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Eigenvalues and ground state eigenfunction of a weird Hamiltonian

  1. Oct 14, 2011 #1
    Hello again everyone!

    I would like to ask a question regarding this Hamiltonian that I encountered. The form is H = Aa^+a + B(a^+ + a). Then there is this question asking for the eigenvalues and ground state wavefunction in the coordinate basis. The only given conditions are, the commutator of a^+ and a is [a^+,a] = 1, and that A > 0 and B are c-number constants. I actually do not understand the meaning of c-number constants. Can anyone suggest how to attack this problem?
  2. jcsd
  3. Oct 14, 2011 #2


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    Are you sure? The canonical convention is that [itex]\hat{a}[/itex] is the destruction and [itex]\hat{a}^{\dagger}[/itex] the creation operator for phonons. Then the commutator should read [itex][\hat{a},\hat{a}^{\dagger}]=1[/itex]. What you have here, is simply a "shifted harmonic oscillator". This problem you find solved in many textbooks on quantum mechanics.
  4. Oct 14, 2011 #3
    Ok, so a and a^+ are the annihilation and creation operators in the harmonic oscillator problem. I thought there are other operators. Thanks for your comment! Anyway, in this shifted harmonic oscillator case, do you expect that the solution for example, the eigenvalues are just shifted by a constant? I think the same is true for the wavefunction.
  5. Oct 27, 2011 #4
    May I know how can I obtain the eigenvalues using the usual eigenvalue problem here? I am quite confused here now.
  6. Oct 28, 2011 #5


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    Define new creation and annihilation operators
    [tex]\tilde a = a+c[/tex]
    [tex]\tilde a^\dagger = a^\dagger + c[/tex]
    where [itex]c[/itex] is a real constant. Choose [itex]c[/itex] so that the hamiltonian is
    [tex]H=A \tilde a^\dagger \tilde a + d[/tex]
    where [itex]d[/itex] is another constant. Note that
    [tex][\tilde a, \tilde a^\dagger]=1[/tex]
  7. Oct 29, 2011 #6
    Hmmm, sounds ok. Thank you for your suggestion. But I am really new to ladder operators, how would you use this translated a+ and a?
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