1
$\begingroup$

I am reading Peskin & Schroeder and I cannot figure out the step between eq. (9.36) and (9.37). They are rewriting the term that appears in the generating functional:

$$\int d^4x \left[\mathcal{L}_0+J\phi\right]=\int d^4x \left[\frac{1}{2}\phi(-\partial^2-m^2+i\epsilon)\phi+ J\phi\right]$$

They introduce $\phi'(x)=\phi(x)-i\int d^4y D_F(x-y)J(y)$. Inserting this I get:

$$\int d^4x \left[\mathcal{L}_0+J\phi\right]=\int d^4x \left[\frac{1}{2}\left(\phi'(x)+i\int d^4y D_F(x-y)J(y)\right)(-\partial^2-m^2+i\epsilon)\left(\phi'(x)+i\int d^4y' D_F(x-y')J(y')\right)+ J(x)\left(\phi'(x)+i\int d^4y D_F(x-y)J(y)\right)\right]~~(*)$$

Now I use

$$(-\partial^2-m^2+i\epsilon)D_F(x-y)=i\delta^{(4)}(x-y).$$

Then $(*)$ reduces to:

$$\int d^4x\left[\frac{1}{2}\phi'(x)\left(-\partial^2-m^2+i\epsilon\right)\phi'(x)+\frac{i}{2}J(x)\int d^4y D_F(x-y)J(y)+\frac{i}{2}\int d^4yD_F(x-y)J(y)(-\partial^2-m^2+i\epsilon)\phi'(x)+\frac{1}{2}J(x)\phi'(x)\right]$$

According to Peskin and Schroeder the last two terms should cancel, but I cannot see that this is the case?

$\endgroup$

1 Answer 1

1
$\begingroup$

Integrate by parts twice (for $x$). For simplicity, I omit $m$ and $\epsilon$: \begin{align} & \frac{i}{2} \int d^4y \, \int d^4x \, D_F(x-y) J(y) (-\partial_x^2) \phi'(x) \\ &= - \frac{i}{2} \int d^4y \, \int d^4x \, D_F(x-y) J(y) \partial_x (\partial_x \phi'(x)) \\ &= \frac{i}{2} \int d^4y \, \int d^4x \,(\partial_x D_F(x-y)) J(y) (\partial_x \phi'(x)) \\ &= - \frac{i}{2} \int d^4y \, \int d^4x \, (\partial_x^2 D_F(x-y)) J(y) \phi'(x) \\ &= \frac{i}{2} \int d^4y \, \int d^4x \, (-\partial_x^2) D_F(x-y)) J(y) \phi'(x) \\ &= \frac{i}{2} \int d^4y \, \int d^4x \, i \mathop{\delta^{(4)}}(x-y) J(y) \phi'(x) \\ &= - \frac{1}{2} \int d^4x \, J(x) \phi'(x) . \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.