# Euclidean action and Hamiltonian

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• ShayanJ
In summary, the conversation discusses the concept of Euclidean action in a quantum field theory and its relationship to the Hamiltonian. The speaker explains that the Euclidean action is obtained by applying a Wick rotation to the Lagrangian or Hamiltonian action in the path integral. The conversation also mentions the need to properly choose the time component in relativistic theories and provides an example calculation from Maggiore's book. The conclusion is that the Euclidean action has a similar structure to the Hamiltonian, but is not exactly equal to it due to the use of Euclidean time.
ShayanJ
Gold Member
Yesterday I was asking questions from someone and in between his explanations, he said that the Euclidean action in a QFT is actually equal to its Hamiltonian. He had to go so there was no time for me to ask more questions. So I ask here, is it true? I couldn't find anything on google. If its true, how can I prove it?
Thanks

The only meaning of "Euclidean action" I know of is the Wick rotation (from Minkowski spacetime to an Euclidean spacetime with complex time) applied to any of the Lagrangian and Hamiltonian actions you put in the path integral. In order to properly perform the calculation of the path integral (just as in ordinary QM in 3D), you must perform this Wick rotation. So the correct wording is: "Euclidean Hamiltonian action".

I don't know the answer, but it should be simple to work out if you just use a theory of a scalar field with a potential:

$$S = \int d^4 x \; \big( \partial_\mu \phi \partial^\mu \phi - V(\phi) \big)$$
The reason to throw in a potential is to make sure the Lagrangian ##T - V## is not just the same as the Hamiltonian ##T + V##.

Note that in relativistic theories, the "Hamiltonian" you would get just by doing the standard Legendre transformation comes out to be identically zero. This is because the Hamiltonian (much like the energy) is only sensible after you make a choice of frame. So you need to single out the time component, re-write your integral as

$$S = \int dt \int d^3 x (\text{stuff})$$
and now do the standard Legendre transformation using the time coordinate ##t##.

There is some chance of this working, since the Wick rotation also requires singling out the time component. You may find that it only works in certain cases, though.

ShayanJ
dextercioby said:
The only meaning of "Euclidean action" I know of is the Wick rotation (from Minkowski spacetime to an Euclidean spacetime with complex time) applied to any of the Lagrangian and Hamiltonian actions you put in the path integral. In order to properly perform the calculation of the path integral (just as in ordinary QM in 3D), you must perform this Wick rotation. So the correct wording is: "Euclidean Hamiltonian action".

What do you mean by Lagrangian and Hamiltonian actions?
The action is defined by the equation ## S=\int_{t_1}^{t_2} L dt=\int_{t_1}^{t_2} (p\dot q-H) dt ##. Although in one of the integrals is the Hamiltonian and in the other there is the Lagrangian, but these are no two different kinds of action, just two different expressions for it. So I don't understand what you're talking about!

Anyway, I saw the calculation below in Maggiore's book:

## S=\int d^4x \left[ \frac 1 2 \left( \partial^\mu \phi\partial_\mu \phi-m^2\phi^2 \right)-V(\phi) \right]=\int d^4x \left[ \frac 1 2 (\partial_t\phi)^2-\frac 1 2 (\partial_i \phi)^2-m^2\phi^2-V(\phi) \right]=-i\int (d^4x)_E \left[-\frac 1 2 (\partial_{t_E}\phi)^2-\frac 1 2 (\partial_i \phi)^2-m^2\phi^2-V(\phi) \right]\equiv iS_E ##

So we have ## S_E=\frac 1 2 (\partial_{t_E}\phi)^2+\frac 1 2 (\partial_i \phi)^2+m^2\phi^2+V(\phi) ##. Its not equal to the Hamiltonian exactly because the time derivative is w.r.t. the Euclidean time but it has the same structure. So I think this is what he was talking about.

## 1. What is the Euclidean action in physics?

The Euclidean action is a mathematical concept used in physics to describe the evolution of a system over time. It is a function of the system's state and its derivatives, and it is used to calculate the probability of a system transitioning from one state to another.

## 2. How is the Euclidean action related to the Hamiltonian?

The Euclidean action is related to the Hamiltonian through the path integral formalism. The Hamiltonian is the classical limit of the Euclidean action, and it determines the dynamics of a system by calculating the rate of change of the system's state over time.

## 3. What is the significance of the Euclidean action in quantum mechanics?

In quantum mechanics, the Euclidean action is used to calculate the probability of a quantum system transitioning from one state to another. It is a key component of the path integral formalism, which is used to calculate the quantum amplitude for a system to evolve from one state to another.

## 4. How is the Euclidean action used in the path integral formalism?

In the path integral formalism, the Euclidean action is used to calculate the quantum amplitude for a system to evolve from one state to another. It is integrated over all possible paths that the system could take between the initial and final states, and the paths with the highest probability contribute the most to the overall amplitude.

## 5. What are some applications of the Euclidean action and Hamiltonian in physics?

The Euclidean action and Hamiltonian have a wide range of applications in physics, including quantum field theory, statistical mechanics, and cosmology. They are used to calculate the dynamics of systems and to study the behavior of particles and fields at the quantum level.

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