Why doesn't the Schrödinger equation contain a term of $mc^2$? Admitting the ansatz
$$ψ=e^{i(kx-ω t)} \tag{1}$$
then
$$k^2=-ψ^{-1} \frac {∂^2ψ}{∂x^2} \tag{2}$$
and
$$ω=iψ^{-1} \frac {∂ψ}{∂t} \tag{3}$$
If one admits that the total energy ($E$) is related to momentum ($p$) as $E=\frac{p^2}{2m}+U$, admiting also the De Broglie relations $E=ħω$; $p=ħk$ it follows that
$$\frac {-ħ^2}{2m} \frac {∂^2ψ}{∂x^2}+Uψ= iħ \frac {∂ψ}{∂t} \tag{4}$$
This is Schrödinger's equation. This equation is said to be non relativistic because of its use of $E= \frac{p^2}{2m}+U$ (rigorously speaking though, it is non relativistic because it is not Lorentz invariant).
However, starting from the relativistic total energy equation
$$E=\frac{1}{ \sqrt{1- \frac{v^2}{c^2}}}mc^2=T+mc^2 \tag{5}$$
Where, $T$ is the kinetic energy and $mc^2$ the particle’s self energy. Now, using the expansion of $\frac{1}{ \sqrt{1- \frac{v^2}{c^2}}}mc^2$ 
$$E=mc^2 + \frac{mv^2}{2} + \frac{3mv^4}{8c^2} + \frac{5mv^6}{16c^4}+...$$
and ignoring members dividing by $c$ (because we are considering $v\ll c$). It becomes
$$E=\frac{1}{2} mv^2+mc^2=T+mc^2 = \frac{p^2}{2m}+mc^2 \tag{6}$$
or
$$\frac{p^2}{2m}+mc^2=ħω=\frac{ħ^2k^2}{2m}+mc^2 \tag{7}$$
So, $mc^2$ do not vanishes even under classical approximation.
Admitting that Planck's and De Broglie's equations holds in every situation and that $E$ in Planck equation is the total energy, substituting equation (2) and (3) into (7) the Schrödinger equation “would” have the form
$$\frac {-ħ^2}{2m} \frac {∂^2ψ}{∂x^2}+mc^2 ψ = iħ \frac {∂ψ}{∂t} \tag{9}$$
Now we could postulate this equation, making the steps of getting it less fundamental then the end result.
I tried to consider that $T \ll mc^2$ in Schrödinger equation, but I realize that an electron in hydrogen atom moving with half the speed of light (using classical equations as we are analyzing Schrödinger’s equation) it would have less than $\rm 100keV$ ($\rm ≈64keV$ if my math is not wrong) of kinetic energy, but $\rm 511keV$ of self energy.
So, my question is: why Schrödinger equation do not have an $mc^2$ term, if $ħω$ is supposed to be the total energy and not just the kinetic energy.
 A: In nonrelativistic quantum mechanics, particles cannot be created or destroyed, and each particle has constant mass $m$. That means the extra $E = mc^2$ energy is just a constant, so it can be subtracted out by adding a constant to the Hamiltonian; only energy differences matter.
The $E = mc^2$ can play a role in quantum field theory, since particles can be created or destroyed there; for example, it is released in pair annihilation, giving the products extra energy.
A: Just because to introduce the constant added term $mc^2I$ to the Hamiltonian operator  would be equivalent to redefine $\psi \to \psi'= e^{imc^2 t/\hbar}\psi$. This sort of phases do not matter  in QM. You cannot see them by measuring any observable. Pure states are actually operators of the form $|\psi \rangle \langle \psi|$ and you see that these phases cancel each other.
Instead, if the mass were replaced by a mass operator with discrete spectrum the picture would change. In the classical limit the rapid temporal oscillations of the phases (I am assuming that the mass is big if compared with the typical energies of the system), would destroy the coherence of superpositions of different masses giving dynamically rise to superselection rule of the mass Bargmann's superselection rule (see here or here).
A: Let's say we start with this equation of yours :

$$-\frac {\hbar^2}{2m} \frac {\partial^2\psi}{\partial x^2}+V\psi +mc^2\psi=i\hbar \frac {\partial \psi}{\partial t}$$

Now a simple transform reduces it to the original form :
$$\psi = e^{Wt}\phi$$
$$-\frac {\hbar^2}{2m} e^{Wt} \frac {\partial^2\phi}{\partial x^2}+Ve^{Wt}\phi+mc^2e^{Wt}\phi=i\hbar e^{Wt} \frac {\partial \phi}{\partial t} + i\hbar  W e^{Wt}\phi$$
Which reduces to :
$$-\frac {\hbar^2}{2m} \frac {\partial^2\phi}{\partial x^2}+V\phi+mc^2\phi=i\hbar  \frac {\partial \phi}{\partial t} + i\hbar W \phi $$
And it is easy to see that $i\hbar W := mc^2$ recovers the original form of the equation.
So as we did in adding the rest mass term was add a rather pointless phase term that does nothing for us :
$$\psi = \phi \,\, exp\left( i\frac{mc^2}{\hbar} t\right)$$
This is the phase change that has no net effect that is discussed in the answers by Valter Moretti and Ruslan
A: Consider a Schrödinger equation:
$$-\frac {\hbar^2}{2m} \frac {\partial^2\psi}{\partial x^2}+V(x,t)\psi=i\hbar \frac {\partial \psi}{\partial t}.$$
Let some wave function $\psi(x,t)$ be its solution. Let's now replace $V(x,t)\to V(x,t)+\hbar W$, where $W=\mathrm{const}$. Corresponding solution of the new equation will change: $\psi(x,t)\to\psi(x,t)\exp(-iWt).$
Consider now an observable $K$, with corresponding operator $\hat K$. Its expected value, calculated for the solution of the original equation, will be
$$\overline K(t)=\int\limits_{-\infty}^{\infty}dx\,\psi(x,t)^*\hat K\psi(x,t).$$
Now let's replace $\psi$ in the above integral with the solution of the modified equation where we shifted the potential energy:
$$\int\limits_{-\infty}^{\infty}dx\,(\psi(x,t)\exp(-iWt))^*\hat K(t)(\psi(x,t)\exp(-iWt))=\\
=\int\limits_{-\infty}^{\infty}dx\,\psi(x,t)^*\exp(iWt)\exp(-iWt)\hat K(t)\psi(x,t)=\\
=\int\limits_{-\infty}^{\infty}dx\,\psi(x,t)^*\hat K(t)\psi(x,t)=\overline K(t).$$
You can see that regardless of the global phase, the observable $K$ appears the same. Similarly can check that matrix elements of an operator will also be independent of the global phase of the basis functions you calculate the matrix elements in.
Any physically-relevant calculation ultimately is about observables, not about particular values of abstract functions like a wavefunction. Thus you shouldn't worry too much about gaining an extra phase factor when adding or removing a constant term to the potential.
