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The quark spectral functions and the Hadron Vacuum Polarization from application of DSEs in Minkowski space

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Abstract

The hadronic vacuum polarization function \(\Pi _h\) for two light flavors is computed on the entire domain of spacelike and timelike momenta using a framework of Dyson–Schwinger equations. The analytical continuation of the function \(\Pi _h\) is based on the utilization of the Gauge Technique with the entry of QCD Green’s functions determined from generalized quark spectral functions. For the first time, the light quark spectral functions are extracted from the solution of the gap equation for the quark propagator. The scale is set up by the phenomena of dynamical chiral symmetry breaking, which is a striking feature of low energy QCD.

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Appendices

Appendix A: Rainbow Ladder Quark Self-Energy in Arbitrary Linear Gauge

The Dyson–Schwinger equation for the quark propagator can be converted into the Unitary equations for Nakanishi weights, by comparing of imaginary and real parts of assumed integral representation (2.15), i.e.

$$\begin{aligned} \sigma _{v,s}(p^2)=-\frac{\mathfrak {I}S_{v,s}(p^2)}{\pi }\qquad \mathfrak {R}S_{v,s}(p^2)= P.\int ds \frac{\sigma _{v,s}(x)}{p^2-x} , \end{aligned}$$
(A1)

and by the integral representation for the inverse of the propagator, which is readily derivable from quark gap Eq. (2.5). After the renormalization it reads

$$\begin{aligned} S^{-1}=\not p -m(\mu )-\Sigma (p) , \end{aligned}$$
(A2)

where the self-energy functions \(\Sigma =\Sigma _{V}+\Sigma _{\xi }\) are evaluated in details in this “Appendix”.

Let us start with unrenormalized self-energy (hence index 0), which comes from the product of a gauge term and the quark propagator expressed through the Hilbert transformation. The first line in (2.5) reads

$$\begin{aligned} \Sigma _{\xi }^0(q)=-i\xi g^2\int \frac{d^4k}{(2\pi )^4} \int d o \, \gamma _{\mu } \frac{\sigma _v(o)\not k+\sigma _s(o)}{(k^2-o+i\epsilon )}\gamma _{\nu } \frac{(k-q)^{\mu }(k-q)^{\nu }}{{((k-q)^2+i\epsilon )}^2} . \end{aligned}$$
(A3)

After a standard treatment and a little algebra it can be written into the following form

$$\begin{aligned} \Sigma _{\xi }^0(q)&=-\xi g^2\int \nolimits _0^1 dx \int \frac{d^4 k_E}{(2\pi )^4}\int d o\nonumber \\&\quad \times \left[ \frac{4(1-x)\sigma _v(o) k.q \not k}{D^3} +\frac{\sigma _v(o)(-2-x) \not q+\sigma _s(o)}{D^2}+ \frac{4(1-x)x^2\sigma _v(o) q^2 \not q}{D^3}\right] , \end{aligned}$$
(A4)

where the denominator \(D=-k^2_E-q_E^2(1-x)x-o x\) is strictly negative, noting the Wick rotation is working for positive as well as for the negative variable o. Thus to go to the Euclidean space is what we only need here in order to integrate over the momenta, as the result we get

$$\begin{aligned} \Sigma _{\xi }^0(q)= & {} -\frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx \int d o ((1-x)\not q\sigma _v(o)\left[ c(d)+\ln {(\Omega /\mu ^2)}\right] \nonumber \\&+\,[\sigma _v(o)(-2-x) \not q+\sigma _s(o)][c(d)+\ln {(\Omega /\mu ^2)}] +\frac{(1-x)x \sigma _v(o) q^2 \not q}{q^2(1-x)-o+i\epsilon } , \end{aligned}$$
(A5)

where \( \Omega =q^2x(1-x)-o x +i\epsilon \) and \(\mu \) is the spacelike renormalization scale (\(\mu ^2<0\) in our metric convention).

