The unit of the cosmological constant In natural units, it’s known that the unit of the cosmological constant is $\text{eV}^2$. I don‘t get why the paper arXiv: 2201.09016, p. 1, it is said the value of $ \Lambda \sim \text{meV}^4$, this means $\Lambda \sim (10^6 ~ \text{eV})^4 \sim 10^{24} \text{eV}^4 $, should not the unit be $\text{eV}^2$ instead?
 A: 
Dr. phy asked: "What is the unit of the cosmological constant?"

In SI units it is $\rm [1/m^2]$, since in the Friedmann equation $\rm (\dot{a}/a)^2+...=\Lambda c^2/3$ with dimensionless $\rm a=r/r_0=[m/m]=[1]$, the square of the Hubble parameter $\rm H^2=(\dot{a}/a)^2=(da/dt/a)^2$ has units of $\rm [1/s^2]$, which is the same as $\rm \Lambda c^2$ when $\rm \Lambda$ is in $\rm [1/m^2]$.
In the covariant field equation $\Lambda g_{\mu \nu}+...=\rm T_{\mu \nu} 8 \pi G/c^4$ with $g_{\rm t t}$ in $\rm c^2=[m^2/s^2]$ and ${\rm T_{t t}}/g_{\rm t t}$ and ${\rm T^{t t}}/g^{\rm t t}$ in units of $\rm [J/m^3]$ the $\Lambda$ also has to be in $\rm [1/m^2]$ in order for the units to match.
The energy stress momentum tensor component for the dark energy density is $\rm T^t_t = c^4 \Lambda/(4 \pi G)$ so for that to have the units $\rm [J/m^3]=[kg/m/s^2]$ you need $\Lambda$ in $\rm [1/m^2$].
In our ΛCDM universe the value is around $\rm \Lambda= 10^{-52}/m^2$. In a de Sitter universe containing no matter and radiation so that the $+...$ term vanishes, the relation to the Hubble parameter is simply $\rm H=c\sqrt{\Lambda/3}$, and $\rm H$ has the units $\rm [m/m/s]=[1/s]$ (the frequency of one e-fold).
As you can see in the other answer above, sometimes it is given already multiplied with $\rm c^2$, then the numeric value is around $\rm 10^{-36}/s^2$. That is the case in sources where it simply says $\rm H^2=\Lambda/3$, but in that case they should mention that they set $\rm c=1$ and then you also should not see any other $\rm c$s in the equation.
So if you read different numbers in different references and they don't mention which constants they omitted, just take the $\rm \Lambda=10^{-52}/m^2$ and multiply the number in your source with the constants in question until the number matches the units, then you know for sure by what's the conversion factor between your different references.
A: with Friedman equation
$$\left(\frac{\dot a}{a}\right)^2+\ldots =\frac{1}{3}\Lambda\,c^2$$
with $~c=[1]~$ you obtain that the unit of $~\Lambda~$ is $\left[\frac{1}{s^2}\right]=[eV]^2$

with
$$E=m\,c^2~,c=1\quad\text{the unit of mass and energy is $~[eV]$}$$
with
$$E^2-p^2\,c^2=m^2\,c^4,c=1\quad\text{the unit of p is $~[eV]$}$$
with
$$\Delta p\,\Delta  L\ge h,h=1\quad \text{the unit of L is $~[eV]^{-1}$}$$
with
$$c=\frac{[L]}{[t]},c=1\quad\text{the unit of t is $~[eV]^{-1}$}$$
