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What is going on with this generating functional? (QFT)

Physics Asked on December 19, 2020

I am reading Peskin and Schroeder’s chapter on functional methods and they compute the following correlation function:
begin{equation*}
begin{split}
langle 0| Tphi_1phi_2phi_1phi_3 |0rangle
&=
frac{delta}{delta J_1 }
frac{delta}{delta J_2 }
frac{delta}{delta J_3 }[-J_xD_{x4}]e^{-frac{1}{2}J_xD_{xy}J_y}bigg|_{J=0}
&=
frac{delta}{delta J_1 }
frac{delta}{delta J_2 }
[-D_{34}+J_xD_{x4}J_yD_{y3}]
e^{-frac{1}{2}J_xD_{xy}J_y}bigg|_{J=0}
&=
frac{delta}{delta J_1 }
[-D_{34}J_xD_{x2}+D_{24}J_yD_{y3}+J_xD_{x4}D_{23}]
e^{-frac{1}{2}J_xD_{xy}J_y} bigg|_{J=0}
&=D_{34}D_{12}+D_{24}D_{13}+D_{14}D_{23}
end{split}
end{equation*}

However, I don’t understand why in the third line the term
begin{equation*}
J_xD_{x4}J_yD_{y3} frac{delta}{delta J_2} e^{-frac{1}{2}J_xD_{xy}J_y}
=
J_xD_{x4}J_yD_{y3}(-J_zD_{z2})
end{equation*}

doesn’t appear. I would expect this term from the product rule acting on $J_xD_{x4}J_yD_{y3}e^{-frac{1}{2}J_xD_{xy}J_y}$ in the second line.

What am I missing?

Note on notation: I am using Peskin’s short notation where:
$$ -frac{1}{2}J_xD_{xy}J_y=-frac{1}{2} int d^4x’,d^4y’, J(x’) D_F(x’-y’)J(y’) $$

One Answer

Since all the $J_u$ are set to zero at the end, any term in which they are still present after the final functional derivative has acted will die anyway.

That's why Peskin and Schroeder elide them - but of course this term arises due to the product rule (and is included in the derivation in e.g. Schwartz)

Correct answer by Nihar Karve on December 19, 2020

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