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Use units $\hslash =c=1$ and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, so $p\cdot x=E_{p}t-\mathbf{p}\cdot \mathbf{x}$. The canonical momentum of the real scalar field is $\pi =\dot{\varphi }$, and the Hamiltonian operator isCanonical quantization of a real scalar field promotes $\varphi$ and $\pi$ to operator-valued distributions satisfying the equal-time canonical commutation relationProducts and operator commutators can first be interpreted after spatial smearing. An additive vacuum constant in $H$ does not affect the following equations. In the Heisenberg picture, $\dot{O}=i[H,O]$. Applying the canonical commutation relation to the quadratic Hamiltonian operator, and integrating the gradient term by parts, givesThus the Klein-Gordon equation is an operator equation. The spatial Fourier transform reduces it to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, with $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches, together with the Hermitian adjoint condition ${\varphi }^{†}=\varphi$, give the mode expansion of a free fieldThe measure $d\mu (p)$ is Lorentz invariant; its normalization fixes that of the creation and annihilation operators rather than imposing any extra physical assumption.

For invariant-normalized scalar mode extraction, take the Klein-Gordon inner product of $f_p$ with the field. The positive-frequency contribution has coefficient $E_p+E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$. The negative-frequency contribution has coefficient $E_p-E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}+\mathbf{q})$, so it vanishes because $E_{-p}=E_p$. ConsequentlyTaking the Hermitian adjoint, with the order of the scalar factors immaterial, yieldsEquivalently $a(\mathbf{p})=\int d^{3}x{f}_p^{\ast }(E_p\varphi +i\pi )$ and $a^{†}(\mathbf{q})=\int d^{3}y{f}_q(E_q\varphi -i\pi )$. At a common time their operator commutator therefore isThe same calculation gives $[a(\mathbf{p}),a(\mathbf{q})]=[a^{†}(\mathbf{p}),a^{†}(\mathbf{q})]=0$: the surviving factor is an energy difference supported at opposite spatial momenta. These are the bosonic canonical commutation relations in relativistic normalization of a one-particle state.

Choose the Fock vacuum by $a(\mathbf{p})|0⟩=0$ and $⟨0|0⟩=1$. Only the annihilator-creator product contributes to the Wightman function, givingUsing time ordering and the reverse product for $y^0>x^0$, the Feynman propagator, with the convention $D_F=i{\Delta }_F$ in this calculation, isIn particular the denominator in the spatial measure is $2E_p$ in both terms.

To verify the four-dimensional Fourier transform representation, let $\tau =x^0-y^0$ and integrate the energy variable first:The Feynman i-epsilon prescription puts the positive-energy pole at $E_p-i0$ and the negative-energy pole at $-E_p+i0$. For $\tau >0$, the contour integral closes in the lower half-plane with clockwise orientation. The residue theorem multiplies the residue $i{e}^{-i{E}_p\tau }/(2E_p)$ by $-i$, giving $e^{-i{E}_p\tau }/(2E_p)$. For $\tau <0$, it closes counterclockwise above: multiplying the residue $-i{e}^{i{E}_p\tau }/(2E_p)$ by $i$ gives $e^{i{E}_p\tau }/(2E_p)$. Thus $I(\tau )=e^{-i{E}_p|\tau |}/(2E_p)$. After inserting the spatial exponential, reversing $\mathbf{p}$ in the negative-time term gives exactly the time ordering expression above. HenceThis equality is understood as a distribution limit; the same prescription gives $(\square +m^2)D_F(x-y)=-i{\delta }^4(x-y)$.

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.

The operator commutator is $[A,B]=AB-BA$, acting on a common domain where both products are defined. Use the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, with positions mutually commuting and momenta mutually commuting. On smooth test functions the only nonzero terms in the expansion areThus the orbital angular momentum commutation relations giveThe product rule for a operator commutator givesbecause the last operator expression is symmetric in $i,k$, whereas ${ϵ}_{iwk}$ is antisymmetric. Hence $\boxed{[L^2,L_w]=0}$ for every component.

