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A Operation of Hamiltonian roots on wave functions

  1. May 16, 2017 #1
    How come a+a- ψn = nψn ? This is eq. 2.65 of Griffith, Introduction to Quantum Mechanics, 2e. I followed the previous operation from the following analysis but I cannot get anywhere with this statement. Kindly help me with it. Thank you for your time.
     
  2. jcsd
  3. May 16, 2017 #2

    BvU

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    HI,

    I only have 1st ed PDF and there, in the text "Algebraic method" he shows that the Schroedinger eqn

    [ ..2 + .. 2] ##\psi = E\psi## ​

    can be factored into
    ##[a_+ a_- ] + {1\over 2 } \hbar \omega \psi = E\psi##
    Or ##[a_- a_+ ] - {1\over 2 } \hbar \omega \psi = E\psi##.


    Then he proves that if ##\psi## satisfies the Schroedinger eqn, with energy ##E##, then ##a_+\psi## satisfies the Schroedinger eqn, with energy ##E+\hbar\omega##
    and - same way - ##a_-\psi## satisfies the Schroedinger eqn, with energy ##E-\hbar\omega##

    (Walking down the ladder there is a ground state (##\ {1\over 2 } \hbar \omega\ ##) and in n steps down you have subtracted n times ##\hbar\omega##.)

    The eigenfunction part has now been shown. The normalization coefficient is left as an exercise. That's where the ##i\sqrt{\left (n+1\right) \hbar\omega\,} ## and ##-i\sqrt{n \hbar\omega\,} ## pop up.

     
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