Introduction
Hamiltonians that are not self-adjoint in the Dirac sense may still organize a consistent spectrum when an antilinear symmetry is present, and the dividing line between the two regimes of such a symmetry is sharp. In the unbroken regime the eigenvalues are real, a positive definite metric exists and the evolution is unitary with respect to it [8 , 9 , 10 , 12 ] . In the broken regime the eigenvalues appear in complex conjugate pairs, no positive definite metric exists, and the discussion usually stops there: for a finite dimensional system the broken phase is the end of the line, two complex eigenvalues and no spectrum left to speak of. Our interest is in what happens to that statement in a quantum field theory, where a conjugate pair of elementary excitations still admits a two-particle sector, and where composite operators built out of such constituents are known to possess real Källén–Lehmann representations [15 , 17 , 18 ] — a pattern first met in the indefinite-metric sector of Lee and Wick [13 ] and later in the infrared of confining gauge theories, where the Gribov–Zwanziger gluon propagator factorizes into the i i i ± i γ 2 \pm i\gamma^2 ± i γ 2 [15 , 16 ] .
In a recent work [1 ] we have constructed a model in which this question can be posed concretely. The field content is two fermions ψ \psi ψ χ \chi χ a μ a_\mu a μ a ^ μ \hat{a}_\mu a ^ μ ϕ \phi ϕ a a a b b b a b ab ab a b > 0 ab>0 ab > 0 a b < 0 ab<0 ab < 0 a = − b a=-b a = − b a b = 0 ab=0 ab = 0 [1 ] established that the one-loop fermion contribution to ⟨ ϕ † ( p ) ϕ ( − p ) ⟩ \langle\phi^\dagger(p)\phi(-p)\rangle ⟨ ϕ † ( p ) ϕ ( − p )⟩ ( m + m ∗ ) 2 = 4 N 2 (m+m^*)^2=4N^2 ( m + m ∗ ) 2 = 4 N 2 p 2 p^2 p 2 2 N 2N 2 N
However, a scalar two-point function is not the quantity an experiment would address, and the model has a feature that invites a sharper question. One combination of its two Abelian gauge fields remains massless after the scalar acquires its expectation value, so that the model has a well-defined notion of an external electromagnetic probe and of the current it drives. Our aim in the present work is to answer what the conductivity of such a medium looks like, both as a function of frequency and, what turns out to be the more revealing question, as a function of time; and to use the vector channel, where the Dirac algebra is simpler and the residues are positive definite, to settle the question of positivity that the scalar channel left open.
Our strategy is to compute the current-current correlator at one loop in the basis that diagonalizes the mass matrix, keeping the temperature and the chemical potential finite from the start, and to follow which pairs of complex masses enter each channel. The fact that the two conserved combinations of the original currents organize themselves, in that basis, one as a diagonal and the other as a mixed bilinear, is what separates the analysis into two problems with opposite analytic structures, and it is the origin of every result reported below. At this point we make a brief interruption to state that the thermal part of the calculation requires the Fermi–Dirac occupation numbers to be evaluated at complex energies, a step whose validity is not a matter of taste and which we therefore establish rather than assume; the discussion occupies Section IV and underlies the positivity bound of § 2 .
In the next section we fix the notation, exhibit the exceptional point of the family, the decomposition of the currents, the indefinite metric of the broken phase and the exact Z 2 \mathbb{Z}_2 Z 2 § 1.4 and VI treat the unbroken and the broken channel respectively, Section VII obtains the response kernel in time, and § 2 contains the positivity results. Section IX applies the Osterwalder–Schrader criterion in the Euclidean region, where it settles both channels without any choice of boundary value, and § 2.2 treats the longitudinal part of the mixed channel and the field-strength composite that removes it. The conclusions gather the results and indicate what we believe should be examined next. Four appendices collect the Dirac algebra, the dispersion relation of the complex-mass excitations, the reading of the channel separation in terms of the balance of the cuts, and the covariant vacuum computation of the two spectral densities.
The family, its exceptional point, and the two current channels
The one-parameter family
We begin with the usual definitions. After the spontaneous breaking the fermionic mass matrix of the model of [1 ] reads, in the ( ψ , χ ) (\psi,\chi) ( ψ , χ )
M ( a , b ) = ( N a v b v N ) , m ± = N ± v a b , (1) M(a,b)=\begin{pmatrix} N & av \\ bv & N \end{pmatrix}, \qquad m_{\pm}=N\pm v\sqrt{ab}, \tag*{(1)} M ( a , b ) = ( N b v a v N ) , m ± = N ± v ab , (1) with a a a b b b v v v a b > 0 ab>0 ab > 0 a = b a=b a = b [1 ] . For a b < 0 ab<0 ab < 0 a = − b a=-b a = − b
m = N + i a v , m ∗ = N − i a v . (2) m=N+iav,\qquad m^{*}=N-iav. \tag*{(2)} m = N + ia v , m ∗ = N − ia v . (2) Between the two lies a b = 0 ab=0 ab = 0 m + = m − = N m_{+}=m_{-}=N m + = m − = N M − N 1 M-N\mathbf{1} M − N 1 M M M
it carries is not the ordinary one. In fact, at b = 0 b=0 b = 0
M_{\mathrm{fermi}}(k)=\begin{pmatrix}\slashed{k}-N & -av \\ 0 & \slashed{k}-N\end{pmatrix}, \tag*{(3)}
and the off-diagonal entry of its inverse is
G(\psi\overline{\chi})=\frac{av}{(\slashed{k}-N)^2}=av\frac{(\slashed{k}+N)^2}{(k^2-N^2)^2}, \tag*{(4)}
a double pole. Put in another way, the exceptional point of the family is a dipole field, the non-diagonalizability of the mass matrix appearing in the propagator as a pole of second order. Dipole structures of this kind are familiar from the Gribov–Zwanziger horizon and from the BRST doublets of algebraic renormalization [15 , 23 ] , where they carry zero norm and no positive spectral weight, and Section IIC establishes that this is exactly what happens here: the state that survives the collision has zero norm under every admissible metric. In the sequence of this work we remain on the broken side, a = − b a=-b a = − b
The biorthogonal basis and the currents
Since M M M
ψ = 1 2 ( Ψ m + Ψ m ∗ ) , χ = i 2 ( Ψ m − Ψ m ∗ ) , \psi=\frac{1}{\sqrt{2}}(\Psi_m+\Psi_{m^*}),\qquad\chi=\frac{i}{\sqrt{2}}(\Psi_m-\Psi_{m^*}), ψ = 2 1 ( Ψ m + Ψ m ∗ ) , χ = 2 i ( Ψ m − Ψ m ∗ ) , ψ ‾ = 1 2 ( Ψ ‾ m + Ψ ‾ m ∗ ) , χ ‾ = − i 2 ( Ψ ‾ m − Ψ ‾ m ∗ ) , (5) \overline{\psi}=\frac{1}{\sqrt{2}}(\overline{\Psi}_m+\overline{\Psi}_{m^*}),\qquad\overline{\chi}=-\frac{i}{\sqrt{2}}(\overline{\Psi}_m-\overline{\Psi}_{m^*}), \tag*{(5)} ψ = 2 1 ( Ψ m + Ψ m ∗ ) , χ = − 2 i ( Ψ m − Ψ m ∗ ) , (5) so that the quadratic form becomes diagonal and Ψ m , Ψ m ∗ \Psi_m,\Psi_{m^*} Ψ m , Ψ m ∗ m m m m ∗ m^* m ∗
G(\Psi_m\overline{\Psi}_m)=\frac{i}{\slashed{k}-m},\qquad G(\Psi_{m^*}\overline{\Psi}_{m^*})=\frac{i}{\slashed{k}-m^*}, \tag*{(6)}
the mixed components vanishing.
The gauge couplings of the model are carried by the two covariant derivatives, so that the interaction reads e a μ ψ ‾ γ μ ψ − e a ^ μ χ ‾ γ μ χ e\,a_\mu\overline{\psi}\gamma^\mu\psi-e\,\widehat{a}_\mu\overline{\chi}\gamma^\mu\chi e a μ ψ γ μ ψ − e a μ χ γ μ χ A μ = a μ + a ^ μ A_\mu=a_\mu+\widehat{a}_\mu A μ = a μ + a μ B μ = a μ − a ^ μ B_\mu=a_\mu-\widehat{a}_\mu B μ = a μ − a μ e 2 v 2 e^2v^2 e 2 v 2
e 2 [ A μ ( J ψ μ − J χ μ ) + B μ ( J ψ μ + J χ μ ) ] , (7) \frac{e}{2}\left[A_\mu\left(J_\psi^\mu-J_\chi^\mu\right)+B_\mu\left(J_\psi^\mu+J_\chi^\mu\right)\right], \tag*{(7)} 2 e [ A μ ( J ψ μ − J χ μ ) + B μ ( J ψ μ + J χ μ ) ] , (7) with J ψ μ = ψ ‾ γ μ ψ J_\psi^\mu=\overline{\psi}\gamma^\mu\psi J ψ μ = ψ γ μ ψ J χ μ = χ ‾ γ μ χ J_\chi^\mu=\overline{\chi}\gamma^\mu\chi J χ μ = χ γ μ χ ϕ \phi ϕ A μ A_\mu A μ B μ B_\mu B μ B μ B_\mu B μ J B = J ψ + J χ J_B=J_\psi+J_\chi J B = J ψ + J χ
It is at this point that the biorthogonal expansion (5) produces the separation on which everything else rests. Inserting it into the two conserved combinations one finds
J B μ = Ψ ‾ m γ μ Ψ m + Ψ ‾ m ∗ γ μ Ψ m ∗ , J A μ = Ψ ‾ m γ μ Ψ m ∗ + Ψ ‾ m ∗ γ μ Ψ m . (8) \begin{aligned}
J_B^\mu&= \overline{\Psi}_m\gamma^\mu\Psi_m+\overline{\Psi}_{m^*}\gamma^\mu\Psi_{m^*},\\
J_A^\mu&= \overline{\Psi}_m\gamma^\mu\Psi_{m^*}+\overline{\Psi}_{m^*}\gamma^\mu\Psi_m. \tag*{(8)}
\end{aligned} J B μ J A μ = Ψ m γ μ Ψ m + Ψ m ∗ γ μ Ψ m ∗ , = Ψ m γ μ Ψ m ∗ + Ψ m ∗ γ μ Ψ m . (8) The current of the unbroken U ( 1 ) U(1) U ( 1 ) ( m , m ) (m,m) ( m , m ) ( m ∗ , m ∗ ) (m^*,m^*) ( m ∗ , m ∗ ) U ( 1 ) U(1) U ( 1 ) Ψ ‾ + Ψ − \overline{\Psi}_+\Psi_- Ψ + Ψ − [1 ] , and its two-point function is built out of the mixed pair ( m , m ∗ ) (m,m^*) ( m , m ∗ ) m − m ∗ m-m^* m − m ∗
The metric of the broken phase
Before computing anything we construct the metric of the broken phase, which anticipates the two results of § 1.4 and VI and explains them. No positive definite metric exists, since a pseudo-Hermitian Hamiltonian admits one if and only if its spectrum is real [10 , 11 ] . What does exist is an indefinite η \eta η η H η − 1 = H † \eta H\eta^{-1}=H^\dagger η H η − 1 = H † (1) with a = − b a=-b a = − b ( ψ , χ ) (\psi,\chi) ( ψ , χ )
η = ( 1 0 0 − 1 ) , η M η − 1 = M † , (9) \eta=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad\eta M\eta^{-1}=M^\dagger, \tag*{(9)} η = ( 1 0 0 − 1 ) , η M η − 1 = M † , (9) Hermitian, with eigenvalues ± 1 \pm1 ± 1
It is instructive to follow the metric along the whole family (1) and not only at a = − b a=-b a = − b b ≠ 0 b\ne0 b = 0 η M η − 1 = M † \eta M\eta^{-1}=M^\dagger η M η − 1 = M †
η = ( 1 0 0 a / b ) , (10) \eta=\begin{pmatrix}1&0\\0&a/b\end{pmatrix}, \tag*{(10)} η = ( 1 0 0 a / b ) , (10) which is positive definite for a b > 0 ab>0 ab > 0 a = b a=b a = b a b < 0 ab<0 ab < 0 (9) , and diverges as b → 0 b\to0 b → 0 b = 0 b=0 b = 0 η M = M † η \eta M=M^\dagger\eta η M = M † η
η = ( 0 q q r ) , q , r ∈ R , (11) \eta=\begin{pmatrix}0&q\\q&r\end{pmatrix},\qquad q,r\in\mathbb{R}, \tag*{(11)} η = ( 0 q q r ) , q , r ∈ R , (11) so that η 11 \eta_{11} η 11 u = ( 1 , 0 ) u=(1,0) u = ( 1 , 0 ) every admissible metric. The vanishing norms of (14) are thus not a peculiarity of the broken side. They are already present at the exceptional point, where the two zero-norm modes of the broken phase have merged into one, and it is the dipole propagator (4) that carries the property in the field theory, a double pole admitting no positive Källén–Lehmann weight. Positive norms exist only on the Hermitian side a b > 0 ab>0 ab > 0 (9) is a discrete flavor transformation, and is therefore implemented on the Fock space by the χ \chi χ
η = ( − 1 ) N χ , N χ = ∫ d 3 x χ † χ , (12) \eta=(-1)^{N_\chi},\qquad N_\chi=\int d^3x\,\chi^\dagger\chi, \tag*{(12)} η = ( − 1 ) N χ , N χ = ∫ d 3 x χ † χ , (12) and one verifies η H η − 1 = H † \eta H\eta^{-1}=H^\dagger η H η − 1 = H † χ → − χ \chi\to-\chi χ → − χ χ ‾ → − χ ‾ \overline{\chi}\to-\overline{\chi} χ → − χ χ ‾ χ \overline{\chi}\chi χ χ (7) are invariant, while the mixed mass term a v ( ψ ‾ χ − χ ‾ ψ ) a v(\overline{\psi}\chi-\overline{\chi}\psi) a v ( ψ χ − χ ψ ) ( ψ ‾ χ ) † = χ ‾ ψ (\overline{\psi}\chi)^\dagger=\overline{\chi}\psi ( ψ χ ) † = χ ψ a v av a v
The operator (12) is Hermitian and unitary with η 2 = 1 \eta^2=1 η 2 = 1 χ \chi χ [13 ] , here not an analogy but an explicit operator. Since η \eta η χ \chi χ (5) ,
η : Ψ m ⟷ Ψ m ∗ , (13) \eta:\ \Psi_m\longleftrightarrow\Psi_{m^*}, \tag*{(13)} η : Ψ m ⟷ Ψ m ∗ , (13) as it must, H † H^\dagger H † ∣ Ψ m ⟩ ∝ ∣ ψ ⟩ + i ∣ χ ⟩ |\Psi_m\rangle\propto|\psi\rangle+i|\chi\rangle ∣ Ψ m ⟩ ∝ ∣ ψ ⟩ + i ∣ χ ⟩ η ∣ Ψ m ⟩ ∝ ∣ ψ ⟩ − i ∣ χ ⟩ \eta|\Psi_m\rangle\propto|\psi\rangle-i|\chi\rangle η ∣ Ψ m ⟩ ∝ ∣ ψ ⟩ − i ∣ χ ⟩
⟨ Ψ m ∣ Ψ m ⟩ η = ⟨ Ψ m ∗ ∣ Ψ m ∗ ⟩ η = 0 , ⟨ Ψ m ∣ Ψ m ∗ ⟩ η ≠ 0. (14) \langle\Psi_m|\Psi_m\rangle_\eta=\langle\Psi_{m^*}|\Psi_{m^*}\rangle_\eta=0,\qquad\langle\Psi_m|\Psi_{m^*}\rangle_\eta\ne0. \tag*{(14)} ⟨ Ψ m ∣ Ψ m ⟩ η = ⟨ Ψ m ∗ ∣ Ψ m ∗ ⟩ η = 0 , ⟨ Ψ m ∣ Ψ m ∗ ⟩ η = 0. (14) Each mode of complex mass has vanishing norm, and the only pairing that carries norm is the one between conjugate partners. The same holds one level up, where the structure is sharper than it first appears. In the two-particle sector, with the metric acting as η ⊗ η \eta\otimes\eta η ⊗ η
⟨ Ψ r Ψ s ∣ Ψ r Ψ s ⟩ η = 0 , r , s ∈ { m , m ∗ } , (15) \langle\Psi_r\Psi_s|\Psi_r\Psi_s\rangle_\eta=0,\qquad r,s\in\{m,m^*\}, \tag*{(15)} ⟨ Ψ r Ψ s ∣ Ψ r Ψ s ⟩ η = 0 , r , s ∈ { m , m ∗ } , (15) the norm residing entirely in the off-diagonal overlaps, which are the same in the two possible pairings,
⟨ Ψ m Ψ m ∗ ∣ Ψ m ∗ Ψ m ⟩ η = ⟨ Ψ m Ψ m ∣ Ψ m ∗ Ψ m ∗ ⟩ η = 1. (16) \langle\Psi_m\Psi_{m^*}|\Psi_{m^*}\Psi_m\rangle_\eta
=\langle\Psi_m\Psi_m|\Psi_{m^*}\Psi_{m^*}\rangle_\eta=1. \tag*{(16)} ⟨ Ψ m Ψ m ∗ ∣ Ψ m ∗ Ψ m ⟩ η = ⟨ Ψ m Ψ m ∣ Ψ m ∗ Ψ m ∗ ⟩ η = 1. (16) The conjugate pairings ( m , m ∗ ) (m,m^*) ( m , m ∗ ) ( m , m ) (m,m) ( m , m ) + 1 +1 + 1 − 1 -1 − 1 η \eta η i i i [15 , 17 ] in the present model: what carries physical content is not the constituent of complex mass, whose norm vanishes, but the symmetrized pair.
It is worth being accurate about what (15) and (16) do and do not decide. They establish that no single mode of complex mass is a state, and that in each channel the current reaches the combination of norm +1 and is η \eta η − 1 -1 − 1 § 2 that the positivity obtained for the mixed channel is the reflection of that fact. They do not by themselves separate the two channels, the two planes having the same signature. What separates them is a symmetry, and it is exact. Neither do they settle the positivity of a current–current correlator, and on this we must be explicit, because the η \eta η J A J_A J A § 2 ; and, more importantly, the η \eta η η \eta η J B ∣ 0 ⟩ J_B|0\rangle J B ∣0 ⟩ (16) and (37) , and the channel it belongs to may still admit no positive-norm realization, which is what Section IX proves of that very channel. An indefinite metric that is conserved is not a Hilbert space in disguise.
