Calculation of the Vacuum Energy Density using Zeta Function Regularization
aa r X i v : . [ phy s i c s . g e n - ph ] F e b Calculation of the Cosmological Constant from VacuumEnergy Density using Zeta Function Regularization
Siamak Tafazoli ∗ Abstract - This paper presents a theoretical calculation and estimate of the Cosmolog-ical Constant (CC) from the vacuum energy density by summing the contributions ofall quantum fields’ vacuum states which turns out to match the predictions of currentcosmological models and all observational data to date. The basis for this calculation isa Zeta function regularization method used to tame the infinities present in the improperintegrals of power functions.
Much has been written on the Cosmological Constant problem and the vacuum catastrophe,refer to [1] for a pedagogical overview. In this paper, we attack the heart of the problem whichis the divergent sums and integrals that have also plagued many other areas of Quantum FieldTheory (QFT), using as our tool, a Zeta regularization technique to tame these infinities [2].Refer also to [3] for a pedagogical overview of the Zeta function and the associated Zeta func-tion regularization techniques with applications to physics.In [4], the vacuum energy density was shown to be within the upper bound set by the cur-rent cosmological models [5] and observational data available in year 2000, however the effectof the particle masses and the quantum states of the different quantum fields were not prop-erly taken into account. This paper remedies these issues and also compares with more recentobservational data from year 2020 [6].The total vacuum (zero-point) energy of a free (non-interacting) quantum field in flat space-time, can be modeled as the sum of zero-point energies of a set of infinite number of quantumharmonic oscillators, one for each normal mode and given simply as + P k ~ ω k for bosonicfields and − P k ~ ω k for fermionic fields [7], and where ~ ω k represents the eigenvalues of thefree Hamiltonian, ω k = √ m + k in Natural units, k is the wave number (with units of 1/m), m is the particle mass associated with a specific field, and ~ is the reduced Planck constant.The negative sign, in front of the infinite sum, is due to the negative energy solutions allowedin the fermionic fields which must follow the Pauli Exclusion principle by obeying the canonicalanti-commutation relations.Consequently, the vacuum energy density of a single state of a given field (i.e. energy E pervolume V ) can be derived [7] (section 10.8) to beˆ ρ vac = EV = ± V X k ~ ω k = ... = ± π Z ∞ √ m + k k dk (1) ∗ email: [email protected] ~ and c in (1), the total vacuum energy density of a field is given byˆ ρ vac = g π Z ∞ p ( mc ) + ( ~ kc ) k dk (2)where g is the degeneracy factor which includes, a sign factor ( − j , a spin factor 2 j + 1 formassive fields and 2 for massless fields, a factor of 3 for fields with color charge, and a factor of2 for fields that have an antiparticle distinct from their particle. Note also that c is the speedof light in vacuum and j is the spin.Equation (2) can be written asˆ ρ vac = g ~ c π Z ∞ r(cid:16) mc ~ (cid:17) + k k dk (3)and for a massless field (i.e. photon and gluon with m = 0), we can further simplify (3) toˆ ρ vac = g ~ c π Z ∞ k dk (4)For massive fields however, we need to obtain the Maclaurin series expansion of the inte-grand in equation (3) with respect to the wave number k . Let’s define f ( k ) = k √ a + k (5)where a = mc ~ . Now the Maclaurin series expansion of (5) can be obtained as f ( k ) = √ a h k + k a − k a + k a + O ( k , k , k , ... ) i (6)At first glance, when substituting (6) as the integrand in (3), the integral in (3) looks highlydivergent, essentially a sum of divergent even powered polynomial integrals. However, finitevalues for these integrals can be assigned using a Zeta regularization technqiue (see appendixA). These integral values are R ∞ k dk = − , R ∞ k dk = − , R ∞ k dk = − and R ∞ k dk = − , etc. Furthermore, given the large value of a = +2 . × m in (6), the only dominating term is the first one in this series so we get the following simpleexpression for massive fields from (3)( ˆ ρ vac ) m =0 = g ~ c π (cid:16) √ a Z ∞ k dk (cid:17) = g ~ c π (cid:16) a − (cid:17) = − g ~ c π (cid:16) mc ~ (cid:17) = − g mc π × (7)2ow, the sign of the vacuum energy density in (2) is represented by the ( − j sign factorwhich dictates that the vacuum energy contribution is positive for bosonic fields (e.g. j = 1)and negative for fermionic fields (i.e. j = 1 / k which is what is typically done in literature). However,this is not what we obtain using our Zeta regularization technique where we assign a negative value to the otherwise divergent integral R ∞ k dk in (7) and hence a negative value for theintegral in (2), and by doing so we reverse the vacuum energy density signs for the fermionicfields (positive) and massive bosonic fields (negative). This is indeed an unexpected turn ofevents but one that will result in calculating the observed positive CC!For the vacuum energy density contributions of the massless photon and gluon bosonicfields, given by (4), the value for the integral R ∞ k dk can be calculated to be ( − (3+1)(3+2) = with units of 1/m and so we obtain the following( ˆ ρ vac ) m =0 = g ~ c π × (8)which (just by visual inspection) indicates negligible contribution (in the order of 10 − J/m ) towards the vacuum energy density and hence will not be further considered in the cal-culations that follows.The expectation value of vacuum energy densities for the 15 free massive quantum fieldsof the Standard Model (with their associated particle masses [8]) are tabulated below and onecan see that the total expectation value of the vacuum energy density does not diverge , ispositive and nearly zero. In fact, we obtain a positive value of ˆ ρ vac = 4 . × − J/m and acorresponding CC value of Λ = 3 . × − eV using [1] (Eqn 14) ρ vac = Λ8 πG (9)where G is Newton’s gravitational constant.This result agrees very well with the value obtained from the large scale cosmological ob-servations of approximately 5 . × − J/m (Λ = (4 . ± . × − eV ) [6]. This closeagreement between theory and observational data essentially confirms that vacuum has verylittle energy content leading to an almost vanishing CC, addressing the ”vacuum catastrophe”and the ”cosmological constant problem”.The small discrepancy of 9 . × − J/m between the theoretical and the observationalvalues may be due to:(1) the vacuum energy contribution of the gravitational field: If we assume that a quantum the-ory of gravity will be eventually formulated with a graviton mass of zero then its contributionto vacuum energy density will also be negligible based on equation (8), however this remainsto be seen; 3ields Degeneracy factor g sign × spin × color × antiparticle Mass (eV/ c ) ˆ ρ vac (J/m )Quarks (x 6): fermionic − × (2 × × × −
