Eigenvalues of disturbed Hamiltonian

[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.

Homework Statement

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.

Homework Equations

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

 

[mfb]

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.