Debye Mass of Massless Φ4-Theory to Order g6 at Weak Coupling


The Debye mass is calculated to order g6 at weak coupling of massless ϕ4-theory. Debye mass gets contributions from the hard and soft momentum scale. We use dimensional reduction and the effective field theory to distinct two scales of order T and gT. Three dimensional effective theory is used to calculate contributions of soft scale whereas hard scale is evaluated by applying a perturbation theory, in power series of g. For the Debye mass to g6-order corrections, we calculate the hard part to g6 order and the soft part through mass parameter to g6 and calculate four loop self energy diagrams. The convergence of series and loop expansion are discussed.

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The authors would like to thank Jens O. Andersen for useful discussions.

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Correspondence to Rashid Khan.

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Appendix A: Three-Dimensional Integrals

The infrared and ultraviolet divergences appearing in three-dimensional integrals over momenta, can be regularized by dimensional regularization. The integral is performed in d-dimensions (real number) for which the integral is convergent and then it is analytically continued back to the physical dimension, d = 3. The divergence will be isolated in an expansion parameter which goes to zero, and can be removed by renormalization. The momentum space integration measure is,

$$ {\int}_{p} \equiv \left( \frac{e^{\gamma}\mu^{2}}{4\pi}\right)^{\epsilon} \int {d^{3-2\epsilon}p \over (2 \pi)^{3-2\epsilon}} . $$

A.1 One Loop

For general loop calculations the integral is given below. It is used to calculate I1, I2, I3 and I4,

$$ \begin{array}{@{}rcl@{}} I_n&\equiv&{\int}_p{1\over(p^{2}+m^{2})^n}\\ &=&{1\over8\pi}(e^{\gamma_E}\mu^{2})^{\epsilon} {\Gamma(n-\frac{3}{2}+\epsilon) \over{\Gamma}(\frac{1}{2}) {\Gamma}(n)}m^{3-2n-2\epsilon} . \end{array} $$

A.2 Two Loop

Two-loop integrals that are required for the calculations are given below,

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}(p=im_{D})&=& {1\over4(4\pi)^{2}} \left( {\mu\over2m}\right)^{4\epsilon}\\ && \times\left[ {1\over\epsilon}+4\log2+6 +6{m\over m_{D}}\log{3m-m_{D}\over3m+m_{D}}-2\log{9m^{2}-m_{D}^{2}\over m^{2}} +\mathcal{O}(\epsilon) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} =\\ I_{\text{sun}}^{\prime}(p=im)&=& {\log2\over4m^{2}(4\pi)^{2}}\left( {\mu\over2m}\right)^{4\epsilon} \left[1 +\mathcal{O}(\epsilon) \right]\\ && +{{\Pi}_{1}\over32m^{4}(4\pi)^{2}} \left( {\mu\over2m}\right)^{4\epsilon} \left[ 3-4\log2 +\mathcal{O}\left( \epsilon\right) \right] , \end{array} $$
$$ \begin{array}{@{}rcl@{}} J_{\text{sun}}(p=im)&=& {1\over32m^{4}(4\pi)^{2}}\left( {\mu\over2m}\right)^{4\epsilon} \left[1+\mathcal{O}(\epsilon)\right] , \end{array} $$
$$ \begin{array}{@{}rcl@{}} K_{\text{sun}}(p=im)&=& {5\over96m^{4}(4\pi)^{2}}\left( {\mu\over2m}\right)^{4\epsilon} \left[1+\mathcal{O}(\epsilon)\right] . \end{array} $$

The sunset diagram is evaluated in Ref. [17] to order 𝜖0. We calculate the remaining two-loop integrals in Appendix B.