The third term in the Eq. (A5) is UV finite and can be rewritten as

$$\begin{aligned} -\frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx \int d o \frac{(1-x)x \sigma _v(o) q^2 \not q}{q^2(1-x)-o+i\epsilon } =-\frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx \int d o (1-x)x S_v(\omega ^{,}) q^2 \not q , \end{aligned}$$
(A6)

where we have employed the Hilbert transformation (2.16) and label \(\omega ^{,}=q^2(1-x)\).

In addition to the usual Minimal Subtraction counter-terms we will sent also the following terms

$$\begin{aligned} \delta Z_{\psi }= & {} \int \nolimits _0^1 dx\frac{g^2\xi }{(4\pi )^2}\int d o \sigma _v(o) (1+2x) ln x =\frac{3g^2\xi }{2(4\pi )^2}\int d o \sigma _v(o)\nonumber \\ \delta Z_{m}= & {} -\int \nolimits _0^1 dx\frac{g^2\xi }{(4\pi )^2}\int d o \sigma _s(o) ln x =-\frac{g^2\xi }{(4\pi )^2}\int d o \sigma _s(o) \end{aligned}$$
(A7)

into the renormalization constant \(Z_2(Z_{\psi })\) and \(Z_4(Z_{m})\).

For the first two terms in (A5) we thus have

$$\begin{aligned} -\frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx \int d o \left[ \not q (-1-2x)\sigma _v(o)+\sigma _s(o)] ln \frac{q^2(1-x)-o+i\epsilon }{\mu ^2}\right] , \end{aligned}$$
(A8)

where we have drop out all renormalization constants. Using per-partes integration and sending momentum independent boundary terms into renormalization constants again one finally gets for the rest of (A8):

$$\begin{aligned} q^2 \frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx \int d o \frac{-\not q x(1+x)\sigma _v(o)+x \sigma _s(o)}{q^2(1-x)-o+i\epsilon } \end{aligned}$$
(A9)

Summing this with (A6 )one finally gets

$$\begin{aligned} \Sigma _{\xi }(q)= & {} q^2 \frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx \int d o \frac{-2x \sigma _v(o) \not q+x\sigma _s(o)}{q^2(1-x)-o+i\epsilon }\nonumber \\= & {} q^2\frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^1 dx x [ -2 S_v(\omega ^{,}) \not q+S_s(\omega ^{,})] . \end{aligned}$$
(A10)

where we have employed the Hilbert transformation once again. To end, we make the substitution \(s=\omega ^{'}\) and arrive into the desired renormalized result:

$$\begin{aligned} \Sigma _{\xi }(q)=-\frac{\xi g^2}{(4\pi )^2}\int \nolimits _0^{q^2} d s \left( 1-\frac{s}{q^2}\right) [\not q 2 S_v(s)-S_s(s)] +C_A \not q +C_B \end{aligned}$$
(A11)

where \(C_{A,B}\) are in principle arbitrary finite constants specifying a given renormalization scheme. In this paper we will use modified MS for which we take \(C_A=C_B=0\) (i.e. the renormalization follow the derivation above). Fully equivalently we employ momentum renormalization scheme, which defines self-energy at given renormalization scheme with renormalization constants given by renormalized self-energy at given point. Numerically, both methods work equally.

The expression (A11) is valid either for the timelike or the spacelike momenta \(q^2\), however for the spacelike arguments \(q^2<0\) one should keep in mind that the integration runs over the negative values of variable s. Thus for an aesthetic reasoning, for spacelike q, we rather write

$$\begin{aligned} \Sigma _{\xi }(q,q^2<0)=\frac{\xi g^2}{(4\pi )^2}\int ^0_{q^2} d s \left( 1-\frac{s}{q^2}\right) \left[ \not q 2 S_v(s)-S_s(s)\right] +C_A \not q +C_B , \end{aligned}$$
(A12)

which keeps the lower integral boundary smaller then the upper one.

