I was following along with these notes, and just above equation (32) on page 3, the author makes the claim that, "for a Bose condensate, the ground state boson creation operator acquires a finite expectation value, $\langle b^\dagger \rangle = \sqrt{n} e^{i\phi}$". How can one arrive at this conclusion? Naively, it seems to me that the expectation value for any isolated creation or annihilation operator should be zero because $$ \langle n | b^\dagger | n\rangle \propto \langle n | n+1\rangle = 0$$ Thanks for the help!
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$\begingroup$ check this out en.wikipedia.org/wiki/Coherent_state $\endgroup$– Ryan ThorngrenMay 11, 2022 at 23:54
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$\begingroup$ You are right: for any vector of a Hilbert space of $N$ particles, this expectation value is zero. If you want a non-zero expectation value, you have to go to Fock spaces. $\endgroup$– Tobias FünkeMay 12, 2022 at 8:22
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Since we are considering BEC, a ground state has a macroscopic occupation. It means that we can replace the field operators $a_{0}$ and $a^{\dagger}_{0}$ into c-numbers. Hence the ground state boson creation operator acquires a finite expectation value.
