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How to show only multiples of unit energy are allowed in QHO by ladder operator

  1. Sep 7, 2011 #1
    I'd like to know how to do it without solving via Hermite polynomials, so that I can check by both methods when I solve other problems. I have tried to figure out myself but I need some help.

    Let's say [itex] H:=x^2+p^2 [/itex], and [itex] a:= x-ip [/itex].

    So that [itex] [a,a^+]=2\hbar [/itex], [itex] H a \varphi_i = (E_i - 2\hbar)\varphi_i [/itex], and [itex] H a^+ \varphi_i = (E_i + 2\hbar)\varphi_i [/itex], where [itex] \varphi_i [/itex] is an eigenstate.

    Further, [itex] \langle i|[H,a^+]|j\rangle=(E_i-E_j)\langle i|a^+|j\rangle=2\hbar\langle i|a^+|j\rangle[/itex] .

    My text says "the matrix element [itex] \langle i|a^+|j\rangle [/itex] vanishes, unless the difference [itex] E_i-E_j [/itex] between the energies of the two eigenstates is the constant [ [itex] 2\hbar [/itex] ]."
    ( I have put ' [ ] ' where I changed due to simplification of H. )

    I understand the statement. My question is: How does it suggest that there won't be any energy eigenvalues between those multiples? ( Or is there any other way to show it? ) Note that the text doesn't explicitly say that, but it is followed by

    "This fact implies that we may sequentially order the eigenstates connected by [itex] a^+ [/itex], the difference between two consecutive energies being [ [itex] 2\hbar [/itex] ]. Another consequence is that we may assign an integer number n to each eigenstate."

    Any help would be appreciated.

    Source: p.35, Quantum Mehcanics: A Modern and Concise Introductory Course
     
    Last edited: Sep 7, 2011
  2. jcsd
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