12 178 . × +7 . × − Leptons (x 6): fermionic − × (2 × × × − . × +2 . × − W: bosonic 1 × (2 × × × . × − . × − Z : bosonic 1 × (2 × × × . × − . × − Higgs : bosonic 1 × (2 × × × . × − . × − Total +4 . × − (2) the effect of interacting quantum fields on vacuum energy density;(3) the vacuum energy contribution of an unknown fermionic field (e.g. the elusive dark matterfield): An exciting possibility would indeed be the discovery of a new (dark matter) fermionicfield and in this case, a plausible dark matter fermionic field could be a spin field, with nocolor charge, and with or without a distinct antiparticle. Using (7), its associated particlemass would be about 35 . c (Dirac fermion, g = −
8) or 71 GeV/ c (Majorana fermion, g = − g W = 4 and g Z = 2) closes this energy density gapand allows to arrive within the range of the measured observational CC value. In this case,ˆ ρ vac = 5 . × − J/m and Λ = (4 . ± . × − eV ! It can be argued that the removalof these 2 RH particle states may indeed be justified since they have not been detected at largefractions in high energy collision experiments [9] [10], and do not seem to play a role in particleinteractions in our universe. References [1] S. M. Carroll, “The Cosmological Constant”, 2000. https://arxiv.org/abs/astro-ph/0004075 [2] F. Aghili and S. Tafazoli, “Analytical Solution to Improper Integral ofDivergent Power Functions Using The Riemann Zeta Function”, 2018. https://arxiv.org/abs/1805.10480 [3] E. Elizalde, “Zeta Functions and the Cosmos—A Basic Brief Review”. Universe 2021, 7,5. https://doi.org/10.3390/universe7010005
44] S. Tafazoli, “A Resolution to the Vacuum Catastrophe”, 2019. https://vixra.org/pdf/1910.0266v1.pdf [5] A. G´omez-Valent, “Vacuum Energy in Quantum Field Theory and Cosmology”, 2017. https://arxiv.org/abs/1710.01978 [6] Planck Collaboration, “Planck 2018 results. VI. Cosmological parameters”, 2020. https://arxiv.org/abs/1807.06209 [7] R. D. Klauber, “The Student Friendly Quantum Field Theory”, 2013. [8] P.A. Zyla et al., (Particle Data Group), Prog. Theor. Exp. Phys. , 083C01, (2020). https://pdglive.lbl.gov [9] D0 Collaboration: V.M. Abazov et al., “Model-independent measurement of the W bosonhelicity in top quark decays”, 2007. https://arxiv.org/abs/0711.0032 [10] CMS Collaboration, “Measurements of the W boson rapidity, helicity, double-differentialcross sections, and charge asymmetry in pp collisions at √ s = 13 TeV”, 2020. https://arxiv.org/abs/2008.04174 [11] E. W. Weisstein, “Binomial Theorem.” From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/BinomialTheorem.html [12] E. W. Weisstein “Bernoulli Number.” From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/BernoulliNumber.html [13] J. Sondow and E. W. Weisstein, “Riemann Zeta Function.” From MathWorld–A WolframWeb Resource. https://mathworld.wolfram.com/RiemannZetaFunction.html A Improper integrals of power functions with non-negativeinteger exponents
Let’s define the µ function as the improper integral of a power function with exponent pµ ( p ) = Z ∞ x p dx (10)with p ∈ N , where N = { , , , · · · } .The mu function can be interpreted as the natural extension of the Riemann zeta functionwhere the discrete sum is replaced by a continuous integral. The above improper integral canbe equivalently written as the following infinite series by splitting the limits of integration intosuccessive integer numbers. That is µ ( p ) = ∞ X n =1 ∆( p, n ) (11a)5here ∆( p, n ) = Z nn − x p dx = 1 p + 1 (cid:16) n p +1 − ( n − p +1 (cid:17) (11b)is the definite integral of the power function over limits n − n . In what follows, we willshow that the divergent series in the RHS of (11a) can converge to a finite result!One can write the binomial polynomial expansion of ( n − p +1 in (11b) as [11] (Eqn 2)( n − p +1 = p +1 X k =0 (cid:18) p + 1 k (cid:19) n k ( − p +1 − k , (12a)where (cid:18) ts (cid:19) = t ! s !