Let us make short digression here and remind that the dynamical symmetry breaking and massless pion are attributed to chiral limit \(m=0\) in QCD. This is a more complicated issue when one is dealing with spectral representation and as we have sent (irrespective of their UV finiteness) scalar piece of the self-energy into the renormalized constant, we cannot use our scheme directly for the calculation in the chiral limit. To set the mass exactly to zero one must also require

$$\begin{aligned} \int d o \rho _s(o)=0 , \end{aligned}$$
(A13)

which, at least at the formal level allows us to skip the mass renormalization at all. The sum rule condition (A13) could be explicitly used before the momentum integration in the chiral limit. otherwise we are facing the ambiguity \(c(d)\int \sigma _s=\infty .0\) and the result turns to be ordering dependent (not well defined). In this paper we will deal with the physical pion and we leave the question of solution in exact chiral limit \(m=0\) unanswered for future task.

For the combination of the vector interaction \(V_v\) with the spectral part of the quark propagator we get

$$\begin{aligned} \Sigma _{V}=-i\int \frac{d^4k}{(2\pi )^4} \int d o \frac{\gamma ^{\mu }(\not k \sigma _v(o)+\sigma _s(o))\gamma _{\mu }}{k^2-o+i\epsilon }\left[ \frac{c_V}{(k-q)^2-m_g^2+i\epsilon } -\frac{c_V}{(k-q)^2-\Lambda _g^2+i\epsilon } \right] , \end{aligned}$$
(A14)

which is the standard one loop expression integrated over the continuous mass o giving us the known result:

$$\begin{aligned} \Sigma _{V}=c_v \int ^1_0 dx \int d o \frac{-2 \not q (1-x) \sigma _v(o)+4 \sigma _s(o)}{(4\pi )^2} log\left( \frac{q^2(1-x)-o-m_g^2\frac{1-x}{x}+i\epsilon }{q^2(1-x)-o-\Lambda _g^2\frac{1-x}{x}+i\epsilon }\right) . \end{aligned}$$
(A15)

For numerical purpose it is suited to further proceed by per partes integration

$$\begin{aligned} \Sigma _{V}= & {} \frac{c_v}{(4\pi )^2} \int ^1_0 dx \int d o \frac{2 \not q (1-x/2) \sigma _v(o)-4 \sigma (o)}{q^2(1-x)-o-m_g^2\frac{1-x}{x}+i\epsilon }\left( -q^2+\frac{m_g^2}{x}\right) -(m_g\rightarrow \Lambda _g)\nonumber \\= & {} \frac{c_v}{(4\pi )^2} \int ^1_0 dx [2 \not q (1-x/2) S_v(\hat{a})-4 S_s(\hat{a})] \left[ -q^2+\frac{m_g^2}{x}\right] -(m_g\rightarrow \Lambda _g) , \end{aligned}$$
(A16)

where the argument in the first term of the second line reads \(\hat{a}=q^2(1-x)-m_g^2\frac{1-x}{x}\), which can be seen by virtue of Hilbert transformation again.

The last step advantageous for numerical solution is the introduction of the following functional identities

$$\begin{aligned} 1=\int \nolimits _{-\infty }^{\infty } da \delta (a-\hat{a}) \, , \delta (f(x))=\Sigma _i \frac{\delta (x-x_i)}{|\frac{df}{dx}(x_i)|} \end{aligned}$$
(A17)

in the previous equation. Interchanging the order of integration and integrating over the variable x one gets

$$\begin{aligned} \Sigma _{V}=\Sigma _{i=\pm }\frac{c_v}{(4\pi )^2}\int \nolimits _{-\infty }^{\infty } da \frac{2 \not q (x_i-x_i/2) S_v(a)-4 x_i S_s(a)}{sgn(-q^2 x_i^2+m_g^2)}\Theta (x_i)\Theta (1-x_i)\Theta (D)-(m_g\rightarrow \Lambda _g) ,\nonumber \\ \end{aligned}$$
(A18)

where the roots are

$$\begin{aligned} x_{\pm }= & {} \frac{-(m_g^2+q^2-a)\pm \sqrt{D}}{-2q^2} \nonumber \\ D= & {} (m_g^2+q^2-a)^2-4q^2m_g^2 \end{aligned}$$
(A19)

for the first term. The expression (A18) has been actually used in our numerical code.