A.3 Three Loop

Three-loop integrals needed are given below,

$$ \begin{array}{@{}rcl@{}} I_{\text{ball}}^{\prime} &=&{1\over8m(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon}\left[ {1\over\epsilon}+2-4\log2 +\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} J_{\text{ball}} &=&{1\over16m^3(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon}\left[ 1+\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} K_{\text{ball}} &=&{1\over32m^3(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon}\left[ {1\over\epsilon}+5-4\log2 +\mathcal{O}\left( \epsilon\right) \right] , \end{array} $$
$$ \begin{array}{@{}rcl@{}} I_{\text{c3rung}} &=& {7\over16m(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon} \left[\zeta(3) +\mathcal{O}\left( \epsilon\right) \right]\\ && +{{\Pi}_{1}\over96m^3(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon} \left[ \pi^{2}+24\log2-21\zeta(3) +\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} J_{\text{c3rung}} &=&-{1\over96m^3(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon} \left[\pi^{2}-24\log2+\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} K_{\text{c3rung}} &=&{1\over192m^3(4\pi)^3}\left( {\mu\over2m}\right)^{6\epsilon} \left[\pi^{2}+\mathcal{O}\left( \epsilon\right) \right] . \end{array} $$

Iball, massive basketball diagram is evaluated to order 𝜖0 by Ref. [17] , and to order 𝜖 by Ref. [27]. Iball can be differentiated with respect to m to get \(I_{\text {ball}}^{\prime }\). In Appendix B the J and K loop integral are calculated.

A.4 Four Loop

Four-loop integrals needed are given below,

$$ \begin{array}{@{}rcl@{}} I_{\text{c4rung}} &=&{1\over32m^{2}(4\pi)^{4}}\left( {\mu\over2m}\right)^{8\epsilon}\left[ -2\pi^{2}\log2+ 15\zeta(3)+\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} I_{\text{triangle}}^{\prime} &=&{\pi^{2}\over48m^{2}(4\pi)^{4}}\left( {\mu\over2m}\right)^{8\epsilon}\left[ 1+\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} I &=& I_{\text{cballsun}}(p=im)-{1\over4(4\pi)^{2}\epsilon}I_{\text{sun}}^{\prime}(p=im)\\ &=& {\log2\over4m^{2}(4\pi)^{4}}\left[ \log{\mu\over2m} -0.01353 +\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} I_{\text{fish1}}(p=im) &=&{1\over16m^{2}(4\pi)^{4}}\left( {\mu\over2m}\right)^{8\epsilon} \left[4.6029+\mathcal{O}\left( \epsilon\right) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} I_{\text{fish2}}(p=im) &=&{1\over16m^{2}(4\pi)^{4}}\left( {\mu\over2m}\right)^{8\epsilon} \left[1.778+\mathcal{O}\left( \epsilon\right) \right] . \end{array} $$

In Ref. [25], Itriangle, the triangle diagram was calculated, so by differentiation Itriangle with respect to m2 we get \(I_{\text {triangle}}^{\prime }\). The fish diagrams are calculated in Appendix Appendix.

Appendix B: Explicit Calculations

B.1 Two Loop Diagrams

The sunset diagram Isun(p = im) is known to order 𝜖0 in Ref. [17]. However, we need to evaluate Isun(p) at p = imD. By going to coordinate space, we can write

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}(p)&=& {\int}_{qr}{1\over (q^{2}+m^{2})(r^{2}+m^{2})} {1\over(\mathbf{p}+\mathbf{q}+\mathbf{r})^{2}+m^{2}} \end{array} $$
$$ \begin{array}{@{}rcl@{}} &=& {\int}_{R}e^{i\mathbf{p}\cdot\textbf{R}}V^3(R), \end{array} $$

where the coordinate-space integral is

$$ \begin{array}{@{}rcl@{}} {\int}_{R}&=&\left( {e^{\gamma_E}\mu^{2}\over4\pi}\right)^{-\epsilon} \int d^{dR} , \end{array} $$

the Fourier transform of momentum-space propagator is represented by V (R) and given as,

$$ \begin{array}{@{}rcl@{}} V(R)&=&{\int}_q{e^{i\mathbf{q}\cdot\textbf{R}}\over q^{2}+m^{2}}\\ &=& \left( {e^{\gamma_E}\mu^{2}\over4\pi}\right)^{-\epsilon} {1\over2\pi^{3/2-\epsilon}}\left( {m\over R}\right)^{1/2-\epsilon} K_{{1\over2}-\epsilon}(mR). \end{array} $$

Here, Kν(x) is Bessel function of the second kind. After integrating over angles, (36) reduces to

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}(p)&= &\left( {e^{\gamma_E}\mu^{2}\over2p}\right)^{-\epsilon} {(2\pi)^{3\over2}\over\sqrt{p}} {\int}_{0}^{\infty}dRR^{{3\over2}-\epsilon}J_{{1\over2}-\epsilon}(pR) V^3(R). \end{array} $$

