= Solution
Use $x^\mu=(t,x)$ and <Minkowski metric> $\eta^{\mu\nu}=\operatorname{diag}(1,-1)$, with the wave speed equal to one. Regarding $\phi$ as a <scalar field>, the symmetric <canonical stress-energy tensor> is
$$
T^{\mu\nu}=\partial^\mu\phi\,\partial^\nu\phi-\eta^{\mu\nu}\mathcal L
=\begin{pmatrix}\mathcal E&p\\p&\mathcal E\end{pmatrix},\qquad
\mathcal E=\frac12(\phi_t^2+\phi_x^2),\quad p=-\phi_t\phi_x.
$$
It obeys $\partial_\mu T^{\mu\nu}=0$ on the <wave equation>. For a <scalar field> the intrinsic spin current vanishes. The total orbital <Lorentz current> and its integrated angular-momentum tensor are
$$
\boxed{\mathcal J^{\lambda\mu\nu}=x^\mu T^{\lambda\nu}-x^\nu T^{\lambda\mu},\qquad
J^{\mu\nu}(t)=\int_{\mathbb R}\mathcal J^{0\mu\nu}\,dx.}
$$
Because there is only one spatial dimension, its only independent possibly nonzero component is
$$
\boxed{J^{01}=\int_{\mathbb R}(t p-x\mathcal E)\,dx=tP-\int_{\mathbb R}x\mathcal E\,dx,\qquad J^{10}=-J^{01},}
$$
where $P=\int p\,dx$. The corresponding flux is $\mathcal J^{101}=t\mathcal E-xp$. Symmetry and conservation of $T$ give
$$
\partial_\lambda\mathcal J^{\lambda\mu\nu}=T^{\mu\nu}-T^{\nu\mu}=0.
$$
Thus $J^{01}$ is conserved provided the boundary flux vanishes and the integrals exist. For example, localized wave packets on an infinite line have this property. One can also check directly that $d\int x\mathcal E\,dx/dt=P$, so the two terms in $J^{01}$ have cancelling derivatives.
The significance is the <boost charge of a one-dimensional scalar field>. There is no independent spatial rotation or ordinary spatial angular-momentum component in one dimension. Instead, the action has the mathematical <Lorentz invariance> of a massless wave field with invariant speed $c_s=1$, and $J^{01}$ generates its boost. Its conservation says that the center of energy moves uniformly. This is a symmetry of the ideal wave-field model, rather than an assertion of fundamental vacuum Lorentz symmetry for a material rod. Fixed boundaries can break it, and a general solution with nonvanishing boundary flux need not conserve the integrated charge.
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