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.

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.