Can somebody help me in deriving the Hamiltonian of system starting from Euclidean Lagrangian?

Say we are given the Minkowski Lagrangian

$$L_m = \frac{\dot{\phi}^2}{2} - V(\phi).$$

The Hamiltonian can then be found by Legendre transformation

$$H = \dot{\phi}\frac{\partial L_m}{\partial\dot{\phi}} - L_m,$$

which equals $$H = \frac{1}{2}\dot{\phi}^2 + V(\phi)$$ which was not hard.

Now consider the corresponding Euclidean Lagrangian

$$L_e = -\frac{\dot{\phi}^2}{2} - V(\phi).$$

How do I calculate the Hamiltonian in this formalism? The above way applied naively will not give the correct result.


1 Answer 1


The only physical interesting correspondence is between a Lagrangian $L$ (if you are considering a standard time) , and the Hamiltonian $H$ (if you are considering euclidean (imaginary) time) . For this, you have to consider quantum mechanics.

In quantum mechanics, a amplitude can be expressed as as sum about paths :

$$\mathcal A = \int [D\phi]~e^{iS(\phi)} = \int [D\phi]~e^{\large i\int dtL(\phi,t)} = \int [D\phi]~ e^{\large i\int dt (\frac{1}{2} (\large \frac{d\phi}{dt})^2 - V(\phi))} \tag{1}$$

Now, defining an imaginary time $t' = it$, you get :

$$\mathcal A = \int [D\phi]~ e^{\large -\int dt' (\frac{1}{2} (\large \frac{d\phi}{dt'})^2 + V(\phi))} = \int [D\phi]~e^{\large -\int dt'H(\phi,t')}\tag{2}$$

So, when you go from a standard time to an euclidean (imaginary time), you have the correspondence :

$$ L(\phi) \to - H(\phi)\tag{3}$$


Your Answer

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

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