# Path Integral in QM with a position-dependent kinetic energy

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).

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:

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

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: $$ = N \int Dq \exp \left( \frac{i}{\hbar} \int dt L \right)$$

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

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