# Integration by parts in derivation of LSZ reduction formula

This is something that every text book or notes skips to explain in the derivation of the LSZ reduction formula

Suppose we have $$a_{1}^{\dagger} \equiv \int d^{3} k f_{1}(\mathbf{k}) a^{\dagger}(\mathbf{k})\tag{5.6}$$ where $$f_{1}(\mathbf{k}) \propto \exp \left[-\left(\mathbf{k}-\mathbf{k}_{1}\right)^{2} / 4 \sigma^{2}\right]\tag{5.7}$$ is an appropriate wave packet, and $$\sigma$$ is its width in momentum space.

In the derivation of LSZ reduction formula at certain point in this notes Quantum Field Theory by Mark Srednicki at page 50 they have \begin{aligned} &-i \int d^{3} k f_{1}(\mathbf{k}) \int d^{4} x e^{i k x}\left(\partial_{0}^{2}-\overleftarrow{\nabla}^{2}+m^{2}\right) \phi(x) \\ &=-i \int d^{3} k f_{1}(\mathbf{k}) \int d^{4} x e^{i k x}\left(\partial_{0}^{2}-\overrightarrow{\nabla}^{2}+m^{2}\right) \phi(x) \end{aligned}\tag{5.10} He claims that here the wave packet is needed to avoid a surface term but I am not seeing how. I am using this to proof it \begin{align*} \int f(y)g''(y)dy &= f(y)g'(y)| - \int g'(y)f'(y)dy\\ \int g(y)f''(y)dy &= g(y)f'(y)| - \int f'(y)g'(y)dy\quad\text{so}\\ \int f(y)g''(y)dy - \int g(y)f''(y)dy &= f(y)g'(y)| - g(y)f'(y)|.\\ \end{align*}

If we take the $$x$$ part of the Laplacian we would have \begin{align*} \int^{+\infty}_{-\infty}{dr^3 \left(\partial_x^2 e^{-ikr}\right)\phi}=\int^{+\infty}_{-\infty}{dr^3 \ e^{-ikr} \partial_x^2\phi}+\left[e^{-ik_xx}\right]^{\infty}_{-\infty}\int^{+\infty}_{-\infty}{dydze^{-ik_yy-ik_zz}\partial_x\phi}+\int^{+\infty}_{-\infty}{dydzik_xe^{-ikr}\left[\phi\right]^{x=\infty}_{x=-\infty}}. \end{align*}

Now if we assume that $$\lim_{r\rightarrow \infty}{\phi}=\lim_{r\rightarrow -\infty}{\phi}=0$$ we have that $$\int^{+\infty}_{-\infty}{dydzik_xe^{-ikr}\left[\phi\right]^{x=\infty}_{x=-\infty}}=0.$$

$$\left[e^{-ik_xx}\right]^{\infty}_{-\infty}\int^{+\infty}_{-\infty}{dydze^{-ik_yy-ik_zz}\partial_x\phi}.$$

why it is zero?

• How quickly does $\phi_{,x}$ vanish at $\pm\infty$?
– J.G.
Sep 5, 2021 at 10:55
• Crossposted from math.stackexchange.com/q/3887424/11127 Sep 9, 2021 at 14:32

Such terms are set to zero by remembering that the $$S$$-matrix is strictly defined using wave-packets and then using the Riemann-Lebesque Lemma.
Multiply the boundary term by $${\tilde f}(k_x,k_y,k_z)$$ and then integrate over $$\vec{k}$$ so you have get a wave-packet state. The boundary term then simplifies to $$\lim_{x \to \pm\infty} \int dy dz f(x,y,z) \partial_x \phi(x,y,z)$$ where $$f(\vec{x})$$ is the wave-packet.
The wave-packet is localized at $$\vec{x}=\vec{v}t$$ where $$\vec{v}$$ is the velocity of the particle. So, if I take $$x \to \infty$$ keeping $$y,z,t$$ fixed, then $$f \to 0$$ and such terms can be neglected.