FIeld configuration with just one particle For a simple real-valued field $$(\partial^2+m^2)\phi=0$$ I would like to get its configuration in explicit form for the case of one particle.
Let's go through the well-known steps:
1) $\phi(x)=\sum_{\vec{p}}{c(t)\exp(i\vec{p}\vec{r})}$
2) $\frac{d^2}{dt^2}c(t)+(\vec{p}^2+m^2)c(t)=0$
3) $c(t) = a_\vec{p}*\exp(-iEt) + b_\vec{p}*\exp(+iEt)$
4) $\phi(x)=\sum_{\vec{p}}{[a_\vec{p}\exp(-iEt) + b_\vec{p}*\exp(+iEt)]\exp(i\vec{p}\vec{r})}$
5) change $\vec{p} \to -\vec{p}$ in second term and apply $\phi(x)=\phi^*(x)$
6) $a^*_\vec{p} = b_{-\vec{p}}$
7) $\phi(x)=\sum_{\vec{p}}{[a_\vec{p}*\exp(-i(px)) + a^*_{\vec{p}}*\exp(+i(px))]}$
Next, I find field's energy and momentum (by computing energy-momentum tensor) and get the well-known result:
$$E = \sum_{\vec{p}}E_{\vec{p}}a_{\vec{p}}a^*_{\vec{p}} = \sum_{\vec{p}}E_{\vec{p}}|a_{\vec{p}}|^2$$
$$\vec{P} = \sum_{\vec{p}}\vec{p}a_{\vec{p}}a^*_{\vec{p}} = \sum_{\vec{p}}\vec{p}|a_{\vec{p}}|^2$$
Next I make the claim: "the field configuration corresponds to one particle in the system with energy $\omega$ and momentum $\vec{k}$". 
Therefore, I make conclusion that all $|a_{\vec{p}}|^2$ are zero excpet for the one with $\vec{p}=\vec{k}$.
And so, field configuration is:
$$\phi(x) = exp(-if)\exp(-i(kx)) + exp(+if)\exp(+i(kx))$$
$$\phi(x) = \exp(-i(kx)-if)) + \exp(+i(kx)+if))$$
$$\phi(x) = 2\cos((kx)+f)$$
where $f$ is a phase factor.
So, indeed, it's REAL-valued. Why do we use COMLEX exponentials in canonical quantization though?
P.S. Please, save my time (and yours) not telling me to read canonical quantization chapter in QFT books :) I've been doing this for last 5 years or so and still hate the whole approach. Therefore, the best answer to this question would be an answer to THIS question. Next questions regarding fundamentals of QFT will follow in separate topics, just starting with this one. Thanks you.
 A: If I got that right - the core of the question is really - why complex numbers. I am also not fond of canonical quantization, in particular because any complex field is representable as two real fields, anyway. 
So, my answer is that complex numbers come in because if you start with a classical field theory then the requirement to be Lorentz invariant means that there has to be a factor isomorphic to the complex numbers in the algebra - because you need something that squares to negative unity to satisfy the relativistic energy equation. 
But, there could be multiple such things, and as such I feel that saying they are the complex numbers is misleading, because they are distinct. The root of negative unity in one is not the same as in the other. This is why Dirac in derivation of electron spin ended up with what are effectively the biquaternians. Witten said as much in stating that quantum field theory is mathematically the special case of classical field theory in which we have roots of negative unity.
A: The main reason is that one generally prefers to work with momentum eigenstates and these are complex exponentials. This is so because we have free fields and the relation $p^2=m^2$ is satisfied anyway. One has
$$
p_\mu=i\partial_\mu
$$
and then
$$
i\partial_\mu\phi_p=p_\mu\phi_p
$$
yields complex solutions that are also solutions of the original equation. This choice can be different as can be seen in the very famous paper by Enrico Fermi on quantum electrodynamics that is anyway a masterpiece. To have a complete orthonormal basis on a Hilbert space, the following normalization condition should hold
$$
\int d^4x\phi_p(x)^*\phi_p'(x)=\delta^4(p-p').
$$
This is so because the momentum operator is part of a $C^*$ algebra of operators acting on a Hilbert space. 
A: Using $c=\hbar=1$ ...
The original motive of Schroedinger was to find a wave analogue of particle dynamics. He actually started with the relativistic energy-momentum equation, and using an analogy to classical wave theory, found the equation (the Klein Gordon) he needed. In doing so, he came up with the idea of replacing $E$ with $i\frac{\partial}{\partial t}$ and $P$ with $-i\frac{\partial}{\partial x}$, and suggested this as a generic process. 
Dirac worked more with abstract operators. The property $[x,p]=i$ is an abstraction of the differential operators used by Schroedinger, but does not really change any of the algebra. 
So in both cases, we see that to get the algebra to work, that is, to produce field equations that project to particle equations we need the root of negative unity in the algebra somewhere.
Now, in dealing with the (nothing specifically to do with quantum) equation $\ddot{x}=-x$ a basis for the solution space over the complex numbers is $\exp(\pm i t)$. This uses simple spectral analysis. But, if we wish to have a real valued solution, we can add up $a \exp(+it) + a^*\exp(-it)$. This has been obtained by using spectral analysis over the complex numbers using a simple uniform method, and then (in a sense) a trick to force a real valued solution by working with a function and its complex conjugate.
Skipping any discussion of "second quantization" (2nd breakfast?) and so on, and going to the field equation $(\partial^2+m^2)\phi=0$, we have a very similar situation with the differential equation. Using complex analysis means that we can use simple uniform methods to solve the equation, but then we use the sum of an expression and its complex conjugate to ensure that we select a real solution.
In other words - it would be possible to recast the argument to use real numbers alone, but doing so would add no intuition to the the process. It would just come out as an awkward rephrasing of statements that are cleaner using the complex numbers. And since we are using complex numbers, we are using the complex exponential.
Note: the definition of the complex exponential is essentially the power series, and does not come from $x^2=x \times x$. 
