# Hamiltonian from Euclidean lagrangian?

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

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 :

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