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Ladder operators in quantum mechanics

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data

    This is problem 2.11 from Griffith's QM textbook under the harmonic oscillator section.

    Show that the lowering operator cannot generate a state of infinite norm, ie, [tex]\int | a_{-} \psi |^2 < \infty [/tex]

    2. Relevant equations

    This isn't so hard, except that I consistently get the wrong "sign". I have worked based on Griffith's a- operator

    [tex]a_{-} = \frac{1}{\sqrt{2m}} (\frac{h}{i} \frac{d}{dx} - im\omega x)[/tex].

    As suggested by the question, I am to find that [tex]\int (a_{-}\psi)^{"*"} (a_{-}\psi) dx = \int \psi^{"*"} (a_{+} a_{-} \psi) dx[/tex]. It's easy to get the answer after getting this intermediate step, but I get the wrong sign, ie, [tex]- \int \psi^{"*"} (a_{+} a_{-} \psi) dx[/tex] instead of with a positive sign.

    [tex]\int (a_{-}\psi)^{"*"} (a_{-}\psi) dx = \int (\frac{1}{\sqrt{2m}}\frac{h}{i} \frac{\partial \psi^{"*"}}{\partial x} - im\omega x \psi^{"*"}) (a_{-} \psi) dx[/tex].

    Once I multiply out the two brackets (which is ok because there is only one operator involved here right?) I get

    [tex]\int \frac{h}{i} \frac{1}{\sqrt{2m}} (a_{-}\psi) \frac{\partial \psi^{"*"}}{\partial x} - im \omega x a_{-} \psi \psi^{"*"} dx[/tex]

    If I apply integration by parts to the first term, I get an expression along the lines of [tex]-\int \psi^{"*"} \frac{\partial \psi}{\partial x} dx[/tex]. This would transform the initial expression to [tex]-a_{+}[/tex] and leave a minus sign in the final expression!

    I do know that due to the way Griffiths derives the operators, there is an alternative operator, ie [tex]a_{-} = - \frac{h}{i} \frac{d}{dx} + m\omega x[/tex]. This would turn out the correct answer I think.

    But am I missing something here? Why wouldn't both methods check out equally as well? Surely the sign is important in these types of derivations.
     
  2. jcsd
  3. Mar 28, 2009 #2
    Maybe you have a different edition than me, but in my book he states:
    [tex]a_{\pm} = \frac{1}{\sqrt{2\hbar m \omega}}\left(\mp i p + m \omega x\right)[/tex]

    This would lead to:
    [tex]a_- = \frac{1}{\sqrt{2\hbar m \omega}}\left(+ ip + m\omega x\right)[/tex]

    Furthermore, it is much easier to show this by noting that [itex]a_-[/itex] and [itex]a_+[/itex] are hermitian conjugates. So you can write:
    [tex]\int \left| a_- \psi \right|^2 \, dx = \int (a_- \psi)^*(a_- \psi) \, dx = \int \psi^*(a_+a_- \psi) \, dx[/tex]
    Then use equation 2.54:
    [tex]a_+a_- = \frac{1}{\hbar \omega} H - \frac{1}{2}[/tex]
    And you should be on your way to get the answer:
    [tex]\int \left| a_- \psi \right|^2 \, dx = \frac{1}{\hbar \omega} E - \frac{1}{2} < \infty[/tex]
     
  4. Mar 28, 2009 #3
    I've got an older edition (ca 1995!!) Nick. It's weird because Griffiths does a few examples (ie, expand a+ a-, etc) with the operator that I stated at the beginning.

    I didn't know about the Hermitian part. But this particular question, which I don't think is in newer editions (I checked), requires me to do out the whole thing with integration by parts.
     
  5. Mar 28, 2009 #4

    turin

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    Homework Helper

    Nick's method is the way that I would show this to myself, but, perhaps the exercise is to show this using the expressions for a± in terms of x and p.

    You have to be very careful about which operators are acting on which functions.
    You also need to be careful about the operator definitions.

    You need to be careful how you are doing the complex conjugation.

    You need to be careful about what the a- is operating on (i.e. keeping parenthesis would be a good idea).

    Momentum (the partial derivative operator) does not commute with a-.

    I don't have Griffiths QM text, but I suspect that this is a typo. This gives a Hermitian operator, but a- should not be Hermitian if it is a lowering operator.
     
    Last edited: Mar 28, 2009
  6. Aug 5, 2011 #5
    There's not a typo in Griffiths. Remember to negate the first trm of the lowering operator that contains an i, as well as the second when you take the complex conjugate. I got hung up at first trying to go too far with the math. For the steps I finally used to get it correct see:

    http://copaseticflow.blogspot.com/2011/08/its-obvious-not-knowing-when-not-to-do.html" [Broken]

    Hamilton
     
    Last edited by a moderator: May 5, 2017
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