One should not read this as the construction of a physical subspace in the strong sense. Since η H η − 1 = H † \eta H\eta^{-1}=H^\dagger η H η − 1 = H † H H H η \eta η H H H χ \chi χ (12) are not preserved by the evolution, and the partner of norm − 1 -1 − 1 (16) sits in the spectrum beside the physical one. In the broken phase this is forced rather than accidental, and the removal of such a partner by a cohomological mechanism of the kind familiar from BRST doublets [23 ] is, with η \eta η (12) is the metric of the quadratic sector, and that it is discrete rather than of the form e − Q e^{-Q} e − Q P T PT P T [9 , 11 ] , precisely because the non-Hermiticity of the model is a discrete flavor structure; whether it survives the Yukawa fluctuations of ϕ \phi ϕ
The unbroken Z₂ and the separation of the channels
The symmetry that separates the two channels is not the metric, and it is not a flavor symmetry either. Consider first the exchange
C : ψ ↔ χ , ϕ → − ϕ † , a μ ↔ − a ^ μ , (17) C:\quad\psi\leftrightarrow\chi,\qquad\phi\to-\phi^\dagger,\qquad a_\mu\leftrightarrow-\hat a_\mu, \tag*{(17)} C : ψ ↔ χ , ϕ → − ϕ † , a μ ↔ − a ^ μ , (17) which leaves the kinetic terms, the diagonal masses and the Yukawa coupling of the a = − b a=-b a = − b
R : α = 0 , α ^ = π / e , ψ → ψ , χ → − χ , ϕ → − ϕ , (18) \mathcal{R}:\quad\alpha=0,\qquad\hat{\alpha}=\pi/e,\qquad\psi\to\psi,\qquad\chi\to-\chi,\qquad\phi\to-\phi, \tag*{(18)} R : α = 0 , α ^ = π / e , ψ → ψ , χ → − χ , ϕ → − ϕ , (18) which is the element of the Higgsed U ( 1 ) U(1) U ( 1 ) η \eta η (9) on the flavor doublet. Neither is a symmetry of the broken vacuum: each reverses the sign of ϕ \phi ϕ v v v M ( a v ) → M ( − a v ) M(av)\to M(-av) M ( a v ) → M ( − a v )
Z 2 = C ⋅ R : ( ψ χ ) ⟶ ( 0 − 1 1 0 ) ( ψ χ ) , v → v , (19) Z_2=\mathcal{C}\cdot\mathcal{R}:\quad
\begin{pmatrix}\psi\\\chi\end{pmatrix}
\longrightarrow
\begin{pmatrix}0&-1\\1&0\end{pmatrix}
\begin{pmatrix}\psi\\\chi\end{pmatrix},
\qquad v\to v, \tag*{(19)} Z 2 = C ⋅ R : ( ψ χ ) ⟶ ( 0 1 − 1 0 ) ( ψ χ ) , v → v , (19) and this is the only combination of the three that preserves the mass matrix and the vacuum at once. It is the Z 2 Z_2 Z 2 [1 ] , which is thus not an accidental flavor symmetry but the diagonal of a flavor exchange and a discrete element of the broken gauge group, each of which is separately broken by ⟨ ϕ ⟩ \langle\phi\rangle ⟨ ϕ ⟩
The statement is in fact stronger than the mass matrix, and holds before the shift. On the Yukawa terms a ψ ˉ ϕ χ + b χ ˉ ϕ † ψ a\bar{\psi}\phi\chi+b\bar{\chi}\phi^\dagger\psi a ψ ˉ ϕ χ + b χ ˉ ϕ † ψ (19) together with ϕ → ϕ † \phi\to\phi^\dagger ϕ → ϕ †
ψ ˉ ϕ χ → − χ ˉ ϕ † ψ , χ ˉ ϕ † ψ → − ψ ˉ ϕ χ , (20) \bar{\psi}\phi\chi\to-\bar{\chi}\phi^\dagger\psi,\qquad
\bar{\chi}\phi^\dagger\psi\to-\bar{\psi}\phi\chi, \tag*{(20)} ψ ˉ ϕ χ → − χ ˉ ϕ † ψ , χ ˉ ϕ † ψ → − ψ ˉ ϕ χ , (20) so that a → − b a\to-b a → − b b → − a b\to-a b → − a b = − a b=-a b = − a ϕ \phi ϕ D μ ϕ → ( D μ ϕ ) † D_\mu\phi\to(D_\mu\phi)^\dagger D μ ϕ → ( D μ ϕ ) † A μ → − A μ A_\mu\to-A_\mu A μ → − A μ ϕ → ϕ † \phi\to\phi^\dagger ϕ → ϕ † ϕ † ϕ \phi^\dagger\phi ϕ † ϕ ϕ = ( v + σ + i π ) / 2 \phi=(v+\sigma+i\pi)/\sqrt{2} ϕ = ( v + σ + iπ ) / 2
σ → σ , π → − π , (21) \sigma\to\sigma,\qquad\pi\to-\pi, \tag*{(21)} σ → σ , π → − π , (21) so that the Higgs mode is even and the Goldstone odd, as it must be, π \pi π A μ A_\mu A μ (7) it reads
A μ → − A μ , B μ → B μ , (22) A_\mu\to-A_\mu,\qquad B_\mu\to B_\mu, \tag*{(22)} A μ → − A μ , B μ → B μ , (22) that is, the residual charge conjugation of the Higgsed U ( 1 ) U(1) U ( 1 ) ξ = ξ ^ \xi=\hat{\xi} ξ = ξ ^
The consequence we need follows at once from (8) and (22) :
Z 2 : J B μ → + J B μ , J A μ → − J A μ . (23) Z_2:\qquad J_B^\mu\to+J_B^\mu,\qquad J_A^\mu\to-J_A^\mu. \tag*{(23)} Z 2 : J B μ → + J B μ , J A μ → − J A μ . (23) The two currents carry opposite parity under an exact symmetry, so that ⟨ J A J B ⟩ \langle J_AJ_B\rangle ⟨ J A J B ⟩ (23) . The diagonal and the mixed channel are therefore two independent spectral problems, built on the two hyperbolic planes of (16) , with no amplitude connecting them. In particular the question of boundary value which § 1.4 raises for Π B \Pi_B Π B Π A \Pi_A Π A § 2 are protected from it by symmetry rather than by the numerical check we also report. The same Z 2 Z_2 Z 2
The polarization at finite temperature and density
We proceed to the computation of the retarded current-current correlator [21 ] at vanishing external three-momentum, which is the kinematics of the optical conductivity, in the imaginary-time formalism [22 ] . It is convenient to write the propagator in Hamiltonian form. With α i = γ 0 γ i \alpha^{i} = \gamma^{0}\gamma^{i} α i = γ 0 γ i β = γ 0 \beta= \gamma^{0} β = γ 0
H ( k , M ) = α ⋅ k + β M , H 2 = k 2 + M 2 ≡ E 2 , (24) H(\mathbf{k},M) = \boldsymbol{\alpha} \cdot\mathbf{k} + \beta M,\qquad H^{2} = \mathbf{k}^{2} + M^{2} \equiv E^{2}, \tag*{(24)} H ( k , M ) = α ⋅ k + β M , H 2 = k 2 + M 2 ≡ E 2 , (24) the Matsubara propagator is S = γ 0 ( i ω ~ n − H ) − 1 S = \gamma^{0}(i\widetilde{\omega}_{n} - H)^{-1} S = γ 0 ( i ω n − H ) − 1 ω ~ n = ω n + i μ \widetilde{\omega}_{n} = \omega_{n} + i\mu ω n = ω n + i μ
P s ( k , M ) = 1 2 ( 1 + s H ( k , M ) E ) , s = ± 1 , (25) P_{s}(\mathbf{k},M) = \frac{1}{2}\left(1+s\frac{H(\mathbf{k},M)}{E}\right),\qquad s=\pm1, \tag*{(25)} P s ( k , M ) = 2 1 ( 1 + s E H ( k , M ) ) , s = ± 1 , (25) remain idempotent and complete for complex M M M
T s s ′ ( k ) ≡ 1 3 ∑ i tr [ α i P s ( k , M 1 ) α i P s ′ ( k , M 2 ) ] = 1 − s s ′ k 2 / 3 + M 1 M 2 E 1 E 2 , (26) \begin{aligned}
T_{ss^{\prime}}(\mathbf{k}) \equiv\frac{1}{3}\sum_{i}\operatorname{tr}\left[\alpha^{i}P_{s}(\mathbf{k},M_{1})\alpha^{i}P_{s^{\prime}}(\mathbf{k},M_{2})\right] \\
&= 1-ss^{\prime}\frac{\mathbf{k}^{2}/3+M_{1}M_{2}}{E_{1}E_{2}}, \tag*{(26)}
\end{aligned} T s s ′ ( k ) ≡ 3 1 i ∑ tr [ α i P s ( k , M 1 ) α i P s ′ ( k , M 2 ) ] = 1 − s s ′ E 1 E 2 k 2 /3 + M 1 M 2 , (26) with E 1 , 2 = k 2 + M 1 , 2 2 E_{1,2} = \sqrt{\mathbf{k}^{2}+M_{1,2}^{2}} E 1 , 2 = k 2 + M 1 , 2 2 i q l → ω + i 0 iq_{l}\to\omega+i0 i q l → ω + i 0
Π ( ω ) = ∑ s , s ′ ∫ d 3 k ( 2 π ) 3 T s s ′ ( k ) n F ( s E 1 − μ ) − n F ( s ′ E 2 − μ ) ω + i 0 − s E 1 + s ′ E 2 , (27) \Pi(\omega)=\sum_{s,s^{\prime}}\int\frac{d^{3}k}{(2\pi)^{3}}\,T_{ss^{\prime}}(\mathbf{k})\frac{n_{F}(sE_{1}-\mu)-n_{F}(s^{\prime}E_{2}-\mu)}{\omega+i0-sE_{1}+s^{\prime}E_{2}}, \tag*{(27)} Π ( ω ) = s , s ′ ∑ ∫ ( 2 π ) 3 d 3 k T s s ′ ( k ) ω + i 0 − s E 1 + s ′ E 2 n F ( s E 1 − μ ) − n F ( s ′ E 2 − μ ) , (27) with n F ( z ) = ( e z / T + 1 ) − 1 n_{F}(z)=(e^{z/T}+1)^{-1} n F ( z ) = ( e z / T + 1 ) − 1 (40) . The dissipative part of the conductivity is
σ 1 ( ω ) = Im Π ( ω ) ω . (28) \sigma_{1}(\omega)=\frac{\operatorname{Im}\Pi(\omega)}{\omega}. \tag*{(28)} σ 1 ( ω ) = ω Im Π ( ω ) . (28) The two channels of (8) correspond to the two assignments
Π B = Π ( m , m ) + Π ( m ∗ , m ∗ ) , Π A = Π ( m , m ∗ ) + Π ( m ∗ , m ) , (29) \Pi_{B}=\Pi(m,m)+\Pi(m^{*},m^{*}),\qquad
\Pi_{A}=\Pi(m,m^{*})+\Pi(m^{*},m), \tag*{(29)} Π B = Π ( m , m ) + Π ( m ∗ , m ∗ ) , Π A = Π ( m , m ∗ ) + Π ( m ∗ , m ) , (29) and in both of them the two terms are exchanged by complex conjugation, since m m ∗ = ∣ m ∣ 2 mm^{*}=|m|^{2} m m ∗ = ∣ m ∣ 2 Π B \Pi_{B} Π B Π A \Pi_{A} Π A [1 ] , and it holds here at every point of the momentum integration and for every temperature and chemical potential. What distinguishes the two channels is not the reality of the result but the position of the singularities that the denominators of (27) produce.
Writing E k = k 2 + m 2 = ε k + i Γ k E_{k}=\sqrt{\mathbf{k}^{2}+m^{2}}=\varepsilon_{k}+i\Gamma_{k} E k = k 2 + m 2 = ε k + i Γ k
ε k Γ k = N a v , ε k 2 = 1 2 [ Δ k + Δ k 2 + 4 N 2 a 2 v 2 ] , (30) \varepsilon_{k}\Gamma_{k}=Nav,\qquad
\varepsilon_{k}^{2}=\frac{1}{2}\left[\Delta_{k}+\sqrt{\Delta_{k}^{2}+4N^{2}a^{2}v^{2}}\right], \tag*{(30)} ε k Γ k = N a v , ε k 2 = 2 1 [ Δ k + Δ k 2 + 4 N 2 a 2 v 2 ] , (30) with Δ k = k 2 + N 2 − a 2 v 2 \Delta_{k}=\mathbf{k}^{2}+N^{2}-a^{2}v^{2} Δ k = k 2 + N 2 − a 2 v 2 ε 0 = N \varepsilon_{0}=N ε 0 = N Γ 0 = a v \Gamma_{0}=av Γ 0 = a v
Γ k = a v N ε k ≤ a v . (31) \Gamma_{k}=av\frac{N}{\varepsilon_{k}}\le av. \tag*{(31)} Γ k = a v ε k N ≤ a v . (31) The imaginary part of the mass therefore acts as a rest-frame rate a v av a v ε k \varepsilon_{k} ε k (31) , which follows from ε k ≥ ε 0 = N \varepsilon_{k}\ge\varepsilon_{0}=N ε k ≥ ε 0 = N § 2 sharp. The fact that the two members of the conjugate pair carry opposite signs of Γ k \Gamma_{k} Γ k Ψ m \Psi_{m} Ψ m Ψ m ∗ \Psi_{m^{*}} Ψ m ∗ P T PT P T [9 , 19 ] and, in the present model, is enforced by the Z 2 \mathbb{Z}_{2} Z 2 [19 ] .
The thermal continuation is exact
The occupation numbers in (27) are evaluated at the complex energies ± ( ε k + i Γ k ) \pm(\varepsilon_{k}+i\Gamma_{k}) ± ( ε k + i Γ k ) n F n_{F} n F z = i ( 2 n + 1 ) π T z=i(2n+1)\pi T z = i ( 2 n + 1 ) π T Γ k \Gamma_{k} Γ k π T \pi T π T (27) is exact for arbitrary complex masses.
The identity under discussion is
T ∑ n 1 ( i ω n + i q l + μ − u ) ( i ω n + μ − w ) = n F ( w − μ ) − n F ( u − μ ) i q l − u + w . (32) T\sum_{n}\frac{1}{(i\omega_{n}+iq_{l}+\mu-u)(i\omega_{n}+\mu-w)}
=\frac{n_{F}(w-\mu)-n_{F}(u-\mu)}{iq_{l}-u+w}. \tag*{(32)} T n ∑ ( i ω n + i q l + μ − u ) ( i ω n + μ − w ) 1 = i q l − u + w n F ( w − μ ) − n F ( u − μ ) . (32) with u , w ∈ C u,w\in\mathbb{C} u , w ∈ C f ( z ) f(z) f ( z ) z = i ω n z=i\omega_{n} z = i ω n n F ( z ) n_{F}(z) n F ( z ) n F n_{F} n F − T -T − T f f f z − 2 z^{-2} z − 2
T ∑ n f ( i ω n ) = − ∑ j Res z j [ n F ( z ) f ( z ) ] , (33) T\sum_{n}f(i\omega_{n})=-\sum_{j}\operatorname{Res}_{z_{j}}\left[n_{F}(z)f(z)\right], \tag*{(33)} T n ∑ f ( i ω n ) = − j ∑ Res z j [ n F ( z ) f ( z ) ] , (33) the sum running over the poles z j z_{j} z j f f f f f f u u u w w w (32) being meromorphic in ( u , w ) (u,w) ( u , w ) f f f n F n_F n F
We have verified (32) by direct summation, with symmetric truncation at 4 × 10 6 4 \times10^6 4 × 1 0 6 n − 2 n^{-2} n − 2 Γ / π T = 0.3 \Gamma/\pi T = 0.3 Γ/ π T = 0.3 0.9 0.9 0.9 1.1 1.1 1.1 1.7 1.7 1.7 2.4 2.4 2.4 3.3 3.3 3.3 n F n_F n F ε \varepsilon ε μ \mu μ q l ≠ 0 q_l \ne0 q l = 0 q l = 0 q_l = 0 q l = 0
It remains to locate the obstruction. A pole of f f f n F n_F n F
ε k = μ and Γ k = ( 2 n + 1 ) π T (34) \varepsilon_k = \mu\qquad\text{and} \qquad\Gamma_k = (2n+1)\pi T \tag*{(34)} ε k = μ and Γ k = ( 2 n + 1 ) π T (34) hold simultaneously, two real conditions on one integration variable. However, (31) ties the two together: at ε k = μ \varepsilon_k = \mu ε k = μ N a v / μ N av/\mu N a v / μ (34) becomes a single condition on the parameters,
a v = ( 2 n + 1 ) μ π T N , (35) av = (2n+1)\frac{\mu\pi T}{N}, \tag*{(35)} a v = ( 2 n + 1 ) N μ π T , (35) a locus of codimension one. On the n = 0 n=0 n = 0 ω = 2 μ \omega= 2\mu ω = 2 μ ( ω − 2 μ ) − 1 (\omega-2\mu)^{-1} ( ω − 2 μ ) − 1 σ 1 A ( 2 μ ) \sigma_1^A(2\mu) σ 1 A ( 2 μ )
The diagonal channel and the choice of boundary value
We examine first the channel of the massless gauge field, where the two terms of (29) are complex conjugates of one another and the question of which boundary value to take becomes acute. In Π ( m , m ) \Pi(m,m) Π ( m , m ) E 1 = E 2 E_1=E_2 E 1 = E 2 s = s ′ s=s' s = s ′ (27) have an identically vanishing numerator, while the interband terms have the denominators ω − 2 s E k \omega-2sE_k ω − 2 s E k (30) the quantity 2 E k = 2 ε k + 2 i Γ k 2E_k=2\varepsilon_k+2i\Gamma_k 2 E k = 2 ε k + 2 i Γ k k k k Π ( m , m ) \Pi(m,m) Π ( m , m ) ω \omega ω Π ( m ∗ , m ∗ ) \Pi(m^*,m^*) Π ( m ∗ , m ∗ ) Π B \Pi_B Π B
Im Π B ( ω ) = 0 ⟹ σ 1 B ( ω ) = π D B δ ( ω ) , σ 1 B ( ω ≠ 0 ) ≡ 0 , (36) \operatorname{Im}\Pi_B(\omega)=0
\qquad\Longrightarrow\qquad
\sigma_1^B(\omega)=\pi D_B\delta(\omega),\qquad\sigma_1^B(\omega\ne0)\equiv0, \tag*{(36)} Im Π B ( ω ) = 0 ⟹ σ 1 B ( ω ) = π D B δ ( ω ) , σ 1 B ( ω = 0 ) ≡ 0 , (36) the remaining weight D B D_B D B (36) numerically by evaluating (27) with a finite regulator i Δ i\Delta i Δ i 0 i0 i 0 Im Π B \operatorname{Im}\Pi_B Im Π B Δ \Delta Δ Im Π A \operatorname{Im}\Pi_A Im Π A P T \mathcal{PT} P T [20 ] ; there it arises between a generalized commutator and the norm corrections of a non-unitary evolution, and here between the two members of the conjugate pair, which we return to in Section VII.
The two settings are not interchangeable, however, and the difference decides whether the norm corrections of [20 ] ought to have appeared in (27) . Their non-Hermitian Hamiltonian generates a postselected evolution: the density matrix obeys i ℏ ∂ t ρ = H ρ − ρ H † i\hbar\partial_t\rho=H\rho-\rho H^\dagger i ℏ ∂ t ρ = H ρ − ρ H † tr ρ \operatorname{tr}\rho tr ρ (12) gives
d d t ⟨ ψ ∣ η ∣ ψ ⟩ = i ⟨ ψ ∣ ( H † η − η H ) ∣ ψ ⟩ = 0 , (37) \frac{d}{dt}\langle\psi|\eta|\psi\rangle
= i\langle\psi|(H^\dagger\eta-\eta H)|\psi\rangle=0, \tag*{(37)} d t d ⟨ ψ ∣ η ∣ ψ ⟩ = i ⟨ ψ ∣ ( H † η − η H ) ∣ ψ ⟩ = 0 , (37) exactly, so that what the evolution preserves is the indefinite norm and no trace decays. A decaying positive norm and a conserved indefinite one are different structures, and the absence of a norm correction in (27) is a consequence of the second rather than an omission.
