3 Using T (0) = 4 πa$_{s}$/m and the one-loop result for δ$_{c}$ , we get By definition, the low-energy physics of the system with a wide Feshbach resonance at zero density is parametrized by only one parameter a$_{s}$ . Hence, for wide Feshbach resonances, we may simply take λ (0) = λ$_{∗}$ if we are only interested in the physics in the unitary regime $^{14}$, and thus we obtain in d = 3 ( ϵ = 1) where b = 4 πK$_{3}$/λ 2 ∗ = 2 / ( πλ 2 $_{∗}$). Since λ ( l ) = λ$_{∗}$ for wide Feshbach resonances, Eq. (8) can now be solved easily. Using Eq. (7) to eliminate l$_{∗}$ , we get The free energy density f is given by where sgn( x ) = 1 , − 1 for x > 0 and x < 0, respectively. where ˜ f ( l$_{∗}$ ) is the free energy density for the renormal- ized Hamiltonian at l = l$_{∗}$ . Since λ ( l$_{∗}$ ) = O ( ϵ ) and u$_{α}$ ( l$_{∗}$ ) ≈ 0, we may calculate ˜ f ( l$_{∗}$ ) in terms of the per- turbation theory in the coupling constants. The leading order results will be given by the mean-field theory of the renormalized Hamiltonian. Moreover, the prefactor in front of ˜ f ( l$_{∗}$ ) in Eq. (12) is always positive, the phases of this system can be determined directly from ˜ f ( l$_{∗}$ ). To proceed, we make a change of variables: the action S [Eq. (1)] can be written as The point that the molecule-molecule repulsion is cru- cial to stabilize this system was also emphasized by the previous mean-field stability analysis $^{6}$. In the µ, δ < 0 re- gion, the mean-field theory predicts that the ASF state is stable for u$_{2}$ > 0 $^{6}$. Our RG analysis, however, indi- cates that in this region this system is thermodynam- ically unstable as | µ | smaller than some critical value. We notice that this conclusion is consistent with an ear- lier RPA calculation $^{15}$. Within our approach, the reason behind this instability is that the molecule-molecule in- teraction becomes attractive at low energy even when the bare value of u$_{2}$ is positive. Equation (16) suggests that the FR Bose gas is stable either at moderate density or with strong dimer-dimer repulsions. In other words, u$_{2}$ must be strong enough to guarantee the global stability of this system in the thermodynamical limit. For given u$_{2}$ > 0, the constraint u$_{2}$ > U$_{c}$ can also be expressed as a lower bound for µ , i.e. in the dilute limit. The stability conditions require that u$_{1}$ > 0 and u$_{2}$ > U$_{c}$ where In Eq. (14), we have neglected the r$_{1}$ and r$_{2}$ dependence, and thus they are valid only in the dilute limit and in the unitary regime. Moreover, we do not consider the loop corrections due to the u$_{1}$$_{,}$$_{2}$ terms because they are irrelevant operators. For the wide Feshbach resonances, we may simply set λ = λ$_{∗}$ , and thus we get The one-loop RG equations for s$_{1}$$_{,}$$_{2}$ are given by where s ∗ α = s$_{α}$ ( l$_{∗}$ ). In order that the renormalized Hamil- tonian is bounded from below, we must require that s ∗ 1 , 2 > 0. A. Stability conditions III. THE PHASE DIAGRAM AT ZERO TEMPERATURE m 4 πa$_{s}$ = − mν 4 π = − δ 2 g 2 0 . δ = ν bma$_{0}$ , (10) r$_{2}$ ( l$_{∗}$ ) = 2sgn( µ ) − (2 ma 2 $_{0}$) ϵ/ $^{2}$δ | µ | 1 − d/ 2 , (11) f = e − ( d +2) l$_{∗}$ 2 ma d +2 0 ˜ f ( l$_{∗}$ ) , (12) x = a$_{0}$ ˜ x , τ = 2 ma 2 $_{0}$˜ τ , Ψ$_{a,m}$ = a − d/ 2 0 ˜ Ψ$_{a,m}$ , S = ∫ ˜ β 0 d ˜ τ ∫ d $^{d}$˜ x ˜ Ψ † $_{a}$( ∂$_{˜}$ τ − ∇ 2 − r$_{1}$ ) ˜ Ψ$_{a}$ + ∫ ˜ β 0 d ˜ τ ∫ d $^{d}$˜ x ˜ Ψ † m ( ∂$_{˜}$ τ − ∇ 2 2 − r$_{2}$ ) ˜ Ψ$_{m}$ + ∫ ˜ β 0 d ˜ τ ∫ d $^{d}$˜ x U , 2 m | µ | a 2 0 > [ 16 ϵ $^{2}$a d − 2 0 ( d − 2) K$_{d}$mu$_{2}$ ] 2 / ( d − 2) . (16) U$_{c}$ = 16 ϵ 2 ( d − 2) K$_{d}$m (2 m | µ | ) 1 − d/ 2 . s ∗ 1 = K$_{d}$mu$_{1}$ 16 a d − 2 0 e (2 − d ) l$_{∗}$ , s ∗ 2 ≈ K$_{d}$mu$_{2}$ 16 a d − 2 0 e (2 − d ) l$_{∗}$ − ϵ 2 d − 2 , (15) ds$_{1}$ dl = (2 − d ) s$_{1}$ , (13) ds$_{2}$ dl = (2 − d ) s$_{2}$ − λ 4 . (14) 16 s ∗ 1 K$_{d}$ | ˜ Ψ$_{a}$ | 4 + 16 s ∗ 2 K$_{d}$ | ˜ Ψ$_{m}$ | 4 , At large values of | ˜ Ψ$_{a,m}$ | , the renormalized potential U at l = l$_{∗}$ is of the form where ˜ β = β/ (2 ma 2 $_{0}$), s$_{1}$$_{,}$$_{2}$ (0) = K d 16 mu$_{1}$$_{,}$$_{2}$a 2 − d 0 , and U = √ 2 K$_{d}$ λ ( ˜ Ψ † $_{m}$˜ Ψ 2 $_{a}$+H .c. ) + 16 s$_{1}$ K$_{d}$ | ˜ Ψ$_{a}$ | $^{4}$+ 16 s$_{2}$ K$_{d}$ | ˜ Ψ$_{m}$ | 4 .