Eigenfunctions of an Harmonic Oscillator perturbed with an Electric Field Knowing that a particle of mass $m$ and electric charge $q$ is under an uni dimensional harmonic potential of frequency $\omega$ perturbed with and electric field $\vec{E}= E_f \hat{x} $, $\hspace{0.2cm}$ I would like to find the eigenfunctions and e
eigenvalues of $\hat{H} = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 - q E_f x $.
I know that I can find the eigenvalues by noticing:
\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + \frac{1}{2} m \omega^2 \bigg( x^2 - \frac{2 q E_f}{m\omega²}x + \frac{q^2 E_f^2}{m^2 \omega^4}x - \frac{q^2 E_f^2}{m^2 \omega^4} \bigg) = E \psi \Leftrightarrow
\end{equation}
\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + \frac{1}{2} m \omega^2 \bigg( x  - \frac{q E_f}{m \omega^2} \bigg)^2 = \bigg(E  +  \frac{q^2 E_f^2}{2 m \omega^2}\bigg)\psi 
\end{equation}
So I use the change of variables: $x' = x - \frac{q E_f}{m \omega^2}$.
Using this, I get that the new eigenvalues are of $\hat{H}$ are: $E = \hbar \omega \bigg( n+ \frac{1}{2} \bigg) - \frac{q^2 E_f^2}{m \omega^2}$.
My question is: If I use the change of variables I found in the expression of the eigenfunctions of the harmonic oscillator, do I get the orthonormal eigenspace for this specific potential and how I could prove it?
Furthermore I would like to have a physical explanation to understand why this trick works?
Thanks in advance
 A: 
My question is: If I use the change of variables I found in the expression of the eigenfunctions of the harmonic oscillator, do I get the orthonormal eigenspace for this specific potential and how I could prove it?

Yes. First, we can see it directly: if $\psi_n(x)$ and $\psi_m(x)$ are eigenfunction of the original harmonic oscillator (before the shift in variables) then we know they are orthogonal i.e. $$\int_{-\infty}^{\infty}dx \psi^*_m(x)\psi_n(x) = \delta_{m,n}$$ so a constant shift $\tilde{\psi}_n(x)=\psi_n(x-x_0)$ will leave this relationship intact
$$\int_{-\infty}^{\infty}dx \tilde{\psi}^*_m(x)\tilde{\psi}_n(x) = \int_{-\infty}^{\infty}dx \psi^*_m(x-x_0)\psi_n(x-x_0) = \int_{-\infty}^{\infty}dx \psi^*_m(x)\psi_n(x) = \delta_{m,n}$$
because it is just a shift in the variable of integration, equally amont both function.
Another way to see it is a bit more formally, without resorting to the representation of the eigenfunction in position space. The shift is done by a unitary transformation $|\tilde{n} \rangle = T_{x_0} |n\rangle$ where $T_{x_0}=\exp(-ipx_0/\hbar)$ is the generator of translations. Since this is unitary $T_{x_0}T^{\dagger}_{x_0} = 1 = T^{\dagger}_{x_0}T_{x_0}$ then it doesn't change inner product between states $\langle \tilde{m} | \tilde{n} \rangle = \langle m | T^{\dagger}_{x_0}T_{x_0}|n\rangle = \langle m | n \rangle$. This is useful to remember as it is applicable to all such unitary transformations, regardless of the specific problem at hand.

Furthermore I would like to have a physical explanation to understand why this trick works?

The reason is similar to the classic case of a spring under gravity, for example. The equilibrium point of the the harmonic oscillator was shifted due to the force. So now, instead of having $x=0$ as the equilibrium, you have $x=x_0 = qE_f/m\omega^2$ as the equilibrium, but the particle will oscillate about this new equilibrium just as it has oscaillated about $x=0$ before hand. As long as we don't add terms with higher orders of $x$ than $x^2$, the harmonic nature of the potential is preserved, just around a new point (or, if we add a term $\propto x^2$, a new frequency). So this shift in variables just translates the problem to its true equilibrium.
