0
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ check this out en.wikipedia.org/wiki/Coherent_state $\endgroup$ May 11, 2022 at 23:54
  • $\begingroup$ 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. $\endgroup$ May 12, 2022 at 8:22

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.