# How to judge constitution boson "BEC" from the dispersion of bosonic quasi-particle?

We know the spin-1/2 anti-ferromagneitc (AFM) Heisenberg model can be expressed as Schwinger boson $$\begin{array}{l}{S_{i}^{+}=b_{i \uparrow}^{\dagger} b_{i \downarrow}} \\ {S_{i}^{-}=b_{i \downarrow}^{\dagger} b_{i \uparrow}} \\ {S_{i}^{z}=b_{i \uparrow}^{\dagger} b_{i \uparrow}-b_{i \downarrow}^{\dagger} b_{i \downarrow}}\end{array}$$ with the constraint: $$1=n_{i \uparrow}+n_{i \downarrow}=b_{i \uparrow}^{\dagger} b_{i \uparrow}+b_{i \downarrow}^{\dagger} b_{i \downarrow}$$ under the mean field level, the Heisenberg Hamiltonian can be expressed as: $$H_{J}^{\mathrm{MF}} \simeq-\frac{J \Delta_{0}^{s}}{4} \sum_{i, j, \alpha} \alpha b_{i, \alpha}^{\dagger} b_{j,-\alpha}^{\dagger}+h . c .+\lambda \sum_{i}\left(\sum_{\sigma} b_{i, \sigma}^{\dagger} b_{i, \sigma}-1\right)$$ where $$\Delta_0^s$$ is the mean-field RVB (resonating valance bond) order parameter: $$\Delta_0^s=\langle\Delta_{i, j}\rangle=\langle b_{i \uparrow} b_{j \downarrow}-b_{i \downarrow} b_{j \uparrow}\rangle=\langle\sum_{\sigma} \sigma b_{i, \sigma} b_{j,-\sigma}\rangle$$ After Bogoliubov transform, we can obtain the quasiparticle $$\gamma$$ and its dispersion $$E_k$$: $$H^{MF}=\sum_{k, \sigma} E_{k} \gamma_{k, \sigma}^{\dagger} \gamma_{k, \sigma}$$ where $$E_{k}=\sqrt{\lambda^{2}-J^{2}\left|\Delta_{0}^{s}\right|^{2}\left(\cos k_{x}+\cos k_{y}\right)^{2}}$$ The magnitude of gap is $$\sqrt{\lambda^2-4J^2\Delta_0}$$ Also, the constraints $$1=n_{i \uparrow}+n_{i \downarrow}$$ as well as order parameter will give us two self-consistent equation. Assume that under $$T=T_c^{\Delta}$$, $$\Delta_0\neq0$$, which means the RVB singlet pairing has been built. Besides, at $$T=T_c^{con}$$, gap closes, and the constraints $$1=n_{i \uparrow}+n_{i \downarrow}$$ cannot be satisfied unless we add a condense term $$n_s$$. Now, if we calculate the expression of AFM order, we will find the non-zero spontaneous stagger magnetization, which means the system has long-range AFM order.

## My question

1. I can just understand the relation between the "Bose-Einstein condense(BEC)" and the gapless dispersion in mathematical level, but not in physical picture. Because $$E_k$$ describes the dispersion of quasiparticle $$\gamma$$, which has no definite particle number and $$\mu$$ is always zero. But the "BEC" happens for constitution boson $$b$$. It seems different from the picture of naive BEC for free Bosonic gas, i.e. chemical potential will increase as $$T$$ increase until $$\mu=0$$.
2. We firstly use the mean field $$\Delta$$ as the order parameter, i.e. RVB pairing. Then we study another order parameter, i.e. AFM magnetization, under the initial mean-field, I am confused that will this lead to conflicts?

In summary, my question may be generalized to :How can judge BEC of constitution boson with definite number from the dispersion of bosonic quasi-particle with non-definite number?

## Reference

• Ch.18, Auerbach, Interacting electrons and quantum magnetism
Technically, you can solve the self-consistent equations first, if there is a solution $$(\lambda, \Delta_{0}^{s})$$, then it means no Bose condensation.
If there is no solution, then introduce the Bose condensation part (a constant term $$m_0$$ in both self-consistent equations), now there are 3 unknown term $$(\lambda, \Delta_{0}^{s},m_0)$$, another equation is the spectrum should be gapless, i.e. $$\lambda^{2}=4J^{2}\Delta_{0}^{s}$$, then solve these equations. For the 2D magnets, the Bose condensation occurs at zero temperature (Merlin-Wagner theorem).
A simple case for antiferromagnetic Heisenberg model on square lattice, the Bose condensation part $$m_{0}=0.305$$ and the mean field around 0.579 at $$T=0$$.