The content of (36) is that the absorption which the loss mode would produce is exactly compensated by the emission of the gain mode, at every frequency and for any temperature and density, and its structural counterpart is the one anticipated in Section II C: this is the Z 2 \mathbb{Z}_2 Z 2 (16) out of modes which, by (14) , are individually null. Put in another way, the pair-creation threshold which in an ordinary Dirac medium opens at twice the mass has here been carried into the complex ω \omega ω Π B \Pi_B Π B ω = ± 2 E k \omega=\pm2E_k ω = ± 2 E k 2 Γ k = 4 N a v / ω 2\Gamma_k=4N av/\omega 2 Γ k = 4 N a v / ω ω \omega ω η \eta η η ≳ 2 Γ k \eta\gtrsim2\Gamma_k η ≳ 2 Γ k a v → 0 av\to0 a v → 0
The mixed channel
On the other hand, in Π ( m , m ∗ ) \Pi(m,m^*) Π ( m , m ∗ )
E 1 − E 2 = 2 i Γ k , E 1 + E 2 = 2 ε k , (38) E_1-E_2=2i\Gamma_k,\qquad E_1+E_2=2\varepsilon_k, \tag*{(38)} E 1 − E 2 = 2 i Γ k , E 1 + E 2 = 2 ε k , (38) the first purely imaginary and the second real. The interband denominators are therefore ω − 2 s ε k \omega-2s\varepsilon_k ω − 2 s ε k ε k ≥ ε 0 = N \varepsilon_k\ge\varepsilon_0=N ε k ≥ ε 0 = N
ω = 2 ε 0 = 2 N , that is p 2 = ( m + m ∗ ) 2 = 4 N 2 , (39) \omega=2\varepsilon_0=2N,\qquad\text{that is}\qquad p^2=(m+m^*)^2=4N^2, \tag*{(39)} ω = 2 ε 0 = 2 N , that is p 2 = ( m + m ∗ ) 2 = 4 N 2 , (39) which is precisely the threshold of the mixed pair found in [1 ] , here recovered as the optical gap of the broken channel. Carrying out the momentum integration with the help of (30) we obtain the closed form
σ 1 A ( ω ) = − k ω T + − F 2 π ω ( ω 2 + 8 N 2 a 2 v 2 ω 3 ) , (40) \sigma_1^A(\omega)=-\frac{k_\omega T_{+-}F}{2\pi\omega}\left(\frac{\omega}{2}+\frac{8N^2a^2v^2}{\omega^3}\right), \tag*{(40)} σ 1 A ( ω ) = − 2 π ω k ω T +− F ( 2 ω + ω 3 8 N 2 a 2 v 2 ) , (40) where T + − T_{+-} T +− (26) at s = + s=+ s = + s ′ = − s'=- s ′ = − ( M 1 , M 2 ) = ( m , m ∗ ) (M_1,M_2)=(m,m^*) ( M 1 , M 2 ) = ( m , m ∗ ) ε = ω / 2 \varepsilon=\omega/2 ε = ω /2 Γ = 2 N a v / ω \Gamma=2Nav/\omega Γ = 2 N a v / ω
k ω 2 = ω 2 4 − N 2 + a 2 v 2 − 4 N 2 a 2 v 2 ω 2 , T + − = 1 + k ω 2 / 3 + N 2 + a 2 v 2 ω 2 / 4 + 4 N 2 a 2 v 2 / ω 2 , F = Re [ n F ( ε + i Γ − μ ) − n F ( − ε + i Γ − μ ) ] . (41) \begin{aligned}
k_\omega^2&=\frac{\omega^2}{4}-N^2+a^2v^2-\frac{4N^2a^2v^2}{\omega^2},\\
T_{+-}&=1+\frac{k_\omega^2/3+N^2+a^2v^2}{\omega^2/4+4N^2a^2v^2/\omega^2},\\
F&=\operatorname{Re}\left[n_F(\varepsilon+i\Gamma-\mu)-n_F(-\varepsilon+i\Gamma-\mu)\right]. \tag*{(41)}
\end{aligned} k ω 2 T +− F = 4 ω 2 − N 2 + a 2 v 2 − ω 2 4 N 2 a 2 v 2 , = 1 + ω 2 /4 + 4 N 2 a 2 v 2 / ω 2 k ω 2 /3 + N 2 + a 2 v 2 , = Re [ n F ( ε + i Γ − μ ) − n F ( − ε + i Γ − μ ) ] . (41) It is straightforward to confirm that k ω k_\omega k ω ω = 2 N \omega=2N ω = 2 N a v → 0 av\to0 a v → 0 (40) reduces to twice the free Dirac result ω 12 π ( 1 + 2 N 2 ω 2 ) 1 − 4 N 2 ω 2 \frac{\omega}{12\pi}\left(1+\frac{2N^2}{\omega^2}\right)\sqrt{1-\frac{4N^2}{\omega^2}} 12 π ω ( 1 + ω 2 2 N 2 ) 1 − ω 2 4 N 2 (27) . We have also checked (40) against a direct numerical evaluation of (27) , with agreement to four significant figures.
The intraband sector of this channel deserves a separate comment, since it is the one in which the width (31) becomes visible. From (38) the four intraband terms of the two orderings combine into
Σ i n t r a = − 8 Γ k ( ζ k + ζ ˉ k ) ω 2 + 4 Γ k 2 , ζ k = Im n F ( ε k + i Γ k − μ ) , ζ ˉ k = Im n F ( − ε k + i Γ k − μ ) , (42) \Sigma_{\mathrm{intra}}=-\frac{8\Gamma_k(\zeta_k+\bar{\zeta}_k)}{\omega^2+4\Gamma_k^2},\qquad
\zeta_k=\operatorname{Im}n_F(\varepsilon_k+i\Gamma_k-\mu),\qquad
\bar{\zeta}_k=\operatorname{Im}n_F(-\varepsilon_k+i\Gamma_k-\mu), \tag*{(42)} Σ intra = − ω 2 + 4 Γ k 2 8 Γ k ( ζ k + ζ ˉ k ) , ζ k = Im n F ( ε k + i Γ k − μ ) , ζ ˉ k = Im n F ( − ε k + i Γ k − μ ) , (42) a Lorentzian of half-width 2 Γ k 2\Gamma_k 2 Γ k 2 Γ F = 2 N a v / μ 2\Gamma_F=2Nav/\mu 2 Γ F = 2 N a v / μ (42) is real, and therefore contributes to the reactive part of the conductivity and not to σ 1 \sigma_1 σ 1 q = 0 q=0 q = 0 0 / 0 0/0 0/0
The response kernel in time
We proceed to answer the question which we regard as the more revealing one. The kernel of the linear response, J ( t ) = ∫ d t ′ σ ( t − t ′ ) E ( t ′ ) J(t)=\int dt'\,\sigma(t-t')E(t') J ( t ) = ∫ d t ′ σ ( t − t ′ ) E ( t ′ ) σ ( ω ) = i K ( ω ) / ( ω + i 0 ) \sigma(\omega)=iK(\omega)/(\omega+i0) σ ( ω ) = i K ( ω ) / ( ω + i 0 ) K = Π d i a + Π K=\Pi_{\mathrm{dia}}+\Pi K = Π dia + Π Π d i a \Pi_{\mathrm{dia}} Π dia Π d i a θ ( t ) \Pi_{\mathrm{dia}}\,\theta(t) Π dia θ ( t ) (42) , whose poles sit at ω = ± 2 i Γ k \omega=\pm2i\Gamma_k ω = ± 2 i Γ k
σ i n t r a ( t ) σ i n t r a ( ∞ ) = { 1 − 1 2 e − 2 Γ k t , t > 0 , 1 2 e + 2 Γ k t , t < 0 , (43) \frac{\sigma_{\mathrm{intra}}(t)}{\sigma_{\mathrm{intra}}(\infty)}=
\begin{cases}
1-\frac{1}{2}e^{-2\Gamma_k t},&t>0,\\
\frac{1}{2}e^{+2\Gamma_k t},&t<0,
\end{cases} \tag*{(43)} σ intra ( ∞ ) σ intra ( t ) = { 1 − 2 1 e − 2 Γ k t , 2 1 e + 2 Γ k t , t > 0 , t < 0 , (43) a continuous function which rises from zero at t → − ∞ t\to-\infty t → − ∞ t → + ∞ t\to+\infty t → + ∞
τ p r e c = 1 2 Γ F = μ 2 N a v . (44) \tau_{\mathrm{prec}}=\frac{1}{2\Gamma_F}=\frac{\mu}{2Nav}. \tag*{(44)} τ prec = 2 Γ F 1 = 2 N a v μ . (44) In the vacuum the intraband sector is absent, there being no carriers, and the precursor is then generated by the branch point at ω = 2 N + 2 i a v \omega=2N+2iav ω = 2 N + 2 ia v 1 / 2 a v 1/2av 1/2 a v μ \mu μ T T T [13 , 14 ] , here exhibited in a transport coefficient rather than in a scattering amplitude, and it is the price of the exact cancellation reported in Section V: a response function that is analytic in the upper half plane and has no imaginary part on the real axis is necessarily a constant, so that the vanishing of the absorptive part of Section V and the precursor of (43) are two readings of the same fact. One should add that (43) follows from the analytic continuation of the Euclidean correlator, which is the definition adopted in [1 ] , and that the paragraphs below show this to be the wrong assignment of boundary value for a retarded response, so that the precursor is an artifact of that assignment and not a property of the model. We note, in contrast with the Lee–Wick case, that (31) would make the window (44) grow with the energy of the mode, so that the acausality carried by the common boundary value is a phenomenon of long times and low frequencies rather than a short-distance disease.
Before the alternative, a word on what the choice of contour is doing, because the precursor is a property of that choice and not of the model, and the alternative to it is not innocuous either. Take a single intraband mode, with meromorphic coefficient K k ( z ) = C k / ( z 2 + a k 2 ) K_k(z)=C_k/(z^2+a_k^2) K k ( z ) = C k / ( z 2 + a k 2 ) a k = 2 Γ k > 0 a_k=2\Gamma_k>0 a k = 2 Γ k > 0 σ k ( z ) = − i K k ( z ) / z \sigma_k(z)=-iK_k(z)/z σ k ( z ) = − i K k ( z ) / z σ k = i σ 0 k ( z − 1 − z / ( z 2 + a k 2 ) ) \sigma_k=i\sigma_{0k}(z^{-1}-z/(z^2+a_k^2)) σ k = i σ 0 k ( z − 1 − z / ( z 2 + a k 2 )) σ 0 k = − C k / a k 2 \sigma_{0k}=-C_k/a_k^2 σ 0 k = − C k / a k 2 (43) : bounded, and noncausal. Inverting instead along Im z = γ > a k \operatorname{Im} z = \gamma> a_k Im z = γ > a k t > 0 t > 0 t > 0 t < 0 t < 0 t < 0
σ k c a u s a l ( t ) σ 0 k = θ ( t ) [ 1 − cosh ( a k t ) ] , (45) \frac{\sigma_k^{\mathrm{causal}}(t)}{\sigma_{0k}} = \theta(t)\left[1-\cosh(a_k t)\right], \tag*{(45)} σ 0 k σ k causal ( t ) = θ ( t ) [ 1 − cosh ( a k t ) ] , (45) which is causal and unstable, with no finite late-time limit. The two differ by the homogeneous growing solution,
σ k a x i s ( t ) − σ k c a u s a l ( t ) σ 0 k = 1 2 e a k t , (46) \frac{\sigma_k^{\mathrm{axis}}(t)-\sigma_k^{\mathrm{causal}}(t)}{\sigma_{0k}} = \frac{1}{2}e^{a_k t}, \tag*{(46)} σ 0 k σ k axis ( t ) − σ k causal ( t ) = 2 1 e a k t , (46) both satisfying f ′ ′ − a k 2 f = − a k 2 f^{\prime\prime}-a_k^2 f=-a_k^2 f ′′ − a k 2 f = − a k 2 t > 0 t>0 t > 0 f ( 0 ) = f ′ ( 0 ) = 0 f(0)=f^\prime(0)=0 f ( 0 ) = f ′ ( 0 ) = 0 f ( 0 ) = 1 / 2 f(0)=1/2 f ( 0 ) = 1/2 f ′ ( 0 ) = a k / 2 f^\prime(0)=a_k/2 f ′ ( 0 ) = a k /2
With that said, the alternative for the diagonal channel. The cancellation of § 1.4 rests on assigning the same boundary value ω + i 0 \omega+i0 ω + i 0 (29) , although each has its singularities on a different side of the real axis. Continuing each term instead from the half plane in which it is analytic replaces 2 Re Π ( m , m ) 2\operatorname{Re}\Pi(m,m) 2 Re Π ( m , m ) 2 Π ( m ∗ , m ∗ ) 2\Pi(m^*,m^*) 2Π ( m ∗ , m ∗ ) 2 Γ k 2\Gamma_k 2 Γ k
σ 1 B , r e t ( ω ) = 2 ω Im Π ( m ∗ , m ∗ ) ( ω ) ≠ 0. (47) \sigma_1^{B,\mathrm{ret}}(\omega)=\frac{2}{\omega}\operatorname{Im}\Pi(m^*,m^*)(\omega)\ne0. \tag*{(47)} σ 1 B , ret ( ω ) = ω 2 Im Π ( m ∗ , m ∗ ) ( ω ) = 0. (47) The two assignments do not differ by a small correction. They differ by the whole of the Hermitian absorption, as Table I shows, and (47) carries the ordinary ultraviolet behaviour of the massless theory, σ 1 → ω / 6 π \sigma_1\to\omega/6\pi σ 1 → ω /6 π σ 1 B , r e t / ω = 0.05304 \sigma_1^{B,\mathrm{ret}}/\omega=0.05304 σ 1 B , ret / ω = 0.05304 ω = 100 N \omega=100\,N ω = 100 N a v = 0.40 N av=0.40\,N a v = 0.40 N 1 / 6 π = 0.05305 1/6\pi=0.05305 1/6 π = 0.05305
Table I. The two assignments of boundary value, in units N = 1 N=1 N = 1 T = 0.15 N T=0.15N T = 0.15 N μ = 1.4 N \mu=1.4N μ = 1.4 N a v = 0.40 N av=0.40N a v = 0.40 N ω + i 0 \omega+i0 ω + i 0 (36) ; the third continues each term from its own half plane, Eq. (47) ; the fourth is the same integral at a v → 0 av\to0 a v → 0
It should be said plainly that this second assignment does not produce a retarded function either. The interband sum contains both signs of frequency, and with the surviving mass m ∗ m^* m ∗
ω = + 2 ε k − 2 i Γ k , ω = − 2 ε k + 2 i Γ k , (48) \omega=+2\varepsilon_k-2i\Gamma_k,\qquad\omega=-2\varepsilon_k+2i\Gamma_k, \tag*{(48)} ω = + 2 ε k − 2 i Γ k , ω = − 2 ε k + 2 i Γ k , (48) the second of which lies in the upper half plane for every k \mathbf{k} k
One would want to settle the matter with the optical sum rule, and it cannot be done, for a reason that Section IX establishes as a theorem rather than an expectation. The rule in its familiar form, and with the requirement that a deformation by a mass leave the ultraviolet asymptotics alone, is a statement about a Källén–Lehmann representation with a positive measure . The diagonal channel has none: its Euclidean correlator (69) changes sign in τ \tau τ (71) at which the moment sequence first fails. Every repair one might attempt is closed in Table II , the choice of boundary value among them — that test is performed in the Euclidean region and never continues to real ω \omega ω
Table II. The ways out of (69) , and why each is closed.
We are careful about what this does and does not say. The current J B J_B J B η \eta η + 1 +1 + 1 (37) . But the η \eta η η \eta η (36) and the precursor of (43) are therefore statements about a correlator with no positive spectral representation, and the choice of boundary value for Π B \Pi_B Π B Table I retains its full interest as the measure of what the two assignments differ by. We regard the question as disposed of rather than answered, and prefer to say so.
The physical content sits in the mixed channel, and there the discussion cannot reach, for two independent reasons. The first is kinematic: the interband singularities sit at ω = 2 ε k \omega=2\varepsilon_k ω = 2 ε k Π A \Pi_A Π A Im Π A \operatorname{Im}\Pi_A Im Π A (23) the two currents have opposite Z 2 Z_2 Z 2 ⟨ J A J B ⟩ = 0 \langle J_AJ_B\rangle=0 ⟨ J A J B ⟩ = 0
One further question belongs here, and § 2 answers it: where the ultraviolet weight of the theory actually is. Doing the vacuum integral of the mixed channel in closed form shows that the whole of the free two-flavor asymptotics ω / 6 π \omega/6\pi ω /6 π Table I measures is not missing from the model but sits where no prescription is in question.
Positivity of the spectral density
It remains to examine the sign of the spectral density, a question that the scalar computation of [1 ] left open.
The vacuum
For T → 0 T \to0 T → 0 μ = 0 \mu= 0 μ = 0 (41) reduces to F = − 1 F = -1 F = − 1 T + − T_{+-} T +− (26) is
T + − = 1 + k 2 / 3 + ∣ m ∣ 2 ∣ E k ∣ 2 > 0 , (49) T_{+-} = 1 + \frac{\mathbf{k}^{2}/3 + \lvert m\rvert^{2}}{\lvert E_k\rvert^{2}} > 0, \tag*{(49)} T +− = 1 + ∣ E k ∣ 2 k 2 /3 + ∣ m ∣ 2 > 0 , (49) strictly positive, since E 1 E 2 = ∣ k 2 + m 2 ∣ E_1 E_2 = \lvert\mathbf{k}^{2} + m^{2}\rvert E 1 E 2 = ∣ k 2 + m 2 ∣ k \mathbf{k} k
σ 1 A ( ω ) = 2 π ω ∫ d 3 k ( 2 π ) 3 T + − ( k ) δ ( ω − 2 ε k ) > 0 , (50) \sigma_1^{A}(\omega) = \frac{2\pi}{\omega}\int\frac{\mathrm{d}^{3}k}{(2\pi)^{3}}\,T_{+-}(\mathbf{k})\,\delta(\omega- 2\varepsilon_k) > 0, \tag*{(50)} σ 1 A ( ω ) = ω 2 π ∫ ( 2 π ) 3 d 3 k T +− ( k ) δ ( ω − 2 ε k ) > 0 , (50) so that in the vector channel the transverse spectral density is not merely real but positive, and manifestly so. To our knowledge this is the first positivity statement available for a composite correlator of this model; Section IX will promote it from a computed outcome to a theorem, by showing that the Euclidean correlator of this channel is completely monotone. The qualifier is not decorative: at q = 0 \mathbf{q} = 0 q = 0
The delta function can be removed and the integral done in closed form. With D k = ∣ E k ∣ 2 = ε k 2 + Γ k 2 D_k = \lvert E_k\rvert^{2} = \varepsilon_k^{2} + \Gamma_k^{2} D k = ∣ E k ∣ 2 = ε k 2 + Γ k 2 ε k Γ k = N a v \varepsilon_k\Gamma_k=N av ε k Γ k = N a v
k 2 = ( ε k 2 − N 2 ) ( ε k 2 + a 2 v 2 ) ε k 2 , k d k d ε k = D k ε k , (51) \mathbf{k}^2=\frac{(\varepsilon_k^2-N^2)(\varepsilon_k^2+a^2v^2)}{\varepsilon_k^2},\qquad k\frac{dk}{d\varepsilon_k}=\frac{D_k}{\varepsilon_k}, \tag*{(51)} k 2 = ε k 2 ( ε k 2 − N 2 ) ( ε k 2 + a 2 v 2 ) , k d ε k d k = ε k D k , (51) so that in the variable s = 4 ε k 2 s=4\varepsilon_k^2 s = 4 ε k 2
λ ( s ) = ( s − 4 N 2 ) ( s + 4 a 2 v 2 ) , k = λ ( s ) 2 s , (52) \lambda(s)=(s-4N^2)(s+4a^2v^2),\qquad k=\frac{\sqrt{\lambda(s)}}{2\sqrt{s}}, \tag*{(52)} λ ( s ) = ( s − 4 N 2 ) ( s + 4 a 2 v 2 ) , k = 2 s λ ( s ) , (52) one has d 3 k / ( 2 π ) 3 = ( 2 π 2 ) − 1 k ( D k / ε k ) d ε k d^3k/(2\pi)^3=(2\pi^2)^{-1}k(D_k/\varepsilon_k)d\varepsilon_k d 3 k / ( 2 π ) 3 = ( 2 π 2 ) − 1 k ( D k / ε k ) d ε k (49) collapses algebraically: writing out D k D_k D k k 2 \mathbf{k}^2 k 2 ε k \varepsilon_k ε k
D k T + − = D k + 1 3 k 2 + ∣ m ∣ 2 = 2 3 ( 2 ε k 2 + N 2 ) ( 1 + a 2 v 2 ε k 2 ) = ( s + 2 N 2 ) ( s + 4 a 2 v 2 ) 3 s , (53) \begin{aligned}
D_kT_{+-}=D_k+\frac{1}{3}\mathbf{k}^2+|m|^2 \\
&=\frac{2}{3}(2\varepsilon_k^2+N^2)\left(1+\frac{a^2v^2}{\varepsilon_k^2}\right) \\
&=\frac{(s+2N^2)(s+4a^2v^2)}{3s}, \tag*{(53)}
\end{aligned} D k T +− = D k + 3 1 k 2 + ∣ m ∣ 2 = 3 2 ( 2 ε k 2 + N 2 ) ( 1 + ε k 2 a 2 v 2 ) = 3 s ( s + 2 N 2 ) ( s + 4 a 2 v 2 ) , (53) all four terms combining into a product. Evaluating (50) at ε k = ω / 2 \varepsilon_k=\omega/2 ε k = ω /2 ω > 2 N \omega>2N ω > 2 N
σ 1 A ( ω ) = λ ( ω 2 ) ( ω 2 + 2 N 2 ) ( ω 2 + 4 a 2 v 2 ) 6 π ω 5 , (54) \sigma_1^A(\omega)=\frac{\sqrt{\lambda(\omega^2)}(\omega^2+2N^2)(\omega^2+4a^2v^2)}{6\pi\omega^5}, \tag*{(54)} σ 1 A ( ω ) = 6 π ω 5 λ ( ω 2 ) ( ω 2 + 2 N 2 ) ( ω 2 + 4 a 2 v 2 ) , (54) and zero below. Two limits check it. At a v → 0 av\to0 a v → 0 ( ω / 6 π ) ( 1 + 2 N 2 / ω 2 ) 1 − 4 N 2 / ω 2 (\omega/6\pi)(1+2N^2/\omega^2)\sqrt{1-4N^2/\omega^2} ( ω /6 π ) ( 1 + 2 N 2 / ω 2 ) 1 − 4 N 2 / ω 2 ρ 1 ( s ) = s / 6 π 2 + a 2 v 2 / π 2 + O ( s − 1 ) \rho_1(s)=s/6\pi^2+a^2v^2/\pi^2+O(s^{-1}) ρ 1 ( s ) = s /6 π 2 + a 2 v 2 / π 2 + O ( s − 1 ) (56) below,
σ 1 A ( ω ) = ω 6 π + a 2 v 2 π ω + O ( ω − 3 ) . (55) \sigma_1^A(\omega)=\frac{\omega}{6\pi}+\frac{a^2v^2}{\pi\omega}+O(\omega^{-3}). \tag*{(55)} σ 1 A ( ω ) = 6 π ω + π ω a 2 v 2 + O ( ω − 3 ) . (55) This settles the question raised in Section VII. The leading term is exactly the free value for the two flavors, the same number that (47) reproduces for the diagonal channel under the second assignment of boundary value. The
ultraviolet weight that Table I measures is therefore not missing from the model: it is carried in full by the mixed channel, whose singularities lie on the real axis and whose prescription is not in question. What the cancellation of (36) says is the narrower thing, that the massless probe does not see it.
