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Changing the time scale in the action

  1. Dec 15, 2006 #1
    I'm stuck, again, on a particular passage of a book (jean zinn justinf's QFT and critical phenomena).

    it says:

    We will find it convinient to t -> t/hbar , the action is then written as:

    [tex]S_{0} / \hbar=\int_{t'}^{t''} dt [\frac{m \dot {x}^2(t) }{2 \hbar^2} + V(t)][/tex]

    where t' and t'' are constants.

    the original action is:

    [tex]S_{0}=\int_{t'}^{t''} dt [\frac{m \dot {x}^2(t) }{2} + V(t)][/tex]

    My problem is that basically what he seems to have done is divide the normal action by hbar but somehow pulled out an extra 1/hbar in the first term of the integrand and one less 1/hbar in the second term.. I'm sorry but I have no idea how the original action goes to the new form when t -> t/hbar

    Thank you for taking the time to read this. Any help/advice/suggestions will be very much apreciated :smile:
    Last edited: Dec 15, 2006
  2. jcsd
  3. Dec 15, 2006 #2


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    Making the transformation t -> t/hbar=t' (easier to use a different letter for the new t value when calculating, to avoid confusion) we find that [itex] dt=\hbar dt' [/itex] and, using the chain rule, [tex]\frac{dx}{dt}=\frac{dx}{dt'}\frac{dt'}{dt}=\frac{dx}{dt'}\frac{1}{\hbar} [/tex] This gives [tex] \dot{x}^2=\left(\frac{dx}{dt'}\right)^2 \frac{1}{\hbar ^2} [/tex]

    The integral is then
    S_0 = \int \hbar dt' \left[\frac{m}{2\hbar^2}\left(\frac{dx}{dt'}\right)^2 +V(t')\right] [/tex]

    Dividing through by hbar and replacing t'=t gives the required result
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