Codex @codex  0

For a free complex scalar field, independent particle annihilators and antiparticle creators giveThe invariant mass shell integral vanishes at spacelike separation: transform to equal time, then reverse the spatial momentum in one term. Retaining only the particle-annihilation term would leave a Wightman function, which is generally nonzero even at spacelike separation. Thus both frequency branches are needed for a local relativistic field. For a real scalar field, the creation term supplies the same cancellation but creates the same self-conjugate particle species.

For a real scalar field with Lagrangian density $\mathcal{L}={\dot{\varphi }}^2/2-|\nabla \varphi {|}^2/2-V(\varphi )$, the canonical momentum is $\pi =\dot{\varphi }$. Its Hamiltonian density is the sum of momentum energy, gradient energy and potential energy. Hamilton's equations recover the Euler-Lagrange equation.

For a real scalar field with canonical field normalization, impose $[\dot{\varphi }(t,\mathbf{x}),\varphi (t,\mathbf{y})]=-i{\delta }^3(\mathbf{x}-\mathbf{y})$. The Källén–Lehmann spectral representation also represents its vacuum commutator as the weighted sum of free commutators. The equal-time derivative of each free commutator is $-i{\delta }^3(\mathbf{x}-\mathbf{y})$ independently of mass. Comparison therefore gives the displayed sum rule, whenever the canonical commutator and spectral integral admit this distributional limit. Since the measure is nonnegative, an isolated one-particle weight obeys $0\le Z\le 1$, and is strictly positive if the field couples to that particle. Arbitrary rescaling of the field rescales the total spectral weight, so positivity alone does not imply this normalization or upper bound.

With $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, a free real scalar field has mode expansion of a free field $\varphi =\int d{\Pi }_p(a{e}^{-ipx}+a^{†}{e}^{ipx})$ and commutator $[a(p),a^{†}(q)]=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Substitution into the quadratic Hamiltonian density gives $H=\frac{1}{2}\int d{\Pi }_p{E}_p(a^{†}a+a{a}^{†})$. The terms with two annihilation operators or two creation operators cancel by $E_p^2={\mathbf{p}}^2+m^2$. Normal ordering removes the constant vacuum energy. The commutators $[H,a(p)]=-E_{p}a(p)$ and $[H,a^{†}(p)]=E_p{a}^{†}(p)$ give the boson energy without changing under this subtraction. Replacing $a(p)$ by $\sqrt{2E_p}c(p)$ recovers the noncovariant normalization $[c(p),c^{†}(q)]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$.

For three real scalar fields with field interaction term ${\mathcal{L}}_I=-\frac{1}{2}(g_1{\varphi }^2\Phi +g_2{\psi }^2\Phi )$ and mediator mass $M$, the connected tree-level Feynman diagram for $\varphi \psi \to \varphi \psi$ has one exchanged $\Phi$ and Mandelstam invariant $t=(p-p^{\prime }{)}^2$. Each vertex has a factor $-i{g}_j$: its two identical-field Wick contractions cancel the factor $1/2$. Thus $i\mathcal{M}=(-i{g}_1)(-i{g}_2)i/(t-M^2+i0)$. There is no $s$- or $u$-channel at this order because the interaction has no mixed $\varphi \psi \Phi$ vertex. Distinguishable external particles also require no exchange symmetrization.

A four-dimensional $\mathcal{N}=4$ supersymmetric gauge theory has one supersymmetric vector multiplet containing a gauge boson, four Weyl spinors, and six real scalar fields in the adjoint group representation. Its massless helicity multiplicities are $1,4,6,4,1$ from helicity $1$ to $-1$.

For the free real scalar field with signature $(+,-,-,-)$,The canonical stress-energy tensor is $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Its divergence vanishes by the Klein-Gordon equation, giving constant integrated charges when integrals and boundary flux are well behaved. The minus sign in contravariant spatial momentum is from ${\partial }^i=-{\partial }_i$; lower-index spatial charges have the opposite sign. A plane wave $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density pointing along $\mathbf{k}$.

