Eigenvalues of disturbed Hamiltonian

  • #1
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 [itex]V(x)[/itex] is governed by the Hamiltonian [itex]H_0=p^2/2m+V(x)[/itex], where [itex]p=-i\hbar d/dx[/itex] is the momentum operator. Let [itex]E_n^{(0)}, n=1,2,3,...,[/itex] be the eigenvalues of [itex]H_0[/itex]. Now consider a new Hamiltonian [itex]H=H_0+\lambda p/m[/itex], where [itex]\lambda[/itex] is a given parameter. Given [itex]\lambda, m [/itex] and [itex]E_n^{(0)}[/itex], find the eigenvalues of [itex]H[/itex].

Homework Equations




The Attempt at a Solution


The new Hamiltonian is
[itex]H=H_0+\lambda p/m=p^2/2m+\lambda p/m+V(x)=(p+\lambda)^2/2m+V(x)-\lambda^2/2m, [/itex]
or
[itex]H'=p'^2/2m+V(x),[/itex]
where [itex]H'=H+\lambda^2/2m, p'=p+\lambda[/itex]
The eigenfunctions and eigenvalues of [itex]H'[/itex] are respectively [itex]E_n^{(0)}[/itex] and [itex]\psi_n^{(0)}[/itex]

Why does [itex]H'[/itex] have the same eigenfunctions and eigenvalues as [itex]H_0[/itex]?


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

Kind regards
Alex
 

Answers and Replies

  • #2
35,441
11,874
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.
 

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