The following argument is quite common in the QFT books
1. A term like $ \int \mathrm{d}^4 x \partial_\mu M $ can be transformed to be a surface integration in the space-time infinity.
2. $ M $ is some arithmetic combination of the field. The field should vanish at space-time infinity and thus this surface integration should also vanish.
But, the following examples make me wonder whether the second point is completely correct.
1. On P. 120 of Greiner's Field Quantization, he discusses the possibility of the Dirac Lagrangian being non-hermitian. But, it would be quite obvious that $ \mathcal{L}_\mathrm{D} $ (5.27) and $ \mathcal{L}^\dagger_\mathrm{D} $ (5.3) differ from each other only by a similar surface term $ \partial^\mu \left(\bar\psi\gamma_\mu\psi\right) $ (see also Eq. (1) of the exercise on the next page). So why is this surface term not zero here? As long as those Lagrangians lead to the same action, I think at least from the field-theoretic viewpoint, they should be considered the same.
2. Coleman's QFT Lectures have a very elegant formula of the conserved charge as Eq. (6.59) (on P. 116). But if we start from this and follow the above surface term argument, we would get a zero charge no matter what symmetry we are talking about.
$$
\begin{aligned}
Q&=\int\mathrm{d}^3 \mathbf{x} J^0(\mathbf{x},0) \\
&=\int\mathrm{d}^4x \delta(n\cdot x) n\cdot J(x) \\
&=\int\mathrm{d}^4x\left[ \partial_\mu \theta(n\cdot x) \right] J^\mu(x) \\
&\text{(here is the end of Coleman's deduction and the following is mine)} \\
&=\int\mathrm{d}^4x\partial_\mu \left[ \theta(n\cdot x) J^\mu(x)
\right] - \int\mathrm{d}^4x\left[ \partial_\mu J^\mu(x) \right] \theta(n\cdot x) \\
&=\int\mathrm{d}^4x\partial_\mu \left[ \theta(n\cdot x) J^\mu(x)
\right] \\
&=0\text{, if the surface-term argument is correct.}
\end{aligned}
$$
In the second last line, I use the fact that we have a conserved current $ \partial\cdot J=0 $. Here $ n_\mu=(1,0,0,0) $ is the vector to pick out the time component.
3. The LSZ formula contains many limits to the time infinity when it comes to the definition of in and out states/fields, like
$$
\lim_{t\rightarrow +\infty}\left<\psi\middle|\phi^{\prime f_2}(t)\middle|f_1\right>=\left<\psi\middle|f_1,f_2\right>^{\mathrm{in}}\quad\text{Eq. (14.12) in $\mathit{QFT\;Lectures}$} \\
\lim_{x_0\rightarrow -\infty}\left<b\middle|\hat{\phi}(x)\middle|a\right> = \sqrt{Z} \left<b\middle|\hat{\phi}_{\mathrm{in}}(x)\middle|a\right>\quad\text{Eq. (9.35a) in $\mathit{Field\;Quantization}$} \\
$$
If we require that the field vanishes also at the time infinity, the above definitions are obviously non-sense.
4. During the derivation of a lemma useful for LSZ formula, Eq (14.21) of Coleman's QFT Lectures, the author writes
$$\begin{aligned}
i\int\mathrm{d}^4 x f(x) (\partial^2+\mu^2)A(x)&=i\int\mathrm{d}^4 x f (\partial_0^2 A-\nabla^2 A+\mu^2 A) \\
&=i\int\mathrm{d}^4 x \left[f (\partial_0^2 A)+A(-\nabla^2 A+\mu^2 )f\right] \\
&=i\int\mathrm{d}^4 x \left[f (\partial_0^2 A)-A(\partial_0^2 f)\right]
\end{aligned}$$
Here, $ f $ is a classical solution to the KG equation that behaves well enough for any of our operations and $ A $ is some arbitrary scalar quantum field. In the second line, the author performs two integration-by-part for the space-derivative. If now the integration-by-part and zero-at-infinity argument is also true for the time-derivative, then we can obviously further get
$$\begin{aligned}
i\int\mathrm{d}^4 x f(x) (\partial^2+\mu^2)A(x) &= i\int\mathrm{d}^4 x \left[f (\partial_0^2 A)-A(\partial_0^2 f)\right] \\
&=i\int\mathrm{d}^4 x \left[(\partial_0^2 f)A-A(\partial_0^2 f)\right] = 0
\end{aligned}$$
Since the above deduction require that $ A $ field disappears in the time infinity, this can also be seen from the right-handed side of Eq. (14.21) of Coleman's QFT Lectures.
5. If one looks carefully at the calculus of variation textbook, which I believe is the formal math behind the field theory and concepts like E-L equation. One would find that they would only require that the variation of the field vanish at the time endpoint, but not the field itself. For example, one can see this from Eqs. (1, 2) and surrounding discussions of the second chapter of Gelfand's Calculus of Variations.
6. Let's also check the expansion of the free uncharged scalar field
$$\begin{aligned}
\hat{\varphi}(x) &= \int \frac{\mathrm{d}^3 \mathbf{p}}{\sqrt{2E_{\mathbf{p}}(2\pi)^3}} \left[e^{-ip\cdot x} a_{\mathbf{p}}^\dagger + e^{ip\cdot x } a_{\mathbf{p}} \right] \\
&= e^{-iE_{\mathbf{p}}t} \int \mathrm{d}^3 \mathbf{p} e^{i\mathbf{p}\cdot \mathbf{x}} \frac{a_{\mathbf{p}}^\dagger}{\sqrt{2E_{\mathbf{p}}(2\pi)^3}} + e^{iE_{\mathbf{p}}t} \int \mathrm{d}^3 \mathbf{p} e^{-i\mathbf{p}\cdot \mathbf{x}} \frac{a_{\mathbf{p}}}{\sqrt{2E_{\mathbf{p}}(2\pi)^3}}
\end{aligned}$$
For the space infinity, one can argue that some form of the Riemann Lebesgue lemma is working here and thus we would reach zero when we take $ |\mathbf{x}|\rightarrow\infty $. But there exists no integral for time and then how do we get a zero when $ |t|\rightarrow\infty $?
My guess is that actually it is not required that the field (quantum or classical) should vanish at time-infinity and actually one cannot use the surface-zero argument for the time derivative. But I don't know whether I'm right. If this is right, then we have to conclude that the Dirac Lagrangian is non-hermitian. And then how should we deal with this?
I have to stress that the above discussions don't concern the gauge field since we already know that the gauge field is not required to vanish even at the space-infinity.
-----
References
1. Chen, B. G., Derbes, D., Griffiths, D., Hill, B., Sohn, R., & Ting, Y. Sen. (2018). Lectures of Sidney Coleman on quantum field theory. World Scientific. https://doi.org/10.1080/00107514.2019.1606046
2. Greiner, W., Bromley, D. A., & Reinhardt, J. (2013). Field Quantization. Springer Berlin Heidelberg. https://books.google.com.hk/books?id=C-DVBAAAQBAJ
3. Fomin, S. V., Gelfand, I. M. (2012). Calculus of Variations. Dover Publications. https://www.google.com.hk/books/edition/Calculus_of_Variations/CeC7AQAAQBAJ?hl=zh-CN&gbpv=0
Q&A (4871)
