Changing the time scale in the action

1. Dec 15, 2006

alfredblase

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:

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

where t' and t'' are constants.

the original action is:

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

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

Last edited: Dec 15, 2006
2. Dec 15, 2006

cristo

Staff Emeritus
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 $dt=\hbar dt'$ and, using the chain rule, $$\frac{dx}{dt}=\frac{dx}{dt'}\frac{dt'}{dt}=\frac{dx}{dt'}\frac{1}{\hbar}$$ This gives $$\dot{x}^2=\left(\frac{dx}{dt'}\right)^2 \frac{1}{\hbar ^2}$$

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

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