The same reduction delivers the other form factor, and there the sign is the other one. The covariant vacuum calculation of Appendix D decomposes the mixed correlator into a transverse and a longitudinal part, with spectral densities
ρ 1 ( s ) = λ ( s ) ( s + 2 N 2 ) ( s + 4 a 2 v 2 ) 6 π 2 s 2 , \rho_1(s)=\frac{\sqrt{\lambda(s)}(s+2N^2)(s+4a^2v^2)}{6\pi^2s^2}, ρ 1 ( s ) = 6 π 2 s 2 λ ( s ) ( s + 2 N 2 ) ( s + 4 a 2 v 2 ) , ρ 0 ( s ) = − a 2 v 2 λ ( s ) ( s − 4 N 2 ) π 2 s 2 , (56) \rho_0(s)=-\frac{a^2v^2\sqrt{\lambda(s)}(s-4N^2)}{\pi^2s^2}, \tag*{(56)} ρ 0 ( s ) = − π 2 s 2 a 2 v 2 λ ( s ) ( s − 4 N 2 ) , (56) both supported on s ≥ 4 N 2 s\ge4N^2 s ≥ 4 N 2 (54) again through σ 1 A ( ω ) = ( π / ω ) ρ 1 ( ω 2 ) \sigma_1^A(\omega)=(\pi/\omega)\rho_1(\omega^2) σ 1 A ( ω ) = ( π / ω ) ρ 1 ( ω 2 ) J A J_A J A m − m ∗ m-m^* m − m ∗ (8) , and it is negative . That this is the non-Hermiticity and not a convention follows from an identity. For any two masses, with E i = k 2 + m i 2 E_i=\sqrt{k^2+m_i^2} E i = k 2 + m i 2
( E 1 E 2 ) 2 − ( k 2 + m 1 m 2 ) 2 = k 2 ( m 1 − m 2 ) 2 , (57) (E_1E_2)^2-(k^2+m_1m_2)^2=k^2(m_1-m_2)^2, \tag*{(57)} ( E 1 E 2 ) 2 − ( k 2 + m 1 m 2 ) 2 = k 2 ( m 1 − m 2 ) 2 , (57) as one sees by expanding both sides. The longitudinal residue is T 0 = 1 − ( k 2 + ∣ m ∣ 2 ) / D k T_0=1-(k^2+|m|^2)/D_k T 0 = 1 − ( k 2 + ∣ m ∣ 2 ) / D k E 1 E 2 − k 2 − m 1 m 2 E_1E_2-k^2-m_1m_2 E 1 E 2 − k 2 − m 1 m 2 m 1 = m m_1=m m 1 = m m 2 = m ∗ m_2=m^* m 2 = m ∗ real masses the right-hand side of (57) is non-negative, so that T 0 ≥ 0 T_0\ge0 T 0 ≥ 0 m 1 = m 2 m_1=m_2 m 1 = m 2 ( m − m ∗ ) 2 = − 4 a 2 v 2 (m-m^*)^2=-4a^2v^2 ( m − m ∗ ) 2 = − 4 a 2 v 2 E 1 E 2 = D k E_1E_2=D_k E 1 E 2 = D k k 2 + ∣ m ∣ 2 k^2+|m|^2 k 2 + ∣ m ∣ 2
D k 2 − ( k 2 + ∣ m ∣ 2 ) 2 = − 4 a 2 v 2 k 2 < 0 , (58) D_k^2-(k^2+|m|^2)^2=-4a^2v^2k^2<0, \tag*{(58)} D k 2 − ( k 2 + ∣ m ∣ 2 ) 2 = − 4 a 2 v 2 k 2 < 0 , (58) so D k < k 2 + ∣ m ∣ 2 D_k<k^2+|m|^2 D k < k 2 + ∣ m ∣ 2 T 0 < 0 T_0<0 T 0 < 0 k ≠ 0 k\ne0 k = 0 D k T 0 = − 2 a 2 v 2 ( s − 4 N 2 ) / s D_kT_0=-2a^2v^2(s-4N^2)/s D k T 0 = − 2 a 2 v 2 ( s − 4 N 2 ) / s (56) . The negative longitudinal weight is thus the square of the mass splitting, and the splitting is imaginary.
Two things follow. The first is that the positivity of (50) is a statement about the transverse channel, which is the one conductivity probes: at q = 0 q=0 q = 0 ρ 0 \rho_0 ρ 0 diag ( ρ 0 , ρ 1 , ρ 1 , ρ 1 ) \operatorname{diag}(\rho_0,\rho_1,\rho_1,\rho_1) diag ( ρ 0 , ρ 1 , ρ 1 , ρ 1 )
What secures (50) , at the level of the computation, is that the residue of the mixed pair is a sum of positive terms, the non-Hermiticity having been confined to the position of the pole and not to its weight; what secures it structurally is (16) , the intermediate state of this channel being the symmetrized conjugate pair, of norm + 1 +1 + 1 η \eta η − 1 -1 − 1 (50) is in that sense not an accident of the Dirac algebra but the norm of the i i i Z 2 \mathbb{Z}_2 Z 2 (23) , not the signature of the metric. Nor does the argument from (16) extend to ρ 0 \rho_0 ρ 0 (58) is a statement about its Lorentz structure, which the one- and two-particle norms of Section II C do not see.
Finite temperature: a sharp bound
In a medium the sign of F F F
t = e ( ε − μ ) / T , c = cos ( Γ / T ) , (59) t=e^{(\varepsilon-\mu)/T}, \qquad c=\cos(\Gamma/T), \tag*{(59)} t = e ( ε − μ ) / T , c = cos ( Γ/ T ) , (59) a short computation gives the closed form
Re n F ( ε + i Γ − μ ) = c t + 1 t 2 + 2 c t + 1 , (60) \operatorname{Re} n_F(\varepsilon+i\Gamma-\mu)=\frac{ct+1}{t^2+2ct+1}, \tag*{(60)} Re n F ( ε + i Γ − μ ) = t 2 + 2 c t + 1 c t + 1 , (60) whose denominator is strictly positive for ∣ c ∣ < 1 |c|<1 ∣ c ∣ < 1
∂ ∂ ε Re n F = − t ( c t 2 + 2 t + c ) T ( t 2 + 2 c t + 1 ) 2 . (61) \frac{\partial}{\partial\varepsilon}\operatorname{Re} n_F=-\frac{t(ct^2+2t+c)}{T(t^2+2ct+1)^2}. \tag*{(61)} ∂ ε ∂ Re n F = − T ( t 2 + 2 c t + 1 ) 2 t ( c t 2 + 2 t + c ) . (61) For c > 0 c>0 c > 0 (61) is strictly positive on t > 0 t>0 t > 0 Re n F \operatorname{Re} n_F Re n F ε \varepsilon ε Γ \Gamma Γ F F F Γ \Gamma Γ ε \varepsilon ε − ε -\varepsilon − ε F < 0 F<0 F < 0 σ 1 A ( ω ) > 0 \sigma_1^A(\omega)>0 σ 1 A ( ω ) > 0 c < 0 c<0 c < 0 Γ > π T / 2 \Gamma>\pi T/2 Γ > π T /2 (31) we arrive at
a v < π T 2 ⟹ σ 1 A ( ω ) > 0 for every ω , (62) a v<\frac{\pi T}{2}\quad\Longrightarrow\quad\sigma_1^A(\omega)>0\ \text{for every }\omega, \tag*{(62)} a v < 2 π T ⟹ σ 1 A ( ω ) > 0 for every ω , (62) a positivity statement at finite temperature and density with an explicit bound, the non-Hermitian scale being measured against the first Matsubara frequency. The bound is sharp: a numerical determination of the true onset a v c a v_c a v c σ 1 A \sigma_1^A σ 1 A T ∈ [ 0.08 , 0.35 ] N T\in[0.08,0.35]\,N T ∈ [ 0.08 , 0.35 ] N μ ∈ [ 1.1 , 4.2 ] N \mu\in[1.1,4.2]\,N μ ∈ [ 1.1 , 4.2 ] N π T / 2 \pi T/2 π T /2 μ / N \mu/N μ / N Figure 2 .
Figure 2. The positivity bound (62) . (a) The real part of the occupation number (60) as a function of ( ε − μ ) / T (\varepsilon-\mu)/T ( ε − μ ) / T Γ \Gamma Γ π T / 2 \pi T/2 π T /2 F < 0 F<0 F < 0 σ 1 A > 0 \sigma_1^A>0 σ 1 A > 0 cos ( Γ / T ) \cos(\Gamma/T) cos ( Γ/ T ) [ 0 , 1 ] [0,1] [ 0 , 1 ] a v c a_{\mathrm{vc}} a vc σ 1 A \sigma_1^A σ 1 A μ / N \mu/N μ / N μ ≫ N \mu\gg N μ ≫ N
Above the bound the medium amplifies
For c < 0 c<0 c < 0 (61) vanishes at t ∓ = ( 1 ∓ 1 − c 2 ) / ∣ c ∣ t_{\mp}=(1\mp\sqrt{1-c^2})/|c| t ∓ = ( 1 ∓ 1 − c 2 ) /∣ c ∣ Re n F \operatorname{Re} n_F Re n F ε \varepsilon ε t < t − t<t_- t < t − t > t + t>t_+ t > t + σ 1 A \sigma_1^A σ 1 A Figure 1 (a) shows the effect inside the Pauli-blocked window 2 N < ω < 2 μ 2N<\omega<2\mu 2 N < ω < 2 μ Figure 2 (a) shows the mechanism, the real part of the occupation number leaving the interval [ 0 , 1 ] [0,1] [ 0 , 1 ] Γ \Gamma Γ π T / 2 \pi T/2 π T /2
Figure 1. Linear response of the complex-mass fermions, in units N = 1 N = 1 N = 1 T = 0.15 N T = 0.15N T = 0.15 N μ = 1.4 N \mu= 1.4N μ = 1.4 N ω = 2 N \omega= 2N ω = 2 N 2 N < ω < 2 μ 2N < \omega< 2\mu 2 N < ω < 2 μ § 2 . (b) The intraband Lorentzian of the mixed channel, normalized at the origin, with the half-height points at ω = 2 Γ F = 2 N a v / μ \omega= 2\Gamma_F = 2Nav/\mu ω = 2 Γ F = 2 N a v / μ (43) , showing the precursor at negative times. (d) The singularities of the polarization in the complex ω \omega ω ω = 2 N \omega= 2N ω = 2 N
The onset can be located exactly. Away from a Matsubara collision the denominator of (61) is positive, so the continued interband coefficient is negative precisely when
cos ( Γ / T ) < − cosh ( ε / T ) cosh ( μ / T ) , (63) \cos(\Gamma/T)<-\frac{\cosh(\varepsilon/T)}{\cosh(\mu/T)}, \tag*{(63)} cos ( Γ/ T ) < − cosh ( μ / T ) cosh ( ε / T ) , (63) which requires ∣ μ ∣ > ε ≥ N |\mu|>\varepsilon\geq N ∣ μ ∣ > ε ≥ N ∣ μ ∣ ≤ N |\mu|\leq N ∣ μ ∣ ≤ N a v a v a v (63) at equality gives a v = ( ε T / N ) arccos [ − cosh ( ε / T ) / cosh ( ∣ μ ∣ / T ) ] a v=(\varepsilon T/N)\arccos[-\cosh(\varepsilon/T)/\cosh(|\mu|/T)] a v = ( εT / N ) arccos [ − cosh ( ε / T ) / cosh ( ∣ μ ∣/ T )] ε \varepsilon ε N ≤ ε < ∣ μ ∣ N\leq\varepsilon<|\mu| N ≤ ε < ∣ μ ∣ ε = N \varepsilon=N ε = N
( a v ) c ( T , μ ) = T arccos [ − cosh ( N / T ) cosh ( ∣ μ ∣ / T ) ] . (64) (a v)_c(T,\mu)=T\arccos\left[-\frac{\cosh(N/T)}{\cosh(|\mu|/T)}\right]. \tag*{(64)} ( a v ) c ( T , μ ) = T arccos [ − cosh ( ∣ μ ∣/ T ) cosh ( N / T ) ] . (64) The spectral weight vanishes at that endpoint, but for a v a v a v ( a v ) c (a v)_c ( a v ) c (63) , and we find the onset numerically at ( a v ) c (a v)_c ( a v ) c ∣ μ ∣ / T → ∞ |\mu|/T\to\infty ∣ μ ∣/ T → ∞ ( a v ) c → π T / 2 (a v)_c\to\pi T/2 ( a v ) c → π T /2 (62) is uniform in the chemical potential and that it is saturated, not merely respected. At larger a v a v a v (64) describes the first onset and not a monotonic phase boundary.
This is a novelty, as, up to now, the discussion of complex conjugate poles has been carried out in the vacuum, where the conjugation pairing is enough to secure reality and, as (50) shows, positivity as well. One should notice that an occupation number outside [ 0 , 1 ] [0,1] [ 0 , 1 ] (14) the diagonal modes of complex mass have vanishing η \eta η F F F
The Euclidean criterion
The two results so far — the positivity of (50) and the vanishing of (36) — are statements on the real frequency axis, and each was argued with a boundary value already chosen. There is a test that needs none, and it settles both channels at once. At vanishing spatial momentum, performing the frequency integral of the one-loop bubble and transforming back in Q 4 Q_4 Q 4 τ > 0 \tau>0 τ > 0
C ( τ ) = ∫ d 3 k ( 2 π ) 3 2 Re { [ 1 + k 2 / 3 + m 1 m 2 E 1 E 2 ] e − ( E 1 + E 2 ) τ } , (65) C(\tau)=\int\frac{d^3k}{(2\pi)^3}\,2\operatorname{Re}\left\{\left[1+\frac{\mathbf{k}^2/3+m_1m_2}{E_1E_2}\right]e^{-(E_1+E_2)\tau}\right\}, \tag*{(65)} C ( τ ) = ∫ ( 2 π ) 3 d 3 k 2 Re { [ 1 + E 1 E 2 k 2 /3 + m 1 m 2 ] e − ( E 1 + E 2 ) τ } , (65) E i = k 2 + m i 2 E_i=\sqrt{\mathbf{k}^2+m_i^2} E i = k 2 + m i 2 on the principal branch. Everything turns on whether E 1 + E 2 E_1+E_2 E 1 + E 2
The criterion is the Osterwalder–Schrader one [6 ] . A field with a positive-norm realization has a reflection-positive Euclidean correlator, and at zero spatial momentum this reduces to the statement that τ ↦ C ( τ ) \tau\mapsto C(\tau) τ ↦ C ( τ ) ( − 1 ) n C ( n ) ( τ ) ≥ 0 (-1)^nC^{(n)}(\tau)\geq0 ( − 1 ) n C ( n ) ( τ ) ≥ 0 n n n τ > 0 \tau>0 τ > 0 ρ ≥ 0 \rho\geq0 ρ ≥ 0 ω \omega ω
The mixed channel: positivity as a theorem
With m 1 = m m_1=m m 1 = m m 2 = m ∗ m_2=m^* m 2 = m ∗ E 1 E 2 = ∣ E k ∣ 2 = D k E_1E_2=\lvert E_k\rvert^2=D_k E 1 E 2 = ∣ E k ∣ 2 = D k E 1 + E 2 = 2 ε k E_1+E_2=2\varepsilon_k E 1 + E 2 = 2 ε k (65) is the residue T + − T_{+-} T +− (49) , and the real part is taken of something already real:
C A ( τ ) = ∫ d 3 k ( 2 π ) 3 T + − ( k ) e − 2 ε k τ . (66) C_A(\tau)=\int\frac{d^3k}{(2\pi)^3}\,T_{+-}(\mathbf{k})e^{-2\varepsilon_k\tau}. \tag*{(66)} C A ( τ ) = ∫ ( 2 π ) 3 d 3 k T +− ( k ) e − 2 ε k τ . (66) This is a positive superposition of decaying exponentials, so that
( − 1 ) n C A ( n ) ( τ ) = ∫ d 3 k ( 2 π ) 3 ( 2 ε k ) n T + − ( k ) e − 2 ε k τ > 0 (67) (-1)^nC_A^{(n)}(\tau)=\int\frac{d^3k}{(2\pi)^3}(2\varepsilon_k)^nT_{+-}(\mathbf{k})e^{-2\varepsilon_k\tau}>0 \tag*{(67)} ( − 1 ) n C A ( n ) ( τ ) = ∫ ( 2 π ) 3 d 3 k ( 2 ε k ) n T +− ( k ) e − 2 ε k τ > 0 (67) for every n n n τ > 0 \tau>0 τ > 0 C A C_A C A
ρ 1 ( s ) ≥ 0. (68) \rho_1(s)\geq0. \tag*{(68)} ρ 1 ( s ) ≥ 0. (68) The positivity of the mixed channel is therefore a theorem and not a computed outcome, and the proof uses only two facts: that the residue T + − T_{+-} T +− (54) is then a corollary one may check, as we have.
