Klein-Gordon quantization and SHO analogy I understand that the procedure to quantize Klein-Gordon's field is to manipulate in a such a way to bring up the simple harmonic oscillator behavior of the field. This is done by Fourier transforming the space variable of the field $\phi\left(\vec{x},t\right)$ and plugging back into KG's equation.
The result of this is to obtain a SHO equation of motion for each modes,
$$
\left(\frac{d^2}{dt^2}+\omega_p^2\right)\phi\left(\vec{p},t\right)=0.
$$
The conjugate momentum given by $\pi\left(\vec{p},t\right)=\dot{\phi}\left(\vec{p},t\right)$ is also the Fourier transform of the space variable of the conjugate momentum $\pi\left(\vec{x},t\right)=\dot{\phi}\left(\vec{x},t\right)$.
Now, to quantize the SHO in non-relativistic quantum mechanics, we impose commutation relations. Since what is behaving like an oscillator are the modes, we should impose
$$
\left[\phi\left(\vec{p},t\right),\pi\left(\vec{p}',t\right)\right]=i\hbar\delta\left(\vec{p}-\vec{p}'\right).
$$
But this is not what's done in textbooks. The commutation relations are instead imposed on the actual fields
$$
\left[\phi\left(\vec{x},t\right),\pi\left(\vec{x}',t\right)\right]=i\hbar\delta\left(\vec{x}-\vec{x}'\right),
$$
which in turn implies,
$$
\left[\phi\left(\vec{p},t\right),\pi\left(\vec{p}',t\right)\right]=i\hbar\left(2\pi\right)^3\delta\left(\vec{p}+\vec{p}'\right).
$$
The factor $\left(2\pi\right)^3$ could be included by convention in the first commutation relation. However, the plus sign is what is bugging me. This, of course, also changes the commutation relation between the ladder operators,
$$
\left[a_\vec{p},a^\dagger_{\vec{p}'}\right]=\left(2\pi\right)^3\delta\left(\vec{p}+\vec{p}'\right)
$$
Is it just a convention which doesn't affect the physics or does it have deeper implications?
Thanks
 A: My confusion between the coupled HOs and Klein-Gordon quantization was due to the following.

In coupled HOs we begin, for example, with the Lagrangian
$$
L=\sum_{n=0}^{N+1}\left[\frac{1}{2}m\dot{q}_n^2-\frac{k}{2}\left(q_{n+1}-q_{n}\right)^2\right]
$$
with $q_0=q_{N+1}=0$, and then proceed to uncouple the EOM with the variable transformation $q_n\left(t\right)=\sum_{j=0}^{N+1}Q_j\left(t\right)\sin{n p_j}$ to obtain
$$
L=\sum_{n=0}^{N+1}\left[\frac{1}{2}m\dot{Q}_n^2-\frac{1}{2}m \omega_n^2Q^2_n\right].
$$
In analogy to the SHO we now can impose the commutation relations on the mode coordinates:
$$
\left[Q_m,P_n\right]=i\delta_{mn},
$$
and
$$
\left[Q_m,Q_n\right]=\left[P_m,P_n\right]=0.
$$

In the same way, with KG, we would do a change of variables using the Fourier transform
$$
\phi\left(\vec{x},t\right)=\int\frac{d^3p}{\left(2\pi\right)^3}\phi\left(\vec{p},t\right)e^{-i\vec{p}\cdot\vec{x}},
$$
and get the uncoupled Lagrangian
$$
L=\frac{1}{2}\dot{\phi}^2-\frac{1}{2}\omega^2_p\phi^2.
$$
where now $\phi=\phi\left(\vec{p},t\right)$. Immediate analogy would lead to
$$
\left[\phi\left(\vec{p},t\right),\pi\left(\vec{p}',t\right)\right]=i\left(2\pi\right)^3\delta\left(\vec{p}-\vec{p}'\right).
$$
and this would lead to the inconsistencies I mentioned above.
The problem is that this commutation relation is wrong. The reason being that the original variable transformations from $q\rightarrow Q$ made the $Q$'s real while the modes of the field $\phi\left(\vec{p},t\right)$ are not. Thus the right commutation relations are 
$$
\left[\phi\left(\vec{p},t\right),\pi^\dagger\left(\vec{p}',t\right)\right]=i\left(2\pi\right)^3\delta\left(\vec{p}-\vec{p}'\right).
$$
This corrects the sign of all other commutation relations. Btw, this is supported by the fact that the Lagrangian should be a real function and thus the uncoupled Lagrangian for the field should actually read
$$
L=\frac{1}{2}\dot{\phi}\dot{\phi}^\dagger-\frac{1}{2}\omega^2_p\phi\phi^\dagger,
$$
where $\phi=\phi\left(\vec{p},t\right)$.
