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For reparameterization ghosts of weights two and minus one, the canonical anticommutators imply $b(z)c(w)\sim 1/(z-w)$. This operator-product expansion fixes the ghost contractions and the ghost stress tensor.

Every positive-energy supermultiplet contains equally many bosonic and fermionic states. A supertrace of a positive supercharge anticommutator proves the equality, provided supersymmetry is exact and the representation is finite-dimensional at fixed momentum.

Use $\{Q_{\alpha }^A,Q_{\beta }^B\}={ϵ}_{\alpha \beta }{Z}^{AB}$. Unitary skew-diagonalization of an antisymmetric matrix gives two-by-two blocks with entries $Z_r=2z_r$. At rest, the paired supercharges have anticommutators $2(M\pm |z_r|)$. The squared norms of an operator and its adjoint sum to the expectation of their anticommutator, so positive norm requires $M\ge |z_r|$ for every block. A convention with $2Z$ in the algebra instead writes $M\ge \max |Z_r|$.

Fermionic creation and annihilation operators satisfy $\{a_i,a_j^{†}\}={\delta }_{ij}$ and vanishing equal-type anticommutators. These relations make each mode either empty or singly occupied.

A real two-mode Bogoliubov transformation preserving the full canonical anticommutation relations. The self anticommutators become $\{a,a^{†}\}=\{b,b^{†}\}=u^2+v^2$, while $\{a,b\}=-uv+uv=0$ and every other cross anticommutator vanishes. Thus its matrix is an orthogonal matrix. For the two-orbital fermionic pairing Hamiltonian, let $E=\sqrt{{\epsilon }^2+{\Delta }^2}$, take $u=\sqrt{(E+\epsilon )/(2E)}$ and $v=\sqrt{(E-\epsilon )/(2E)}$, and obtain $H=E(c^{†}c+d^{†}d)+\epsilon -E$. Inverting gives $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$.

For signature $+---$, use ${\gamma }^0=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ and ${\gamma }^i=\left(\begin{array}{cc}0& {\sigma }^i\\ -{\sigma }^i& 0\end{array}\right)$, where ${\sigma }^i$ are Pauli matrices. Their Clifford algebra follows directly from the Pauli anticommutator. The chirality matrix is ${\gamma }^5=\mathrm{diag}(-I_2,I_2)$, separating the two chiral components of a Dirac spinor.

For a representation $R$ of $SU(N)$ with $N\ge 3$, choose the Hermitian quantum generator convention and normalize the cubic anomaly coefficient byHere the anticommutator symmetrizes the last two generators; cyclic property of the trace makes the whole tensor symmetric. This coefficient multiplies the local four-dimensional gauge anomaly from a chiral fermion in $R$. It adds under direct sums.

The conjugate representation has generators $-(T^a{)}^T$, so transposition and the three minus signs give $A(\overline{R})=-A(R)$. Hence a real representation or pseudoreal representation has zero cubic anomaly coefficient. For $SU(3)$, the two-index antisymmetric representation is the conjugate triplet and has coefficient $-1$. For $SU(2)$ the symmetric cubic invariant itself vanishes, so normalization to a nonzero fundamental cubic coefficient is unavailable; this statement concerns the local cubic anomaly and does not settle global anomalies.^[ref]

With signature $(+,-,-,-)$, one standard representation is ${\tilde{\gamma }}^0=\mathrm{diag}(I_2,-I_2)$ and ${\tilde{\gamma }}^i=\left(\begin{array}{cc}0& {\sigma }^i\\ -{\sigma }^i& 0\end{array}\right)$. Block multiplication and the Pauli matrix multiplication law give time-time anticommutator $2I_4$, mixed anticommutators zero, and spatial anticommutators $-2{\delta }^{ij}{I}_4$. Thus these gamma matrices represent the spacetime Clifford algebra.

A supersymmetric domain wall contributes a boundary or tensorial charge to the extended anticommutator of supercharges. It is not the Lorentz-scalar central charge of an extended particle algebra. The wall's energy per unit area can cancel this charge in half the real charge combinations, giving the BPS tension bound and half-supersymmetry preservation. This does not contradict the rank classification for the ordinary algebra without wall charges.

For $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, summing diagonal components gives $4H$. The expectation of each anticommutator is $\Vert Q_{\alpha }v{\Vert }^2+\Vert Q_{\alpha }^{†}v{\Vert }^2$, so $H$ is a positive semidefinite operator. A vacuum has zero energy exactly when every supercharge annihilates it. Spontaneously broken global supersymmetry with an existing vacuum therefore gives positive vacuum energy density. This statement fixes the energy zero through the algebra and does not apply to the general supergravity F-term potential.

For mode normalization $1/\sqrt{2E_{\mathbf{p}}}$, particle and antiparticle operators satisfy $\{a_{\mathbf{p}}^s,a_{\mathbf{q}}^{r†}\}=\{b_{\mathbf{p}}^s,b_{\mathbf{q}}^{r†}\}=(2\pi {)}^3{\delta }^{sr}{\delta }^3(\mathbf{p}-\mathbf{q})$. The Dirac spinor completeness relations imply ${\sum }_s[u_s(\mathbf{p})u_s^{†}(\mathbf{p})+v_s(-\mathbf{p})v_s^{†}(-\mathbf{p})]=2E_{\mathbf{p}}I$. Inserting this into the mode expansion of a Dirac field gives the displayed local anticommutator. The reversed momentum in the antiparticle completeness term is essential.

