5
$\begingroup$

Assuming we have a Effective Field Theory, for example a Real Scalar Field Theory, defined through a Lagrangian density of the form

$\mathcal{L}_{eff} = \frac{1}{2}\partial_\mu\phi \partial^\mu\phi - \frac{1}{2} m \phi^2 - \frac{\lambda}{4!}\phi^4 + \text{higher dimensional terms}$.

with arbitrary higherdimensional and nonrenormalizable terms such as $\partial_\mu\partial_\nu\phi ~ \partial^\mu\partial^\nu\phi$ etc.

N-point greens functions can be defined through the pathintegral:

$\langle \mathcal{T}\lbrace \phi(x_1) \dots \phi(x_n) \rbrace \rangle = \int \mathcal{D}\phi \ \phi(x_1) \dots \phi(x_n) e^{i\int d^4x \mathcal{L}_{eff}[\phi]}$

which may be evaluated perturbatively.

My Question is:

How can we obtain the Hamiltonian-Operator (as the generator of time-evolution $U(t, t_0) = e^{iH (t - t_0)} $) from the Lagrangian that is used in the Pathintegral ?

$\left \langle \phi_b \right| e^{-iH_{eff}T} \left|\phi_a \right\rangle = \int \mathcal{D}\phi \ e^{i\int_0^T dt \int d^3x \ \mathcal{L}_{eff}[\phi]}$

I would expect the Hamiltonian to be of the form $H_{eff} = \int d^3x \frac{1}{2}\dot\phi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m\phi^2 + \frac{\lambda}{4!}\phi^4 + \text{higher dimensional terms} $,

as a sum of arbitrary higher dimensional Operators (but finitely many at each perturbative order maybe ?).

EDIT: If a Legendre Transformation should be the answer:

What would the justification be for doing a Legendre Transformation ?

    In contrast to the case where time derivatives appear quadratic in the Lagrangian we have arbitrary higher power of time derivatives, terms in the Lagrangian containing $\frac{d^n}{d^n t}\phi$. The usual way of casting the Pathintegral from the Lagrangian form into the Hamiltonian form: $\int\mathcal{D}\phi e^{i\int \ \mathcal{L}[\phi]} \rightarrow \int\mathcal{D}\phi\mathcal{D}\pi\ e^{i\int (\pi\dot\phi - H[\phi, \pi])}$, and deriving the Hamilton Operator from it or vice versa doesn't work !
$\endgroup$
  • 2
    $\begingroup$ What is stopping us from performing the usual Legendre transform to obtain the Hamiltonian? $\endgroup$ – ACuriousMind Sep 3 '14 at 22:27

Your Answer

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

Browse other questions tagged or ask your own question.