The integral (39) is ultraviolet divergent is three dimensions for small values of R, i.e. for large values of the momenta. We therefore split the integral into an integral from R = 0 to R = r and one from R = r to \(R=\infty \). The first integral is calculated in d = 3 − 2𝜖 dimensions and expanded in powers of 𝜖, while the second is calculate directly in d = 3. This yields

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}(p)&=& \left( {e^{\gamma_E}\mu^{2}\over4\pi}\right)^{-\epsilon} {(2\pi)^{3\over2}\over\sqrt{p}} {\int}_{0}^{r} dR R^{{3\over2}-\epsilon}J_{{1\over2}-\epsilon}(pR) V^3(R)\\ && +{4\pi\over p}{\int}_{R}^{\infty} dR R\sin(pR)V_{0}^3(R), \end{array} $$

where we have used that \(J_{1\over 2}(x)=\sqrt {2/\pi x} \sin \limits (x)\). The subscript on V0(R) indicates that we use the expression for V (R) with 𝜖 = 0, i.e. the Coulomb potential V0(R) = emR/4πR. In the first term, we use the small-r expansion of the Bessel functions

$$ \begin{array}{@{}rcl@{}} J_{{1\over2}-\epsilon}(pR)&=&{1\over{\Gamma}[{3\over2}-\epsilon]} \left( {{1\over2}pR}\right)^{{1\over2}-\epsilon} \left[ 1+\mathcal{O}(p^{2}R^{2}) \right], \end{array} $$
$$ \begin{array}{@{}rcl@{}} V(R) &=& \left( {e^{\gamma_E}\mu^{2}\over4}\right)^{\epsilon} {\Gamma({1\over2}-\epsilon)\over{\Gamma}{1\over2}}{1\over4\pi} R^{-1+2\epsilon}\left[ 1+{m^{2}R^{2}\over2(1+2\epsilon)}+\mathcal{O}(m^{4}R^{4}) \right] \end{array} $$
$$ \begin{array}{@{}rcl@{}} &&-\left( {e^{\gamma_E}\mu^{2}}\right)^{\epsilon} {\Gamma(-{1\over2}-\epsilon)\over{\Gamma}{-1\over2}}{1\over4\pi} m^{1-2\epsilon} \left[ 1+{m^{2}R^{2}\over2(3-2\epsilon)}+\mathcal{O}(m^{4}R^{4}) \right]. \end{array} $$

The first integral in (40) is denoted by I1 and reduces to

$$ \begin{array}{@{}rcl@{}} I_{1}&=&{1\over4(4\pi)^{2}} \left[ {1\over\epsilon} +4\log{\mu r} +2\gamma_E +\mathcal{O}(\epsilon) \right]. \end{array} $$

The second integral in (40) is denoted by I2 and becomes

$$ \begin{array}{@{}rcl@{}} I_{2}&=& {4\pi\over p}{\int}_{R}^{\infty}dR R \sin(pR)V_{0}^3(R) |_{p=im_{D}}\\ &=& {1\over2(4\pi)^{2}m_{D}} \left [ (3m-m_{D})\log(3m-m_{D})-(3m+m_{D})\log(3m+m_{D}) \right.\\ &&\left. -2m_{D}(-1+\gamma_E+\log r) \right] . \end{array} $$

Adding (44) and (45), the r-dependent terms cancel and we obtain (20).

B.2 Three Loop Diagrams

We can rewrite the diagram Ic3rung(p = im) in (27) as,

$$ \begin{array}{@{}rcl@{}} I_{\text{c3rung}}(p=im) &=&{\int}_q{1\over(\mathbf{p}+\mathbf{q})^{2}+m^{2}} |_{p=im}[I_{\text{bub}}(q)]^{2} , \end{array} $$

where Ibub(q) is given by

$$ \begin{array}{@{}rcl@{}} I_{\text{bubble}}(q)&=& {\int}_{R}{1\over r^{2}+m^{2}}{1\over(\mathbf{r}+\mathbf{q})^{2}+m^{2}}\\ &=& {1\over4\pi q}~\arctan(q/2m). \end{array} $$

Ibub(q) can be calculated by averaging over angles first and then integrating over r. Inserting (47) into (46), averaging over the angle between p and q, and finally integrating over q, we obtain (27).