Write $r=\sqrt{x^2+y^2+z^2}$ and $P=x+y+z$. The differential angular momentum operator $L=-i\hslash x\times \nabla$ annihilates every radial function, so $L_i(P{e}^{-r})=e^{-r}{L}_{i}P$ and the same holds for its square. For example $L_{x}x=0$, $L_y^{2}x={\hslash }^{2}x$ and $L_z^{2}x={\hslash }^{2}x$. Cyclic symmetry gives $L^{2}y=2{\hslash }^{2}y$ and $L^{2}z=2{\hslash }^{2}z$ as well. ThereforeThe wavefunction is an $l=1$ orbital angular momentum state; its nontrivial radial factor does not change its angular eigenvalue. The calculation holds away from the origin and defines the same angular-operator identity in the square-integrable state space.

On a common domain where all products are defined, expansion of the operator commutator givesUse the usual quantum position and momentum operators with $p_j=-i\hslash {\partial }_j$, the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, and wavefunctions with the decay/domain conditions needed for adjoints. The cross-coordinate operators commute, soFor an eigenstate $L\psi =\lambda \psi$, hermiticity gives $\lambda \Vert \psi {\Vert }^2=\overline{\lambda }\Vert \psi {\Vert }^2$, hence its eigenvalue is real.

The product identity gives $[L,x]=i\hslash y$, $[L,y]=-i\hslash x$, $[L,p_x]=i\hslash p_y$, $[L,p_y]=-i\hslash p_x$. Thus the commutators with $x^2,y^2$ cancel, as do those with $p_x^2,p_y^2$, proving $[L,H]=0$. Each oscillator energy eigenspace is finite-dimensional and preserved by $L$; the Hermitian restriction of $L$ can be diagonalized within it. Therefore an energy eigenbasis can be chosen to diagonalize angular momentum as well.

For ${\varphi }_0=e^{-(x^2+y^2)/(2\hslash )}$, differentiation gives $\Delta {\varphi }_0=[(x^2+y^2)/{\hslash }^2-2/\hslash ]{\varphi }_0$, so $H{\varphi }_0=\hslash {\varphi }_0$. Also $[H,x]=-i\hslash p_x$ and $p_x{\varphi }_0=ix{\varphi }_0$, giving $H(x{\varphi }_0)=2\hslash x{\varphi }_0$; the same calculation holds for $y{\varphi }_0$. Directly $L=-i\hslash (x{\partial }_y-y{\partial }_x)$, henceThe simultaneous normalized states areTheir squared unnormalized norm is $\int r^2{e}^{-r^2/\hslash }{d}^{2}x=\pi {\hslash }^2$. Their inner product vanishes because the integral of $(x-iy{)}^2{e}^{-r^2/\hslash }$ is zero: the $xy$ term is odd and the $x^2,y^2$ terms cancel by rotational symmetry.

Put $a=e\mathcal{E}$. The new Hamiltonian hasFor a nonzero applied coupling $a$, a common normalizable eigenstate would be killed by the commutator, so $y\psi =0$. It would have to be supported on the line $y=0$, a set of zero planar measure, and would be the zero vector in $L^2$. Thus there are no nonzero common eigenstates in this case. Merely having a nonzero commutator would not prove that conclusion for arbitrary operators; the multiplication operator here has trivial kernel. If $a=0$, the original simultaneous eigenstates remain.

For the shifted two-dimensional oscillator in a uniform electric field, complete the square:Therefore ${\psi }_1(x^{\prime },y^{\prime })$ and ${\psi }_2(x^{\prime },y^{\prime })$ have the same shifted energyThe modified angular momentum isIt is angular momentum about the displaced centre, commutes with $H^{\prime }$, and has eigenvalues $\pm \hslash$ on the two translated states.

Define ${\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }]$ using the gamma matrix Clifford algebra. Moving ${\gamma }^{\rho }$ through the two factors with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$ proves the displayed operator commutator. For real antisymmetric ${\omega }_{\mu \nu }$ and $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$, it gives $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$. This is the infinitesimal Lorentz covariance identity required by the Dirac equation.