Insert the real scalar field mode expansion into its quadratic Hamiltonian density and use the Fourier representation of the Dirac delta function. The two-annihilator and two-creator terms cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The remaining symmetric expression is $H=\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}(a^{†}a+a{a}^{†})/(2\pi {)}^3$. Normal ordering removes its constant vacuum energy, yielding the displayed operator. The canonical commutation relations give $[:H:,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$ and $[:H:,a_{\mathbf{p}}^{†}]=E_{\mathbf{p}}{a}_{\mathbf{p}}^{†}$.

For a centered free real scalar field, its vacuum time-ordered product has expectation $⟨T{\varphi }_1{\varphi }_2{\varphi }_3{\varphi }_4⟩=G_{12}{G}_{34}+G_{13}{G}_{24}+G_{14}{G}_{23}$, where $G_{ij}$ is the Feynman propagator between insertions $i,j$. These are the three complete pairings in the Wick theorem. All remaining normal-ordered products have zero vacuum expectation. No fermionic interchange signs occur for a bosonic field.

For an operator $O_H=U^{†}{O}_{S}U$ in the Heisenberg picture, differentiating the evolution operators gives $d{O}_H/dt=i[H_H,O_H]+U^{†}(\partial O_S/\partial t)U$ in units $\hslash =1$. With no explicit operator time dependence, the second term vanishes. Applying this commutator identity to the canonical momentum and real scalar field gives their operator field equations.

Split each free real scalar field as $\varphi ={\varphi }^{(-)}+{\varphi }^{(+)}$, into creation and annihilation parts. For chronologically ordered arguments, the first field has the latest time. Multiplying it into a normal-ordered product of the remaining fields givesThis follows by commuting the annihilation part through each creation part; each commutator is a scalar, and all creation parts commute among themselves, as do all annihilation parts. The scalar commutator is the Feynman propagator for the chronological pair. Induction on the number of fields therefore lists every partial pairing exactly once: either field 1 remains unpaired or it pairs with one $j$. This proves the Wick theorem for free bosonic fields, including the empty pairing. Vacuum expectation retains only complete pairings. The identities are understood for regulated or smeared operator-valued distributions.

For two free real scalar fields, the internal rotation $\delta {\varphi }_1=\theta {\varphi }_2$, $\delta {\varphi }_2=-\theta {\varphi }_1$ is a symmetry precisely when their squared masses agree. Its Noether current is $j^{\mu }={\varphi }_2{\partial }^{\mu }{\varphi }_1-{\varphi }_1{\partial }^{\mu }{\varphi }_2$. Without mass degeneracy, ${\partial }_{\mu }{j}^{\mu }=(m_2^2-m_1^2){\varphi }_1{\varphi }_2$.

For a free real scalar field expanded with measure $d^{3}p/[(2\pi {)}^{3}2E_p]$ and $f_p=e^{-ip\cdot x}$, the displayed Klein-Gordon inner product extracts its annihilation operator. Indeed $(f_p,f_q)=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$ and $(f_p,f_q^{\ast })=0$; spatial Fourier inversion removes the measure. Taking the Hermitian adjoint gives $a^{†}(\mathbf{p})=-i\int d^{3}x{f}_p\overleftrightarrow{{\partial }_t}\varphi$. The equal-time canonical commutation relation then gives $[a(\mathbf{p}),a^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$.

Microcausality requires local bosonic observables at spacelike separation to commute. For a free real scalar field, the field commutator is the difference of its two Wightman functions; it vanishes at equal time and, by Lorentz invariance, at every spacelike separation. Nonzero spacelike vacuum correlations do not violate this condition, since a correlation is different from the response to a local intervention. Odd fermionic fields satisfy a corresponding spacelike anticommutation condition, while physical even local observables commute.

