Hi this problem relates directly to path integrals but I imagine it is a maths trick that I am missing. One has an expression such as $$\int dx \exp\left[i\frac{(p)^{2}}{2}-iV(\frac{f}{4})\right] $$ it says simply expand the expression in powers of $(x-y)$ to get $$\int dx \left(i\frac{(y-x)^{2}}{2a}\right)\left[1-iaV(y)+...\right]\left[1+(x-y)\frac{\partial}{\partial y}+\frac{1}{2}(x-y)^{2}\frac{\partial^2}{\partial y^2}\right] $$ I thought it was some sort of Taylor expansion by I can't explain the change in exponential and I don't really understand the derivatives. Can anyone explain why these are equal?
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1$\begingroup$ Please do not vandalize your posts that way. $\endgroup$– dmckee --- ex-moderator kittenApr 19, 2013 at 22:28
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$\begingroup$ Echoing @dmckee's comment, your latest edit (version 5) is less informative than version 4. E.g. now the P&S page number is missing. $\endgroup$– Qmechanic ♦May 22, 2013 at 22:30
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1 Answer
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The derivatives are the Taylor expansion of the propagator, which you have omitted. Notice that on page 278, the propagator under the integral is $U(x_a, x'; T-\epsilon)$, but in equation 9.5 it's $U(x_a,x_b;T-\epsilon)$.
The middle term is the expansion of $\exp[-i/\hbar \epsilon V((x+y)/2)]$, first in terms of powers of $\epsilon$, then in powers of $(x-y)$. Then since both are small, keep terms of first order in both only. this gives $1 - i \epsilon V(y)$. (Here your $a$ is Peskin and Schroeder's $\epsilon$).
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1$\begingroup$ Did you not read my post. The derivatives are not of $V$, they are of $U$. You didn't look at the equations in their entirety. $\endgroup$– nervxxxApr 19, 2013 at 14:57