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I know that if a quantum mechanical particle of mass $m$ is subject to the potential $V(x)=\frac{1}{2}m\omega^2x^2$ then the wave functions $\phi_n(x)=\frac{1}{\sqrt{n!}}(a^t)^n\phi_0(x)$ are eigenfunctions of the Hamiltonian, with eigenvalues $E_n=\hbar \omega(\frac{1}{2}+n)$. Here $a=\frac{1}{\sqrt{2\hbar m\omega}}(m\omega x+ip)$ and $a^t=\frac{1}{\sqrt{2\hbar m\omega}}(m\omega x-ip)$

But what if we modify the potential $V(x)=\frac{1}{2}m\omega^2x^2+kx$ for some constant $k$? How do the eigenvalues change?

I also have the relation $[a,(a^t)^n]=n(a^t)^{n-1}$ and know how to express the position and momentum operators $x$ and $p$ in terms of $a,a^t$. All I can get to is this:

$H\phi_n=E_n\phi_n+kx\phi_n=E_n\phi_n + k\sqrt{\frac{\hbar}{2m\omega}}(a+a^t)\phi_n$

but don't know where to go from here since $a\phi_n=\sqrt{n}\phi_{n-1}$ and $a^t\phi_n=\sqrt{n+1}\phi_{n+1}$ don't seem to be of any use.

Azamat Bagatov
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You are barking up the wrong tree. Note that $$V(x)=\frac{1}{2}m\omega^2x^2+kx=\frac{1}{2}m\omega^2\left (x+\frac{k}{m\omega^2}\right)^2 - \frac{k^2}{2m\omega^2}= \frac{1}{2}m\omega^2y^2- \frac{k^2}{2m\omega^2}~~ ,$$ where the new coordinate is a shift, $y\equiv x + k/m\omega^2 $.

You should know (otherwise check) that the kinetic term, the Laplacian, is unchanged under a coordinate shift. So no need to modify the equations you are solving! You solve the same equations, and find the same solutions, only now in a different variable (y), and so the same energy eigenvalues, only now (down)shifted by the constant $- \frac{k^2}{2m\omega^2}$. The eigenvalues care not about the variable.

Analogously for the creation/annihilator operators. Embrace the new variable.

Cosmas Zachos
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    I assume by the kinetic term you are referring to the $-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi$ part of the Hamiltonian?. Is there a way I can show that this does not affect the eigenvalues from the definition of the Hamiltonian?

    My attempt: I have that $\hat{H}\psi(x)=E\psi(x)$, so from your coordinate change, $\hat{H}\psi(y)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(y)+V_{old}(y)\psi(y)-\frac{k^2}{2m\omega}\psi(y)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial y^2}\psi(y)+V_{old}(y)\psi(y)-\frac{k^2}{2m\omega}\psi(y)=(E-\frac{k^2}{2m\omega})\psi(y)$.

     – Azamat Bagatov May 15 '20 at 22:28
  • Yes, and yes, you got it. The respective spectra are merely shifted by the constant you write down. – Cosmas Zachos May 16 '20 at 01:14