A fermionic operator is odd under fermion parity: if $P=(-1{)}^N$ is the parity operator, then $PAP=-A$. A fermionic creation operator or fermionic annihilation operator is an example. Interchanging odd insertions in a time-ordered product contributes the corresponding fermionic sign. Odd parity alone does not imply that every pair of such operators has zero anticommutator.

A Fermi oscillator has nilpotent lowering operator $B$, anticommutator $B{B}^{†}+B^{†}B=1$, and occupation Hamiltonian $B^{†}B$ with levels zero and one.

An invariant path-integral measure and BRST symmetry action imply this identity when every separated insertion $G_i$ is BRST-closed. The bracket is the graded commutator, so it becomes an anticommutator for an odd $\mathcal{O}$. The arbitrary operator need not itself be closed. Exact insertions therefore decouple from physical correlators.

The bracket of homogeneous elements in a parity-graded associative algebra. It is a commutator if at least one element is even and an anticommutator if both are odd. This is the natural bracket for the action of an odd BRST charge on quantum operators.

A Lie superalgebra representation acts on a parity-graded vector space by linear operators preserving the graded brackets. Odd generators reverse parity, and their brackets are represented by anticommutators.

A Lindblad dissipator is the non-Hamiltonian generator associated with a Lindblad operator $V$. It preserves the trace because cyclicity gives $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, cancelling the two anticommutator terms. It preserves Hermiticity and satisfies $\mathcal{D}[V](I)=[V,V^{†}]$; it need not preserve the identity or the purity of a density operator.

The Lindblad equation describes the generator of a time-homogeneous, completely positive, trace-preserving quantum dynamical semigroup in a finite-dimensional system. In units $\hslash =1$, the Hamiltonian $H$ supplies coherent evolution and the Lindblad operators $L_{\alpha }$ specify dissipative contributions.

A Markovian continuous-time quantum channel obeys

The Lindblad dissipator combines its gain term and two anticommutator terms so that the trace remains one. For $L=\sqrt{\gamma /2}Z$ and $H=0$, diagonal entries in the $Z$ basis stay fixed while an off-diagonal element obeys ${\dot{\rho }}_{01}=-\gamma {\rho }_{01}$: dephasing without population decay. The semigroup assumption is a specific model of environmental memory loss. ^[ref]

For a spinless Bogoliubov transformation $c_p=\cos {\theta }_p{\alpha }_p+i\sin {\theta }_p{\alpha }_{-p}^{†}$, consistency of the doubled Nambu spinor requires $\cos {\theta }_{-p}=\cos {\theta }_p$ and $\sin {\theta }_{-p}=-\sin {\theta }_p$. Thus choose ${\theta }_{-p}=-{\theta }_p$ modulo $2\pi$. The cross anticommutator from the first-row formulas is $\{c_p,c_{-p}\}=i\sin ({\theta }_p+{\theta }_{-p})$; a nontrivial even angle would violate the canonical anticommutation relations. A self-paired zero mode must instead have vanishing mixing in this convention. Real-angle unitarity within each two-by-two block alone does not enforce consistency between duplicated momentum labels.

The proper quark-gluon vertex colour factors come from two one-particle-irreducible Feynman diagrams, shown in the upper-middle and upper-right panels. One has a virtual gluon joining the open quark line on opposite sides of the external gluon insertion. The other has two virtual gluons meeting the external gluon at a three-gluon Feynman vertex, with their other ends joined by an internal quark. A Faddeev-Popov ghost field has no direct quark coupling, so it supplies no additional one-loop Feynman vertex graph here.

Let $T^a$ be the Hermitian generators of the fundamental representation of $SU(N)$, with $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, $[T^a,T^b]=i{f}^{abc}{T}^c$, ${\sum }_b{T}^b{T}^b=C_{F}1$, and $f^{abc}{f}^{dbc}=C_A{\delta }^{ad}$. Here $C_F=(N^2-1)/(2N)$ and $C_A=N$ are quadratic Casimir operators evaluated in the fundamental representation and the Adjoint representation.

For the first graph, the ordered product is ${\sum }_b{T}^b{T}^a{T}^b$. Expand the double commutator:Thus its colour factor isFor the second graph choose the orientation of the three-gluon Feynman vertex so that its group tensor is $f^{abc}$ and the ordered product along the quark line is $T^b{T}^c$. Antisymmetry removes the anticommutator part:The corresponding real group coefficient is $C_A/2=3/2$ after the conventional factor of $i$ is incorporated into the remaining Feynman rules. If the vertex is written with $i{f}^{abc}$, its raw matrix product is instead $-C_A{T}^a/2$. Reversing the ordered product reverses its sign together with the orientation-dependent three-gluon Feynman rule; these conventions describe the same scattering amplitude. External-leg corrections and local counterterms are separate from these two proper one-loop graphs.