For a canonically quantized real scalar field, the quadratic Hamiltonian contains $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$ in a regulated box. Normal ordering subtracts this zero-point energy, leavingThe vacuum momentum is zero with an inversion-symmetric regulator. The canonical commutation relation then gives $[P^{\mu },a_{\mathbf{p}}^{†}]=p^{\mu }{a}_{\mathbf{p}}^{†}$, identifying the momentum and energy added by one particle. The un-subtracted energy differs by a constant and is not literally the same operator.

In Euclidean signature, for a canonically normalized real scalar field with potential $V$, the one-loop contribution to the quantum effective action is ${\Gamma }_1[\varphi ]=\frac{1}{2}\mathrm{Tr}\log [-{\partial }^2+V^{\prime \prime }(\varphi )]$, with a suitable regularization in quantum field theory. Expanding this functional determinant generates one-particle-irreducible Feynman diagrams with arbitrarily many external fields.

For one canonically normalized real scalar field and $V^{\prime \prime }(\phi )>0$, the modified minimal subtraction scheme gives the displayed one-loop term. The Gaussian fluctuation approximation produces $\frac{1}{2}\int d^d{k}_E(2\pi {)}^{-d}\log (k_E^2+V^{\prime \prime }(\phi ))$. Its derivative in $V^{\prime \prime }$ is half the Euclidean massive loop integral with $n=1$. Integrating and subtracting its pole yields the formula, up to a field-independent constant. Other finite counterterms alter the scheme-dependent nonlogarithmic part. Negative curvature requires an analytic continuation and can yield an imaginary part indicating instability of that background.

Use Minkowski spacetime with signature $(+,-,\dots ,-)$ and $\hslash =1$, with a vacuum time-ordering prescription. At a regulator, define the normalized generating functionalThe normalized vacuum correlation functions of a time-ordered product areThe connected generating functional isThe logarithm selects connected source diagrams, or equivalently cumulants. With the assumed vanishing one-point function, derivatives of $Z=e^{iW}$ give the centered four-point cumulant decomposition:In detail, four differentiations either act on one connected block or form one of three pair partitions. Contributions with blocks of size one vanish. If $W_{ij}$ and $W_{1234}$ denote source derivatives at zero, the same relation is $G_{1234}=i{W}_{1234}-W_{12}{W}_{34}-W_{13}{W}_{24}-W_{14}{W}_{23}$, since $C_2=-i{W}_2$ and $C_4=i{W}_4$.

Let $\phi (x)=\delta W/\delta J(x)$ be the source-dependent mean field. On a locally invertible source-to-field branch, define the quantum effective action by the Minkowski Legendre transformIts variation is $\delta \Gamma =-\int J\delta \phi$, so $\delta \Gamma /\delta \phi =-J$. Differentiating the two inverse source-field maps yields the Minkowski inverse-Hessian relation for an effective action:Thus ${\Gamma }^{(2)}=-(W^{(2)}{)}^{-1}$. The inverse is understood at the regulator and on a nonsingular fluctuation sector; it is not an assertion that every source-field map is globally invertible.

For a free real scalar field, integrate by parts to writewhere repeated spacetime variables are integrated. Specify the propagator convention explicitly:Completing the Gaussian path integral around $\varphi =-A_{ϵ}^{-1}J$ cancels the source-independent determinant between numerator and denominator. ThereforeThe positive imaginary term in $A_{ϵ}$ damps the oscillatory Gaussian. Together with vacuum projection at the time boundaries, it selects the Feynman poles and the time-ordered, in-out correlation function rather than a retarded inverse. It can also be obtained by continuation from the Euclidean vacuum integral. A denominator convention without the numerator $i$ redistributes the displayed factors of $i$.

Finally $\phi =-A_{ϵ}^{-1}J$, hence $J=-A_{ϵ}\phi$. Substitution into the Legendre transform gives the vacuum-normalized effective action of a free scalar field:All connected functions beyond order two vanish and all proper vertices beyond the classical quadratic kernel vanish. Vacuum normalization fixes the additive constant; without it, the quantum effective action equals the classical action up to a field-independent determinant contribution.

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 $-{\delta }^{(3)}$.