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Show the functions are eigenfunctions of the hamiltonian

  1. Oct 14, 2012 #1
    Given the hamiltonian in this form: H=[itex]\hbar[/itex][itex]\omega[/itex]([itex]b^{+}[/itex]b+.5)

    b[itex]\Psi_{n}[/itex]=[itex]\sqrt{n}[/itex][itex]\Psi_{n-1}[/itex]
    [itex]b^{+}[/itex][itex]\Psi_{n}[/itex]=[itex]\sqrt{n+1}[/itex][itex]\Psi_{n+1}[/itex]

    Attempt:

    H[itex]\Psi_{n}[/itex]=[itex]\hbar[/itex][itex]\omega[/itex]([itex]b^{+}[/itex]b+.5)[itex]\Psi_{n}[/itex]

    I get to

    H[itex]\Psi_{n}[/itex]=[itex]\hbar[/itex][itex]\omega[/itex][itex]\sqrt{n}[/itex]([itex]b^{+}[/itex][itex]\Psi_{n-1}[/itex]+.5[itex]\Psi_{n-1}[/itex])


    But now I'm stuck. Where can I go from here?
     
  2. jcsd
  3. Oct 14, 2012 #2

    dextercioby

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    It's not correct, you have to split the Hamiltonian as it should be split:

    [tex] H = \hbar\omega b^{\dagger}b + \frac{1}{2}\hbar\omega \hat{1} [/tex]

    then act on an arbitrary vector.
     
  4. Oct 14, 2012 #3
    I still end up with a similar problem though...I will have the raising operator acting on [itex]\Psi_{n-1}[/itex]


    H= ℏω[itex]\sqrt{n}[/itex]([itex]b^{+}[/itex][itex]\Psi_{n-1}[/itex])+[itex]\frac{1}{2}[/itex]ℏω[itex]\Psi_{n}[/itex]
     
  5. Oct 14, 2012 #4

    dextercioby

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    Excelent. You need to do a trick on the relation given, namely realize that the <n> can be replaced by other values. Which substitution is useful ?

    P.S. Always post your HW questions here, in this forum.
     
  6. Oct 14, 2012 #5
    Can I say that if [itex]b^{+}[/itex][itex]\Psi_{n}[/itex]=[itex]\sqrt{n+1}[/itex][itex]\Psi_{n+1}[/itex] then [itex]b^{+}[/itex][itex]\Psi_{n-1}[/itex]=[itex]\sqrt{n}[/itex][itex]\Psi_{n}[/itex]

    which would allow me to write the Hamiltonian as

    H=ℏω(n+[itex]\frac{1}{2}[/itex])[itex]\Psi_{n}[/itex]

    and because ℏω(n+[itex]\frac{1}{2}[/itex]) is just a number, then

    b[itex]\Psi_{n}[/itex]=[itex]\sqrt{n}[/itex][itex]\Psi_{n-1}[/itex] and
    [itex]b^{+}[/itex][itex]\Psi_{n}[/itex]=[itex]\sqrt{n+1}[/itex][itex]\Psi_{n+1}[/itex]

    are eigenfunctions of the Hamiltonian?
     
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