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Zwiebach eqn (17.44) page 395

  1. Nov 8, 2007 #1
    [SOLVED] Zwiebach eqn (17.44) page 395

    1. The problem statement, all variables and given/known data
    Equation (17.44) is:
    [tex]e^{-i\ell x_0/R}pe^{i\ell x_0/R} = p + \frac{\ell}{R}[/tex]



    2. Relevant equations
    Equation (17.43)
    [tex]\ell \in \mathbb{Z}[/tex]



    3. The attempt at a solution
    Equation (17.44) is easy to derive if [itex]\frac{\ell}{R}[/itex] is small. but since [itex]\ell[/itex] can be any integer, there doesn't seem to be any way to insure that. Did the author simply forget to mention some conditions necessary for it to be true?
     
  2. jcsd
  3. Nov 8, 2007 #2

    nrqed

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    It's true to all orders. Just use [itex] [ x_0^n, p] = i n x_0^{n-1} [/itex].
     
  4. Nov 9, 2007 #3
    Thanks for taking a look at this nrqed. If [itex]\frac{\ell}{R}[/itex] were small, then I would expand the exponentials to linear terms. But it isn't small, so it would seem that I need to keep all terms. But if I consider up to quadratic terms I get:
    [tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(x_0^2p + x_0px_0 - px_0^2) + \mathcal{O}({\frac{\ell^3}{R^3}})[/itex]
    Using your suggestion this becomes:
    [tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(2ix_0 + x_0px_0) + \mathcal{O}(\frac{\ell^3}{R^3})[/itex]
    This doesn't seem to improve things. Is there a simpler approach that I am missing?
     
    Last edited: Nov 9, 2007
  5. Nov 9, 2007 #4

    nrqed

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    Hi Jimmy,

    I am in a hurry because I am teaching a class in 15 minutes and have stuff to do. I will get back to you early this afternoon.
     
  6. Nov 9, 2007 #5
    Take your time. When you get back to it, you will find that I erroniously put upper case P's in my equations. They should be lower case. I fixed it in my post.

    Edit - With your help, I have figured this out. Thanks nrqed.
     
    Last edited: Nov 9, 2007
  7. Nov 9, 2007 #6

    nrqed

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    I just got back from my classes. I am glad it worked out! As you see, it works to all orders.

    Glad I could help!


    Patrick
     
  8. Nov 9, 2007 #7
    Just for the record, here is the solution:
    [tex]e^{-ilx_0/R}pe^{ilx_0/R}[/tex]

    [tex]= (\Sigma\frac{1}{n!}(\frac{-il}{R})^nx_0^np)e^{ilx_0/R}[/tex]

    Now using nrqed's suggestion:

    [tex]= (\Sigma\frac{1}{n!}(\frac{-il}{R})^npx_0^n + \Sigma_0\frac{1}{n!}(\frac{-il}{R})^ninx_0^{n-1})e^{ilx_0/R}[/tex]

    [tex]= (p\Sigma\frac{1}{n!}(\frac{-il}{R})^nx_0^n + \Sigma_1\frac{1}{n!}(\frac{-il}{R})^ninx_0^{n-1})e^{ilx_0/R}[/tex]

    [tex]= (pe^{-ilx_0/R} + \frac{\ell}{R}\Sigma_1\frac{1}{(n-1)!}(\frac{-il}{R})^{n-1}x_0^{n-1})e^{ilx_0/R}[/tex]

    [tex]= (pe^{-ilx_0/R} + \frac{\ell}{R}\Sigma_0\frac{1}{n!}(\frac{-il}{R})^n}x_0^n)e^{ilx_0/R}[/tex]

    [tex]= (pe^{-ilx_0/R} + \frac{\ell}{R}e^{-ilx_0/R})e^{ilx_0/R}[/tex]

    [tex]= p + \frac{\ell}{R}[/tex]
     
    Last edited: Nov 9, 2007
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