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In This paper (Kobayashi et al -- Lyman-alpha Constraints on Ultralight Scalar Dark Matter: Implications for the Early and Late Universe) it says, at the beginning of Section 3.1:

A light scalar field stays frozen at its initial field value in the early Universe. Hence, any initial displacement from the potential minimum gives rise to a scalar dark matter density in the later universe.

I don't understand this statement. Can someone explain its meaning? Why would such a configuration give rise to matter later in the universe? Is it due to the fact that later in the universe the scalar field would oscillate and oscillations can be seen as particles?

Sorry if the question is not clear, I studied physics quite a long time ago and study these things in my free time so there are many gaps in my understanding of fundamental physics and Cosmology. Feel free to be as technical as you wish but please remember I'm not expert or anything

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If there is a non-zero particle field in the early universe, this means there are particles present. These particles are driven away so fast from each other (by inflation) that they can't return to the zero-field configuration anymore. So they are "frozen".

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  • $\begingroup$ Thanks, to the oscillation of the field doesn't matter? It's simply the fact that there is a zero particle filed, therefore there are particles. Did I understand it correctly? $\endgroup$ Jun 16 at 9:48
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The answer seems to be much less deep than what OP expects. The field $\phi$ is to be thought of as a purely classical field, its magnitude literally representing the density of matter. So if its value is large, it means there is a lot of matter, and if low, very little.

As the field is locked to its initial value, if this is non-zero, it will represent a non-zero density at all times. Hence the claim: if there is a displacement it leads to matter density at later times.

Note: $\phi$ is not quite the matter density. The density is actually $\rho\propto\phi^2$ but this is irrelevant for this discussion. If $\phi$ is non-zero so is $\rho$, and viceversa.

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  • $\begingroup$ Thank you, I'll award it in a couple of days. it's not about deepness of the answer, it's mostly due to the fact that I am late of understanding. Your answer seems to have clarified my trivial doubts: especially the fact that there is non-zero density at all times really helped, I somehow understood that the we only had non-zero density only later. $\endgroup$ Jul 16 at 9:15
  • $\begingroup$ A minor addendum: the density is not actually $\phi^2$, that would imply the density is constantly oscillating since $\phi \propto \cos(m t)$. Instead the density is roughly $\phi^2 + \dot{\phi}^2$ which is constant since $\cos^2(mt) + \sin^2(mt) = 1$. $\endgroup$
    – knzhou
    Jul 17 at 20:25
  • $\begingroup$ I think this answer also oversimplifies the meaning of $\phi$. Yes, mechanically you can treat it as a classical field in some calculations, but, like everything else, it actually is a quantum field which is occupied by a macroscopic number of particles. In some cases, thinking of the quantum field as a classical field is the right way to go. In other situations, you may want to think of the very same state as a collection of particles. It just depends on what you are coupling to the field. $\endgroup$
    – knzhou
    Jul 17 at 20:27
  • $\begingroup$ @knzhou I agree that the picture is grossly oversimplified but, to be fair, that is a criticism of the original paper, not really my post. It is them who are claiming that $\phi$ should be thought of as a purely classical field (or at least in that specific section, I haven't checked if the discussion is refined in later sections). Yes, $\phi$ (like everything else) should be treated as a quantum object but no, they specifically are not doing that in the quote in the OP. $\endgroup$ Jul 18 at 9:37

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