Appendix B: Determination of Quark Spectral Function

In the “Appendix” above we have determined the selfenergy functions selfconsistently, i.e. assuming the quark propagator satisfies spectral representation

$$\begin{aligned} S(k)=\int \nolimits _0^{\infty }d a\frac{\not p \rho _v(a)+\rho _s(a)}{p^2-a+i\epsilon } \end{aligned}$$
(B1)

we have derived the selfenergy \(\Sigma \), which solely depends on spectral quark functions \(\sigma _v\) and \(\sigma _s\).

In addition we assume that the spectral functions are smooth, the only singularity of the propagator is a cut at timelike real axis of momenta, i.e. it starts at \(p^2=0\). Thus as a consequence, the propagator is real for the spacelike arguments \(p^2\), \(p^2>0\).

Analytically properties are completely exploited in practice in the numerical search. From this it follows, that for a non-trivial solution for propagator evaluated at timelike momentum \(p^2\), there must be a unique ratio \(\mathfrak {R}S_{s,v}(p^2)/ \mathfrak {I}S_{s,v}(p^2)\) such that \(\mathfrak {I}\Sigma _{s,v}(p^2)=-\pi \sigma _{s,v}(p^2)\).

First let us make a projections, writing \(S^{-1}(p)=\not p A(p)-B(p)\) where AB are two scalar function needed to determine the propagator S. Comparing to DSE one gets

$$\begin{aligned} B(p^2)= m(\mu )+\Sigma _B(p^2) ;\quad A(p^2)= 1-\Sigma _A(p^2) \end{aligned}$$
(B2)

where all terms of the selfenergy which are proportional to 4times4times unit matrix are collected in \(\Sigma _B(p)\) and all terms of selfenergy terms listed in previous “Appendix” and proportional to \(\not p\) matrix are collected in the scalar function \(\Sigma _A(p)\)

In what follows we subtract the two equations with itself at some arbitrary timelike scale \(\zeta \).

$$\begin{aligned} B(p^2)= B(\zeta ) +\Sigma _B(p^2)-\Sigma _B(\zeta ) A(p^2)=A(\zeta )-\Sigma _A(p^2)+\Sigma _A(\zeta ) . \end{aligned}$$
(B3)

To get the spectral functions one just needs to inverse the propagator expressed in terms self-energies. From the imaginary parts one gets equation for spectral functions:

$$\begin{aligned} \rho _s(s)=\frac{-1}{\pi }\frac{\mathfrak {I}B(s) R_D(s)+\mathfrak {R}B(s)) I_D(s)}{R_D^2(s)+I_D^2(s)} \end{aligned}$$
(B4)

where \(s=p^2>0\) in our metric, and where the functions \(R_D\) and \(I_D\) stand for the square of the real and the imaginary part of the function \(sA^2(s)-B^2(s)\), i.e.

$$\begin{aligned} R_D(s)= & {} s[\mathfrak {R}A(s)]^2-s [\mathfrak {I}A(s)]^2-\mathfrak {R}B(s)^2+[\mathfrak {I}B(s)]^2 \nonumber \\ I_D(s)= & {} 2 s \mathfrak {R}A(s)\mathfrak {I}A(s) +2 \mathfrak {R}B(s) \mathfrak {I}B(s) , \end{aligned}$$
(B5)

and similarly for the function \(\rho _v\).

$$\begin{aligned} \rho _v(s)=\frac{-1}{\pi }\frac{\mathfrak {I}A(s) I_D(s)-\mathfrak {I}A(s)) R_D(s)}{R_D^2(s)+I_D^2(s)} . \end{aligned}$$
(B6)

To get the solution of quark gap equation consistent with spectral representation one get the solution in two steps. As a first step one solve the coupled system (B3) and (B4, B6) for arbitrary initial choice of four fixed numbers \(\mathfrak {R}B(\zeta ), \mathfrak {I}B(\zeta ), \mathfrak {R}A(\zeta ), \mathfrak {I}A(\zeta )\) assuring the system is convergent, then one changes gradually the imaginary parts in the manner described bellow.

