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If the uncertainty in the age of the Universe is $\Delta t$ then the Uncertainty Principle implies that it has an uncertainty in its energy $\Delta E$ given by

$$\Delta E \ \Delta t \sim h.\tag{1}$$

If this energy fluctuation excites the zero-point electromagnetic field of the vacuum then a photon is created with energy $\Delta E$ and wavelength $\lambda$ given by

$$\Delta E \sim h \frac{c}{\lambda}.\tag{2}$$

Combining Equations $(1)$ and $(2)$ we find that

$$\lambda \sim c\ \Delta t.\tag{3}$$

Now as this characteristic length $\lambda$ is the wavelength of a photon it is a proper length that expands with the Universal scale factor $a(t)$ so that

$$\lambda \sim a(t).\tag{4}$$

Combining Equations $(3)$ and $(4)$, and taking $\Delta t \sim t$, we arrive at a unique linear cosmology with the normalized scale factor $a$ given by

$$a(t) = \frac{t}{t_0}.$$

where $t_0$ is the current age of the Universe.

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The Heisenber uncertainty principle, in the strict mathematical quantum mechanical models is due to the commutation relations of the operators of the specific theory.

When one is dealing with the cosmos, which by necessity has gravity, there is no definitive quantum mechanical theory that includes gravity , thus it is not known what the specific commutation relations will be for the inflation time that could be extrapolated into a Heisenberg uncertainty. Only approximations.

A linear cosmology can also be seen here, there are links outlining the hypotheses assumed. All these are just toy models, imo.

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  • $\begingroup$ I think the graph you link to assumes $a \sim t^{1/2}$ which is a universe comprising of photons. $\endgroup$
    – John Eastmond
    Commented Jun 14, 2018 at 20:29
  • $\begingroup$ I am trying to say that it all depends on assumptionsto come up with some linearity , and taking the HUP is another assumption because we do not know whether at the beginning of time there is a HUP. $\endgroup$
    – anna v
    Commented Jun 15, 2018 at 4:09
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Equation 1 is wrong; the left-hand side should read $\Delta E\Delta t$, i.e. a product of standard deviations.

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  • $\begingroup$ Ok I've tried to be more careful by using $\Delta E$ and $\Delta t$. $\endgroup$
    – John Eastmond
    Commented Jun 14, 2018 at 18:35
  • $\begingroup$ @JohnEastmond Your edit to (2) is wrong; that should have been $E$ all along. There's no way to use equations (1), (2) together in anything like the way you desire. $\endgroup$
    – J.G.
    Commented Jun 14, 2018 at 18:37
  • $\begingroup$ Can't a photon be excited with the energy fluctuation $\Delta E$? $\endgroup$
    – John Eastmond
    Commented Jun 14, 2018 at 19:01
  • $\begingroup$ @JohnEastmond Why would it be? $\endgroup$
    – J.G.
    Commented Jun 14, 2018 at 19:37
  • $\begingroup$ Because there's some energy $\Delta E$ that's available to excite the EM field due to the uncertainty principle. In quantum mechanics if something is permitted it will happen. $\endgroup$
    – John Eastmond
    Commented Jun 14, 2018 at 20:37

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