# Calculating the transition amplitude

Lagrangian of the Lee-Yang model is given by:

$$L=\frac{1}{2}f(q)\dot{q}^2$$

where $f(q)$ is some differentiable function.

I am trying to derive the following expression for the transition amplitude:

$$<q_ft_f|q_it_i>=N \int e^{\frac{i}{\hbar}\int{dt(L(q,\dot{q})-\frac{1}{2}\delta(0)\ln{f(q)})}} Dq$$

EDIT: I almost finished. So, I started with the most general expression for the transition amplitude: $$<q_ft_f|q_it_i>=\int Dq \int Dp e^{\frac{i}{\hbar} \sum \Delta t(p_j\frac{q_{j+1}-q_j}{\Delta t^2}-H(p_j,\overline{q_j}))}$$ where $\overline{q_j}=\frac{q_j+q_{j+1}}{2}$ (I derived this for the Hamiltonian of the form $H=\frac{p^2}{2m}+V(q)$, but I understood it should be valid in general) and managed to obtain:

$$<q_ft_f|q_it_i>= N\int\prod{dq_i} e^{\frac{1}{2}\sum_{j=o}^n (ln(\overline{q_j})-\frac{i}{\hbar}\Delta t\frac{(q_{j+1}-q_j)^2}{\Delta t^2}f(\overline{q_j}))}$$

In limes $n\to\infty$ I can identify the second term as Lagrangian, but I'm still confused with the delta function.

• $\delta(0)$ is e.g. explained in this Phys.SE post. – Qmechanic Nov 10 '16 at 12:06

## 1 Answer

The amplitude equals to the following path integral $$\int Dq(\tau)\int \frac{Dp(\tau)}{2\pi}exp\left[i\int^{t_f}_{t_i}d\tau \tilde{L}[q(\tau),p(\tau)]\right]$$ where $$\tilde{L}=p\dot{q}-\frac{p^2(\tau)}{2f(q(\tau))}$$ which is not the Lagrangian because $p$ is not related to $q$ and $\dot{q}$. The integral over $p$ can be easily performed because it's a Gaussian integral: $$\int\frac{Dp}{2\pi}exp\left[i\int^{t_f}_{t_i}d\tau\int^{t_f}_{t_i}d\tau^\prime \tilde{L}[q(\tau),p(\tau)]\delta(\tau-\tau^\prime)\right] \\=N\frac{1}{\sqrt{detA}}exp\left[i\int^{t_f}_{t_i}d\tau\tilde{L}[q(\tau),\tilde{p}(\tau)]\right]$$ where $\tilde{p}$ is the stationary ponit satisfying the canonical equation $$\dot{q}=\left(\frac{\partial H}{\partial p}\right)_{p=\tilde{p}}$$ so now $\tilde{L}$ can be replaced by the Lagrangian $L$ and we need only do the path integral over $q(\tau)$; and the matrix $A$ is $$A_{\tau,\tau^\prime}=\frac{\delta(\tau-\tau^\prime)}{f(q(\tau))}$$ whose determinant can be expressed as $$detA=e^{trlnA}=exp\left(tr[-\delta(\tau-\tau^\prime)lnf(q(\tau))]\right)\\ =exp\left[-\delta(0)\int d\tau lnf(q(\tau))\right]$$ which can be regarded as a modification to the Lagrangian.