Is the sign of an amplitude in QFT meaningful? Is the sign of an amplitude in QFT physical? On pages 121-126 of Peskin and Shroeder, they use the signs of amplitudes to determine whether an interaction is attractive or repulsive. Whereas in Schwartz's textbook he states on page 229 that the 

"overall sign of the sum of the matrix elements is an unphysical phase...". 

Who is correct?
 A: In Feynman diagram computations, it is certainly important to keep track of the factors of $i$ and $-1$ when dealing with multiple diagrams.
In evaluating the scattering amplitude (e.g. to then compute a decay width), one must write out all the diagrams of the process $A \to B$ to a given order, allowed by the Feynman rules.
If say, we have $\mathcal M_1$ and $\mathcal M_2$ contributing to a process, it would certainly make a difference whether we include $\mathcal M_1 - \mathcal M_2$ or $\mathcal M_1 + \mathcal M_2$ when finding $|\mathcal M_{\mathrm{total}}|^2$ for the computation of interest.
Even when dealing with one diagram, it may also be important if that is the only diagram required to derive a counter-term. The coefficients must match exactly for the right cancellations to occur.
A: Suppose you send a beam of light towards a slab of glass. Part of the beam travels through the glass at a slower speed, then exits and continues in the same direction as before.
Because the glass affects the speed of the light, it produces a phase shift, and the fact that glass slows down light rather than speeds it up is encoded in the sign of the phase shift. But this is a phase shift, not a phase. In general, there is no such thing as "the" phase of a light beam, because the point in space at which you evaluate the phase is arbitrary. The effect of the glass is encoded in a relative phase: it means that the phase of the exiting light beam is different, at a fixed spatial reference point, relative to what it would have been had the glass not been there.
Similarly, the absolute phases of $S$-matrix elements are arbitrary because they can be adjusted by changing the conventions for defining "in" and "out" states. However, once these conventions are fixed, the change in phase due to adding an interaction is meaningful.
A: Eq. 4.73 of Peskin and Schroder fixes the definition of the S-matrix elements
\begin{equation}
\langle \mathbf{p}_1 \cdots \mathbf{p}_n | i T | \mathbf{k}_{\mathcal{A}} \mathbf{k}_{\mathcal{B}} \rangle = (2\pi)^4 \delta^{(4)}(k_{\mathcal{A}}+k_{\mathcal{B}}-\sum p_f) \cdot i \mathcal{M}(k_{\mathcal{A}},k_{\mathcal{B}}\rightarrow p_f)
\end{equation}
where the S-matrix $S$ is related to $T$ by $S=1+iT$.
The quantity $i\mathcal{M}$ for 2 distinguishable fermions interacting via a massive scalar field $\phi$ in the non-relativistic limit is worked out in Eq 4.123
\begin{equation}
i \mathcal{M} = \frac{i g^2}{|\mathbf{p}'-\mathbf{p}|^2+m_\phi^2}
\end{equation}
This is then compared with the S-matrix element for a non-relativistic particle scattering off of a potential in the Born approximation in Eq 4.124
\begin{equation}
\langle \mathbf{p}' | iT | \mathbf{p} \rangle = -i \tilde{V}(\mathbf{q}) (2\pi)\delta(E_{\mathbf{p}'}-E_{\mathbf{p}}), \ \ (\mathbf{q}=\mathbf{p}'-\mathbf{p})
\end{equation}
Now here's the punchline. You can change the overall, but not the relative, phase of the states. But here we are doing a comparison between the same amplitude computed two different ways. If you perform an overall change of phase of $|\mathbf{p}\rangle$ in computing $i \mathcal{M}$, in order to be consistent you must also change the phase of $|\mathbf{p}\rangle$ in $\langle \mathbf{p}' | iT | \mathbf{p} \rangle$. Therefore, it is meaningful to compare the relative sign of these two methods of computing the amplitude and fix the sign of $\tilde{V}(\mathbf{q})$ and then $V(\mathbf{x})$.
Note: I consider this answer to just be adding mathematical details to knzhou's answer, as substantively we are saying the same thing.
A: The phase of a quantum-mechanical probability amplitude is completely arbitrary – this is in fact one of the fundamental postulates of quantum mechanics.
A quantum-field-theoretic scattering amplitude is no exception. You can choose the phase at will, but you have to be self-consistent: relative phases do matter.
In the reference, Peskin and Schröder compute the non-relativistic potential predicted by QFT by comparing the relativistic amplitude with the non-relativistic amplitude. The former is given by a certain Feynman diagram, and the latter by the Born approximation. Both amplitudes carry an arbitrary phase, but it is the same phase: if you want to use a different convention for one of them, you better adjust the phase of the other consistently. The relative phase is not, again, arbitrary.
The book chooses a convention from the get-go, and sticks to it; they are always self-consistent with their choice (or at least one hopes so). The sign in the non-relativistic potential is measurable, and it does not depend on conventions. Had they chosen a different convention for the individual phases, the individual amplitudes would have changed, but the end result would not: the sign in the potential is convention-independent. To reiterate: the book computes the potential by comparing two amplitudes; these both carry an overall arbitrary phase, but the relative phase is fixed.
Signs are always tricky. They are sometimes the hardest part of the computation, and often the most important part of it. Politzer, Gross and Wilczek got a certain prize for it. You want to compute the sign of the potential yourself? Choose a convention, and be very careful that you are always respecting it.
A: very easy question forget about QFT,
assume you have
an Hamiltonian $H$ with
$$H=E_n|n\rangle$$
now you have some state
$$|\alpha\rangle=\frac{1}{\sqrt{2}}(|1\rangle+|2\rangle)$$
and a state $$|\beta\rangle=\frac{1}{\sqrt{2}}(|1\rangle-|2\rangle)$$
those they are orthogonal
$$\langle\beta|\alpha\rangle$$ is zero
now we have
$$\langle 1|\alpha\rangle =\frac{1}{\sqrt{2}}$$
$$\langle 2|\alpha\rangle =\frac{1}{\sqrt{2}}$$
if we just change the phase of the former such that
$$\langle 1|\alpha\rangle =-\frac{1}{\sqrt{2}}$$
keeping
$$\langle 2|\alpha\rangle =\frac{1}{\sqrt{2}}$$
we would have
$$|\alpha\rangle=\frac{1}{\sqrt{2}}(-|1\rangle+|2\rangle)$$
and
$$\langle\beta|\alpha\rangle=1$$
thus you can't change the phase of amplitude blindly you should keep track and be consistent.
