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Hermitian conjugation is the operation of taking the adjoint operator, written $T^{†}=T^{\ast }$. For a matrix in orthonormal bases, it takes the complex conjugate of each entry and exchanges rows with columns:For a real matrix it is the ordinary transpose. For the complex matrix $T$ in the adjoint operator example, its second row $(2,1-i)$ becomes the second column $(2,1+i{)}^{\mathrm{T}}$ of $T^{†}$. This is the same conjugate transpose used in MIT's treatment of complex inner products and Hermitian matrices.^[ref]

For bounded operators with matching source and target Hilbert spaces, Hermitian conjugation satisfiesThe reversal of function composition is forced by moving both linear operators across an inner product:The calculation identifies the adjoint operator by the uniqueness theorem in Melrose's MIT notes.^[ref]

For a computed order check, let $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and reuse $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$. ThenKeeping the original order gives a different matrix. Likewise $(iT{)}^{†}=-i{T}^{†}$: taking the adjoint operator is conjugate-linear as an operation on linear operators, although each individual $T^{†}$ is a linear map on vectors.

The phase in a field transformation $\hat{C}{V}^{\mu }{\hat{C}}^{-1}={\eta }_C{V}^{\mu †}$. For a real vector relativistic quantum field, Hermitian conjugation restricts this phase to $\pm 1$.

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

Set $t_0=1$ and $t_i=-1$ for spatial indices. The complex conjugation of the scalar $i$ supplies a minus sign, while conjugating each of the three spatial gamma matrices supplies three more. In the original order,Consequently the chirality matrix obeysNo rearrangement of the gamma matrices is needed, so no additional anticommutator sign occurs. The star is entrywise complex conjugation, not matrix transpose or Hermitian conjugation.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Taking the Hermitian conjugation of the massive Dirac equation and using the Dirac adjoint givesTaking the matrix transpose and multiplying by the charge conjugation matrix $C$ yieldsThus the charge conjugation of a Dirac field preserves the free massive Dirac equation:The unitarity of $C$ and ${\gamma }^0{C}^{-1}{\gamma }^0=-C^{-1}{\gamma }^{0T}$ also giveConsequentlyHere $\hat{C}$ is the unitary operator acting on the relativistic quantum field, whereas $C$ acts on Dirac spinor indices.

The bosonic annihilation operator $a^{\lambda }(p)$ removes a vector particle of four-momentum $p$ and polarization vector ${\epsilon }^{\mu ,\lambda }(p)$. The bosonic creation operator $c^{\lambda †}(p)$ creates its antiparticle with the same four-momentum and polarization label. The polarization vector describes the corresponding spin-one mode expansion of a free field.

Because charge conjugation acts through a unitary operator, it leaves the numerical polarization vector coefficients unchanged. Applying the specified charge conjugation to the operators givesComparison with the Hermitian conjugation of the original expansion therefore givesThe factor ${\eta }_C$ is the intrinsic charge-conjugation phase.

Write $J_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ for the Dirac current and $J_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ for the axial current. Under parity symmetry in quantum field theory, the Dirac spinor phases cancel andThe extra sign for the axial current follows from ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. Combining these transformations with part (c), both currents transform under CP symmetry as $J_{V,A}^{\mu }(x)\mapsto -{\mathbb{P}}^{\mu }{}_{\nu }{J}_{V,A}^{\nu }(x_P)$. The vector relativistic quantum field transforms as $V^{\mu }(x)\mapsto {\eta }_C{\mathbb{P}}^{\mu }{}_{\nu }{V}^{\nu †}(x_P)$. The two parity matrices cancel in the tensor contraction, givingFor a real vector relativistic quantum field with ${\eta }_C=-1$, the interaction is invariant under CP symmetry. A complex vector relativistic quantum field instead maps to the conjugate-field interaction, which must be compared with its Hermitian conjugation partner.

Let $|1⟩=|D^0⟩$, $|2⟩=|{\bar{D}}^0⟩$. In the rest frame, the CPT theorem interchanges these states up to irrelevant phases. For the weak Hamiltonian operator $H^{\prime }$, with $H^{\prime }=H^{\prime †}$, and its antiunitary CPT symmetry operator $\Theta$,Its diagonal matrix elements are real by Hermitian conjugation, soIn the specified phase convention, the CP symmetry operator on this two-state subspace is $-{\sigma }_x$. If the weak interaction preserves CP symmetry, ${\sigma }_{x}R{\sigma }_x=R$, givingIndependently, Hermitian conjugation gives $R_{21}=R_{12}^{\ast }$ for this $H^{\prime }$. Hence CP symmetry makes the off-diagonal element real in this convention; that conjugation relation alone does not imply CP symmetry.

For neutral meson mixing described by the decay-effective Hamiltonian operator $R=M-i\Gamma /2$, the effective matrix is generally not a Hermitian matrix. The CPT constraints on neutral-meson mixing still give equality of the complex diagonal entries, expressing equal masses and total decay widths, and CP symmetry gives equality of the off-diagonal entries in the specified convention. One must not additionally impose $R_{21}=R_{12}^{\ast }$ on this decay-effective matrix.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

Under spatial reflection, a scalar field obeys $P\varphi (\mathbf{x})P=\varphi (-\mathbf{x})$, and hence each spatial derivative changes sign. The interaction contains three such derivatives, so changing the integration variable from $\mathbf{x}$ to $-\mathbf{x}$ givesThus this is a parity-odd scalar interaction.

Put $O({\mathbf{k}}_a)={\prod }_{a=1}^n{\varphi }_a({\mathbf{k}}_a)$ and $A({\mathbf{k}}_a)=⟨H_{\mathrm{int}}O({\mathbf{k}}_a)⟩$. Hermitian conjugation and commutativity of equal-time scalar fields giveThe parity-invariant vacuum and the odd parity of $H_{\mathrm{int}}$ imply $A(-{\mathbf{k}}_a)=-A({\mathbf{k}}_a)$. ThereforeThere is no conflict with the usual $2i\mathrm{Im}A$ rule for two Hermitian operators: a momentum-space product at fixed ${\mathbf{k}}_a$ is generally not itself Hermitian, since its adjoint carries momenta $-{\mathbf{k}}_a$.