# Is this the correct way to obtain $<f|i>$ term in $\phi^4$ interaction theory? [closed]

Lets first write the expectation value of the fields in the interaction picture;

$$<\Omega|T\phi(x_1)\phi(x_2)\phi(x_1')\phi(x_2')|\Omega>\\=\frac{i\lambda}{4!}<0|T\phi_I(x_1)\phi_I(x_2)\phi_I(x_1')\phi_I(x_2')\int d^4y \phi_I^4(y)|0>\\+\frac{1}{2}(\frac{i\lambda}{4!})^2<0|T\phi_I(x_1)\phi_I(x_2)\phi_I(x_1')\phi_I(x_2')\int d^4y \phi_I^4(y)\int d^4z \phi_I^4(z)|0>+O(\lambda^3)\\ = \frac{i\lambda}{4!}\int d^4y 4\Delta(x_1-y)3\Delta(x_2-y)2\Delta(x_1'-y)\Delta(x_2'-y) \\+\frac{1}{2}(\frac{i\lambda}{4!})^2\int d^4y\int d^4z 16\Delta(y-z)9\Delta(y-z)4\Delta(x_1-y)\{2\Delta(x_2-y)\Delta(x_1'-z)\Delta(x_2'-z)+\Delta(x_2-z)2\Delta(x_1'-y)\Delta(x_2'-z)+\Delta(x_2-z)\Delta(x_1'-z)2\Delta(x_2'-y)\}+O(\lambda^3)\\ = i\lambda\int d^4y \Delta(x_1-y)\Delta(x_2-y)\Delta(x_1'-y)\Delta(x_2'-y) \\+(i\lambda)^2\int d^4y\int d^4z \Delta(y-z)\Delta(y-z)\{\Delta(x_1-y)\Delta(x_2-y)\Delta(x_1'-z)\Delta(x_2'-z)+\Delta(x_1-y)\Delta(x_2-z)\Delta(x_1'-y)\Delta(x_2'-z)+\Delta(x_1-y)\Delta(x_2-z)\Delta(x_1'-z)\Delta(x_2'-y)\}+O(\lambda^3)\\$$

Then we need to write $<f|i>$ such as;

$$<f|i>=(i)^4\int d^4x_1 \int d^4x_2\int d^4x_3\int d^4x_4(-\partial_1^2+m^2)(-\partial_1^2+m^2)(-\partial_1'^2+m^2)(-\partial_2'^2+m^2)<\Omega|T\phi(x_1)\phi(x_2)\phi(x_1')\phi(x_2')|\Omega>$$

My question is that; is the following part right? $$...16\Delta(y-z)9\Delta(y-z)...$$ I mean if I investigate $\phi^5$ theory then this term would contained 3 $\Delta(y-z)$ such as,

$$...25\Delta(y-z)16\Delta(y-z)9\Delta(y-z)...$$

right?

## closed as off-topic by ACuriousMind♦, HDE 226868, Gert, Kyle Kanos, user36790 Nov 3 '15 at 9:08

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