The diagonal channel carries negative norm
With m 1 = m 2 = m m_1=m_2=m m 1 = m 2 = m E 1 + E 2 = 2 E k = 2 ( ε k + i Γ k ) E_1+E_2=2E_k=2(\varepsilon_k+i\Gamma_k) E 1 + E 2 = 2 E k = 2 ( ε k + i Γ k ) (65) becomes
C B ( τ ) = ∫ d 3 k ( 2 π ) 3 2 ∣ W k ∣ e − 2 ε k τ cos ( 2 Γ k τ + φ k ) , (69) C_B(\tau)=\int\frac{d^3k}{(2\pi)^3}\,2\lvert W_k\rvert e^{-2\varepsilon_k\tau}\cos(2\Gamma_k\tau+\varphi_k), \tag*{(69)} C B ( τ ) = ∫ ( 2 π ) 3 d 3 k 2 ∣ W k ∣ e − 2 ε k τ cos ( 2 Γ k τ + φ k ) , (69) W k = ∣ W k ∣ e i φ k = 1 + ( k 2 / 3 + m 2 ) / E k 2 W_k=\lvert W_k\rvert e^{i\varphi_k}=1+(\mathbf{k}^2/3+m^2)/E_k^2 W k = ∣ W k ∣ e i φ k = 1 + ( k 2 /3 + m 2 ) / E k 2 N = 1 N=1 N = 1 a v = 0.6 a_v=0.6 a v = 0.6 C B = 1.82 × 10 − 2 C_B=1.82\times10^{-2} C B = 1.82 × 1 0 − 2 τ = 1 \tau=1 τ = 1 − 1.16 × 10 − 4 -1.16\times10^{-4} − 1.16 × 1 0 − 4 τ = 2 \tau=2 τ = 2 + 2.44 × 10 − 7 +2.44\times10^{-7} + 2.44 × 1 0 − 7 τ = 5 \tau=5 τ = 5 τ ≃ 1.88 \tau\simeq1.88 τ ≃ 1.88 π / 4 a v \pi/4a_v π /4 a v Γ k ≤ Γ 0 = a v \Gamma_k\leq\Gamma_0=a_v Γ k ≤ Γ 0 = a v (31) . Any test function supported where C B < 0 C_B<0 C B < 0
The same conclusion follows from the Euclidean form factor alone, with no reference to τ \tau τ A ( x ) = ∫ ρ ( s ) d s / ( s + x ) A(x)=\int\rho(s)ds/(s+x) A ( x ) = ∫ ρ ( s ) d s / ( s + x ) ρ ≥ 0 \rho\geq0 ρ ≥ 0 A ′ ′ ( x ) = 2 ∫ ρ ( s ) ( s + x ) − 3 d s A''(x)=2\int\rho(s)(s+x)^{-3}ds A ′′ ( x ) = 2 ∫ ρ ( s ) ( s + x ) − 3 d s
c n = ( − 1 ) n A ( n ) ( x 0 ) ( n + 2 ) ! = ∫ ρ ( s ) d s ( s + x 0 ) n + 3 (70) c_n=\frac{(-1)^nA^{(n)}(x_0)}{(n+2)!}=\int\frac{\rho(s)\,ds}{(s+x_0)^{n+3}} \tag*{(70)} c n = ( n + 2 )! ( − 1 ) n A ( n ) ( x 0 ) = ∫ ( s + x 0 ) n + 3 ρ ( s ) d s (70) is a Hausdorff moment sequence of a positive measure, and the Hankel matrix H j k = c j + k H_{jk}=c_{j+k} H j k = c j + k A = ∣ m ∣ 2 = m 2 \mathcal{A}=\lvert m\rvert^2=m^2 A = ∣ m ∣ 2 = m 2 ( m 2 + u u ˉ x ) − ( n + 1 ) (m^2+u\bar{u}x)^{-(n+1)} ( m 2 + u u ˉ x ) − ( n + 1 ) complex denominator, whose real part carries cos [ ( n + 1 ) arg ( m 2 + u u ˉ x ) ] \cos\left[(n+1)\arg(m^2+u\bar{u}x)\right] cos [ ( n + 1 ) arg ( m 2 + u u ˉ x ) ]
n ∗ ≃ π 2 arg m 2 = π 4 arctan ( a v / N ) . (71) n^* \simeq\frac{\pi}{2\arg m^2}=\frac{\pi}{4\arctan(av/N)}. \tag*{(71)} n ∗ ≃ 2 arg m 2 π = 4 arctan ( a v / N ) π . (71) Measured against predicted, the first n n n c n < 0 c_n<0 c n < 0 a v = 0.2 N av=0.2\,N a v = 0.2 N N N N N N N N = 2 N=2 N = 2 a v = 0.7 av=0.7 a v = 0.7 (67) requires.
Nothing repairs it
The obstruction is not an artifact of a choice, and it is worth closing the routes one by one; Table II collects them.
A different boundary value cannot help, because (69) and (70) are computed from the Euclidean correlator and from (D11), where Re ( m 2 + u u ˉ x ) > 0 \operatorname{Re}(m^2+u\bar{u}x)>0 Re ( m 2 + u u ˉ x ) > 0 ω \omega ω [ A ] = 2 [A]=2 [ A ] = 2 A ( 0 ) + x A ′ ( 0 ) A(0)+xA'(0) A ( 0 ) + x A ′ ( 0 ) A ′ ′ A'' A ′′ c n c_n c n n ∗ ≥ 1 n^*\geq1 n ∗ ≥ 1 § 2.2 , does nothing here, because J B J_B J B (57) with m 1 = m 2 m_1=m_2 m 1 = m 2 B P 1 B † BP_1B^\dagger B P 1 B † s > 0 s>0 s > 0 J A J_A J A J A J_A J A (23) the two currents have opposite Z 2 Z_2 Z 2 ⟨ J A J B ⟩ = 0 \langle J_AJ_B\rangle=0 ⟨ J A J B ⟩ = 0 α J A + β J B \alpha J_A+\beta J_B α J A + β J B ∣ α ∣ 2 G A + ∣ β ∣ 2 G B |\alpha|^2G_A+|\beta|^2G_B ∣ α ∣ 2 G A + ∣ β ∣ 2 G B G B G_B G B β ≠ 0 \beta\ne0 β = 0 Θ \Theta Θ C C C (17) both leave J B J_B J B Θ ^ = Θ O S ∘ Θ \hat{\Theta}=\Theta_{\mathrm{OS}}\circ\Theta Θ ^ = Θ OS ∘ Θ Θ ^ C = Θ O S ∘ C ∘ Θ \hat{\Theta}^{C}=\Theta_{\mathrm{OS}}\circ C\circ\Theta Θ ^ C = Θ OS ∘ C ∘ Θ C C C Ψ ˉ m γ Ψ m − Ψ ˉ m ∗ γ Ψ m ∗ \bar{\Psi}_m\gamma\Psi_m-\bar{\Psi}_{m^*}\gamma\Psi_{m^*} Ψ ˉ m γ Ψ m − Ψ ˉ m ∗ γ Ψ m ∗ same two-point function at one loop, the cross term requiring a mixed propagator and vanishing. A single negative direction at q = 0 q=0 q = 0 n ∗ n^* n ∗ (71) is finite for every a v > 0 av>0 a v > 0 a v → 0 av\to0 a v → 0 n ∗ = 17 n^*=17 n ∗ = 17 a v = 0.05 N av=0.05\,N a v = 0.05 N n ∗ = 88 n^*=88 n ∗ = 88 N N N
One line explains why the list is exhaustive. Reflection positivity at q = 0 q=0 q = 0 C ( τ a + τ b ) C(\tau_a+\tau_b) C ( τ a + τ b ) ∑ n f n ( τ a ) f n ( τ b ) \sum_n f_n(\tau_a)f_n(\tau_b) ∑ n f n ( τ a ) f n ( τ b ) (65) the exponent is E 1 + E 2 E_1+E_2 E 1 + E 2 m m m m ∗ m^* m ∗ balanced in the sense of Appendix C. The balanced-cut criterion and reflection positivity are one condition seen twice, and the diagonal channel fails both by construction.
Can a positive subspace be filtered out?
The natural reaction to an indefinite form is to look for a subspace on which it is positive, and then to quotient: this is what Gupta–Bleuler does in electrodynamics and what the quartet mechanism of Kugo and Ojima does in a covariant gauge [23 ] . The question is a sharp one here, and it has an answer in each direction.
A subspace of states is not merely a linear subspace. In the Osterwalder–Schrader construction the Euclidean time translation e − s H e^{-sH} e − sH ∣ f ⟩ |f\rangle ∣ f ⟩ ∑ a c a δ ( τ − τ a ) \sum_a c_a\delta(\tau-\tau_a) ∑ a c a δ ( τ − τ a ) τ a > 0 \tau_a>0 τ a > 0
Q ( s ) = ∑ a , b c a c b C ( τ a + τ b + 2 s ) , (72) Q(s)=\sum_{a,b}c_ac_bC(\tau_a+\tau_b+2s), \tag*{(72)} Q ( s ) = a , b ∑ c a c b C ( τ a + τ b + 2 s ) , (72) and both exponentials in (65) factorize through it. In the mixed channel this gives
Q A ( s ) = ∫ d 3 k ( 2 π ) 3 T + − ( k ) [ ∑ a c a e − 2 ε k τ a ] 2 e − 4 ε k s ≥ 0 , (73) Q_A(s)=\int\frac{d^3k}{(2\pi)^3}T_{+-}(k)\left[\sum_a c_a e^{-2\varepsilon_k\tau_a}\right]^2e^{-4\varepsilon_ks}\geq0, \tag*{(73)} Q A ( s ) = ∫ ( 2 π ) 3 d 3 k T +− ( k ) [ a ∑ c a e − 2 ε k τ a ] 2 e − 4 ε k s ≥ 0 , (73) a manifest square with a positive weight, for every vector and every s s s
Q B ( s ) = ∫ d 3 k ( 2 π ) 3 2 Re { W k [ ∑ a c a e − 2 E k τ a ] 2 e − 4 E k s } , (74) Q_B(s)=\int\frac{d^3k}{(2\pi)^3}2\operatorname{Re}\left\{W_k\left[\sum_a c_a e^{-2E_k\tau_a}\right]^2e^{-4E_ks}\right\}, \tag*{(74)} Q B ( s ) = ∫ ( 2 π ) 3 d 3 k 2 Re ⎩ ⎨ ⎧ W k [ a ∑ c a e − 2 E k τ a ] 2 e − 4 E k s ⎭ ⎬ ⎫ , (74) a complex square. As s s s ε k → N \varepsilon_k\to N ε k → N Γ k → a v \Gamma_k\to av Γ k → a v
Q B ( s ) ∼ ∣ Z ∣ e − 4 N s cos ( 4 a v s + ϕ ) , (75) Q_B(s)\sim|Z|e^{-4Ns}\cos(4av\,s+\phi), \tag*{(75)} Q B ( s ) ∼ ∣ Z ∣ e − 4 N s cos ( 4 a v s + ϕ ) , (75) which changes sign; the bracket is a finite sum of exponentials in E k E_k E k k k k
k = 0 k = 0 k = 0 c a c_a c a N = 1 N = 1 N = 1 Q B Q_B Q B a v = 0.3 av = 0.3 a v = 0.3 2 π / 4 a v = 5.24 2\pi/4av = 5.24 2 π /4 a v = 5.24 No subspace of the diagonal channel that is stable under e − s H e^{-sH} e − sH
That closes the standard devices at once. Quotienting by the radical of the form, which is what the Osterwalder–Schrader reconstruction does, removes null vectors and not negative ones, and leaves a Krein space rather than a Hilbert space. A subsidiary condition of Gupta–Bleuler type would have to cut out an invariant positive subspace, which by the above does not exist. A positive metric e − Q e^{-Q} e − Q P T \mathcal{PT} P T [10 ] , which the broken phase does not have. And no cohomological mechanism removes the offending states, with either of the two operators available. For the ordinary BRST operator the argument is immediate: J B J_B J B s s s s ( X ) s(X) s ( X ) ⟨ s ( X ) J B ⟩ = 0 \langle s(X)J_B\rangle= 0 ⟨ s ( X ) J B ⟩ = 0 A B ′ ′ A^{\prime\prime}_B A B ′′ 4.8 × 10 − 3 4.8 \times10^{-3} 4.8 × 1 0 − 3 x = 0.5 x = 0.5 x = 0.5 a v = 0.3 N av = 0.3\,N a v = 0.3 N s q s_q s q [2 ] and whose transformations are given explicitly in [3 ] ; and here a counting argument disposes of both operators at once, without any appeal to the filtration that relates them. A BRST operator raises the ghost number by one and preserves the dimension , so that J B = s ( X ) J_B = s(X) J B = s ( X ) X X X − 1 -1 − 1 − 1 -1 − 1 [ c ˉ ] ≥ 1 [\bar c] \ge1 [ c ˉ ] ≥ 1 X = c ˉ Y X = \bar c Y X = c ˉ Y [ Y ] ≤ 2 [Y] \le2 [ Y ] ≤ 2 Y Y Y s ( Y ) s(Y) s ( Y ) Y Y Y s ( c ˉ Y ) = b Y − c ˉ s ( Y ) s(\bar cY) = bY - \bar c\,s(Y) s ( c ˉ Y ) = bY − c ˉ s ( Y ) s q s_q s q s s s
The cohomology of this sector can in fact be computed outright, and it is small. The pairs ( c ˉ , b ) (\bar c,b) ( c ˉ , b ) ( c ˉ ^ , b ^ ) (\hat{\bar c},\hat b) ( c ˉ ^ , b ^ ) [23 ] the cohomology may be computed without them; in the reduced complex the only ghosts left carry ghost number + 1 +1 + 1 − 1 -1 − 1 nothing at all is exact at ghost number zero, which is the counting above in a sharper form. What remains is the kernel, and for an abelian theory that is the space of gauge-invariant local operators. At dimension three, parity-even and with one Lorentz index, requiring vanishing charge under both U ( 1 ) U(1) U ( 1 )
even : J B , ∂ μ ( ϕ † ϕ ) , ∂ ν F B μ ν , odd : J A , i ( ϕ † D μ ϕ − h . c . ) , ∂ ν F A μ ν , (76) \begin{aligned}
\text{even}:&\quad J_B,\quad\partial^\mu(\phi^\dagger\phi),\quad\partial_\nu F_B^{\mu\nu},\\
\text{odd}:&\quad J_A,\quad i(\phi^\dagger D^\mu\phi-\mathrm{h.c.}),\quad\partial_\nu F_A^{\mu\nu}, \tag*{(76)}
\end{aligned} even : odd : J B , ∂ μ ( ϕ † ϕ ) , ∂ ν F B μν , J A , i ( ϕ † D μ ϕ − h.c. ) , ∂ ν F A μν , (76) split three and three by the Z 2 \mathbb{Z}_2 Z 2 ψ ˉ γ μ χ \bar\psi\gamma^\mu\chi ψ ˉ γ μ χ ( − e , − e ) (-e,-e) ( − e , − e ) ϕ \phi ϕ ∂ ν F \partial_\nu F ∂ ν F J A J_A J A J B J_B J B
Two things follow that we had obtained separately. In the even sector only J B J_B J B q μ q ν q^\mu q^\nu q μ q ν Π B B \Pi_{BB} Π B B (77) receives the fermion loop and nothing else. In the odd sector there are two transverse classes, J A J_A J A Π A A \Pi_{AA} Π AA [4 ] , and the present model has no such sector. It is worth adding that the pattern found here is the one that construction also reports: the elementary modes complex mass violate positivity, and what is positive is a composite built from them — which is the subject of § 2.2 .
There is, however, a filtering that works, and it is the one the model supplies by itself. The Z 2 \mathbb{Z}_2 Z 2 (19) commutes with the Hamiltonian, so its eigenspaces are stable under the evolution; the two currents sit in different ones by (23) , with no amplitude between them; and on the odd sector (73) is positive for every vector and every s s s
The gauge and scalar sector
Two objects have to be kept apart before the question can be asked. The conductivity of Sections VI and § 2 is built from the correlator of the fermion bilinears (8) ; a gauge line inside that correlator costs e 2 e^2 e 2
The charges settle half of it immediately. The scalar carries charge e e e U ( 1 ) U(1) U ( 1 ) 2 e 2e 2 e A μ A_\mu A μ none under B μ B_\mu B μ
Π B B : fermions only, Π A A : fermions + ( σ , π ) + seagull, Π A B = 0 by x r e f ‘ ( 23 ) ‘ . (77) \begin{aligned}
\Pi_{BB} &: \text{fermions only,}\\
\Pi_{AA} &: \text{fermions} + (\sigma,\pi) + \text{seagull,}\\
\Pi_{AB} &= 0 \text{ by } {xref}`(23)`. \tag*{(77)}
\end{aligned} Π B B Π AA Π A B : fermions only, : fermions + ( σ , π ) + seagull, = 0 by x r e f ‘ ( 23 ) ‘. (77) The first line says that Section IX is not a partial result: the fermion loop is the whole of the one-loop self-energy of the massless field, so the negative norm found there is the complete one-loop statement about the probe an experiment would use.
In Π A A \Pi_{AA} Π AA σ \sigma σ π \pi π m σ m_\sigma m σ m π = ξ m A m_\pi=\sqrt{\xi}\,m_A m π = ξ m A gauge field itself . The last deserves a word, since the absence of abelian self-coupling might suggest otherwise: the scalar kinetic term contains e 2 A 2 ∣ ϕ ∣ 2 ⊃ e 2 v A 2 σ e^2 A^2|\phi|^2 \supset e^2v A^2\sigma e 2 A 2 ∣ ϕ ∣ 2 ⊃ e 2 v A 2 σ A A σ AA\sigma AA σ [5 ] and is not removed by abelianness; tadpoles, or an equivalent consistent shift of the vacuum, belong with it. All of these masses are nevertheless real , hence self-conjugate under the Θ \Theta Θ n m = n m ∗ = 0 n_m=n_{m^*}=0 n m = n m ∗ = 0 ( m σ + m π ) 2 (m_\sigma+m_\pi)^2 ( m σ + m π ) 2 ( m A + m σ ) 2 (m_A+m_\sigma)^2 ( m A + m σ ) 2 (65) with the vertex e A μ ( σ ∂ μ π − π ∂ μ σ ) = e A μ σ ∂ μ ↔ π eA^\mu(\sigma\partial_\mu\pi-\pi\partial_\mu\sigma)=eA^\mu\sigma\overleftrightarrow{\partial_\mu}\pi e A μ ( σ ∂ μ π − π ∂ μ σ ) = e A μ σ ∂ μ π q = 0 \boldsymbol{q}=0 q = 0 1 3 ∑ i ( 2 k i ) 2 = 4 3 k 2 \frac{1}{3}\sum_i(2k_i)^2=\frac{4}{3}\boldsymbol{k}^2 3 1 ∑ i ( 2 k i ) 2 = 3 4 k 2
C s ( τ ) = e 2 3 ∫ d 3 k ( 2 π ) 3 k 2 E σ E π e − ( E σ + E π ) τ , (78) C_s(\tau)=\frac{e^2}{3}\int\frac{d^3k}{(2\pi)^3}\frac{\boldsymbol{k}^2}{E_\sigma E_\pi}e^{-(E_\sigma+E_\pi)\tau}, \tag*{(78)} C s ( τ ) = 3 e 2 ∫ ( 2 π ) 3 d 3 k E σ E π k 2 e − ( E σ + E π ) τ , (78) with E σ E_\sigma E σ E π E_\pi E π (68) . We have checked that the sum of the fermion and this scalar piece stays completely monotone over a range of m σ m_\sigma m σ m π m_\pi m π e 2 e^2 e 2 τ = N − 1 \tau=N^{-1} τ = N − 1
That is as far as we take it, and the limitation should be stated rather than glossed. The positivity just established is that of one term; the complete Π A A \Pi_{AA} Π AA A σ A\sigma A σ m π m_\pi m π ξ \xi ξ 1 2 e 2 A 2 ( 2 v σ + σ 2 + π 2 ) \frac{1}{2}e^2A^2(2v\sigma+\sigma^2+\pi^2) 2 1 e 2 A 2 ( 2 v σ + σ 2 + π 2 ) A μ A_\mu A μ
What the section does settle is the diagonal channel, and that is the one the physical probe couples to: Π B B \Pi_{BB} Π B B q ≠ 0 q\ne0 q = 0 O ( e 2 ) O(e^2) O ( e 2 )
What this costs
It should be said plainly. B μ B_\mu B μ U ( 1 ) U(1) U ( 1 ) (36) and the precursor of (43) are therefore not assertions about physical states, and the optical sum rule, a theorem about a positive measure, has nothing to apply to; Table I keeps its value as the measure of what the two assignments of boundary value differ by, and no more. The physical content of the model sits entirely in the Z 2 \mathbb{Z}_2 Z 2 U ( 1 ) U(1) U ( 1 ) (68) and whose observable is constructed next.
