Addition of a constant potential in the Schrodinger equation A really elementary question:
Solutions of the single particle, non-relativistic, time-dependent Schrödinger equation for a spinless particle of mass $m$ alone in space, with no forces acting on it, with the potential function $V(r,t) = 0$, include all constant frequency, wavelength, and amplitude plane wave functions $\Psi$. Because the similar Schrödinger equation with any constant potential $V(r,t) = P$ describes the same physical situation, since it is only differences in potential that have any physical significance, the Schrödinger equation with the potential $V(r,t) = P$, for any real $P \neq 0$, which also describes a situation with no forces on the particle, should also have as solutions all the ones described above for the Schrödinger equation with $V(r,t) = 0$, but it doesn't.
What is the explanation?
The following reply to Stratiev is too long for a comment, so I am including it here.
If a function $ψ′(x,t) = kψ(x,t)$, for $k$ any complex constant, then $ψ′$ satisfies a given linear partial differential equation, such as the Schrodinger equation, if and only if $ψ$ satisfies that equation. If the wavefunction $ψ′$ that you suggested would satisfy the Schrodinger equation with some non-zero potential $P$ differed by a constant phase factor $k$, which would be a complex number of norm 1, from a solution $ψ$ of the Schrodinger equation with zero $P$, then $ψ$ would also satisfy the same Schrodinger equation for the non-zero $P$ as $ψ′$ satisfied. However, your $ψ′$ for the non-zero $P$ situation differed from the ψ for the zero P by a factor of exp(iPt/(2 π h), which is not a constant phase factor, since it is a non-constant function of t.
[If you will review your differential equations theory, you will see that global
differences in phase which don't affect what linear differential equations a
function satisfies must be differences by a factor of a complex number,  not the value of a function which varies with one of the variables of the differential equation.]
 A: If you solve the time-independent Schrödinger equation first, you'd see that the solutions you get have their energy eigenvalues shifted by a constant $P$. Solving for the time dependence in the eigenstates, you'd get solutions whose frequencies $\omega'_n$ are also shifted by a constant $\omega'_n = \omega_n +\frac{P}{\hbar}$, where $\omega_n$ are the frequencies of the oscillator with $V(r,t)=0$. As is well known, global phases in the wave function are not measurable, so the differences in these phases don't matter.
However, one can think of linear combinations of Hamiltonian eigenstates $\Psi =\sum_{i} e^{i \omega_n t} | n \rangle $. Again, since every eigenvalue is shifted by the same amount, we have $$\Psi' = \sum_n e^{i (\omega_n+\frac{P}{\hbar} t)} |n\rangle=e^{i \frac{P}{\hbar} t} \sum_{i} e^{i \omega'_n t} | n \rangle= e^{i \frac{P}{\hbar} t} \Psi, $$ which is a global shift in phase, which is unobservable. Hence the solutions of the time-dependent Schrödinger equation for a potential shifted by a constant are physically the same, despite differing at first sight.
The above argument can be applied to any time-independent Hamiltonian, not just the free particle case.
A: The solutions aren't the same. In general, the time-independent Schrödinger equation reads
$$\left[\frac{-\hbar^2}{2m}\nabla^2 +V(\vec{r})\right]\psi(\vec{r})=E\psi(\vec{r})$$
I have removed the time dependance of the potential because it is not needed if it is null everywhere or constant. This equation is normally re-written in the form
$$\nabla^2\psi(\vec{r})+k^2\psi(\vec{r})=0$$
where $k^2=\frac{2m(E-V(\vec{r}))}{\hbar^2}$ so the solutions to this equation will vary depending on the value of $V(\vec{r})$. However, as explained in the above answer, adding a constant potential will just phase shift the solution.
