Consider the complex Klein-Gordon field (in finite volume $V$), which can be expanded in terms of plane waves as: $$ \phi\left(\mathbf{x},t\right)=\frac{1}{\sqrt{V}}\sum_{\mathbf{k}}\left(A_{\mathbf{k}}\left(t\right)e^{i\mathbf{k}\cdot\mathbf{x}}+B_{\mathbf{k}}^{\ast}\left(t\right)e^{-i\mathbf{k}\cdot\mathbf{x}}\right) $$ where $A_{\mathbf{k}}\left(t\right)=A_{\mathbf{k}}e^{-i\omega_{k}t}$ and $B_{\mathbf{k}}^{\ast}\left(t\right)=B_{\mathbf{k}}^{\ast}e^{i\omega_{k}t}$ are the Fourier coefficients and $\omega_{k}^{2}=\mathbf{k}^{2}c^{2}+\frac{m^{2}c^{4}}{\hbar^{2}}$.
By promoting these coefficients to operators we obtain the quantized version:
$$ \hat{\phi}\left(\mathbf{x},t\right)=\frac{1}{\sqrt{V}}\sum_{\mathbf{k}}\left(\hat{A}_{\mathbf{k}}\left(t\right)e^{i\mathbf{k}\cdot\mathbf{x}}+\hat{B}_{\mathbf{k}}^{\dagger}\left(t\right)e^{-i\mathbf{k}\cdot\mathbf{x}}\right) $$
and the conjugate field:
$$\hat{\pi}\left(\mathbf{x},t\right)=\frac{i}{\sqrt{V}}\sum_{\mathbf{k}}\frac{\omega_{k}}{c^{2}}\left(\hat{A}_{\mathbf{k}}^{\dagger}\left(t\right)e^{-i\mathbf{k}\cdot\mathbf{x}}-\hat{B}_{\mathbf{k}}\left(t\right)e^{i\mathbf{k}\cdot\mathbf{x}}\right)$$
where $\left[\hat{A}_{\mathbf{k}},\hat{A}_{\mathbf{k}^{\prime}}^{\dagger}\right]=\left[\hat{B}_{\mathbf{k}},\hat{B}_{\mathbf{k}^{\prime}}^{\dagger}\right]=\frac{\hbar c^{2}}{2\omega_{k}}\delta_{\mathbf{k},\mathbf{k}^{\prime}}$.
Typically, we impose periodic boundary conditions such that
$$ \mathbf{k}=2\pi\left(\frac{n_{x}}{L_{x}},\frac{n_{y}}{L_{y}},\frac{n_{z}}{L_{z}}\right),\quad n_{x},n_{y},n_{z}\in\mathbb{Z} $$
Exactly the same approach is used in the quantization of the electromagnetic field. There, the vector potential is expanded as:
$$ \displaystyle \mathbf {A} (\mathbf {r} ,t)=\sum _{\mathbf {k} }\sum _{\mu =\pm 1}\left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)e^{-i\mathbf {k} \cdot \mathbf {r} }\right) $$
However, in that Wikipedia entry they write:
"In this summation $\mathbf{k}$ runs over one side, positive or negative"
I was always under the impression that the summation $\sum_{\mathbf{k}}$ (and in the $V\to\infty$ case - integration) in field expansions is always performed over all possible $\mathbf{k}$ modes - both positive and negative (please correct me if I'm wrong). In other words $\sum_{\mathbf{k}}=\sum_{\mathbf{k}\in S}$ where $S$ is a "symmetric" set in the sense that for every $\mathbf{k} \in S$ we have $-\mathbf{k} \in S$. This, in particular, allows one to deduce things like
$$ \left[\hat{\pi}\left(\mathbf{x},t\right),\hat{\pi}^{\dagger}\left(\mathbf{x}^{\prime},t\right)\right]=\dots=\frac{\hbar}{2c^{2}V}\sum_{\mathbf{k}}\underbrace{\omega_{k}\left(-e^{i\mathbf{k}\cdot\left(\mathbf{x}^{\prime}-\mathbf{x}\right)}+e^{i\mathbf{k}\cdot\left(\mathbf{x}-\mathbf{x}^{\prime}\right)}\right)}_{\text{antisymmetric in }\mathbf{k}}=0 $$
or
$$ \hat{\phi}\left(\mathbf{x},t\right)=\frac{1}{\sqrt{V}}\sum_{\mathbf{k}}\left(\hat{A}_{\mathbf{k}}\left(t\right)+\hat{B}_{\mathbf{-k}}^{\dagger}\left(t\right)\right)e^{i\mathbf{k}\cdot\mathbf{x}} $$
(since $\sum_{\mathbf{k}}=\sum_{-\mathbf{k}}$ due to the symmetric nature of the summation over $S$).
Assuming my understanding is correct, how can it be reconciled with the claim in the cited Wikipedia article? If my reasoning is incorrect, I'd like to known why.
Edit (in response to a comment): I don't see how the condition that the field must be real resolves the contradiction. In the case of the free scalar field the requirement $\phi=\phi^{\dagger}$ implies $\hat{A}_{\mathbf{k}}\left(t\right)=\hat{B}_{\mathbf{k}}\left(t\right)$ which means that
$$\hat{\phi}\left(\mathbf{x},t\right)=\sum_{\mathbf{k}}\left(\hat{A}_{\mathbf{k}}\left(t\right)e^{i\mathbf{k}\cdot\mathbf{x}}+\hat{A}_{\mathbf{k}}^{\dagger}\left(t\right)e^{-i\mathbf{k}\cdot\mathbf{x}}\right)=\sum_{\mathbf{k}}\left(\hat{A}_{\mathbf{k}}\left(t\right)+\hat{A}_{-\mathbf{k}}^{\dagger}\left(t\right)\right)e^{i\mathbf{k}\cdot\mathbf{x}}$$
However:
a) I don't see how this leads to $\hat{A}_{\mathbf{k}}^{\dagger}\left(t\right)=\hat{A}_{-\mathbf{k}}\left(t\right)$
b) Even if $\hat{A}_{\mathbf{k}}^{\dagger}\left(t\right)=\hat{A}_{-\mathbf{k}}\left(t\right)$ holds, I fail to see how it implies that
$$ \hat{\phi}\left(\mathbf{x},t\right)=\sum_{\text{all}\,\mathbf{k}}\left(\hat{A}_{\mathbf{k}}\left(t\right)+\hat{A}_{-\mathbf{k}}^{\dagger}\left(t\right)\right)e^{i\mathbf{k}\cdot\mathbf{x}}=\sum_{\mathbf{k}>0}\left(\hat{A}_{\mathbf{k}}\left(t\right)+\hat{A}_{-\mathbf{k}}^{\dagger}\left(t\right)\right)e^{i\mathbf{k}\cdot\mathbf{x}} $$