The longitudinal part and the field-strength composite
The negative density ρ 0 \rho_0 ρ 0 (56) is attached to the longitudinal projector, and the longitudinal projector is annihilated by an antisymmetric derivative. This suggests the object to consider, and the suggestion works. Let
O μ ν = ∂ μ J A ν − ∂ ν J A μ , O μ ν ( q ) = i B α μ ν J A α ( q ) , (79) \mathcal{O}^{\mu\nu}=\partial^\mu J_A^\nu-\partial^\nu J_A^\mu,\qquad\mathcal{O}^{\mu\nu}(q)=iB^{\mu\nu}_{\alpha}J_A^\alpha(q), \tag*{(79)} O μν = ∂ μ J A ν − ∂ ν J A μ , O μν ( q ) = i B α μν J A α ( q ) , (79) with
B α μ ν = q μ δ α ν − q ν δ α μ , B α μ ν q α = q μ q ν − q ν q μ = 0. (80) B^{\mu\nu}_{\alpha}=q^\mu\delta^\nu_{\alpha}-q^\nu\delta^\mu_{\alpha},\qquad B^{\mu\nu}_{\alpha}q^\alpha=q^\mu q^\nu-q^\nu q^\mu=0. \tag*{(80)} B α μν = q μ δ α ν − q ν δ α μ , B α μν q α = q μ q ν − q ν q μ = 0. (80) Writing the spectral tensor of the mixed current as
ρ J μ ν = ρ 1 ( s ) P 1 μ ν + ρ 0 ( s ) P 0 μ ν , (81) \rho_J^{\mu\nu}=\rho_1(s)P_1^{\mu\nu}+\rho_0(s)P_0^{\mu\nu}, \tag*{(81)} ρ J μν = ρ 1 ( s ) P 1 μν + ρ 0 ( s ) P 0 μν , (81) P 1 μ ν = − η μ ν + q μ q ν s , P 0 μ ν = q μ q ν s , P_1^{\mu\nu}=-\eta^{\mu\nu}+\frac{q^\mu q^\nu}{s},\qquad P_0^{\mu\nu}=\frac{q^\mu q^\nu}{s}, P 1 μν = − η μν + s q μ q ν , P 0 μν = s q μ q ν , the second term drops out of the composite, since B P 0 B † = ( B q ) ( B q ) † / s = 0 BP_0B^\dagger=(Bq)(Bq)^\dagger/s=0 B P 0 B † = ( B q ) ( B q ) † / s = 0 (80) , and
ρ O = B ρ J B † = ρ 1 ( s ) B P 1 B † . (82) \rho_{\mathcal O}=B\rho_JB^\dagger=\rho_1(s)BP_1B^\dagger. \tag*{(82)} ρ O = B ρ J B † = ρ 1 ( s ) B P 1 B † . (82) What survives is positive. For q = ( E , p ) q=(E,\boldsymbol{p}) q = ( E , p ) E 2 = p 2 + s E^2=\boldsymbol{p}^2+s E 2 = p 2 + s p = ∣ p ∣ p=|\boldsymbol{p}| p = ∣ p ∣ e 1 , e 2 e_1,e_2 e 1 , e 2 p \boldsymbol{p} p
e a μ = ( 0 , e a ) , e 3 μ = ( p s , E s p ^ ) . (83) e_a^\mu=(0,e_a),\qquad e_3^\mu=\left(\frac{p}{\sqrt{s}},\frac{E}{\sqrt{s}}\hat{p}\right). \tag*{(83)} e a μ = ( 0 , e a ) , e 3 μ = ( s p , s E p ^ ) . (83) These satisfy q ⋅ e λ = 0 q\cdot e_\lambda=0 q ⋅ e λ = 0 e λ ⋅ e λ ′ = − δ λ λ ′ e_\lambda\cdot e_{\lambda'}=-\delta_{\lambda\lambda'} e λ ⋅ e λ ′ = − δ λ λ ′ q ⋅ e 3 = ( E p − p E ) / s = 0 q\cdot e_3=(Ep-pE)/\sqrt{s}=0 q ⋅ e 3 = ( E p − pE ) / s = 0 e 3 ⋅ e 3 = ( p 2 − e_3\cdot e_3=(p^2- e 3 ⋅ e 3 = ( p 2 − E 2 ) / s = − 1 E^{2})/s=-1 E 2 ) / s = − 1
∑ λ = 1 3 ϵ λ α ϵ λ β ∗ = P 1 α β ( q ) , (84) \sum_{\lambda=1}^{3}\epsilon_{\lambda}^{\alpha}\epsilon_{\lambda}^{\beta*}=P_{1}^{\alpha\beta}(q), \tag*{(84)} λ = 1 ∑ 3 ϵ λ α ϵ λ β ∗ = P 1 α β ( q ) , (84) its 00 component reading p 2 / s = − 1 + E 2 / s p^{2}/s=-1+E^{2}/s p 2 / s = − 1 + E 2 / s 0 i 0i 0 i E p i / s Ep^{i}/s E p i / s i j ij ij δ i j − p ^ i p ^ j + ( E 2 / s ) p ^ i p ^ j = δ i j + p i p j / s \delta^{ij}-\hat{p}^{i}\hat{p}^{j}+(E^{2}/s)\hat{p}^{i}\hat{p}^{j}=\delta^{ij}+p^{i}p^{j}/s δ ij − p ^ i p ^ j + ( E 2 / s ) p ^ i p ^ j = δ ij + p i p j / s f μ ν f_{\mu\nu} f μν
f † ϱ O f = ρ 1 ( s ) ∑ λ = 1 3 ∣ f † B e λ ∣ 2 ≥ 0 , (85) f^{\dagger}\varrho_{\mathcal{O}}f=\rho_{1}(s)\sum_{\lambda=1}^{3}\left|f^{\dagger}Be_{\lambda}\right|^{2}\geq0, \tag*{(85)} f † ϱ O f = ρ 1 ( s ) λ = 1 ∑ 3 f † B e λ 2 ≥ 0 , (85) a sum of squares with a positive coefficient, by (56). In the rest frame q = ( s , 0 ) q=(\sqrt{s},0) q = ( s , 0 ) ( 0 i ) (0i) ( 0 i ) B B B ϱ O = s ρ 1 ( s ) diag ( 1 , 1 , 1 , 0 , 0 , 0 ) \varrho_{\mathcal{O}}=s\rho_{1}(s)\operatorname{diag}(1,1,1,0,0,0) ϱ O = s ρ 1 ( s ) diag ( 1 , 1 , 1 , 0 , 0 , 0 ) ( 01 , 02 , 03 , 12 , 13 , 23 ) (01,02,03,12,13,23) ( 01 , 02 , 03 , 12 , 13 , 23 ) Λ ∧ Λ \Lambda\wedge\Lambda Λ ∧ Λ
The Euclidean counterpart identifies what is being superposed. With D μ ν E ( Q ; s ) D_{\mu\nu}^{E}(Q;s) D μν E ( Q ; s ) s s s
B E D E ( B E ) T = B E ( B E ) T Q 2 + s = K E Q 2 + s , (86) B^{E}D^{E}(B^{E})^{T}=\frac{B^{E}(B^{E})^{T}}{Q^{2}+s}=\frac{K^{E}}{Q^{2}+s}, \tag*{(86)} B E D E ( B E ) T = Q 2 + s B E ( B E ) T = Q 2 + s K E , (86) because the Q μ Q ν / s Q_{\mu}Q_{\nu}/s Q μ Q ν / s D E D^{E} D E (80) , so that the non-local part of the composite correlator is a superposition over s s s ρ 1 \rho_{1} ρ 1 R = diag ( − 1 , 1 , 1 , 1 ) R=\operatorname{diag}(-1,1,1,1) R = diag ( − 1 , 1 , 1 , 1 ) Q 4 Q_{4} Q 4 p p p s s s
M ( p , s ) = ( p 2 / s i E p T / s − i E p / s I + p p T / s ) , (87) M(p,s)=\begin{pmatrix} p^{2}/s & iEp^{T}/s \\ -iEp/s & I+pp^{T}/s \end{pmatrix}, \tag*{(87)} M ( p , s ) = ( p 2 / s − i E p / s i E p T / s I + p p T / s ) , (87) and decomposing f = p ^ f ∥ + f ⊥ f=\hat{p}f_{\parallel}+f_{\perp} f = p ^ f ∥ + f ⊥ s + p 2 = E 2 s+p^{2}=E^{2} s + p 2 = E 2
f † M f = ∣ f ⊥ ∣ 2 + 1 s ∣ E f ∥ − i p f 4 ∣ 2 ≥ 0 , (88) f^{\dagger}Mf=\lvert f_{\perp}\rvert^{2}+\frac{1}{s}\left\lvert Ef_{\parallel}-ipf_{4}\right\rvert^{2}\geq0, \tag*{(88)} f † M f = ∣ f ⊥ ∣ 2 + s 1 E f ∥ − i p f 4 2 ≥ 0 , (88) with M = diag ( 0 , 1 , 1 , 1 ) M=\operatorname{diag}(0,1,1,1) M = diag ( 0 , 1 , 1 , 1 ) p = 0 p=0 p = 0
H 1 = ∫ 4 N 2 ∞ ⊕ ρ 1 ( s ) d s L 2 ( R 3 , d μ s ; C 3 ) , (89) \mathcal{H}_{1}=\int_{4N^{2}}^{\infty}\!{}^{\oplus}\rho_{1}(s)\,ds\,L^{2}(\mathbb{R}^{3},d\mu_{s};\mathbb{C}^{3}), \tag*{(89)} H 1 = ∫ 4 N 2 ∞ ⊕ ρ 1 ( s ) d s L 2 ( R 3 , d μ s ; C 3 ) , (89) d μ s = d 3 p ( 2 π ) 3 2 E p , s , d\mu_{s}=\frac{d^{3}p}{(2\pi)^{3}2E_{p,s}}, d μ s = ( 2 π ) 3 2 E p , s d 3 p , an honest Hilbert space with a positive inner product, and the associated generalized free vector field has W ~ μ ν μ ν ( q ) = 2 π θ ( q 0 ) θ ( q 2 − 4 N 2 ) ρ 1 ( q 2 ) P 1 μ ν ( q ) \widetilde{W}_{\mu\nu}^{\mu\nu}(q)=2\pi\theta(q^{0})\theta(q^{2}-4N^{2})\rho_{1}(q^{2})P_{1}^{\mu\nu}(q) W μν μν ( q ) = 2 π θ ( q 0 ) θ ( q 2 − 4 N 2 ) ρ 1 ( q 2 ) P 1 μν ( q ) ρ 1 ( s ) = O ( s ) \rho_{1}(s)=O(s) ρ 1 ( s ) = O ( s ) Δ s ( x ) = ∫ d μ s ( e − i p x − e i p x ) \Delta_{s}(x)=\int d\mu_{s}(e^{-ipx}-e^{ipx}) Δ s ( x ) = ∫ d μ s ( e − i p x − e i p x ) x 2 < 0 x^{2}<0 x 2 < 0 p → − p p\to-p p → − p x 0 = 0 x^{0}=0 x 0 = 0 supp ∂ T ⊆ supp T \operatorname{supp}\partial T\subseteq\operatorname{supp}T supp ∂ T ⊆ supp T O \mathcal{O} O G R = [ ( ω + i 0 ) 2 − p 2 − s ] − 1 G_{R}=[(\omega+i0)^{2}-p^{2}-s]^{-1} G R = [( ω + i 0 ) 2 − p 2 − s ] − 1
The source
An external source must carry strictly positive ultraviolet dimension. The reason is the usual one, and it is not a matter of taste: if a commuting source has vanishing dimension, power counting places no bound on its powers, every monomial in it is admissible in the counterterm, and the number of insertions is not controlled, so that the extended action is not renormalizable in the algebraic sense [23 ] . We therefore never introduce a source for O μ ν \mathcal{O}^{\mu\nu} O μν [ Ψ ] = 3 / 2 [\Psi]=3/2 [ Ψ ] = 3/2 [ J A ] = 3 [J_{A}]=3 [ J A ] = 3 [ O ] = 4 [\mathcal{O}]=4 [ O ] = 4 O \mathcal{O} O
What the model does admit is the vector source of the current,
S a = ∫ d 4 x a ν ( x ) J A ν ( x ) , [ a ν ] = 1 > 0 , (90) S_{a}=\int d^{4}x\,a_{\nu}(x)J_{A}^{\nu}(x),\qquad[a_{\nu}]=1>0, \tag*{(90)} S a = ∫ d 4 x a ν ( x ) J A ν ( x ) , [ a ν ] = 1 > 0 , (90) unconstrained and of the dimension a gauge field carries, and O μ ν \mathcal{O}^{\mu\nu} O μν W [ a ] W[a] W [ a ]
⟨ O μ ν ( x ) ⟩ a = ( ∂ μ δ δ a ν ( x ) − ∂ ν δ δ a μ ( x ) ) W [ a ] . (91) \langle\mathcal{O}^{\mu\nu}(x)\rangle_{a}=\left(\partial^{\mu}\frac{\delta}{\delta a_{\nu}(x)}-\partial^{\nu}\frac{\delta}{\delta a_{\mu}(x)}\right)W[a]. \tag*{(91)} ⟨ O μν ( x ) ⟩ a = ( ∂ μ δ a ν ( x ) δ − ∂ ν δ a μ ( x ) δ ) W [ a ] . (91) Everything in § 2.2 is of this form. The antisymmetric b μ ν b_{\mu\nu} b μν O \mathcal{O} O
a ν = − ∂ μ b μ ν , ∂ ν a ν = 0 , (92) a_{\nu}=-\partial^{\mu}b_{\mu\nu},\qquad\partial^{\nu}a_{\nu}=0, \tag*{(92)} a ν = − ∂ μ b μν , ∂ ν a ν = 0 , (92) the second relation following from the antisymmetry alone. The map b ↦ a b\mapsto a b ↦ a a ν a_{\nu} a ν (90) to conserved configurations loses nothing; the positivity statements (85) and (88) are exactly the statements about those configurations, with b b b
Two remarks keep this in its place. The first is that a ν a_{\nu} a ν B μ B_{\mu} B μ Z 2 \mathbb{Z}_{2} Z 2 [23 ] , so no conductivity is being redefined here. The second is that we do not claim to have carried out the algebraic renormalization of (90) ; what the dimension count settles is only where such a construction must start, namely from the dimension-one vector source and not from a two-index one.
The four-point test
Everything above is a statement about the two-point function. It exhibits a positive and causal generalized free field with the right vacuum two-point data, and that is all it exhibits: a generalized free field fixes its higher correlators by Gaussian factorization, whereas O μ ν O^{\mu\nu} O μν
The test is the next Osterwalder–Schrader condition. Take N = 1 N=1 N = 1 a v = 3 / 5 av=3/5 a v = 3/5 T = μ = 0 T=\mu=0 T = μ = 0 f , g f,g f , g ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 2 , 0 ) (2,0) ( 2 , 0 )
F = O 41 ( f ) O 41 ( g ) − ω [ O 41 ( f ) O 41 ( g ) ] 1 (93) F=O_{41}(f)O_{41}(g)-\omega\left[O_{41}(f)O_{41}(g)\right]\mathbf{1} \tag*{(93)} F = O 41 ( f ) O 41 ( g ) − ω [ O 41 ( f ) O 41 ( g ) ] 1 (93) is admissible, and reflection positivity demands ω [ ( Θ F ) F ] ≥ 0 \omega[(\Theta F)F]\geq0 ω [( Θ F ) F ] ≥ 0 ( − Y , − X , X , Y ) (-Y,-X,X,Y) ( − Y , − X , X , Y )
Q c r o s s = + 3.3097 × 10 − 9 , Q c o n n = − 3.5033 × 10 − 7 , ω [ ( Θ F ) F ] = − 3.4702 × 10 − 7 < 0. (94) \begin{aligned}
Q_{\mathrm{cross}}&=+3.3097\times10^{-9},\\
Q_{\mathrm{conn}}&=-3.5033\times10^{-7},\\
\omega[(\Theta F)F]&=-3.4702\times10^{-7}<0. \tag*{(94)}
\end{aligned} Q cross Q conn ω [( Θ F ) F ] = + 3.3097 × 1 0 − 9 , = − 3.5033 × 1 0 − 7 , = − 3.4702 × 1 0 − 7 < 0. (94) The magnetic component O 12 O_{12} O 12 − 5.3292 × 10 − 7 -5.3292\times10^{-7} − 5.3292 × 1 0 − 7 § 2.1.5 gives the contractions and the arithmetic; the value is dominated by Q c o n n Q_{\mathrm{conn}} Q conn
This does not contradict (85) , which is a two-point statement and remains true. What it excludes is the identification of the curl algebra of the original model with the positive realization built from its two-point function. An algebra containing both insertions of (93) and their product contains a vector of negative norm, so it cannot be the positive-time part of a positive state. The Z 2 \mathbb{Z}_2 Z 2 X ↦ P X P X\mapsto PXP X ↦ P X P
P X P = ( P X P ) ( P Y P ) + P X ( 1 − P ) Y P , (95) PXP=(PXP)(PYP)+PX(1-P)YP, \tag*{(95)} P X P = ( P X P ) ( P Y P ) + P X ( 1 − P ) Y P , (95) and the second term vanishes only if the retained space is invariant under the operators involved, which would itself have to be proved.
The failure is not uniform in the non-Hermiticity, and the way it switches off can be established rather than merely observed. Two statements do it.
The first is that ω [ ( Θ F ) F ] \omega[(\Theta F)F] ω [( Θ F ) F ] even function of a v av a v a v av a v m m m m ∗ m^* m ∗ m ↔ m ∗ m\leftrightarrow m^* m ↔ m ∗ D m ∗ = D m ‾ D_{m^*}=\overline{D_m} D m ∗ = D m
Q ( − a v ) = Q ( a v ) ‾ = Q ( a v ) , Q ≡ ω [ ( Θ F ) F ] . (96) Q(-av)=\overline{Q(av)}=Q(av),\qquad Q\equiv\omega[(\Theta F)F]. \tag*{(96)} Q ( − a v ) = Q ( a v ) = Q ( a v ) , Q ≡ ω [( Θ F ) F ] . (96) There is no term linear in the non-Hermiticity, and the expansion starts at ( a v ) 2 (av)^2 ( a v ) 2
The second is that the leading coefficient is the Hermitian value and cannot be negative. At a v = 0 av=0 a v = 0 F F F
Q ( 0 ) ≥ 0 for every test configuration. (97) Q(0)\geq0\qquad\text{for every test configuration.} \tag*{(97)} Q ( 0 ) ≥ 0 for every test configuration. (97) Fitting the computed values at the configuration above confirms both statements, the parity to one part in 10 18 10^{18} 1 0 18 10 − 14 10^{-14} 1 0 − 14
Q = 4.6936 × 10 − 7 − 2.9229 × 10 − 6 ( a v ) 2 + 1.4857 × 10 − 6 ( a v ) 4 + ⋯ (98) Q=4.6936\times10^{-7}-2.9229\times10^{-6}(av)^2
+1.4857\times10^{-6}(av)^4+\cdots \tag*{(98)} Q = 4.6936 × 1 0 − 7 − 2.9229 × 1 0 − 6 ( a v ) 2 + 1.4857 × 1 0 − 6 ( a v ) 4 + ⋯ (98) in units of N N N a v ≃ 0.5 N av\simeq0.5N a v ≃ 0.5 N
Together these fix what is and is not established. At a v = 3 / 5 N av=3/5\,N a v = 3/5 N a v av a v a v av a v null direction of the Hermitian theory , where Q ( 0 ) = 0 Q(0)=0 Q ( 0 ) = 0 ( a v ) 2 (av)^2 ( a v ) 2
What survives of this section is then the following. The mixed channel has a positive transverse spectral measure, by (68) ; the curl removes the negative longitudinal piece of (58) ; and the resulting two-point data admit a positive, local, causal generalized free realization. That realization is a legitimate object with which to ask what a composite response with this spectral density would look like. It is not a physical sector of the fermion model: the four-point test says the identification fails. Constructing a smaller algebra that is closed, positive and invariant under the dynamics is left open, and (94) is the obstruction such a construction must avoid.