( t − s )! (12b)is the binomial coefficient, with (cid:0) tt (cid:1) = (cid:0) t (cid:1) = 1. Substituting (12a) into (11b) gives∆( p, n ) = 1 p + 1 (cid:16) n p +1 − p +1 X k =0 (cid:18) p + 1 k (cid:19) n k ( − p +1 − k (cid:17) = 1 p + 1 (cid:16) n p +1 − p X k =0 (cid:18) p + 1 k (cid:19) n k ( − p − k ( − − (cid:18) p + 1 p + 1 (cid:19) n p +1 ( − p +1 − ( p +1) (cid:17) = 1 p + 1 p X k =0 (cid:18) p + 1 k (cid:19) n k ( − p − k (13)Thus (11a) can be written as µ ( p ) = 1 p + 1 ∞ X n =1 p X k =0 (cid:18) p + 1 k (cid:19) ( − p − k n k = 1 p + 1 (cid:16) p X k =0 (cid:18) p + 1 k (cid:19) ( − p − k ∞ X n =1 n k (cid:17) ∀ p ∈ N (14)The last summation in RHS of (14) can be formally written in terms of the Reimann zetafunction ζ ( · ) as ζ ( − k ) = ∞ X n =1 n k (15)and therefore we arrive at µ ( p ) = 1 p + 1 p X k =0 (cid:18) p + 1 k (cid:19) ( − p − k ζ ( − k ) ∀ p ∈ N (16)For integers k ≥
0, the zeta function is related to Bernoulli numbers by [12] (Eqn 64) ζ ( − k ) = ( − k B k +1 k + 1 (17)6oreover, from definition (12b), one can verify that the successive binomial coefficients holdthe following useful identity (cid:18) p + 1 k (cid:19) = k + 1 p + 2 (cid:18) p + 2 k + 1 (cid:19) (18)Finally, upon substituting the relevant terms from (17) and (18) into (16), one can equivalentlyrewrite the latter equation in the following simple form µ ( p ) = 1 p + 1 p X k =0 k + 1 p + 2 (cid:18) p + 2 k + 1 (cid:19) ( − p − k ( − k B k +1 k + 1= ( − p ( p + 1)( p + 2) p X k =0 (cid:18) p + 2 k + 1 (cid:19) B k +1 = ( − p ( p + 1)( p + 2) p +1 X k =1 (cid:18) p + 2 k (cid:19) B k ∀ p ∈ N (19)Furthermore, the Bernoulli numbers satisfy the following property [13] (Eqn 34) l − X k =0 (cid:18) lk (cid:19) B k = 0 (20)Given that B = 1 and using a change of variable l = p + 2, one can equivalently write (20) as p +1 X k =1 (cid:18) p + 2 k (cid:19) B k + (cid:18) p + 20 (cid:19) B = 0 p +1 X k =1 (cid:18) p + 2 k (cid:19) B k = − µ function µ ( p ) = Z ∞ x p dx = ( − p +1 ( p + 1)( p + 2) ∀ p ∈ N (22)This surprising and simple finite result should be viewed as an alternative, that may be usefulin certain cases, to an otherwise divergent solution!As a simple check of (22) for small p (e.g. p = 0 , , µ ( p ) = Z ∞ x p dx = ∞ X n =1 Z nn − x p dx = ∞ X n =1 p + 1 (cid:16) n p +1 − ( n − p +1 (cid:17)
7e get for p = 0: µ (0) = Z ∞ x dx = ∞ X n =1 Z nn − dx = ∞ X n =1 ( n − ( n − ∞ X n =1 ∞ X n =1 n = ζ (0) = −
12 = ( − (0 + 1)(0 + 2)for p = 1: µ (1) = Z ∞ x dx = ∞ X n =1 Z nn − xdx = ∞ X n =1 (cid:16) n − ( n − (cid:17) = 12 ∞ X n =1 (cid:16) n − ( n − n + 1) (cid:17) = 12 ∞ X n =1 ( n − n + 2 n −
1) = 12 ∞ X n =1 (2 n −
1) = 12 (2 ζ ( − − ζ (0))= 12 (cid:16) (cid:16) − (cid:17) − − (cid:17) = 12 (cid:16) − (cid:17) = 16 = ( − (1 + 1)(1 + 2)for p = 2: µ (2) = Z ∞ x dx = ∞ X n =1 Z nn − x dx = ∞ X n =1 (cid:16) n − ( n − (cid:17) = 13 ∞ X n =1 (cid:16) n − ( n − n + 3 n − (cid:17) = 13 ∞ X n =1 (3 n − n + 1) = 13 (cid:16) ∞ X n =1 n − ∞ X n =1 n + ∞ X n =1 n (cid:17) = 13 (3 ζ ( − − ζ ( −
1) + ζ (0))= ζ ( − − ζ ( −
1) + 13 ζ (0) = 0 − (cid:16) − (cid:17) + 13 (cid:16) − (cid:17) = 112 −
16 = −
112 = ( −2+1