The integral Jc3rung in (28) has an extra propagator and we can write it as,

$$ \begin{array}{@{}rcl@{}} J_{\text{c3rung}}(p=im) &=&{\int}_q{1\over[(\mathbf{p}+\mathbf{q})^{2}+m^{2}]^{2}} |_{p=im}[I_{\text{bub}}(q)]^{2} . \end{array} $$

The integral is evaluated in the same manner as Ic3rung(p = im). The last integral of this type, namely Kc3rung(p = im) is,

$$ \begin{array}{@{}rcl@{}} K_{\text{c3rung}}(p=im) &=& {\int}_q{1\over(\mathbf{p}+\mathbf{q})^{2}+m^{2}} |_{p=im}I_{\text{bub}}(q)I_{\text{bub}}^{\prime}(q), \end{array} $$

where \(I_{\text {bub}}^{\prime }\) is given by the derivative of (47) with respect to m2:

$$ \begin{array}{@{}rcl@{}} I_{\text{bubble}}^{\prime}(p)&=&{1\over8\pi m}{1 \over p^{2}+4m^{2}}. \end{array} $$

Inserting (47) and (50) into (49), averaging over the angle between p and q, and finally integrating over q, we obtain (29). The integral Jball is easy to evaluate by noting that it can be written as

$$ \begin{array}{@{}rcl@{}}J_{\text{ball}} &=&{\int}_p[I_{\text{bub}}^{\prime}(q)]^{2}\\ &=& {1\over4(4\pi)^{2}m^{2}}{\int}_p{1\over(p^{2}+4m^{2})^{2}}. \end{array} $$

The integral Kball can be evaluated by noting the relation

$$ \begin{array}{@{}rcl@{}} K_{\text{ball}}&=&{1\over2}\left( 3J_{\text{ball}}-I^{\prime\prime}_{\text{ball}}\right). \end{array} $$

B.3 Four Loop Diagrams

We can rewrite the integral Ic4rung in (30) as,

$$ \begin{array}{@{}rcl@{}} I_{\text{c4rung}} (p=im) &=& {\int}_q{1\over(\mathbf{p}+\mathbf{q})^{2}+m^{2}} |_{p=im}[I_{\text{bub}}(q)]^3 , \end{array} $$

where Ibub(q) is given by (47). Again, the integral can be evaluated by averaging over angles first and then integrating over q.

Let us next calculate Icballsun, which is

$$ \begin{array}{@{}rcl@{}} I&=&I_{\text{sun}}^{\prime}(p=im) \left[I_{\text{sun}}(q) -6{\Delta} m^{2} \right]\\ &=& I_{\text{sun}}^{\prime}(p=im) \left[I_{\text{sun}}(q) -I_{\text{sun}}(q=im)+I_{\text{sun}}(q=im) -6{\Delta} m^{2} \right]. \end{array} $$

The first term is

$$ \begin{array}{@{}rcl@{}}I_{1}&=&I_{\text{sun}}^{\prime}(p=im) \left[I_{\text{sun}}(q)-I_{\text{sun}}(q=im)\right]\\ &=& {\int}_q{1\over q^{2}+m^{2}}I_{\text{bub}}(|\mathbf{p}+\mathbf{q}|) \left[I_{\text{sun}}(q)-I_{\text{sun}}(q=im)\right]|_{p=im} \end{array} $$

The first two terms inside the paranthesis can be written as [29, 30]

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}(q) -I_{\text{sun}}(q=im) &=& -{1\over(4\pi)^{2}} \left[{1\over2}\log\left( {q^{2}+9m^{2}\over64m^{2}} \right)+ {3m\over q}\arctan{q\over3m} \right]. \end{array} $$

The integral (55) can be calculated by averaging over the angles between q and p first and then integrate over q. We get,

$$ \begin{array}{@{}rcl@{}} I_{1}&=&-{\log2\over4m^{2}(4\pi)^{4}}\left[ 0.127234\right]. \end{array} $$

