Dear all of you, I decided to calculate the minima of the potential by myself, to give an independent check. It seems that I do not get eq. (24a) (24b) of the notes, nor what Steve found in his notes. The difference may be due to different initial assumptions, or, of course, my mistake. In any case, I am going to write what I get, and promise to think more about this.

I start with the definitions

ρ_{ϕ} = λ|ϕ|/H  (1)
ρ_{n} = λ|N|/H  (2)
ã = a/H  (3)
m̃_{ϕ,n} = m_{ϕ,n}/H  (4)
Ṽ = \frac{|λ|^{2}}{H^{4}} V  (5)

and the potential

Ṽ = (m̃_{ϕ}^{2} − c_{ϕ})ρ_{ϕ}^{2} + (m̃_{n}^{2} − c_{n})ρ_{n}^{2} − 2|ã + c_{H}|ρ_{ϕ}^{2}ρ_{n} + \frac{1}{4}ρ_{ϕ}^{4} + ρ_{ϕ}^{2}ρ_{n}^{2}  (6)

With these, I get that the minima are given by the same conditions that Steve and Michael found (I use Steve’s notation):

ρ_{n} = \frac{c_{H}ρ_{ϕ}^{2}}{ρ_{ϕ}^{2} − c_{n}}  (7)
ρ_{ϕ}^{2} = 2c_{ϕ} + 4c_{H}ρ_{n} − 2ρ_{n}^{2} ,  (8)
2ρ_{n}^{3} − 6c_{H}ρ_{n}^{2} + (4c_{H}^{2} + c_{n} − 2c_{ϕ})ρ_{n} + 2c_{H}c_{ϕ} = 0  (9)

Now, I assume that |c_{ϕ}| ≪ 1. Under this assumption, eq. (9) has one solution with ρ_{n} ∝ c_{ϕ}. So, consider ρ_{n} to be small too. If I keep only the lowest order, non-zero terms in c_{ϕ}, I get the following solution:

ρ_{n} = −\frac{2c_{ϕ}c_{H}}{4c_{H}^{2} + c_{n}}  (10)

ρ_{ϕ} = \sqrt{\frac{2c_{ϕ}c_{n}}{4c_{H}^{2} + c_{n}}} .  (11)

A check with the second derivatives gives that this point is a minimum if we have c_{ϕ} > 0, c_{H} > 0 and c_{n} < −4c_{H}^{2}. Of course, everything I wrote needs to be reconsidered and checked.

Assuming that this solution is right, we may wonder how natural it is to have small c_{ϕ}. Also, we should keep in mind that, since eq. (9) is cubic, there are other two solutions of course. These solutions may not be real, as Michael wrote. I will think.

Let me say that I studied this scenario, instead than follow Steven and Michael’s work, because I have a difficulty to understand the c_{H} = 1/2 solution that Steven suggested. My problem with it is that if I insert c_{H} = 1/2 in eq. (9), I do not see any particular simplification.... So I am confused. Probably there is something that escapes me at this moment.

Another observation, which may not have practical interest, though, is that if in eq.

(9) both the c_{ϕ} and the c_{N} terms are negligible, but the c_{H} term is not, the equation has a particularly simple set of solutions. These solutions give the following extremes of the potential:

ρ_{n} = 0 , ρ_{ϕ} = 0 ;  (12)
ρ_{n} = c_{H} , ρ_{ϕ} = \sqrt{2}c_{H} ;  (13)
ρ_{n} = 2c_{H} , ρ_{ϕ} = 0 .  (14)

Clearly, this case is very particular because it requires a strong suppression of the Hubble mass terms. Howerver it may be kept in mind if we are interested to know what are the minimal ingredients that we need to have the ADDL mechanism. Also, this particular case may be of some interest in the situation when H ∼ m_{ϕ} ∼ m_{ν}, so that there is an instant of time when the “bare” and the Hubble mass terms cancel.