^{\dagger}_k c_k -1 \end{align}$$

so one would get $\langle [c^{\dagger}_k c_k] \rangle_0=0$.

Maybe it's better if you can give a more concrete example of your question.

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