Why can we neglect surface terms in Field Theory?

I would like to add that in some cases, even if a term is a total derivative -- so a surface term, by Stokes's theorem -- we can't neglect it in QFT! This will also expand a little on Valter Moretti's point that we need to use some sort of boundary conditions.

For this purpose, let us consider $$S_\theta = \int_M \operatorname{tr} F \wedge F.$$ Here $M$ is 4-dimensional spacetime, $F$ is the gauge field strength for a non-Abelian gauge theory, $F = dA + A \wedge A$, so its components are matrices, that's why we have the trace. You may be unfamiliar with this $\wedge$ and $d$. They're called the "wedge product" and "exterior derivative". In index notation, they correspond to taking the anti-symmetric product; and to taking the derivative, then the totally anti-symmetric part, so: $A \wedge A $ corresponds to $A_\mu A_\nu - A_\nu A_\mu$ and $dA$ to $\partial_\mu A_\nu - \partial_\nu A_\nu$. Note that $A \wedge A \neq 0$ because each component of $A$ is a matrix. You can do all of this with Levi-Civitas, e.g., $$S_\theta = \int d^4 x \, \operatorname{tr} \epsilon_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma} \tag{1} $$ but that leads to writing more indices than I want to.

Anyway, it turns out that $$\operatorname{tr} F\wedge F = d \operatorname{tr} (dA\wedge A + \frac{2}{3} A\wedge A \wedge A)$$ (to show this: $d^2 = 0$, the product rule applies to $d$, and $\operatorname{tr} A^{\wedge 4} = 0$ because of the cyclic property of the trace.) So we have that by Stokes's theorem $$S_\theta = \int_{S^3} \operatorname {tr} (dA\wedge A + \frac{2}{3} A \wedge A \wedge A) $$ where $S^3$ is a 3-sphere at infinity, the "boundary of spacetime". As boundary condition, we should use that $F = 0$ at infinity. Then $$S_\theta = -\frac{1}{3} \int_{S^3} \operatorname A\wedge A \wedge A.$$ Now, if $F = 0$, it is possible to find a gauge transformation such that $A = 0$. So is $S_\theta = 0$? No!

But how? It's topological. $A$ defines a function $S^3 \to G$ where $G$ is the gauge group. Such a function belongs to $\pi_3$, which stands for something called the "third homotopy group", which is a generalization of the concept of winding number. You know how if draw a closed curve -- topologically a circle -- in the plane without crossing through the origin, the number of times it winds around the origin cannot be changed by bending stretching the curve smoothly? The third homotopy group is like that, except it's about how 3-spheres wrap around the target space. (I don't know how to visualize it either!)

If $G = SU(n)$ with $n\ge 2$, then $\pi_3$ is just like the winding number: it's an integer. What this means is that unless the "winding number" of $A$ is $0$, a gauge transformation can't make $A = 0$ everywhere on $S^3$.

So we have shown that $S_\theta \neq 0$ in general, even when we use the boundary condition $F =0$ at infinity. In fact $S_\theta = \theta n $ where $n$ is the "winding number" and $\theta$ is some constant. But why should we care? Because it is a surface term, and variations are taking with the boundary condition that the variation vanish on the boundary, $S_\theta$ makes no contribution to the Euler-Lagrange field equations. But this is quantum field theory, and I really want the path integral $$\begin{align} \mathcal Z &= \int D[A, \psi, \ldots]\, \exp(-i(S_0[A, \psi, \ldots] + S_\theta)) \\ & = \sum_n \int D[A, \psi, \ldots]\, \exp(-i S_0[A, \psi, \ldots]) e^{-in\theta} \end{align} $$ where on the second line, each integral is only over $A$ such that the winding number is $n$. So $S_\theta$ determines interference terms between different paths. This means it can -- and does -- have physical effects in quantum field theory. For example, from (1) you can see that because there is a Levi-Civita, $S_\theta$ violates parity symmetry.

I think these topological arguments with winding numbers and other similar concepts are really cool. So does the Nobel committee, because this year's prize was about things like this. If you want to learn more, some keywords to look for are instantons, topological quantum field theory (TQFT), theta terms, strong $CP$ problem, axions...

Without some particular hypotheses on $J^\mu$ the statement is simply false. There are so many elementary examples...Take $$J^\mu = x^\mu$$ in Minkowskian coordinates in Minkowski spacetime, for instance.

It is sufficient that $\partial_\mu J^\mu \geq 0$ with $\partial_\mu J^\mu >0$ in a region with non-vanishing four-volume to make false the statement. I guess your teacher is thinking of some very specific cases.

Perhaps your teacher is referring to currents generated in some bounded region $B$ of the spacetime. In this case if the current is associated to some hyperbolic equation, as those of QFT, the support of $J$ is confined in the causal cone $J(B)$ emanated from $B$ in the past and in the future of $B$.

Choosing a sufficiently large cylinder parallel to the temporal axis of a Minkowskian reference frame and completely including this double cone, you may neglect the contribution of the surface integral over the lateral surface of the cylinder.

However I cannot see any compelling reason to prove that the integration over the (spacelike) bases $\Sigma_{\pm T}$ vanishes for $T \to +\infty$. There is no reason to think it. $J$ vanishes in the remote future/past of $B$ but we cannot say that it vanishes so rapidly to make vanishing the mentioned integrals over $\Sigma_{\pm T}$.

To corroborate this idea against the supposed proof of the statement, think of a conserved current, i.e., $\partial_\mu J^\mu =0$ associated to some causal quantum field. In this case $\int \partial_\mu J^\mu d^4x =0$ as wanted, but the reason is not the one you are adducing.

In this case each integral over the mentioned bases $\Sigma_{\pm T}$ of the cylinder does not vanish for $T \to +\infty$.

What it is true is that the sum of them is zero, because they are evaluated referring to opposite timelike normal unit vectors.

However, each integral separately equals (up to sign) the conserved charge $$Q = \int_{\Sigma_t} J^0 d^3x \neq 0$$ associated with the current.

As an example, consider $J_\mu = \phi \partial_\mu \psi - \psi \partial_\mu \phi$ with $\partial_\mu \partial^\mu \phi + f\phi = \partial_\mu \partial^\mu \psi + f \psi=0$ (for every fixed function $f$). It is easy to chose initial conditions of $\psi$ and $\phi$ over $\Sigma_0$ to obtain $Q \neq 0$. In this case, however, $\phi(t,x)$ and $\psi(t,x)$ vanish as $t\to +\infty$ for every fixed $x$.

My conclusion is that, even in QFT, the statement is generally false: It is better to assume $\partial_\mu J^\mu =0$ from scratch and these currents are not directly associated to (free) quantum fields.