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A Euclidean action and Hamiltonian

  1. Jan 3, 2017 #1

    ShayanJ

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    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
     
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  3. Jan 4, 2017 #2

    dextercioby

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    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".
     
  4. Jan 4, 2017 #3

    Ben Niehoff

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    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.
     
  5. Jan 4, 2017 #4

    ShayanJ

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    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.
     
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