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alfredblase

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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

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

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