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The pullback of the flat Euclidean metric under $x^{\prime }{}^{\mu }=x^{\mu }+ϵ{\omega }^{\mu }(x)$ isFor a conformal transformation it must be a scalar multiple of ${\delta }_{\mu \nu }$. Taking the trace determines that multiple. With $\rho ={\partial }_{\alpha }{\omega }^{\alpha }$, the necessary and sufficient infinitesimal condition is the Conformal Killing equationDifferentiate its $(\alpha ,\beta )$ instance with respect to $x^{\gamma }$, its $(\alpha ,\gamma )$ instance with respect to $x^{\beta }$, and its $(\beta ,\gamma )$ instance with respect to $x^{\alpha }$. Adding the first two and subtracting the last, commuting partial derivatives, leaves $2{\partial }_{\beta }{\partial }_{\gamma }{\omega }_{\alpha }$. Hence flat conformal Killing integrability givesContracting $\beta$ with $\gamma$ gives $\Delta {\omega }_{\alpha }=(2-d){\partial }_{\alpha }\rho /d$. Its divergence is $\Delta \rho =(2-d)\Delta \rho /d$, so $2(d-1)\Delta \rho =0$: thus $\boxed{\Delta \rho =0\text{ for }d>1}$. Taking instead the $\alpha$ divergence of the displayed second-derivative identity givesConsequently $\boxed{{\partial }_{\beta }{\partial }_{\gamma }\rho =0\text{ for }d>2}$.

On a connected flat domain, write $\rho =d\lambda +2db\cdot x$, with constant $\lambda$ and $b$. Integrating the second-derivative identity, and imposing the Conformal Killing equation on its remaining affine part, gives the complete Conformal Killing vector fieldIndeed the quadratic part has precisely the required Hessian matrix, while the symmetric part of the remaining linear term must be $\lambda {\delta }_{\mu \nu }$. The constants generate translations, rotations, dilations and special conformal transformations, respectively. There are $d$, $d(d-1)/2$, $1$ and $d$ independent parameters, so the full conformal group has dimensionHere “full” means local transformations, or global transformations of the conformal compactification of Euclidean space; its connected group is locally $SO(d+1,1)$. Requiring every transformation to be nonsingular on all of uncompactified ${\mathbb{R}}^d$ would exclude the special conformal transformations with finite poles.

For $d\ge 2$, 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 ${\varphi }^{\prime }(x^{\prime })=\Omega (x{)}^{-\Delta }\varphi (x)$. Its infinitesimal fixed-coordinate form isFor $\mathcal{L}=(\partial \varphi {)}^2/2$, contract the Conformal Killing equation with ${\partial }_{\mu }\varphi {\partial }_{\nu }\varphi$. Direct differentiation then givesHere $2(1+\Delta )=d$. Flat conformal Killing integrability gives $\square \sigma =0$, 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 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 $-{\varphi }^2{\omega }^{\prime \prime \prime }/4$, so the free action is invariant under the projective subgroup ${\omega }^{\prime \prime \prime }=0$, rather than arbitrary reparametrizations. The usual field-theory conformal-invariance claim above concerns $d\ge 2$.

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 $\pi =\dot{\varphi }$, and canonical quantization imposes $[\varphi (t,x),\pi (t,y)]=i{\delta }_{\ell }(x-y)$, 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:Strict spatial periodicity excludes a winding term. The canonical commutation relation is equivalent towith ${\alpha }_n^{†}={\alpha }_{-n}$ 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 $i(2\ell {)}^{-1}{\sum }_{n\ne 0}{e}^{2\pi in(x-y)/\ell }$. 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 ${\alpha }_0={\tilde{\alpha }}_0=p/\sqrt{4\pi }$ and the normal ordering quadratic generatorsThese 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 $m>0$,The two contractions of each pair explain the factor $1/2$. Hence the free-boson Virasoro central term yieldsThe tilded modes give a second commuting Virasoro algebra with $\bar{c}=1$. The adjoint condition $L_m^{†}=L_{-m}$ 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 isThe 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.