= Solution
Use the <Minkowski metric> $\eta=\operatorname{diag}(1,-1,-1,-1)$ and the free <real scalar field> Lagrangian density $\mathcal L=\tfrac12\partial_\mu\phi\partial^\mu\phi-\tfrac12m^2\phi^2$. Its conjugate <canonical momentum> is $\pi=\dot\phi$. Spatial <translation symmetry> and <Noether's theorem> give the <canonical stress-energy tensor>
$$
T^{\mu\nu}=\partial^\mu\phi\partial^\nu\phi-\eta^{\mu\nu}\mathcal L,
\qquad
\partial_\mu T^{\mu\nu}=(\Box\phi+m^2\phi)\partial^\nu\phi=0.
$$
Vanishing boundary flux then makes the <four-momentum of a free real scalar field> constant. In particular the contravariant spatial <momentum> is
$$
\boxed{\mathbf P=-\int\pi\,\boldsymbol\nabla\phi\,d^3x}.
$$
The minus sign is from $\partial^i=-\partial_i$; it is essential for the following mode calculation.
In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or <normal ordering>. Write $\int_{\mathbf p}=\int d^3p/(2\pi)^3$. The spatial integral is $(2\pi)^3\delta^{(3)}(\mathbf p\pm\mathbf k)$, leaving
$$
\mathbf P=\frac12\int_{\mathbf p}\mathbf p\left(a_{\mathbf p}a_{\mathbf p}^\dagger+a_{\mathbf p}^\dagger a_{\mathbf p}
-a_{-\mathbf p}a_{\mathbf p}-a_{-\mathbf p}^\dagger a_{\mathbf p}^\dagger\right).
$$
Each last term integrates to zero: replace $\mathbf p$ by $-\mathbf p$ and commute the two <annihilation operators> or the two <creation operators>. The mixed terms reorder by the <canonical commutation relation>. Their vacuum contribution is proportional to the integral of the odd vector $\mathbf p$ and vanishes with an inversion-symmetric regulator; <normal ordering> removes it directly. Thus the <normal-ordered free scalar four-momentum> has spatial part
$$
\boxed{\mathbf P=\int_{\mathbf p}\mathbf p\,a_{\mathbf p}^\dagger a_{\mathbf p}}.
$$
These manipulations can be made first with box modes and a symmetric <momentum> cutoff, then used on finite-particle wave packets.
The <commutator derivation identity> gives
$$
[P_i,a_{\mathbf q}^\dagger]=\int_{\mathbf p}p_i a_{\mathbf p}^\dagger[a_{\mathbf p},a_{\mathbf q}^\dagger]
=q_i a_{\mathbf q}^\dagger,
\qquad [P_i,a_{\mathbf q}]=-q_i a_{\mathbf q}.
$$
For $U(\mathbf y)=e^{-i\mathbf P\cdot\mathbf y}$, the <commutator expansion for exponential conjugation> therefore gives the <translation of a scalar creation operator>
$$
\boxed{U(\mathbf y)a_{\mathbf q}^\dagger U(\mathbf y)^\dagger
=e^{-i\mathbf q\cdot\mathbf y}a_{\mathbf q}^\dagger}.
$$
The vacuum has zero <momentum>, so $U|0\rangle=|0\rangle$. Hence the one-particle <momentum eigenstate> obeys $U|\mathbf q\rangle=e^{-i\mathbf q\cdot\mathbf y}|\mathbf q\rangle$: translation preserves its <momentum> and multiplies it by the associated phase.
The <annihilation operator> receives the opposite phase. Substituting both phases into the field expansion gives
$$
\boxed{U(\mathbf y)\phi(\mathbf x)U(\mathbf y)^\dagger=\phi(\mathbf x+\mathbf y)}.
$$
This is the specified conjugation convention for the <spatial translation operator>. On states, the active translation moves a wave packet by $\mathbf y$, giving a position wavefunction $\psi(\mathbf x-\mathbf y)$. Conjugating the field by $U^\dagger$ instead would give the opposite argument shift. Infinitesimally the displayed result is $[P_i,\phi]=i\partial_i\phi$.
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