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I'm working on the Functional Methods in Peskin (page 280)

enter image description here However, I canʻt obtain Eq.(b) and Eq.(c) from Eq.(a)

Consider Eq.(a)

\begin{align} \left\langle q_{k+1}|f(q)| q_{k}\right\rangle&= f\left(q_{k}\right) \prod_{i} \delta\left(q_{k}^{i}-q_{k+1}^{i}\right)\\&= f\left(q_{k}\right) \prod_{i}\int \frac{dp^i_k}{2 \pi} e^{-i p^i_k \cdot (q_{k}^{i}-q_{k+1}^{i})}\\&=f\left(q_{k}\right)\left(\prod_{i} \int \frac{d p_{k}^{i}}{2 \pi}\right) \exp \left[i \sum_{i} p_{k}^{i}\left(q_{k+1}^{i}-q_{k}^{i}\right)\right] \end{align} It's different from Eq.(b). Is $f\left(q_{k}\right)$ equal to $ f\left(\frac{q_{k+1}+q_{k}}{2}\right) $?

In the Eq.(c), he said "we introduce a complete set of momentum eigenstates", but how to introduce it?

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It's different from Eq. (b). Is $f(q_k)$ equal to $f\left(\frac{q_{k+1}+q_k}{2}\right)$?

Notice in the very first line that $f(q_k)$ multiplies a delta function $\delta(q_k-q_{k+1})$. This delta forces, in anything multipliying it, that $q_k = q_{k+1}$. This is the well-known property:

$$f(x)\delta(x-a)=f(a)\delta(x-a).$$

Now, this means that $$f(q_k)\delta(q_k-q_{k+1})=f\left(\frac{q_k+q_k}{2}\right)\delta(q_k-q_{k+1})=f\left(\frac{q_k+q_{k+1}}{2}\right)\delta(q_k-q_{k+1}).$$

This is, in fact, a convention. Here it may seem weird, but it appears because when you evaluate matrix elements $\langle q'|H|q\rangle$ of the Hamiltonian you need assume one operator ordering convention. The point is that while this seems superflous for a function of just the coordinates $f(q)$, in general $H$ is a function $H(q,p)$ and $q$ and $p$ do not commute.

With the operator ordering that Peskin & Schroeder choose you end up evaluating things at the average of the endpoints of the subintervals.

But again this is just a convention. Check Weinberg's The Quantum Theory of Fields, Vol. 1, Chapter 9. He does the same derivation but he employs another convention. Specifically he orders $H$ such that all position are to the left of all momenta.

In the Eq. (c), he said "we introduce a complete set of momentum eigenstates", but how to introduce it?

Well, now you have a function $f(p)$. What are its matrix elements in coordinate basis $\langle q'|f(P)|q\rangle$? The answer is that by definition you know how to compute $f(P)$ acting on the momentum eigenstates $$f(P)|p\rangle = f(p)|p\rangle,$$ therefore you should use this. Introducing a complete set means using the identity $$\mathbf{1}=\int |p\rangle \langle p| dp,$$ well-known from quantum mechanics.

This will lead you to just evaluate $\langle q|p\rangle$ which is just $\frac{1}{\sqrt{2\pi}}e^{ipq}$.

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  • $\begingroup$ Perfect answer! $\endgroup$ – sky Feb 26 at 16:16

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