Conclusions
We have shown that the complex conjugate masses of the broken phase of the model introduced in [1 ] organize its linear response into two channels with opposite analytic structures, and that the separation is dictated by the biorthogonal basis that diagonalizes the mass matrix: the current of the unbroken U ( 1 ) U(1) U ( 1 ) U ( 1 ) U(1) U ( 1 ) Z 2 Z_2 Z 2 A μ → − A μ A_\mu\to-A_\mu A μ → − A μ U ( 1 ) U(1) U ( 1 ) Z 2 Z_2 Z 2 [1 ] , here identified as the diagonal of a flavor exchange and a discrete element of the broken gauge group, each of which is separately broken by the scalar expectation value. The first channel has a strictly vanishing dissipative conductivity at every real frequency and for any temperature and density, the gain and the loss modes compensating each other exactly, while the second absorbs above the real threshold ω = 2 N \omega= 2N ω = 2 N p 2 = 4 N 2 p^2 = 4N^2 p 2 = 4 N 2 2 Γ k = 2 N a v / ε k 2\Gamma_k = 2Nav/\varepsilon_k 2 Γ k = 2 N a v / ε k μ / 2 N a v \mu/2Nav μ /2 N a v (46) . Finally, the transverse spectral density of the mixed channel was shown to be positive in the vacuum, which settles in the vector channel the question left open in the scalar one, and to remain positive at finite temperature under the bound a v < π T / 2 av < \pi T/2 a v < π T /2 (64) , which shows the bound to be uniform in μ \mu μ ∣ μ ∣ / T → ∞ |\mu|/T \to\infty ∣ μ ∣/ T → ∞ (54) gives σ 1 A \sigma_1^A σ 1 A a v → 0 av \to0 a v → 0 ω / 6 π \omega/6\pi ω /6 π (66) of that channel is a positive superposition of decaying exponentials, hence completely monotone, and Bernstein returns ρ 1 ≥ 0 \rho_1 \ge0 ρ 1 ≥ 0 (35) on which a complex mass pole collides with a Matsubara pole.
Three remarks seem to us worth making. The first is that the vanishing of the absorptive part in the diagonal channel and the precursor of the kernel are not independent findings but the same statement read in the two domains, and that both concern a channel which admits no positive-norm realization at all. This is now a theorem: its Euclidean correlator (69) oscillates, so by Osterwalder–Schrader there is a test function of negative norm, and Table II closes every repair — the choice of boundary value included, that test being carried out in the Euclidean region. Nor can the offending states be filtered away. A subspace of states must be stable under e − s H e^{-sH} e − sH (74) the norm of an evolved vector of the diagonal channel oscillates and changes sign, so no invariant positive subspace exists; that closes Gupta–Bleuler, the quartet mechanism — the states are not s s s ⟨ J B J B ⟩ \langle J_BJ_B\rangle ⟨ J B J B ⟩ e − Q e^{-Q} e − Q (19) , an exact symmetry rather than a condition imposed by hand, and what it selects is the other channel. What the single formula (65) shows is that one condition decides both channels: the sum of the two energies in the loop is real precisely when the cut is balanced, and reflection positivity and the balanced-cut criterion of Appendix C are the same statement. The cost is worth repeating, since B μ B_\mu B μ U ( 1 ) U(1) U ( 1 ) (19) the Z 2 Z_2 Z 2 π \pi π U ( 1 ) U(1) U ( 1 ) A μ → − A μ A_\mu\to-A_\mu A μ → − A μ B μ → B μ B_\mu\to B_\mu B μ → B μ diag ( − i , i ) \operatorname{diag}(-i,i) diag ( − i , i ) Z 2 Z_2 Z 2 (65) are four names for one fact, and the positivity follows from it. The same partition appears in the non-abelian setting, where after spontaneous breaking the generalized Faddeev–Popov operator becomes positive definite along the broken directions [3 ] , and it is tempting to read a mechanism into the coincidence. One should not. The present model is abelian: its Faddeev–Popov operator is − ∂ 2 -\partial^2 − ∂ 2 b = − a b=-a b = − a S L ( 3 , C ) SL(3,\mathbb{C}) S L ( 3 , C ) [2 , 7 ] , it fails in both directions at once: L 8 L 8 L_8L_8 L 8 L 8 i i i s = 0 s=0 s = 0 Tr F 2 \operatorname{Tr} F^2 Tr F 2 i i i Z 2 Z_2 Z 2 S L ( 3 , C ) SL(3,\mathbb{C}) S L ( 3 , C ) Table I retains its interest as the measure of what the two assignments of boundary value differ by, namely the whole of the Hermitian absorption; and by [23 ] neither assignment can affect the mixed channel, which carries the opposite Z 2 \mathbb{Z}_2 Z 2 + 1 +1 + 1 η \eta η − 1 -1 − 1 ρ 0 \rho_0 ρ 0 J A J_A J A (57) the amount by which it is negative is k 2 ( m − m ∗ ) 2 k^2(m-m^*)^2 k 2 ( m − m ∗ ) 2 O μ ν = ∂ μ J A ν − ∂ ν J A μ O^{\mu\nu}=\partial^\mu J_A^\nu-\partial^\nu J_A^\mu O μν = ∂ μ J A ν − ∂ ν J A μ ρ 1 B 1 B 1 † ≥ 0 \rho_1 B_1 B_1^\dagger\ge0 ρ 1 B 1 B 1 † ≥ 0 (89) , a reflection-positive Euclidean form (88) , and a commutator that vanishes outside the light cone. But the scope of that statement is the two-point function, and Section X B shows that it does not extend. The generalized free field fixes its higher correlators by Gaussian factorization, while O μ ν O^{\mu\nu} O μν (94) at a v = 3 / 5 N av=3/5\,N a v = 3/5 N Z 2 \mathbb{Z}_2 Z 2 (98) this witness is positive below a v ≃ 0.5 N av\simeq0.5\,N a v ≃ 0.5 N a v av a v a v av a v (62) measures the non-Hermitian scale a v av a v π T \pi T π T (35) ; we speculate that this is not an accident of the one-loop expression but the sign of a genuine restriction on the regime in which the broken phase may be treated thermally, and that the gain above the bound is the field-theoretic counterpart of the instability which in a finite dimensional system would simply be exponential growth. The third concerns the exceptional point of the family, where the two threads of this paper meet. The metric (10) is positive on the Hermitian side, indefinite on the broken side and divergent at a b = 0 ab=0 ab = 0 (4) being the field-theoretic form of that degeneracy. The zero-norm structure which Lee and Wick met in an indefinite-metric sector [13 ] , which the Gribov–Zwanziger construc-
tion meets in its dipoles [15 ] , and which (14) meets in the complex-mass modes is therefore one and the same structure along the whole range a b ≤ 0 ab\le0 ab ≤ 0 [23 ] to apply to it, although we have not carried that out here. As for the reach of the calculation, Section IX E puts the gauge and scalar sector into the one-loop self-energies and finds that it changes nothing: the massless channel receives no contribution from it, the scalar being neutral there, so the result of Section IX is the complete one-loop statement about the physical probe; and what it adds to the massive channel carries only real masses, hence balanced cuts and, by (78) , a positive weight. The seagull is the London term and has no cut. What is still missing is the Landau damping at q ≠ 0 q\ne0 q = 0 O ( e 2 ) O(e^2) O ( e 2 )
The Dirac algebra
With α i = γ 0 γ i \alpha^i=\gamma^0\gamma^i α i = γ 0 γ i i i i α i ( α ⋅ k ) α i = 2 α i k i − α ⋅ k \alpha^i(\alpha\cdot k)\alpha^i=2\alpha^i k_i-\alpha\cdot k α i ( α ⋅ k ) α i = 2 α i k i − α ⋅ k α i β α i = − β \alpha^i\beta\alpha^i=-\beta α i β α i = − β α i H 1 α i = 2 α i k i − α ⋅ k − β M 1 \alpha^i H_1\alpha^i=2\alpha^i k_i-\alpha\cdot k-\beta M_1 α i H 1 α i = 2 α i k i − α ⋅ k − β M 1 tr [ α a α b ] = 4 δ a b \operatorname{tr}[\alpha^a\alpha^b]=4\delta^{ab} tr [ α a α b ] = 4 δ ab tr [ α β ] = 4 \operatorname{tr}[\alpha\beta]=4 tr [ α β ] = 4 tr [ α β ] = 0 \operatorname{tr}[\alpha\beta]=0 tr [ α β ] = 0
1 3 ∑ i tr [ α i H 1 α i H 2 ] = − 4 k 2 3 − 4 M 1 M 2 (99) \frac{1}{3}\sum_i \operatorname{tr}\left[\alpha^i H_1\alpha^i H_2\right]=-\frac{4k^2}{3}-4M_1M_2 \tag*{(99)} 3 1 i ∑ tr [ α i H 1 α i H 2 ] = − 3 4 k 2 − 4 M 1 M 2 (99) from which (26) follows at once, the terms linear in H H H M 1 = M 2 = M M_1=M_2=M M 1 = M 2 = M T s s = 2 k 2 / 3 E 2 T_{ss}=2k^2/3E^2 T ss = 2 k 2 /3 E 2
The dispersion relation
Writing E k = k 2 + m 2 = ε k + i Γ k E_k=\sqrt{k^2+m^2}=\varepsilon_k+i\Gamma_k E k = k 2 + m 2 = ε k + i Γ k m = N + i a v m=N+iav m = N + ia v E k 2 = k 2 + N 2 − a 2 v 2 + 2 i N a v E_k^2=k^2+N^2-a^2v^2+2iNav E k 2 = k 2 + N 2 − a 2 v 2 + 2 i N a v
ε k 2 − Γ k 2 = k 2 + N 2 − a 2 v 2 , 2 ε k Γ k = 2 N a v (100) \varepsilon_k^2-\Gamma_k^2=k^2+N^2-a^2v^2,\qquad2\varepsilon_k\Gamma_k=2Nav \tag*{(100)} ε k 2 − Γ k 2 = k 2 + N 2 − a 2 v 2 , 2 ε k Γ k = 2 N a v (100) the second of which is the first relation of (30) . Eliminating Γ k \Gamma_k Γ k ε k 4 − Δ k ε k 2 − N 2 a 2 v 2 = 0 \varepsilon_k^4-\Delta_k\varepsilon_k^2-N^2a^2v^2=0 ε k 4 − Δ k ε k 2 − N 2 a 2 v 2 = 0 (30) , whose positive root is the one selected by the principal branch. At k = 0 k=0 k = 0 ε 0 = N \varepsilon_0=N ε 0 = N Γ 0 = a v \Gamma_0=av Γ 0 = a v 2 ε 0 = 2 N 2\varepsilon_0=2N 2 ε 0 = 2 N k k k Γ k ≃ N a v / ∣ k ∣ \Gamma_k\simeq Nav/|k| Γ k ≃ N a v /∣ k ∣ ε k \varepsilon_k ε k ∣ k ∣ |k| ∣ k ∣ Γ k ≤ Γ 0 = a v \Gamma_k\le\Gamma_0=av Γ k ≤ Γ 0 = a v (62) . The regime N 2 > a 2 v 2 N^2>a^2v^2 N 2 > a 2 v 2 [1 ] keeps Δ k > 0 \Delta_k>0 Δ k > 0 k k k m 2 m^2 m 2
The balanced-cut criterion and the two channels
The separation obtained in Section II D by a symmetry argument has an equivalent formulation in terms of the analytic continuation from the Euclidean region, which we record here because it reaches the same conclusion from a different direction and places the boundary-value question of § 1.4 where it belongs.
Let Θ \Theta Θ M M M
Θ : u m ⟶ u m ∗ , m ⟶ m ∗ , (101) \Theta:\quad u_m \longrightarrow u_{m^*},\qquad m\longrightarrow m^*, \tag*{(101)} Θ : u m ⟶ u m ∗ , m ⟶ m ∗ , (101) the mass matrix (1) being real, so that its complex eigenvalues necessarily come in conjugate pairs. It is worth noting that it is this antilinear map, and not the linear Z 2 \mathbb{Z}_2 Z 2 (19) , that implements the exchange m ↔ m ∗ m\leftrightarrow m^* m ↔ m ∗ (19) acts diagonally on Ψ m \Psi_m Ψ m Ψ m ∗ \Psi_{m^*} Ψ m ∗
For a one-loop two-point function the Landau equations place a normal threshold on the first sheet at
s C = ( ∑ j ∈ C m j ) 2 , s_C=\left(\sum_{j\in C}m_j\right)^2, s C = j ∈ C ∑ m j 2 , the sum running over the lines crossed by the cut C C C (2) ,
Im ∑ j ∈ C m j = a v [ n m ( C ) − n m ∗ ( C ) ] , (102) \operatorname{Im}\sum_{j\in C}m_j=av\left[n_m(C)-n_{m^*}(C)\right], \tag*{(102)} Im j ∈ C ∑ m j = a v [ n m ( C ) − n m ∗ ( C ) ] , (102) so that the threshold is real if and only if the cut is balanced , crossing as many lines of mass m m m m ∗ m^* m ∗ Θ \Theta Θ [7 ] , which sits inside the procedure of [2 ] : one first selects operators invariant under the full BRST operator, and then uses the nilpotent operator s q s_q s q [2 , 3 ] , whose cohomology classifies the observables there, to decompose them into sectors; the balance of the cuts is the condition that decides which of those sectors admits a particle representation. It is a criterion of identification and does not by itself assert anything about the sectors it rejects, which is why Section IX is needed: what the criterion predicts for the diagonal channel, the Euclidean correlator proves. In [7 ] the same analysis is carried out for the sectors of the S L ( 3 , C ) SL(3,\mathbb{C}) S L ( 3 , C )
The two channels of (8) are precisely the two classes. The cut of Π A \Pi_A Π A n m = n m ∗ = 1 n_m=n_{m^*}=1 n m = n m ∗ = 1
s C = ( m + m ∗ ) 2 = 4 N 2 , (103) s_C=(m+m^*)^2=4N^2, \tag*{(103)} s C = ( m + m ∗ ) 2 = 4 N 2 , (103) real and, remarkably, independent of a v av a v p 2 = 4 N 2 p^2=4N^2 p 2 = 4 N 2 [1 ] and the ω = 2 N \omega=2N ω = 2 N (39) . The cut of Π B \Pi_B Π B s C = 4 ( N ± i a v ) 2 s_C=4(N\pm iav)^2 s C = 4 ( N ± ia v ) 2 ω = 2 ε k ± 2 i Γ k \omega=2\varepsilon_k\pm2i\Gamma_k ω = 2 ε k ± 2 i Γ k § 1.4 . The Z 2 \mathbb{Z}_2 Z 2 (23) is thus the algebraic label of the balance: the odd channel is balanced and has a particle representation, the even one is not and does not.
Two consequences are worth stating. The first is that the ambiguity of § 1.4 arises exactly, and only, in the unbalanced sector — the sector which by the criterion would have no particle representation under any prescription — while the sector that does have one is free of it, its singularities sitting on the real axis. The second is that the criterion is itself prescription dependent: it is for the analytic continuation of the Euclidean correlator that balanced sectors acquire a real cut with positive weight, and the positivity of (50) is the statement, in the vector channel of the present model, that they do.
