Canonical quantisation: How to find the scalar product? I am trying to understand the canonical quantisation procedure. I understood that one takes the classical field equation and replaces the field by an operator Φ which solves the field equations. Then one imposes the commutation relations on Φ and Π.
Since Φ solves the field equation one can write it as 
    Φ = Σk uk ak + uk* bk
.
where uk are solutions to the classical field equation and form a orthonormal basis. My question is now how to find the scalar product to which they are orthonormal? Is there a definite way how to find this scalar product? Or do I need to guess one?
 A: The general answer to your question is given by holomorphic quantization (a good treatment of which can be found in Woodhouse, "Geometric Quantization", especially chapter 5 & section 9.21). The idea is that you can build a scalar product from two ingredients:


*

*a symplectic form $Ω$, which is a non-degenerate antisymmetric real-valued form, related to the Poisson brackets via:
$$df = Ω(X_f,\,\cdot\,)$$
$$\{f,g\} = d_{X_f} g$$
(where $X_f$ denotes the Hamiltonian vector field generated by $f$): this you get for free from the commutation relations;

*a complex structure2, which is a (real-)linear operator $J$ on your phase space (seen as a real vector space) satisfying $J^2 = - \mathbf{1}$: given such a complex structure, any phase space can be turned into a complex vector space, defining the complex scalar multiplication by:
$$\lambda u := \text{Re}(\lambda)\, u + \text{Im}(\lambda)\, J u.$$
If $J$ preserves the symplectic structure, i.e.
$$Ω(J\cdot,J\cdot) = Ω(\,\cdot\,,\,\cdot\,),$$
then we can define a sesquilinear form by:
$$\langle u|v\rangle := 2Ω(u,Jv) + 2iΩ(u,v).$$
[To Be Continued...]

1 Woodhouse focuses mostly on finite-dimensional phase spaces, but he treats the more general case, namely Kähler manifolds, which requires extra integrability conditions on $Ω$ and $J$. The answer above only considers the linear case of phase spaces as "Kähler vector spaces", aka. Hilbert spaces.
2 A complex structure is a special case of what is called a polarization in the context of geometric quantization.
A: Take the lagrangian of the free spin zero complex particle:
$$\mathcal{L}=\partial_\mu\Phi^*\partial^\mu\Phi-m\Phi^*\Phi$$
You can see that this lagrangian is $U(1)$ invariant:
$$\Phi(x)\to\Phi'(x)=e^{-i\theta}\Phi(x)$$
$$\Phi^*(x)\to\Phi^{*'}(x)=e^{+i\theta}\Phi^*(x)$$
Using the Noether theorem, this yield a conserved current as:
$$\delta\theta J^\mu=\frac{\delta\mathcal{L}}{\delta(\partial_\mu\Phi)}\delta\Phi+\delta\Phi^*\frac{\delta\mathcal{L}}{\delta(\partial_\mu\Phi^*)}=\partial^\mu\Phi^*(-i\delta\theta\Phi)+(i\delta\theta\Phi^*)\partial^\mu\Phi$$
equating the first and the last term:
$$J^\mu=i(\partial^\mu\Phi-\partial^\mu\Phi^*):=i\Phi^*\overleftrightarrow{\partial}\Phi$$
And there is also a conserved charge:
$$Q=\int d\vec{x}J^0=\int d\vec{x}\left(i\Phi^*\overleftrightarrow{\partial}\Phi\right)$$
And for the Klein-Gordon field, when you have its expansion in normal modes:
$$\phi(x)=\int d\vec{k}(f_\vec{k}\,u_\vec{k}(x)+f_\vec{k}^*u_\vec{k}(x)^*)$$
with
$$u_\vec{k}(x)=\frac{1}{(2\pi)^{3/2}}\frac{1}{\sqrt{2\omega_\vec{k}}}e^{-i\omega_\vec{k}t+i\vec{k}\cdot\vec{x}}$$
these $u_\vec{k}(x)$ are orthonormal with respect to the hermitian product:
$$(u_\vec{k}(x),u_\vec{p}(x)):=\int d\vec{x}\left(iu_\vec{k}(x)^*\overleftrightarrow{\partial}u_\vec{p}(x)\right)\overbrace{=}^{\text{it can be checked}}\delta(\vec{k}-\vec{p})$$