As the first, two following functions

$$\begin{aligned} L_s= & {} P. \int \nolimits _0^{\infty }d a\frac{\rho _s(a)}{p^2-a} \nonumber \\ R_s= & {} \frac{-\mathfrak {I}B(s) I_D(s)+\mathfrak {R}B(s) R_D(s)}{R_D^2(s)+I_D^2(s)} \end{aligned}$$
(B7)

should agree with each other. The equality obviously implies

$$\begin{aligned} L_s(s)=R_s(s)=\frac{1}{4} \mathfrak {R}Tr S(s)=\mathfrak {R}S_s(s) \end{aligned}$$
(B8)

in this case.

Similarly, defining for resulting solutions

$$\begin{aligned} L_v= & {} P. \int \nolimits _0^{\infty }d a\frac{\rho _v(a)}{p^2-a} \nonumber \\ R_v= & {} \frac{-\mathfrak {R}A(s) R_D(s)+\mathfrak {I}A(s) I_D(s)}{R_D^2(s)+I_D^2(s)} \end{aligned}$$
(B9)

The equality obviously implies

$$\begin{aligned} L_v(s)=R_v(s)=\frac{1}{4p^2} \mathfrak {R}\not p Tr S(s)=\mathfrak {R}S_v(s) . \end{aligned}$$
(B10)

These conditions are not used for purpose of solving the system (B4)+(B3)+(B6). Eqs. (B8), (B10) are obviously not fulfilled for arbitrary values \( \mathfrak {I}B(\zeta )\) and \( \mathfrak {I}A(\zeta )\) and are used as cross check instead.

All the system of equations is then solved iteratively in two enclosed iteration cycles. The inner iteration cycles solve the DSEs (B4)+(B3)+(B6) iteratively. The external cycle scan imaginary parts of A and B at fixed points and search for their correct values by evaluating the norms

$$\begin{aligned} h(\sigma _v,\mathfrak {I}A)= & {} \frac{\int (L_v(s)-R_v(s))^2 ds}{\int (L_v(s)+R_v(s))^2 ds}\nonumber \\ h(\sigma _s,\mathfrak {I}B)= & {} \frac{\int (L_s(s)-R_s(s))^2 ds}{\int (L_s(s)+R_s(s))^2 ds} \end{aligned}$$
(B11)

The solution turns out to be fast convergent and amazingly stable for almost any value of gauge fixing parameters. Few seconds at recent computers are needed for (\(\simeq 30\)) iterations providing the stable solution for fixed masses and renormalization functions and therefore a correct ratio \(\mathfrak {I}B(\zeta )/\mathfrak {R}B(\zeta )\) can be identified with an arbitrary precision. Examples of solutions obtained for different values of the product \(g^2\xi \) is shown in the Fig. 7. The dot-dot-dashed line corresponds with our ladder-rainbow QCD model and is rescaled for purpose of the figure here. All other lines correspond with the renormalized values \(B(\zeta )\) and \(A(\zeta )\) chosen such that \(\mathfrak {R}B(\zeta )=1.0 L\) and \(\mathfrak {I}A(\zeta )=2\) at \(\zeta =0.1L^2\). Other parameters values are identical to those referred in the main text and we show only the real part of renormalization function and the imaginary part of the function B(s). Now, the scale L defines our units wherein \(m_g^2=2L^2\) in this units.

Fig. 7
figure 7

Solutions of the quark gap equation as described in the “Appendix”. Numbers refer various values of the product \(g^2\xi \)

The reader can recognize that the all above procedure remains a well know momentum renormalization scheme, here however performed at the timelike renormalization scale. It can be regarded as, but it should be stresses that it has nothing to do with the renormalization of UV infinities presented in the theory, noting that the renormalization constants stay real since they were identified by imposing the renormalization conditions at spacelike t’Hooft renormalization scale \(\mu \) which is a must for theory with Hermitian Lagrangian.

Providing the equalities (B8) and (B10) hold, the propagator evaluated at spacelike momentum is equivalent to the solution of the fermion DSE solved directly in the momentum Euclidean space. The uniqueness of analytical continuation ensures the statement exactly, while mutual numerical comparison provides independent check (at least of numerical accuracy).

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Šauli, V. The quark spectral functions and the Hadron Vacuum Polarization from application of DSEs in Minkowski space. Few-Body Syst 61, 23 (2020). https://doi.org/10.1007/s00601-020-01555-3

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