- #1

ChrisVer

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The Lagrangian is:

$$ L = \frac{1}{2} (\partial \phi)^2 - c \phi - \frac{m^2}{2} \phi^2 + \frac{g}{3!} \phi^3$$

When I want to draw all the 1-loop 1PI diagrams, should I make something like these:

?

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- Thread starter ChrisVer
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- #1

ChrisVer

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The Lagrangian is:

$$ L = \frac{1}{2} (\partial \phi)^2 - c \phi - \frac{m^2}{2} \phi^2 + \frac{g}{3!} \phi^3$$

When I want to draw all the 1-loop 1PI diagrams, should I make something like these:

?

- #2

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P.S. Are you sure about that linear term in [itex]\phi[/itex] it's quite unusual. Also you should be aware that the [itex]\phi^3[/itex] is an unstable theory when quantized since the energy is unbounded from below.

- #3

ChrisVer

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Interesting... how would it contribute to the Feynman diagrams? A tadpole?

- #5

ChrisVer

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I don;t think it would contribute to Feynman diagram... :(

- #6

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Well it can definitely generate a 1-point function. I'm wordering what does that mean...

- #7

ChrisVer

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[itex] [ \partial^2 + m^2 ] \phi = c + \frac{g}{2} \phi^2 [/itex]

- #8

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If there is a linear term you are not expanding the theory around the metastable point ...

- #9

ChrisVer

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Well I am not sure, I'm using what I read in the exercise I was trying to work on...

- #10

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What I was wondering is what is the role of a linear term in the various n-point functions.

- #11

ChrisVer

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What I was wondering is what is the role of a linear term in the various n-point functions.

The constant ##c## can play the role of a constant-everywhere existing current with which the field interacts with, as for example in a term [itex] \phi \partial_\mu J^\mu [/itex]?

To get the n-point function (if there exists such a thing) I think you have to start acting on the generating functional with functional derivatives wrt to the current J?

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ChrisVer

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