In the section 2.1 of QFT book written by Peskin and Schroeder, the amplitudes for a free particle to propagate from $x_0$ to $x$ are obtained. In a relativistic theory, the amplitude is obtained by using relation $E=\sqrt{p^2+m^2}$, which reads $$U(t)=\frac{1}{2\pi^2}\frac{(i t)m^2}{|x-x_0|^2-t^2}K_2(|m|\sqrt{|x-x_0|^2-t^2}),$$ where $K_\nu(z)$ is modified Bessel functions. It can be found that this amplitude is imaginary because $K_\nu(x\rightarrow\infty)\sim\sqrt{\pi/(2 x)}e^{-x}$ for large space-like interval. Thus, my question is that wether or not the imaginary amplitude can be understood physically.
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$\begingroup$ Amplitudes $(\mathcal{A})$ can be complex no? and it's the amplitude squared $|\mathcal{A}|^2$ that's interpreted as a probability? $\endgroup$– snultyCommented Jan 20, 2017 at 16:25
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Amplitudes are in general complex and even purely imaginary. This is absolutely fine since the amplitude itself is not a physical observable. Usually, the square modulus of the amplitude is taken to be the probability of probability density.
If we think of a system in superposition, then we can think of the amplitude as the amount that the wavefunction of the system favours one particular state, perhaps after we have acted on it with some operator $\mathcal{O}$.
That being said, since it's not a physical obervable, in a general quantum system this very much depends on your interpretation of quantum mechanics: the only sensible thing to do is to interpret the probability you get from the amplitude.
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$\begingroup$ No, if they are purely imaginary then the square moduli are positive (They are always positive) since you define: $|A|^2 = AA^*$. For a purely imaginary amplitude, say $i\times s$, the square modulus is $is\times-is = s^2$ $\endgroup$– AkobenCommented Jan 23, 2017 at 16:13