The vacuum spectral densities of the mixed channel
We record the covariant vacuum computation that produces (56) . Work in Euclidean space with { γ μ E , γ ν E } = 2 δ μ ν \{\gamma_\mu^E,\gamma_\nu^E\}=2\delta_{\mu\nu} { γ μ E , γ ν E } = 2 δ μν S m ( k ) = ( − i ̸ k E + m ) / ( k E 2 + m 2 ) S_m(k)=(-i\not{k}_E+m)/(k^2_E+m^2) S m ( k ) = ( − i k E + m ) / ( k E 2 + m 2 ) j μ E = i ( Ψ ‾ m γ μ E Ψ m ∗ + ( m ↔ m ∗ ) ) j_\mu^E=i(\overline{\Psi}_m\gamma_\mu^E\Psi_{m^*}+(m\leftrightarrow m^*)) j μ E = i ( Ψ m γ μ E Ψ m ∗ + ( m ↔ m ∗ )) i 2 ( − 1 ) = + 1 i^2(-1)=+1 i 2 ( − 1 ) = + 1 p = k + Q p=k+Q p = k + Q ∣ m ∣ 2 = N 2 + a 2 v 2 |m|^2=N^2+a^2v^2 ∣ m ∣ 2 = N 2 + a 2 v 2
N μ ν = − 4 [ k μ p ν + k ν p μ − δ μ ν k ⋅ p ] + 4 ∣ m ∣ 2 δ μ ν , (104) \mathcal{N}_{\mu\nu}=-4\left[k_\mu p_\nu+k_\nu p_\mu-\delta_{\mu\nu}\,k\cdot p\right]+4|m|^2\delta_{\mu\nu}, \tag*{(104)} N μν = − 4 [ k μ p ν + k ν p μ − δ μν k ⋅ p ] + 4∣ m ∣ 2 δ μν , (104) the odd traces vanishing. Current conservation fails, so two form factors survive,
C μ ν E ( Q ) = ( δ μ ν − Q μ Q ν x ) A ( x ) + Q μ Q ν x L ( x ) , x = Q 2 . (105) C_{\mu\nu}^E(Q)=\left(\delta_{\mu\nu}-\frac{Q_\mu Q_\nu}{x}\right)A(x)+\frac{Q_\mu Q_\nu}{x}L(x),\qquad x=Q^2. \tag*{(105)} C μν E ( Q ) = ( δ μν − x Q μ Q ν ) A ( x ) + x Q μ Q ν L ( x ) , x = Q 2 . (105) Take Q E = ( Q , 0 ) Q_E=(Q,0) Q E = ( Q , 0 ) k E = ( z , k ) k_E=(z,\mathbf{k}) k E = ( z , k ) A A A L L L
1 3 ∑ i N i i = 4 z ( z + Q ) + 4 1 3 k 2 + 4 ∣ m ∣ 2 , \frac{1}{3}\sum_i\mathcal{N}_{ii}=4z(z+Q)+4\frac{1}{3}\mathbf{k}^2+4|m|^2, 3 1 i ∑ N ii = 4 z ( z + Q ) + 4 3 1 k 2 + 4∣ m ∣ 2 , N 44 = − 4 z ( z + Q ) + 4 k 2 + 4 ∣ m ∣ 2 , (106) \mathcal{N}_{44}=-4z(z+Q)+4\mathbf{k}^2+4|m|^2, \tag*{(106)} N 44 = − 4 z ( z + Q ) + 4 k 2 + 4∣ m ∣ 2 , (106) and both need the same two frequency integrals. For Re E , Re F > 0 \operatorname{Re}E,\operatorname{Re}F>0 Re E , Re F > 0 z = i E z=iE z = i E z = − Q + i F z=-Q+iF z = − Q + i F z = − i E z=-iE z = − i E z = − Q − i F z=-Q-iF z = − Q − i F
I = ∫ d z 2 π 1 ( z 2 + E 2 ) [ ( z + Q ) 2 + F 2 ] = E + F 2 E F [ ( E + F ) 2 + Q 2 ] , (107) \begin{aligned}
I=\int\frac{dz}{2\pi}\frac{1}{(z^2+E^2)[(z+Q)^2+F^2]} \\
&=\frac{E+F}{2EF\left[(E+F)^2+Q^2\right]}, \tag*{(107)}
\end{aligned} I = ∫ 2 π d z ( z 2 + E 2 ) [( z + Q ) 2 + F 2 ] 1 = 2 E F [ ( E + F ) 2 + Q 2 ] E + F , (107) the apparent pole at Q = − i ( E − F ) Q=-i(E-F) Q = − i ( E − F ) 2 z ( z + Q ) = 2z(z+Q)= 2 z ( z + Q ) = ( z 2 + E 2 ) + [ ( z + Q ) 2 + F 2 ] − ( Q 2 + E 2 + F 2 ) (z^2+E^2)+[(z+Q)^2+F^2]-(Q^2+E^2+F^2) ( z 2 + E 2 ) + [( z + Q ) 2 + F 2 ] − ( Q 2 + E 2 + F 2 )
K = ∫ d z 2 π z ( z + Q ) ( z 2 + E 2 ) [ ( z + Q ) 2 + F 2 ] = E + F 2 [ ( E + F ) 2 + Q 2 ] (108) K=\int\frac{dz}{2\pi}\frac{z(z+Q)}{(z^2+E^2)[(z+Q)^2+F^2]}=\frac{E+F}{2[(E+F)^2+Q^2]} \tag*{(108)} K = ∫ 2 π d z ( z 2 + E 2 ) [( z + Q ) 2 + F 2 ] z ( z + Q ) = 2 [( E + F ) 2 + Q 2 ] E + F (108) since 1 4 E + 1 4 F − 1 2 ( Q 2 + E 2 + F 2 ) I \frac{1}{4E}+\frac{1}{4F}-\frac{1}{2}(Q^2+E^2+F^2)I 4 E 1 + 4 F 1 − 2 1 ( Q 2 + E 2 + F 2 ) I ( E + F ) 2 + Q 2 − Q 2 − E 2 − F 2 = 2 E F (E+F)^2+Q^2-Q^2-E^2-F^2=2EF ( E + F ) 2 + Q 2 − Q 2 − E 2 − F 2 = 2 E F F = E ∗ F=E^* F = E ∗ E + F = 2 ε k E+F=2\varepsilon_k E + F = 2 ε k E F = D k EF=D_k E F = D k K = ε k / ( x + 4 ε k 2 ) K=\varepsilon_k/(x+4\varepsilon_k^2) K = ε k / ( x + 4 ε k 2 ) I = K / D k I=K/D_k I = K / D k
A ( x ) = ∫ d 3 k ( 2 π ) 3 8 ε k T + − x + 4 ε k 2 , L ( x ) = − ∫ d 3 k ( 2 π ) 3 8 ε k T 0 x + 4 ε k 2 (109) A(x)=\int\frac{d^3k}{(2\pi)^3}\frac{8\varepsilon_kT_{+-}}{x+4\varepsilon_k^2},\qquad L(x)=-\int\frac{d^3k}{(2\pi)^3}\frac{8\varepsilon_kT_0}{x+4\varepsilon_k^2} \tag*{(109)} A ( x ) = ∫ ( 2 π ) 3 d 3 k x + 4 ε k 2 8 ε k T +− , L ( x ) = − ∫ ( 2 π ) 3 d 3 k x + 4 ε k 2 8 ε k T 0 (109) with T + − T_{+-} T +− (49) and T 0 = 1 − ( k 2 + ∣ m ∣ 2 ) / D k T_0=1-(k^2+|m|^2)/D_k T 0 = 1 − ( k 2 + ∣ m ∣ 2 ) / D k
Passing to s = 4 ε k 2 s=4\varepsilon_k^2 s = 4 ε k 2 (51) and (52) , the measure is d 3 k / ( 2 π ) 3 = ( 2 π 2 ) − 1 k ( D k / ε k ) d ε k d^3k/(2\pi)^3=(2\pi^2)^{-1}k(D_k/\varepsilon_k)d\varepsilon_k d 3 k / ( 2 π ) 3 = ( 2 π 2 ) − 1 k ( D k / ε k ) d ε k d s = 8 ε k d ε k ds=8\varepsilon_kd\varepsilon_k d s = 8 ε k d ε k 1 / ( s + x ) 1/(s+x) 1/ ( s + x )
ρ 1 , 0 ( s ) = k D k T + − , 0 2 π 2 ε k ∣ ε k = s / 2 . (110) \rho_{1,0}(s)=\left.\frac{kD_kT_{+-,0}}{2\pi^2\varepsilon_k}\right|_{\varepsilon_k=\sqrt{s}/2}. \tag*{(110)} ρ 1 , 0 ( s ) = 2 π 2 ε k k D k T +− , 0 ε k = s /2 . (110) The two residues factorize. The first is (53) ; the second collapses even more simply, every term in ε k 2 \varepsilon_k^2 ε k 2
D k T 0 = D k − k 2 − ∣ m ∣ 2 = − 2 a v 2 ε k 2 − N 2 ε k 2 = − 2 a v 2 ( s − 4 N 2 ) s . (111) \begin{aligned}
D_kT_0=D_k-k^2-|m|^2=-2a_v^2\frac{\varepsilon_k^2-N^2}{\varepsilon_k^2} \\
&=-\frac{2a_v^2(s-4N^2)}{s}. \tag*{(111)}
\end{aligned} D k T 0 = D k − k 2 − ∣ m ∣ 2 = − 2 a v 2 ε k 2 ε k 2 − N 2 = − s 2 a v 2 ( s − 4 N 2 ) . (111) which with k = λ / 2 s k=\sqrt{\lambda/2}\sqrt{s} k = λ /2 s (56) . The threshold is at s = 4 N 2 s=4N^2 s = 4 N 2 a v a_v a v λ \lambda λ s = − 4 a v 2 s=-4a_v^2 s = − 4 a v 2
The representations need subtracting. Since ρ 1 ( s ) = s / ( 6 π 2 ) + a v 2 / π 2 + O ( s − 1 ) \rho_1(s)=s/(6\pi^2)+a_v^2/\pi^2+O(s^{-1}) ρ 1 ( s ) = s / ( 6 π 2 ) + a v 2 / π 2 + O ( s − 1 ) ρ 0 ( s ) = − a v 2 / π 2 + O ( s − 1 ) \rho_0(s)=-a_v^2/\pi^2+O(s^{-1}) ρ 0 ( s ) = − a v 2 / π 2 + O ( s − 1 ) L L L A A A ( s + x ) − 1 − s − 1 + x s − 2 = x 2 s − 2 ( s + x ) − 1 (s+x)^{-1}-s^{-1}+xs^{-2}=x^2s^{-2}(s+x)^{-1} ( s + x ) − 1 − s − 1 + x s − 2 = x 2 s − 2 ( s + x ) − 1
A ^ ( x ) = x 2 ∫ 4 N 2 ∞ ρ 1 ( s ) d s s 2 ( s + x ) , L ^ ( x ) = − x 2 ∫ 4 N 2 ∞ ρ 0 ( s ) d s s 2 ( s + x ) (112) \widehat A(x)=x^2\int_{4N^2}^{\infty}\frac{\rho_1(s)\,ds}{s^2(s+x)},\qquad\widehat L(x)=-x^2\int_{4N^2}^{\infty}\frac{\rho_0(s)\,ds}{s^2(s+x)} \tag*{(112)} A ( x ) = x 2 ∫ 4 N 2 ∞ s 2 ( s + x ) ρ 1 ( s ) d s , L ( x ) = − x 2 ∫ 4 N 2 ∞ s 2 ( s + x ) ρ 0 ( s ) d s (112) with A ^ = A − A ( 0 ) − x A ′ ( 0 ) \widehat A=A-A(0)-xA'(0) A = A − A ( 0 ) − x A ′ ( 0 ) L L L
A ′ ′ ( x ) = 2 ∫ 4 N 2 ∞ ρ 1 ( s ) d s ( s + x ) 3 , L ′ ′ ( x ) = − 2 ∫ 4 N 2 ∞ ρ 0 ( s ) d s ( s + x ) 3 . (113) A''(x)=2\int_{4N^2}^{\infty}\frac{\rho_1(s)\,ds}{(s+x)^3},\qquad L''(x)=-2\int_{4N^2}^{\infty}\frac{\rho_0(s)\,ds}{(s+x)^3}. \tag*{(113)} A ′′ ( x ) = 2 ∫ 4 N 2 ∞ ( s + x ) 3 ρ 1 ( s ) d s , L ′′ ( x ) = − 2 ∫ 4 N 2 ∞ ( s + x ) 3 ρ 0 ( s ) d s . (113) As a check independent of the whole radial route, the same two quantities follow from (D2) evaluated with Feynman parameters in dimensional regularization. Writing A ( u ) = ( 1 − u ) m 2 + u ( m ∗ ) 2 = N 2 − a v 2 + 2 i N a v ( 1 − 2 u ) \mathcal{A}(u)=(1-u)m^2+u(m^*)^2=N^2-a_v^2+2iNa_v(1-2u) A ( u ) = ( 1 − u ) m 2 + u ( m ∗ ) 2 = N 2 − a v 2 + 2 i N a v ( 1 − 2 u ) I 2 ( Δ ) = − d − 2 Δ I 0 ( Δ ) I_2(\Delta)=-\frac{d-2}{\Delta}I_0(\Delta) I 2 ( Δ ) = − Δ d − 2 I 0 ( Δ )
A ′ ′ ( x ) = 1 2 π 2 ∫ 0 1 d u ( u u ˉ ) 2 ( 3 A + ∣ m ∣ 2 + 2 u u ˉ x ) ( A + u u ˉ x ) 2 , (114) A''(x)=\frac{1}{2\pi^2}\int_0^1du\frac{(u\bar u)^2(3\mathcal{A}+|m|^2+2u\bar ux)}{(\mathcal{A}+u\bar ux)^2}, \tag*{(114)} A ′′ ( x ) = 2 π 2 1 ∫ 0 1 d u ( A + u u ˉ x ) 2 ( u u ˉ ) 2 ( 3 A + ∣ m ∣ 2 + 2 u u ˉ x ) , (114) with u ˉ = 1 − u \bar u=1-u u ˉ = 1 − u L ′ ′ L'' L ′′ ( ∣ m ∣ 2 − A ) ( u u ˉ ) 2 (|m|^2-\mathcal{A})(u\bar u)^2 ( ∣ m ∣ 2 − A ) ( u u ˉ ) 2 u u u 1 − u 1-u 1 − u 3 × 10 − 7 3\times10^{-7} 3 × 1 0 − 7 N N N a v a_v a v x x x (56) we know.
The four-point reflection test
We record the computation behind (94) . Coordinates are ordered ( x 4 , x 1 , x 2 , x 3 ) (x_4,x_1,x_2,x_3) ( x 4 , x 1 , x 2 , x 3 ) N = 1 N=1 N = 1 a v = 3 / 5 a_v=3/5 a v = 3/5 T = μ = 0 T=\mu=0 T = μ = 0 Re m 2 = 16 / 25 > 0 \operatorname{Re}m^2=16/25>0 Re m 2 = 16/25 > 0 Ψ = ( Ψ m , Ψ m ∗ ) T \Psi=(\Psi_m,\Psi_{m^*})^T Ψ = ( Ψ m , Ψ m ∗ ) T V a = i γ τ E x V_a=i\gamma^E_\tau x V a = i γ τ E x τ x \tau_x τ x
S ( x ) = diag ( S m ( x ) , S m ∗ ( x ) ) , S(x)=\operatorname{diag}(S_m(x),S_{m^*}(x)), S ( x ) = diag ( S m ( x ) , S m ∗ ( x )) , S m ( x ) = ( m − γ E ⋅ ∂ ) D m ( x ) , D m ( x ) = m K 1 ( m r ) 4 π 2 r , (115) S_m(x)=(m-\gamma^E\cdot\partial)D_m(x),\qquad D_m(x)=\frac{mK_1(mr)}{4\pi^2r}, \tag*{(115)} S m ( x ) = ( m − γ E ⋅ ∂ ) D m ( x ) , D m ( x ) = 4 π 2 r m K 1 ( m r ) , (115) r = ∣ x ∣ > 0 r=|x|>0 r = ∣ x ∣ > 0 z = m r z=mr z = m r ( K 1 / z ) ′ = − K 2 / z (K_1/z)'=-K_2/z ( K 1 / z ) ′ = − K 2 / z ( − K 2 / z ) ′ = − K 2 / z 2 + K 3 / z (-K_2/z)'=-K_2/z^2+K_3/z ( − K 2 / z ) ′ = − K 2 / z 2 + K 3 / z D D D D = ( m 2 / 4 π 2 ) f D=(m^2/4\pi^2)f D = ( m 2 /4 π 2 ) f D ′ = ( m 3 / 4 π 2 ) f ′ D'=(m^3/4\pi^2)f' D ′ = ( m 3 /4 π 2 ) f ′ f = K 1 / z f=K_1/z f = K 1 / z f ′ ′ = 2 K 2 / z 3 − K 1 ′ / z 2 + K 1 ′ / z − K 3 / z 2 f''=2K_2/z^3-K_1'/z^2+K_1'/z-K_3/z^2 f ′′ = 2 K 2 / z 3 − K 1 ′ / z 2 + K 1 ′ / z − K 3 / z 2 K 1 ′ = − ( K 1 + K 3 ) / 2 K_1'=-(K_1+K_3)/2 K 1 ′ = − ( K 1 + K 3 ) /2 K 3 ′ = − ( K 2 + K 4 ) / 2 K_3'=-(K_2+K_4)/2 K 3 ′ = − ( K 2 + K 4 ) /2
∂ 4 D = D ′ ς , ∂ 4 2 D = D ′ ′ , ∂ i ∂ j D = D ′ r δ i j , \partial_4D=D'\varsigma,\qquad\partial_4^2D=D'',\qquad\partial_i\partial_jD=\frac{D'}{r}\delta_{ij}, ∂ 4 D = D ′ ς , ∂ 4 2 D = D ′′ , ∂ i ∂ j D = r D ′ δ ij , ∂ 4 3 D = D ′ ′ ′ ς , ∂ 4 ∂ i ∂ j D = ( D ′ ′ r − D ′ r 2 ) δ i j , (116) \partial_4^3D=D'''\varsigma,\qquad\partial_4\partial_i\partial_jD=\left(\frac{D''}{r}-\frac{D'}{r^2}\right)\delta_{ij}, \tag*{(116)} ∂ 4 3 D = D ′′′ ς , ∂ 4 ∂ i ∂ j D = ( r D ′′ − r 2 D ′ ) δ ij , (116) with ς = sgn ( x 4 ) \varsigma=\operatorname{sgn}(x_4) ς = sgn ( x 4 )
This is more than bookkeeping: it collapses the Dirac content of the whole calculation into a finite table. Since S = ( m − γ ⋅ ∂ ) D S=(m-\gamma\cdot\partial)D S = ( m − γ ⋅ ∂ ) D ∂ 4 \partial_4 ∂ 4
P = A 1 + B γ 4 + C γ 1 , (117) P=A1+B\gamma_4+C\gamma_1, \tag*{(117)} P = A 1 + B γ 4 + C γ 1 , (117) TABLE III. The three-term reduction ((E3)): every propagator that occurs, with any admissible derivative content, in terms of D D D r r r ς = sgn ( x 4 ) \varsigma=\operatorname{sgn}(x_4) ς = sgn ( x 4 ) Δ ≡ D ′ ′ / r − D ′ / r 2 \Delta\equiv D''/r-D'/r^2 Δ ≡ D ′′ / r − D ′ / r 2
with the scalars of Table III . The Dirac algebra is therefore exhausted by the four traces Tr [ γ b 1 P 1 γ b 2 P 2 ] \operatorname{Tr}[\gamma_{b_1}P_1\gamma_{b_2}P_2] Tr [ γ b 1 P 1 γ b 2 P 2 ] Tr [ γ b 1 P 1 ⋯ γ b 4 P 4 ] \operatorname{Tr}[\gamma_{b_1}P_1\cdots\gamma_{b_4}P_4] Tr [ γ b 1 P 1 ⋯ γ b 4 P 4 ] b i ∈ { 4 , 1 } b_i \in\{4,1\} b i ∈ { 4 , 1 } A 1 … C 4 A_1\ldots C_4 A 1 … C 4
Table III. The three-term reduction ((E3)): every propagator that occurs, with any admissible derivative content, in terms of D D D r r r ς = sgn ( x 4 ) \varsigma=\operatorname{sgn}(x_4) ς = sgn ( x 4 ) Δ ≡ D ′ ′ / r − D ′ / r 2 \Delta\equiv D''/r-D'/r^2 Δ ≡ D ′′ / r − D ′ / r 2
Tr [ γ 4 P 1 γ 4 P 2 ] = 4 ( A 1 A 2 + B 1 B 2 − C 1 C 2 ) , \operatorname{Tr}[\gamma_4P_1\gamma_4P_2] = 4(A_1A_2+B_1B_2-C_1C_2), Tr [ γ 4 P 1 γ 4 P 2 ] = 4 ( A 1 A 2 + B 1 B 2 − C 1 C 2 ) , Tr [ γ 1 P 1 γ 1 P 2 ] = 4 ( A 1 A 2 − B 1 B 2 + C 1 C 2 ) , \operatorname{Tr}[\gamma_1P_1\gamma_1P_2] = 4(A_1A_2-B_1B_2+C_1C_2), Tr [ γ 1 P 1 γ 1 P 2 ] = 4 ( A 1 A 2 − B 1 B 2 + C 1 C 2 ) , Tr [ γ 4 P 1 γ 1 P 2 ] = 4 ( B 1 C 2 + B 2 C 1 ) , (118) \operatorname{Tr}[\gamma_4P_1\gamma_1P_2] = 4(B_1C_2+B_2C_1), \tag*{(118)} Tr [ γ 4 P 1 γ 1 P 2 ] = 4 ( B 1 C 2 + B 2 C 1 ) , (118) and the four-point ones carry twenty or twenty-one monomials each. Assembled with the combinatorics below they reproduce (94) exactly, which is how we have checked the reduction.
The flavor trace is where the structure of the model reappears. The vertices carry τ x \tau_x τ x
Tr f l [ τ x S ( 1 ) τ x S ( 2 ) ⋯ ] = S m ∗ ( 1 ) S m ( 2 ) S m ∗ ( 3 ) ⋯ + ( m ↔ m ∗ ) , (119) \operatorname{Tr}_{\mathrm{fl}}\left[\tau_x S^{(1)}\tau_x S^{(2)}\cdots\right]
= S_{m^*}^{(1)}S_m^{(2)}S_{m^*}^{(3)}\cdots+(m\leftrightarrow m^*), \tag*{(119)} Tr fl [ τ x S ( 1 ) τ x S ( 2 ) ⋯ ] = S m ∗ ( 1 ) S m ( 2 ) S m ∗ ( 3 ) ⋯ + ( m ↔ m ∗ ) , (119) since τ x diag ( S m , S m ∗ ) \tau_x\operatorname{diag}(S_m,S_{m^*}) τ x diag ( S m , S m ∗ ) alternate around every surviving loop, which is the balanced-cut condition of Appendix C read off the trace rather than off the Landau equations. And since the separations are real, D m ∗ = D m D_{m^*}=D_m D m ∗ = D m ω [ ( Θ F ) F ] \omega[(\Theta F)F] ω [( Θ F ) F ]
For separated points Wick contraction gives
C 2 ( i , j ) = − Tr ( V i S i j V j S j i ) , C_2(i,j)=-\operatorname{Tr}(V_iS_{ij}V_jS_{ji}), C 2 ( i , j ) = − Tr ( V i S ij V j S j i ) , C 4 = ∑ pairings C 2 C 2 − ∑ ( i , j , k ) ∈ Perm ( 2 , 3 , 4 ) L ( 1 , i , j , k ) , C_4=\sum_{\text{pairings}}C_2C_2-\sum_{(i,j,k)\in\operatorname{Perm}(2,3,4)}L(1,i,j,k), C 4 = pairings ∑ C 2 C 2 − ( i , j , k ) ∈ Perm ( 2 , 3 , 4 ) ∑ L ( 1 , i , j , k ) , L ( 1 , i , j , k ) = Tr ( V 1 S 1 i V i S i j V j S j k V k S k 1 ) , (120) L(1,i,j,k)=\operatorname{Tr}(V_1S_{1i}V_iS_{ij}V_jS_{jk}V_kS_{k1}), \tag*{(120)} L ( 1 , i , j , k ) = Tr ( V 1 S 1 i V i S ij V j S j k V k S k 1 ) , (120) the trace running over Dirac and flavor indices, with three disconnected pairings and six oriented four-cycles; cycles of odd length vanish because each vertex carries τ x \tau_x τ x S S S O a b = ∂ a j b − ∂ b j a O_{ab}=\partial_a j_b-\partial_b j_a O ab = ∂ a j b − ∂ b j a
Time reflection acts with R = diag ( − 1 , 1 , 1 , 1 ) R=\operatorname{diag}(-1,1,1,1) R = diag ( − 1 , 1 , 1 , 1 ) Θ O 4 i ( x ) = − O 4 i ( θ x ) \Theta O_{4i}(x)=-O_{4i}(\theta x) Θ O 4 i ( x ) = − O 4 i ( θ x ) Θ O i j ( x ) = + O i j ( θ x ) \Theta O_{ij}(x)=+O_{ij}(\theta x) Θ O ij ( x ) = + O ij ( θ x ) + 1 +1 + 1 ( − Y , − X , X , Y ) (-Y,-X,X,Y) ( − Y , − X , X , Y ) X = ( 1 , 0 ) X=(1,0) X = ( 1 , 0 ) Y = ( 2 , 0 ) Y=(2,0) Y = ( 2 , 0 ) (93) , and the two pairings that cross the reflection together with the connected cycles give the numbers quoted in (94) . Smearing against non-negative normalized bumps supported in disjoint neighbourhoods of X X X Y Y Y
References 
