= Solution
For $d\ge2$, let $\omega$ be a <Conformal Killing vector field>, put $\sigma=\partial\cdot\omega/d$, and assign the massless <scalar field> <scaling dimension> $\Delta=(d-2)/2$. In a finite <conformal transformation> with metric scale $\Omega^2$, the transformation law is $\phi'(x')=\Omega(x)^{-\Delta}\phi(x)$. Its infinitesimal fixed-coordinate form is
$$
\delta_0\phi=-\omega^\mu\partial_\mu\phi-\Delta\sigma\phi.
$$
For $\mathcal L=(\partial\phi)^2/2$, contract the <Conformal Killing equation> with $\partial_\mu\phi\partial_\nu\phi$. Direct differentiation then gives
$$
\begin{aligned}
\delta_0\mathcal L
&=-\omega\cdot\partial\mathcal L-(1+\Delta)\sigma(\partial\phi)^2-\frac\Delta2\partial_\mu(\phi^2)\partial^\mu\sigma\\
&=-\partial_\mu(\omega^\mu\mathcal L)-\frac\Delta2\partial_\mu(\phi^2\partial^\mu\sigma)+\frac\Delta2\phi^2\Box\sigma.
\end{aligned}
$$
Here $2(1+\Delta)=d$. <Flat conformal Killing integrability> gives $\Box\sigma=0$, with the Lorentzian version obtained by replacing the flat metric consistently. Thus the action variation is a boundary integral, vanishing for suitable <boundary conditions>. This proves \b[conformal invariance with field dimension $(d-2)/2$], without using the field equation. The connected finite symmetry follows by integrating the infinitesimal generators. Equivalently its <conformal currents> follow from the <improved stress-energy tensor of a free massless scalar>.
In one dimension the metric conformality condition is vacuous. The same action-variation calculation contains $-\phi^2\omega'''/4$, so the free action is invariant under the projective subgroup $\omega'''=0$, rather than arbitrary reparametrizations. The usual field-theory conformal-invariance claim above concerns $d\ge2$.
In $d=2$, $\Delta=0$, so the field transforms simply as a scalar. There is no improvement term and no derivative of the conformal scale in the variation. All local independent reparametrizations of the two null coordinates preserve the action; after Euclidean continuation they are holomorphic and antiholomorphic maps. The conformal algebra is therefore infinite-dimensional, in contrast to $d>2$.
Now use Lorentzian coordinates with spatial period $\ell$. The <canonical momentum> is $\pi=\dot\phi$, and <canonical quantization> imposes $[\phi(t,x),\pi(t,y)]=i\delta_\ell(x-y)$, with the periodic delta distribution $\delta_\ell(s)=\ell^{-1}\sum_{n\in\mathbb Z}e^{2\pi ins/\ell}$. Solving the wave equation and imposing reality gives a convenient oscillator normalization:
$$
\phi(t,x)=q+\frac{pt}{\ell}+\frac{i}{\sqrt{4\pi}}\sum_{n\ne0}\frac1n\left[\alpha_ne^{-2\pi in(t-x)/\ell}+\widetilde\alpha_ne^{-2\pi in(t+x)/\ell}\right].
$$
Strict spatial periodicity excludes a winding term. The <canonical commutation relation> is equivalent to
$$
[q,p]=i,\qquad [\alpha_m,\alpha_n]=m\delta_{m+n,0},\qquad
[\widetilde\alpha_m,\widetilde\alpha_n]=m\delta_{m+n,0},\qquad [\alpha_m,\widetilde\alpha_n]=0,
$$
with $\alpha_n^\dagger=\alpha_{-n}$ and the corresponding tilded relation. Both oscillator families commute with $q,p$. To verify the normalization, the zero-mode commutator supplies $i/\ell$, while each oscillator family supplies $i(2\ell)^{-1}\sum_{n\ne0}e^{2\pi in(x-y)/\ell}$. Their sum is exactly $i\delta_\ell(x-y)$. The positive-index modes are <annihilation operators> and the negative-index modes are <creation operators>, giving two independent <bosonic Fock spaces> together with the free-particle zero mode.
Define $\alpha_0=\widetilde\alpha_0=p/\sqrt{4\pi}$ and the <normal ordering> quadratic generators
$$
L_m=\frac12\sum_{n\in\mathbb Z}:\alpha_{m-n}\alpha_n:,
\qquad \widetilde L_m=\frac12\sum_{n\in\mathbb Z}:\widetilde\alpha_{m-n}\widetilde\alpha_n:.
$$
These are the Fourier modes of the two chiral components of the <stress-energy tensor>. The oscillator <operator commutators> immediately give $[L_m,\alpha_n]=-n\alpha_{m+n}$. Applying this to both factors of $L_n$ gives $(m-n)L_{m+n}$, except for a scalar term when $m+n=0$, produced by reordering creators and annihilators. Its value is independent of the momentum sector. Evaluate it in the formal zero-momentum oscillator vacuum. For $m>0$,
$$
L_{-m}|0\rangle=\frac12\sum_{r=1}^{m-1}\alpha_{-r}\alpha_{-(m-r)}|0\rangle,
\qquad
\langle0|L_mL_{-m}|0\rangle=\frac12\sum_{r=1}^{m-1}r(m-r)=\frac{m^3-m}{12}.
$$
The two contractions of each pair explain the factor $1/2$. Hence the <free-boson Virasoro central term> yields
$$
\boxed{[L_m,L_n]=(m-n)L_{m+n}+\frac{m^3-m}{12}\delta_{m+n,0}},\qquad \boxed{c=1}.
$$
The tilded modes give a second commuting <Virasoro algebra> with $\bar c=1$. The adjoint condition $L_m^\dagger=L_{-m}$ and the positive oscillator norms make this a <unitary highest-weight Virasoro representation> in each momentum sector. In a fixed momentum sector, $L_0=p^2/(8\pi)+\sum_{n>0}\alpha_{-n}\alpha_n$, so the oscillator ground state is a highest-weight state of weight $p^2/(8\pi)$ in either sector. For a noncompact field the zero mode is a free particle: definite momentum ground states are generalized states, rather than a normalizable zero-mode vacuum. The oscillator representation and central term remain well defined. With the standard cylinder vacuum-energy subtraction, the Hamiltonian is
$$
\boxed{H=\frac{2\pi}{\ell}\left(L_0+\widetilde L_0-\frac1{12}\right)}.
$$
The constant is the <Casimir energy of a two-dimensional conformal field theory>; omitting it amounts to a different additive Hamiltonian convention, not a change of the plane <Virasoro algebra>.
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