I Physical Meaning of the Imaginary Part of a Wave Function

[deuteron]

TL;DR

As far as I've encountered, the imaginary part of functions describing physical phenomena have a physical meaning too. What is the physical meaning of the imaginary part of the wave function for the plane wave?

We know the wave function:
$$ \frac {\partial^2\psi}{\partial t^2}=\frac {\partial^2\psi}{\partial x^2}v^2,$$

where the function $\psi(x,t)=A\ e^{i(kx-\omega t)}$ satisfies the wave function and is used to describe plane waves, which can be written as:

$$ \psi(x,t)=A\ [\cos(kx-\omega t)+i\sin(kx-\omega t)]$$

Here, the real part of the equation alone, $\Re(\psi)=A\cos(kx-\omega t)$, also describes a plane wave, however what is the physical meaning of the imaginary part? I know that in QM, since $|\psi|$ depends on the imaginary part too, it has some physical relevance, but my question is not necessarily limited to quantum mechanics. I have seen other similar questions, but I unfortunately haven't seen a satisfying answer
The motivation behind my question is that so far the complex parts of physical variables I have encountered also have a physical meaning: The complex part of the refraction index corresponds to the absorption, the complex part of the scattering amplitude indicates the existence of inelastic processes; that's why I am curious

If it doesn't have a meaning, why don't we say that $A\sin(kx-\omega t),\ A\cos(kx-\omega t)$ and $A\exp[i(kx-\omega t)]$ all satisfy the wave equation, where we don't know the physical meaning of the exponential one?

 

[vanhees71]

It depends of course on the physics you consider. If your field, $\psi$, is a real quantity you look of course only for real solutions. Since it's a linear differential equation with real coefficients for any complex solution you get two real solutions by taking $\mathrm{Re} \psi$ and $\mathrm{Im} \psi$.

As an initial-value problem the solution is uniquely determined by giving initial values $\psi_0(t=0,x)=f(x)$ and $\partial_t \psi_0(t=0,x)=g(x)$.

Sometimes you have in addition also boundary constraints (e.g., if $\psi$ displacement of a string of length $L$ you have $\psi(t,0)=\psi(t,L)=0$).

Note that the general solution of the (1+1)d wave equation is given by
$$\psi(t,x)=\psi_1(x-vt) + \psi_2(x+vt)$$
with arbitrary functions $\psi_1$ and $\psi_2$, i.e., you have enough "freedom" to fulfill the initial and boundary conditions.