1
$\begingroup$

I have been studying the book on Quantum Field Theory by Lewis H. Ryder and I am finding a Gaussian integration a little bit confusing. In the book, the transition amplitude (Eq. $(5.15)$) is given as $$\langle q_ft_f|q_it_i \rangle = \int \frac{\mathscr{D}q\mathscr{D}p}{h}\exp \left [\frac{i}{\hbar} \int_{t_i}^{t_f}dt \left [p \dot q - H(p,q) \right ] \right ]. \tag{5.13}$$ If we use the Hamiltonian to be of the form $H = p^2/2m \: + \:V(q)$, then we get the form of the transition amplitude as $$\langle q_ft_f|q_it_i \rangle = N \int \mathscr{D}q \exp \left [\frac{i}{\hbar} \int_{t_i}^{t_f}L(q, \dot q)dt\right ]. \tag{5.15}$$ Further, it introduces to an Lagrangian of a system with a velocity-dependent potential, given by $$L = \frac{{\dot q}^2}{2}f(q)\tag{p.160}$$ and we get the Hamiltonian corresponding to this Lagrangian as $$H=\frac{1}{2}\frac{p^2}{f(q)}.\tag{p.160}$$ We substitute this Hamiltonian into the equation $(5.13)$ and doing the $p-$integration, we get the transition amplitude to be $$\langle q_ft_f|q_it_i \rangle = N \int \mathscr{D}q \exp \left (\frac{i}{\hbar}S_{\text{eff}}\right )\tag{p.160}$$ where $$S_{\text{eff}} = \int dt \left [L(q, \dot q) - \frac{i}{2}\delta(0)\ln f(q)\right ].\tag{p.160}$$ But when I did the $p-$integration, I got $S = \int L\text{d}t$, where $L$ is the same Lagrangian I defined above, I am not getting the $\ln f(q)$ term in the effective action equation. What am I doing wrong? Any hint and help is appreciated.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

If I am not mistaken, you simply forgot that $e^{\frac{1}{2}\delta(0)\ln(f(q))} \leadsto \det(f(q))^{1/2}$. Indeed: \begin{align} &\int \mathcal{D}p \mathcal{D}q\,e^{i\int dt (p\dot{q}-H)},\,\,\, H=\frac{p^2}{2f(q)} \\ \sim&\int \mathcal{D}q \det(f(q))^{1/2}\,e^{i\int dt L},\,\,\, L = \frac{\dot{q}{}^2}{2}f(q) \end{align} But we have $\det(\cdots) = e^{\text{tr}(\ln(\cdots))}$. The trace should give the $\delta(0)$ because inside of the determinant, $f(q)$ is in fact $f(q)\times \delta(\cdots)$. Putting all the pieces together, we finally have: \begin{equation} \int \mathcal{D}q \,e^{i\int dt (L-\frac{i}{2}\delta(0) \ln f(q))} \end{equation} Which is the desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.