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Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

Use natural units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Legendre transform in mechanics givesFor canonical quantization of a real scalar field, impose the equal-time canonical commutation relationsThe Heisenberg picture equation $\dot{O}=i[H,O]$ then gives $\dot{\varphi }=\pi$. For the other equation, the momentum term commutes with $\pi$; commuting the gradient term past $\pi$ and integrating the derivative of the Dirac delta function by parts gives $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$. ConsequentlyThe Fourier transform reduces this Klein-Gordon equation to oscillators of frequency $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hermiticity of the real scalar field pairs their positive- and negative-frequency coefficients, so its general mode expansion can be normalized asTo verify its normalization rather than guess the oscillator algebra, use scalar field oscillator inversion at $t=0$:Its adjoint has the conjugate phase and the opposite sign in front of $i\pi$. The two mixed field-momentum commutators giveThe Fourier representation of the Dirac delta function sets the energies equal in the last step. Similarly, $[a(\mathbf{p}),a(\mathbf{q})]$ has a factor $E_{\mathbf{q}}-E_{\mathbf{p}}$ multiplying ${\delta }^3(\mathbf{p}+\mathbf{q})$ and vanishes; the two-creator commutator vanishes as well.

Substituting the expansion in $H$ and integrating the spatial phases, the two-annihilator and two-creator coefficients cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The free real scalar Hamiltonian in oscillator variables isIt differs from $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}a/(2\pi {)}^3$ by its constant vacuum energy; normal ordering removes that constant. Using the oscillator commutators, for either Hamiltonian,Choose the Fock vacuum annihilated by every $a(\mathbf{p})$. Each creation operator adds a particle of energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$; its annihilation operator removes one. Their mutual commutativity gives bosonic Fock space and symmetric many-particle states. Thus the scalar mass-shell frequency has a particle interpretation, rather than merely describing a classical oscillation.

For $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$, vacuum contraction and time ordering give the Feynman propagatorIn the scalar Feynman propagator pole prescription, the integral of $i{e}^{-i{p}^{0}t}/((p^0{)}^2-E_{\mathbf{p}}^2+i0)$ has poles at $E_{\mathbf{p}}-i0$ and $-E_{\mathbf{p}}+i0$. Close clockwise below for $t>0$ and counterclockwise above for $t<0$. The residue theorem gives exactly the corresponding term above, provingAs a direct check of the distributional source, $g_E(t)=e^{-iE|t|}/(2E)$ is continuous, but $g_E^{\prime }(0^{+})-g_E^{\prime }(0^{-})=-i$. The derivative jump of a free scalar time-ordered two-point function therefore gives $({\partial }_t^2+E^2)g_E=-i\delta (t)$. Spatial Fourier inversion yields

With the mode normalization given, the oscillator canonical commutation relations are $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, with the other two commutators zero. Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p^0=E_{\mathbf{p}}$. The convention compatible with the requested numerator iswithout an additional factor of $i$ outside the vacuum expectation value. Put $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. The Fock vacuum is annihilated by $a_{\mathbf{p}}$, so only $a{a}^{†}$ contributes to the two Wightman functions. Consequently,In the second term we changed $\mathbf{p}\mapsto -\mathbf{p}$ to give the same spatial exponential.

Now perform the energy contour integralHere the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared distribution limit, not as an isolated normalized oscillator. The Feynman i-epsilon prescription puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau >0$, close in the lower half-plane, clockwise; the residue theorem gives $-i$ times the residue $i{e}^{-iE\tau }/(2E)$, hence $e^{-iE\tau }/(2E)$. For $\tau <0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-i{e}^{iE\tau }/(2E)$, again giving a positive $e^{iE\tau }/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the scalar Feynman propagator pole prescription:The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a distribution limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the derivative jump of a free scalar time-ordered two-point function gives $(\square +m^2){\Delta }_F=-i{\delta }^4(x-y)$, an independent check of both the numerator and the sign.

The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the time ordering of the Feynman propagator.