I have not encountered a great deal about the path integral but I am trying to learn it, I would like to derive by hand each step in Peskin and Schroeder however I am hitting a wall at equation $9.5$ on page $279$:

$$U(x_a,x_b;T) = \int_{-\infty}^{+\infty}\frac{dx'}{C}\exp\left(\frac{i}{\hbar}\frac{m}{2\epsilon}(x_b-x')^2\left[1-\frac{i\epsilon}{\hbar}V(x_b)+\text{ }...\right]\right)$$ $$\times\left[ 1+(x'-x_b)\frac{\partial}{\partial x_b}+\frac{1}{2}(x'-x_b)^2\frac{\partial ^2}{\partial x_b^2}+\text{ } ...\right] U(x_a,x_b;T-\epsilon) \tag{9.5}.$$

We obtain this from the previous expression:

$$U(x_a,x_b;T)=\int_{-\infty}^{+\infty}\frac{dx'}{C}\exp\left[\frac{i}{\hbar}\frac{m}{2\epsilon}(x_b-x')^2-\frac{i}{\hbar}\epsilon V\left(\frac{x_b+x'}{2}\right)\right]U(x_a,x';T-\epsilon).$$

The guidance provided is that we "expand the above expression in powers of $(x'-x_b)$" to arrive at $9.5$.

It seems clear that the first step is to split up the exponentials and Taylor expand the potential term in $\epsilon$, however we then appear to Taylor expand the propagator, $U(x_a,x';T-\epsilon)$ in $(x'-x_b)$. It is strange to me that we can even do this, is the propagator a function of the quantity $(x'-x_b)$?

I understand that this exact equation has been asked about here, and it has an answer, but I cannot follow the answer that is given. I would ideally like if someone could clearly write out the necessary steps, that being said I would appreciate any help.


1 Answer 1


I'm not sure why you think that the propagator can't be a function of $(x'-x_b)$, it lies within the domain for that parameter and has the effect of shifting the endpoint of the propagator by a fixed distance. Nevertheless, this is actually immaterial since we don't encounter the object $U(x_a, \ x'{-}x_b, \ T{-}\epsilon)$ at any point during the Taylor expansion.

As mentioned, $\epsilon$ acts as a Lagrange multiplier, constraining the value of $x'$ to be close to $x_b$ - this means that during the analysis, we should only keep terms that are first order in $(x-x_b)$ or $\epsilon$ (but not first-order in both, since that would be a second-order term overall). Recall the standard definition for a Taylor series: $$ f(x) = f(a) + \frac{1}{1!}f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2+\frac{1}{3!}f'''(a)(x-a)^3 + \dots $$

So $U(x_a, x')$ in terms of $U(x_a, x_b)$ is simply (now with partial derivatives to account for the multivariability of $U$) $$ U(x_a, x') = U(x_a,x_b)+\frac1{1!}(x'-x_b)\frac{\partial}{\partial x_b}U(x_a,x_b)+\frac{1}{2!}(x'-x_b)^2\frac{\partial ^2}{\partial x_b^2}U(x_a,x_b)+\dots \\ = \left(1+(x'-x_b)\frac{\partial}{\partial x_b}+\frac{1}{2}(x'-x_b)^2\frac{\partial ^2}{\partial x_b^2}+\dots \right)U(x_a,x_b), $$ precisely the second factor.

The first factor similarly comes from expanding the exponential in $\epsilon$, and then Taylor expanding $V\left(\frac{x_b+x'}{2}\right)$ in the resultant first-order term $-\frac {i\epsilon}{\hbar} V\left(\frac{x_b+x'}{2}\right)$:

$$ V\left(\frac{x_b+x'}{2}\right) = V(x_b) + \frac{1}{1!}\frac{x-x_b}{2}V'(x_b)+\frac{1}{2!}\left(\frac{x-x_b}{2}\right)^2 V''(x_b) + \dots \\ \therefore\exp\left[-\frac {i\epsilon}{\hbar} V\left(\frac{x_b+x'}{2}\right)\right] = 1 -\frac {i\epsilon}{\hbar} V\left(\frac{x_b+x'}{2}\right) + \dots \\ = 1 -\frac {i\epsilon}{\hbar}V(x_b) -\underbrace{\frac {i\epsilon}{\hbar}\frac{1}{1!}\frac{x-x_b}{2}V'(x_b)-\frac {i\epsilon}{\hbar}\frac{1}{2!}\left(\frac{x-x_b}{2}\right)^2 V''(x_b)}_\text{Neglect, since $\epsilon$ and $(x-x_b)$ are both small} + \dots\\ = 1 -\frac {i\epsilon}{\hbar}V(x_b) + \dots $$

which is the first factor (which, incidentally, is not supposed to be inside the exponential, as the first expression in the question seems to suggest). Thus the final expression for the propagator is

$$ U(x_a,x_b;T) = \int_{-\infty}^{+\infty}\frac{dx'}{C}\exp\left(\frac{i}{\hbar}\frac{m}{2\epsilon}(x_b-x')^2\right)\left[1-\frac{i\epsilon}{\hbar}V(x_b)+\text{ }...\right]$$ $$\times\left[ 1+(x'-x_b)\frac{\partial}{\partial x_b}+\frac{1}{2}(x'-x_b)^2\frac{\partial ^2}{\partial x_b^2}+\text{ } ...\right] U(x_a,x_b;T-\epsilon) $$


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