The second term is

$$ \begin{array}{@{}rcl@{}} I_{2}&=&I_{\text{sun}}^{\prime}(p=im) \left[ I_{\text{sun}}(q=im) -6{\Delta} m^{2} \right]\\ \end{array} $$

The first factor is \(I_{\text {sun}}^{\prime }(p=im)\) and can be evaluated by taking the derivative of Isun(p = imD) with respect to m and then setting m = mD. This yields

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}^{\prime}(p=im) &=&{\log2\over4m^{2}(4\pi)^{2}} . \end{array} $$

The second integral is simply given by the finite part of Isun(p = im):

$$ \begin{array}{@{}rcl@{}} I_{\text{sun}}(q=im) -6{\Delta} m^{2} &=&{1\over4m^{2}(4\pi)^{2}} \left[ \log{\mu\over2m}+{3\over2}-2\log2 \right] . \end{array} $$

Multiplying (59) and (60), and adding the result to (57), we obtain (32), where we have defined Cballsun = − 0.01353.

Let us next discuss the first fish-like diagram in (33). We can write it as,

$$ \begin{array}{@{}rcl@{}} I_{\text{fish1}}(p=im)&=& {\int}_{qr} {1\over(\mathbf{p}+\mathbf{r})^{2}+m^{2}} |_{p=im} [I_{\text{tri}}(\mathbf{q},\mathbf{r})]^{2} , \end{array} $$


$$ \begin{array}{@{}rcl@{}} I_{\text{tri}}(\mathbf{q},\mathbf{r}) &=&{\int}_{s}{1\over s^{2}+m^{2}}{1\over(\mathbf{q}+\mathbf{s})^{2}+m^{2}} {1\over(\mathbf{r}+\mathbf{s})^{2}+m^{2}}. \end{array} $$

The integral (62) is finite and we can be write it as, as [28]

$$ \begin{array}{@{}rcl@{}} I_{\text{tri}}(\mathbf{q},\mathbf{r})&=&{\arctan(\sqrt{D}/C)\over8\pi\sqrt{D}}, \end{array} $$


$$ \begin{array}{@{}rcl@{}} C&=&{q^{2}+r^{2}+\mathbf{q}\cdot \mathbf{r}+4m^{2}\over m^{2}}, \end{array} $$
$$ \begin{array}{@{}rcl@{}} D&=&{q^{2}r^{2}(\mathbf{q}-\mathbf{r})^{2}+4m^{2}[q^{2}r^{2}-(\mathbf{q}\cdot \mathbf{r})^{2}]\over4m^6} . \end{array} $$

The integral (61) can be calculated numerically by averaging over the angles first and then integrating over p and q. We get,

$$ \begin{array}{@{}rcl@{}} I_{\text{fish1}}(p=im)&=& {1\over16m^{2}(4\pi)^{4}}\left( {\mu\over2m}\right)^{8\epsilon} \left[ 4.6029 \right]. \end{array} $$

We denote the numerical constant inside the paranthesis by Cfish1 = 4.6029. Finally, we consider the second fish-like diagram and rewriting

$$ \begin{array}{@{}rcl@{}} I_{\text{fish2}}(p=im)&=& {\int}_{qr} {1\over(\mathbf{q}+\mathbf{p})^{2}+m^{2}}{1\over(\mathbf{r}+\mathbf{p})^{2}+m^{2}} |_{p=im}I_{\text{bub}}(q)I_{\text{tri}}(\mathbf{q},\mathbf{r}) \end{array} $$

The integral is finite in three dimensions but we must average over the angles between p and q, p and r, q and r. This is simplified somewhat since the direction of the external momentum p is fixed. All the angular averages as well the integrals over q and r must be done numerically [26]. This yields

$$ \begin{array}{@{}rcl@{}} I_{\text{fish2}}(p=im)&=& {1\over 16 m^{2}(4\pi)^{4}}\left( {\mu\over2m}\right)^{8\epsilon} \left[ 1.778 \right] . \end{array} $$

We denote the numerical constant inside the paranthesis by Cfish2 = 1.778.

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Khan, R., Shahzeb, B. Debye Mass of Massless Φ4-Theory to Order g6 at Weak Coupling. Int J Theor Phys 59, 1523–1536 (2020).

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  • Effective filed theory
  • Quark-gluon plasma
  • Dimensional reduction