The Dilute Bose Gas has quartic Hamiltonian

$$H=\sum_{k}\epsilon_k b_k^\dagger b_k+u\sum_{k\,k'q}b_{k+q}^\dagger b_{k'-q}^\dagger b_kb_{k'}.$$

It is said in a reference that

Since the lowest energy state is occupied by macroscopic number of particles in BEC, one can neglect the quantum fluctuation of this state and replace the operator $b_0$ with a c-number": $$b_0=\sqrt N_0$$ http://www.nii.ac.jp/qis/first-quantum/forStudents/lecture/pdf/qis385/QIS385_chap4.pdf

I have heard that the relative fluctuations in $b_0$ go as $1/\sqrt{N_0}$. This, in addition to the quotation above, would yield a solid answer to my question. However, how can I prove that these fluctuations indeed go as specified?

  • $\begingroup$ What you have heard about the relative $1/\sqrt{N_0}$ fluctuation about $b_0$ may be wrong. For coherent state where $\langle b_0\rangle=\sqrt{N_0}$ holds, there is actually no fluctuation of $\langle b_0\rangle$, because coherent state is an eigenstate of $b_0$. $\endgroup$ – Everett You Mar 26 '16 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.