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I am reading Quantum Field Theory, by Mark Srednicki. On page 50, he derives the scattering amplitude for two spinless particles from $\pm \infty$. I am stuck on what I believe is a simple thing. His steps are the following:

\begin{align} a_1^{\dagger}(+\infty)-a_1^{\dagger}(-\infty) &= \int_{-\infty}^{\infty}dt \partial_0 a_1^{\dagger}(t) \\ &= -i\int d^3k f_1(\mathbf{k})\int d^4 x \partial_0 \left( e^{ikx}\overset\leftrightarrow\partial_0 \psi(x) \right) \\ &= -i\int d^3k f_1(\mathbf{k})\int d^4 x e^{ikx}\left( \partial_0^2+\omega^2 \right)\psi(x) \\ &= -i\int d^3k f_1(\mathbf{k})\int d^4 x e^{ikx}\left( \partial_0^2+\mathbf{k}^2 + m^2 \right)\psi(x) \\ &= -i\int d^3k f_1(\mathbf{k})\int d^4 x e^{ikx}\left( \partial_0^2-\overset\leftarrow\nabla^2 + m^2 \right)\psi(x) \\ &= -i\int d^3k f_1(\mathbf{k})\int d^4 x e^{ikx}\left( \partial_0^2-\overset\rightarrow\nabla^2 + m^2 \right)\psi(x) \\ &= -i\int d^3k f_1(\mathbf{k})\int d^4 x e^{ikx}\left( -\partial^2 + m^2 \right)\psi(x) \\ \end{align}

And here is the explanation for these steps. What I don't understand is highlighted in bold.

The first equality is just the fundamental theorem of calculus. To get the second, we substituted the definition of $a^{\dagger}_1(t)$, and combined the $d^3x$ from this definition with the $dt$ to get $d^4x$. The third comes from straightforward evaluation of the time derivatives. The fourth uses $ω^2 = \mathbf{k}^2+m^2$. The fifth writes $k^2$ as $−∇^2$ acting on $e^{ik·x}$. The sixth uses integration by parts to move the $∇^2$ onto the field $\psi(x)$; here the wave packet is needed to avoid a surface term. The seventh simply identifies $∂^2_0 − ∇^2$ as $−∂^2$.

What is the difference between the del with the right arrow and the left arrow? And how does integration by parts change one to the other?

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Before integration by parts: $$ -e^{ikx} (\overleftarrow{\nabla}^2) \psi(x) = -(\nabla^2 e^{ikx}) \psi(x). $$

Integrate by parts once: $$ +(\nabla e^{ikx}) \cdot (\nabla \psi(x)) + \text{boundary terms.} $$

And a second time: $$ -e^{ikx}\, \nabla^2 \psi(x) + \text{boundary terms.} $$

The "left arrow" indicates that the derivative acts on what is left of it (as opposed to the usual notation where derivatives always act on the right). The "right arrow" is just a vector-arrow.

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  • $\begingroup$ Adding to this, the left-arrow del can also be represented as a covector of derivatives, rather than a vector, which is why it acts to the left. $\endgroup$ – probably_someone Jul 14 '17 at 17:41

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