Dear captainjack2000, it's really a good problem, and I wish I could do some help.
Firstly, I'm to re-express the problem. Generally, Hamiltonian operator is connected with the total energy,
$$
\hat{H}=\frac{\hat{p}^2}{2m}+V=-\frac{\hbar^2}{2m}\nabla^2+V,
$$
hence,
$$
\hat{H}'=\frac{\hat{p}^2}{2m}+(V+C)=-\frac{\hbar^2}{2m}\nabla^2+(V+C).
$$
So, the problem is indeed equal to:\\
\textsf{What are the effects on the eigenvalue spectrum of
Hamiltonian, due to the selection of zero reference level for
potential energy?}\\
Now, let's begin the analyses via Schr$\ddot{o}$dinger Equation (SE):
$$
i\hbar\frac{\partial}{\partial t}\Psi=\hat{H}\Psi, (Eq1)
$$
$$
i\hbar\frac{\partial}{\partial t}\Psi'=\hat{H}'\Psi'=(\hat{H}+C)\Psi', (Eq2)
$$
where $C$ is the difference resulting from the change of reference level for potential energy. Next let's formulate how $\Psi'$ differs from $\Psi$, and we introduce a phase shift factor as an attempt (this is in fact a common method, for example, we turn to this method again when we verify the Galilean invariance of wave equation, or explore the new formalism of wave equation from non-inertial reference under uniform acceleration):
$$
\Psi'=e^{\frac{-i\alpha t}{\hbar}}\Psi. (Eq3)
$$
Eq2 \& Eq3 lead to
$$
i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-i\alpha t}{\hbar}}(\hat{H}+C)\Psi.(Eq4)
$$
On the other hand, time derivative of Eq3 gives rise to
$$
i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-i\alpha t}{\hbar}}(i\hbar\frac{\partial}{\partial t}+\alpha)\Psi. (Eq5)
$$
Comparison of Eq4 and Eq5 results in
$$
(\hat{H}+C)\Psi=(i\hbar\frac{\partial}{\partial t}+\alpha)\Psi. (Eq6)
$$
Combination of Eq6 \& Eq1 immediately lead to
$$
\alpha=C. (Eq7)
$$
Hence, Eq3 and Eq4 are rewritten as
$$
\Psi'=e^{\frac{-iCt}{\hbar}}\Psi, (Eq8)
$$
$$
i\hbar\frac{\partial}{\partial t}\Psi'=e^{\frac{-iCt}{\hbar}}(\hat{H}+C)\Psi, (Eq9)
$$
respectively. Suppose $\Psi$ and $\Psi'$ are for stationary states, then, via separation of variables, we have
$$
\Psi(t,r)=e^{-iEt/\hbar}\phi(r) (Eq10)
$$
$$
\Psi'(t,r)=e^{-iE't/\hbar}\phi'(r) (Eq11)
$$
where $E$ and $E'$ are real constants, and separately represent the energy of the states $\Psi$ and $\Psi'$. From Eq9, Eq10 and Eq11, we
have
$$
i\hbar\frac{\partial}{\partial t}\Psi' =e^{\frac{-iCt}{\hbar}}(\hat{H}+C)\Psi =e^{\frac{-iCt}{\hbar}}(E+C)\Psi =(E+C)\Psi' (Eq12)
$$
where we used
$$
\hat{H}\phi(r)=E\phi(r),\sim \hat{H}\phi(r)e^{-iEt/\hbar}=E\phi(r)e^{-iEt/\hbar},\sim \hat{H}\Psi=E\Psi.
$$
Hence, to sum up, we attained that
$$
\Psi'(t,r)=e^{-iEt/\hbar}\Psi(t,r)
$$
$$
E'=E+C
$$
Now, we're ready to get down to the eigenvalue spectrum of the newHamiltonian $\hat{H}'$:
$$
\Psi(r,t)=\sum_n C_n\phi_n(r)e^{-iE'_nt/\hbar}.
$$
After separation of variables, for $\hat{H}'$, $\hat{H}'=\hat{H}+C$:
$$
\left(-\frac{\hbar^2}{2m}\nabla^2+U(r)+C\right)\phi(r)=E\phi(r).
$$
For simplicity, we take one-dimensional case as an example:
$$
\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(r)+C\right)\phi(x)=E\phi(x).
$$
Hence
$$
-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\phi+[(E-C)+U(r)]\phi=0.
$$
Hence
$$
\frac{d^2}{dx^2}\phi+\frac{2m}{\hbar^2}\left[(E-C)+U(r)\right]\phi=0. (Eq13)
$$
This is the usual 2-order ODE, from which the spectra of Hamiltonian will be achieved. And before, with the ODE:
$$
\frac{d^2}{dx^2}\phi+\frac{2m}{\hbar^2}\left[E+U(r)\right]\phi=0, (Eq14)
$$
various special cases (i.e. $U(r)$ is determined in formalism), such as harmonic oscillators, infinite square wells, and we get the corresponding Hamiltonian spectra for each case, which depend on
$U(r)$:
$$
\text{Spectra Set}: {E_n:\quad E=E(n), n=0,1,2\ldots} (Eq15)
$$
Comparing Eq13 with Eq14, we just need to replace $E$ with $E-C$ in Eq15, hence,
$$
\text{Spectra Set}: {E_n: E-C=E(n), n=0,1,2\ldots},
$$
hence,
$$
\text{Spectra Set}: {E_n: E=E(n)+C, n=0,1,2\ldots}.
$$
Hence, we draw the conclusion that:
$$
\hat{H}'=H+C, \sim E'_n=E_n+C
$$
This result holds for arbitrary form of $U(r)$, i.e. for arbitrary potential.
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