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Free lagrangian propagator

  • Thread starter kelly0303
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Homework Statement


I have the Lagrangian $$L=-\frac{1}{2}\phi\Box \phi-\frac{1}{2}m^2\phi^2$$ and I need to show that the propagator in the momentum space I obtain using this lagrangian (considering no interaction) is the same as if I consider the free Lagrangian to be $$L_{free}=-\frac{1}{2}\phi\Box \phi$$ and treat the mass term as an interaction $$L_{int}= -\frac{1}{2}m^2\phi^2$$

Homework Equations




The Attempt at a Solution


So in the normal case the propagator for a mass m scalar particle is $$\frac{i}{p^2-m^2+i\epsilon}$$. For the other approach I get that the propagator looks like this: $$\frac{1}{p^2+i\epsilon}-im^2(\frac{1}{p^2+i\epsilon})^2-m^4(\frac{1}{p^2+i\epsilon})^3+im^6(\frac{1}{p^2+i\epsilon})^4+...$$ But i am not sure how to show they are equal. Can someone help me?
 

Answers and Replies

  • #2
nrqed
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Homework Statement


I have the Lagrangian $$L=-\frac{1}{2}\phi\Box \phi-\frac{1}{2}m^2\phi^2$$ and I need to show that the propagator in the momentum space I obtain using this lagrangian (considering no interaction) is the same as if I consider the free Lagrangian to be $$L_{free}=-\frac{1}{2}\phi\Box \phi$$ and treat the mass term as an interaction $$L_{int}= -\frac{1}{2}m^2\phi^2$$

Homework Equations




The Attempt at a Solution


So in the normal case the propagator for a mass m scalar particle is $$\frac{i}{p^2-m^2+i\epsilon}$$. For the other approach I get that the propagator looks like this: $$\frac{1}{p^2+i\epsilon}-im^2(\frac{1}{p^2+i\epsilon})^2-m^4(\frac{1}{p^2+i\epsilon})^3+im^6(\frac{1}{p^2+i\epsilon})^4+...$$ But i am not sure how to show they are equal. Can someone help me?
It is a geometric series. Alternatively, you can see this as the Taylor expansion around ##m^2=0## of what expression?
 
  • #3
245
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It is a geometric series. Alternatively, you can see this as the Taylor expansion around ##m^2=0## of what expression?
Thank you for this so I have: $$\frac{1}{p^2+i\epsilon}-im^2(\frac{1}{p^2+i\epsilon})^2-m^4(\frac{1}{p^2+i\epsilon})^3+im^6(\frac{1}{p^2+i\epsilon})^4+...$$ $$\frac{1}{p^2+i\epsilon}(1-im^2\frac{1}{p^2+i\epsilon}-m^4(\frac{1}{p^2+i\epsilon})^2+im^6(\frac{1}{p^2+i\epsilon})^3+...)$$ $$lim_{n \to \infty}\frac{1}{p^2+i\epsilon}\frac{1-(-im^2\frac{1}{p^2+i\epsilon})^n}{1+im^2\frac{1}{p^2+i\epsilon}}$$ $$lim_{n \to \infty}\frac{1-(-im^2\frac{1}{p^2+i\epsilon})^n}{p^2+i\epsilon+im^2}$$ I am not sure from here. How does that term behave as n goes to infinity? Also I have a factor of i with that ##m^2##
 
  • #4
nrqed
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Thank you for this so I have: $$\frac{1}{p^2+i\epsilon}-im^2(\frac{1}{p^2+i\epsilon})^2-m^4(\frac{1}{p^2+i\epsilon})^3+im^6(\frac{1}{p^2+i\epsilon})^4+...$$ $$\frac{1}{p^2+i\epsilon}(1-im^2\frac{1}{p^2+i\epsilon}-m^4(\frac{1}{p^2+i\epsilon})^2+im^6(\frac{1}{p^2+i\epsilon})^3+...)$$ $$lim_{n \to \infty}\frac{1}{p^2+i\epsilon}\frac{1-(-im^2\frac{1}{p^2+i\epsilon})^n}{1+im^2\frac{1}{p^2+i\epsilon}}$$ $$lim_{n \to \infty}\frac{1-(-im^2\frac{1}{p^2+i\epsilon})^n}{p^2+i\epsilon+im^2}$$ I am not sure from here. How does that term behave as n goes to infinity? Also I have a factor of i with that ##m^2##
There is a way to write the series in closed form. Consider

$$ S \equiv 1 + x + x^2 + \ldots $$

Then $$ S-1 = x S$$,
right? Then we can can solve for the sum S.
 

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