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Eigenvalues of disturbed Hamiltonian

  1. Sep 25, 2015 #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.
    1. The problem statement, all variables and given/known data
    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].

    2. Relevant equations


    3. 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
     
  2. jcsd
  3. Sep 25, 2015 #2

    mfb

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    2016 Award

    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.
     
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