Eigenvalues of disturbed Hamiltonian

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1. Sep 25, 2015

AwesomeTrains

Hello everyone!
I'm trying to follow a solution to a problem from the book "Problems and Solutions on Quantum Mechanics", it's problem 1017. There's a step where they go on too fast, and I can't follow. I've posted the solution and where my problem is down below.
1. The problem statement, all variables and given/known data
The dynamics of a particle moving one-dimensionally in a potential $V(x)$ is governed by the Hamiltonian $H_0=p^2/2m+V(x)$, where $p=-i\hbar d/dx$ is the momentum operator. Let $E_n^{(0)}, n=1,2,3,...,$ be the eigenvalues of $H_0$. Now consider a new Hamiltonian $H=H_0+\lambda p/m$, where $\lambda$ is a given parameter. Given $\lambda, m$ and $E_n^{(0)}$, find the eigenvalues of $H$.

2. Relevant equations

3. The attempt at a solution
The new Hamiltonian is
$H=H_0+\lambda p/m=p^2/2m+\lambda p/m+V(x)=(p+\lambda)^2/2m+V(x)-\lambda^2/2m,$
or
$H'=p'^2/2m+V(x),$
where $H'=H+\lambda^2/2m, p'=p+\lambda$
The eigenfunctions and eigenvalues of $H'$ are respectively $E_n^{(0)}$ and $\psi_n^{(0)}$

Why does $H'$ have the same eigenfunctions and eigenvalues as $H_0$?

As the wave number is $k'=p'/\hbar=\frac{1}{\hbar}(p+\lambda)$, the new eigenfunctions are
$\psi=\psi^{(0)}e^{-i\lambda x/\hbar}$
and the corresponding eigenvalues are
$E_n=E_0^{(0)}-\lambda^2/2m$

Kind regards
Alex

2. Sep 25, 2015

Staff: Mentor

Shifting p by a constant λ does not change the eigenvalues. You just get eigenfunctions with an additional factor (the exponent you have there) compared to the original ones.