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I'm studying p. 160 in Ryder's book of QFT and there is an example where the standard path integral equation is not valid

$$\langle q_ft_f|q_it_i\rangle = N \int Dq \exp \left( \frac{i}{\hbar}S \right) \tag{5.15} $$

meaning when we have a position-dependent kinetic energy like in the Lagrangian:

$$L=\frac{\dot q^2}{2} f(q). \tag{5.15a} $$

Using the "explicit" expression for $Dq$ I have

$$\langle q_ft_f|q_it_i\rangle = \text{Const} \cdot \lim_{n \rightarrow \infty} \prod_j dq_j \exp \left( \frac{i}{\hbar} \sum_j (q_{j+1}-q_j)^2 \frac{f(q_j)}{\tau} \right) \tag{1} $$

where I separated the path in equal time intervals of $\tau$. The result in the book is

$$\langle q_ft_f|q_it_i\rangle = N \int Dq \exp \left( \frac{i}{\hbar} \int dt (L - \frac{i}{2} \delta(0) \ln f(q)) \right) \tag{5.15d} $$

I don't understand the meaning of $\delta(0)$ and I don't get where the second terms come from (the one with the $\log$), I don't seem to get it from eq. (1).

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It is counter example to derivation Schroedinger equation from path integral. $$ L = \frac{\dot{q}^2}{2}f(q) $$ $$ H = \frac{p^2}{2f(q)} $$ Using (5.13)

And integrate by $p$ to obtain:

$$ <q_ft_f|q_it_i> = N \int Dq \exp \left( \frac{i}{\hbar} \int dt (L - \frac{i}{2} \delta(0) \ln f(q)) \right) $$

To do this one need calculate Jacobian ($p \to p^\prime = \frac{p-f(q)\dot{q}}{\sqrt{f(q)}}$) and take product over initial and final momentum ($p_{in}\in(-\infty, +\infty)$ and ($p_{out}\in(-\infty, +\infty)$, and Jacobian is the same for all momentum. It lead us to take infinite product, which we replace by power $\delta(0)$):

$$ \prod_{in/out} J = \left(\prod_{time} \frac{1}{\sqrt{ f_n}}\right)^{\underbrace{\delta(0)}_{\infty}} = e^{-\frac{1}{2}\int dt \delta(0) \ln(f(q))} $$

In other side: $$ <q_ft_f|q_it_i> = N \int Dq \exp \left( \frac{i}{\hbar} \int dt L \right) $$

So this expressions are different. This is main point of Ryder.

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  • $\begingroup$ Thanks a lot! Could you please elaborate a bit further the piece in which you introduce the $\delta(0)$? $\endgroup$
    – TheoPhy
    Feb 9, 2020 at 22:54
  • $\begingroup$ I updated answer $\endgroup$
    – Nikita
    Feb 9, 2020 at 22:59

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