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squigglywolf
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Path Integral Formalism
Reading through Shankar atm, up to page 232/233.
Reference to pages if interested.
http://books.google.co.nz/books?id=2zypV5EbKuIC&printsec=frontcover&source=gbs_vpt_reviews#v=onepage&q=232&f=false(sorry I am too noob at latex to type all the formulas out..)
It's looking at getting the propagator from the path integral method with potentials of the form [tex]\mbox{V = a + bx + cx^2 +d}\dot{x} \mbox{+ex}\dot{x}[/tex]
I think I understand the steps until 8.6.6 but can't understand why it is assumed L is a quadratic polynomial? Cant V(x) depend on say x^3 and higher?
Also equation 8.6.12, in evaluating the path integral it assumes it has no memory of Xcl, so it can only depend on t, thus it gets replaced by A(t). What exactly does this mean? Is this because we are now taking all paths between 0 --> 0? i.e. the paths no longer have any knowledge of what the classical endpoints are (x --> x' ) ?
This whole section 8.6 has kinda confused me.
Reading through Shankar atm, up to page 232/233.
Reference to pages if interested.
http://books.google.co.nz/books?id=2zypV5EbKuIC&printsec=frontcover&source=gbs_vpt_reviews#v=onepage&q=232&f=false(sorry I am too noob at latex to type all the formulas out..)
It's looking at getting the propagator from the path integral method with potentials of the form [tex]\mbox{V = a + bx + cx^2 +d}\dot{x} \mbox{+ex}\dot{x}[/tex]
I think I understand the steps until 8.6.6 but can't understand why it is assumed L is a quadratic polynomial? Cant V(x) depend on say x^3 and higher?
Also equation 8.6.12, in evaluating the path integral it assumes it has no memory of Xcl, so it can only depend on t, thus it gets replaced by A(t). What exactly does this mean? Is this because we are now taking all paths between 0 --> 0? i.e. the paths no longer have any knowledge of what the classical endpoints are (x --> x' ) ?
This whole section 8.6 has kinda confused me.
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