Singlet Extension of the SM Higgs Sector

Transcription

1 Singlet Extension of the SM Higgs Sector David Sommer KIT December 3, 2015

2 Overview Motivation for an extension of the Higgs sector Addition of a real singlet scalar (xsm Addition of a complex singlet scalar (cxsm 2 / 22

3 Motivation xsm - real singlet addition cxsm - complex singlet addition Cosmological mystery: the missing mass problem Jan Oort ( Fritz Zwicky ( Motion of galaxies and stars in the universe dark matter. Estimated to account for about 85% of the mass in the universe. But where does it come from? 3 / 22

4 MACHOs and WIMPs Massive compact halo object Weakly interacting massive particle 4 / 22

5 MACHOs and WIMPs Properties of WIMPs Little interaction with SM particles. large mass (for a particle. Readily predicted by simple extensions of the SM Higgs sector. We study the addition of a real (xsm as well as a complex (cxsm singlet scalar to the Higgs doublet. 4 / 22

6 Let H be the SM Higgs doublet and s be a single gauge singlet real scalar field. Consider the potential ( V =µ 2 H H + a 1 (H H ( + λ 2 H H ( s + a 2 H H + b 2 2 s2 + b 3 3 s3 + b 4 4 s4. Note: V is Z 2 symmetric in s for a 1 = b 3 = 0 (i.e. symmetric under s s. s 2 5 / 22

7 ( V =µ 2 H H + a 1 (H H What are the conditions on V? ( + λ 2 H H ( s + a 2 H H + b 2 2 s2 + b 3 3 s3 + b 4 4 s4. s 2 It must be bounded from below (existence of a vacuum. It must accomodate electroweak symmetry breaking H 0. It should yield a massive stable scalar s. 6 / 22

8 We write The stationary conditions H = 1 ( 0, with h real (unitary gauge, 2 h and denote the vacuum expectation values of h and s with v and v s. With this, the stationary conditions of V yield V h (h,s=(v,vs = V s µ 2 = λv 2 v s (a 1 + a 2 v s, = 0 (h,s=(v,vs a 1 = a 2 v s 2b 2v s v 2 2b 3v 2 s v 2 2b 4v 3 s v 2. 7 / 22

9 Using the equation for µ 2 we now calculate the mass squared matrix 2 V 2 V M 2 = h 2 h s 2 V 2 V s h s 2 (h,s=(v,v s = 2λv 2 a 1 v + 2a 2 vv s. a 1 v + 2a 2 vv s a 2 v 2 + b 2 + 2b 3 v s + 3b 4 vs 2 Note: A Z 2 symmetry (a 1 = b 3 = 0 is not sufficient to eliminate the mixing terms. 8 / 22

10 This is because the acquisition of a nonzero vev v s 0 of the scalar s breaks the Z 2 symmetry (if imposed spontaneously. unwanted mixing terms. instability of the mass eigenstates. no DM candidate. So, in order to obtain a viable dark matter candidate, we now assume a 1 = b 3 = s = 0. 9 / 22

11 Constraints on the potential After electroweak symmetry breaking, for which we shift h v + h, the potential reads V = µ4 4λ µ2 h 2 + λvh 3 + λ 4 h4 Necessary conditions: (b 2 + a 2 v 2 s 2 + b 4 4 s4 + a 2 vs 2 h + a 2 2 s2 h 2. Existence of a vacuum: λ, b 4 0 and λb 4 a2 2 for negative a 2. The mass squared matrix M 2 = diag(2λv, b 2 + a 2 v 2 must be positive definite. Note: The phenomenological properties of this model are completely determined by a 2 and b 2, or a 2 and m 2 s = b 2 + a 2 v / 22

12 Experimental and theoretical constraints on the parameters Figure: taken from Lei Feng, S. Profumo, L. Ubaldi, [arxiv: ] Highly constrained parameter space for the xsm! 11 / 22

13 Another cosmological mystery: the baryon asymmetry Number of baryons number of antibaryons in the observable universe. Possible Explanations: There is as much antimatter, as there is matter, but its all clunked together far away. The universe began with a small preference for matter. The universe was initially perfectly symmetric, but somehow matter was favoured over time. This requires the electroweak symmetry breaking to be a first order phase transition. In the context of SM, this requires m h 70 GeV. In the context of xsm, this requires S / 22

14 xsm - Conclusive remarks The xsm Model yields either a stable CDM candidate, that doesn t affect EWPT ( S = 0, or generates strong first order EWPT, but only yields unstable mass eigenstates ( S 0. So, it is impossible to explain both these mysteries in the context of a single xsm. Unsatisfactory? 13 / 22

15 Let S = S + ia be a single gauge singlet complex scalar field. Consider the U(1 and Z 2 symmetric Potential V = m2 2 ( H H + λ ( 2 H δ 2 H H H S 2 + b 2 2 S 2 + d 2 4 S 4 14 / 22

16 Let S = S + ia be a single gauge singlet complex scalar field. Consider the U(1 and Z 2 symmetric Potential V = m2 2 ( H H + λ ( 2 H δ 2 H H H S 2 + b 2 2 S 2 + d 2 4 S 4 Glodstone s theorem: S 0 massless particle (Spontaneous breaking of the U(1 symmetry. 14 / 22

17 Let S = S + ia be a single gauge singlet complex scalar field. Consider the Z 2 symmetric Potential V = m2 ( H H λ ( 4 ( b1 4 eiφ b 1 S 2 + c.c. 2 H δ 2 H + 2 H H S 2 + b 2 2 S 2 + d 2 4 S 4 Glodstone s theorem: S 0 massless particle (Spontaneous breaking of the U(1 symmetry. We therefore break the U(1 symmetry explicitly. 14 / 22

18 Let S = S + ia be a single gauge singlet complex scalar field. Consider the Potential V = m2 ( H H λ ( 2 H δ 2 H + 4 ( b1 4 eiφ b 1 S 2 + a 1 e iφa 1 S + c.c. 2 H H S 2 + b 2 2 S 2 + d 2 4 S 4 In the same fashion, we explicitly break the Z 2 symmetry. 14 / 22

19 Let S = S + ia be a single gauge singlet complex scalar field. Consider the Potential V = m2 ( H H λ ( 2 H δ 2 H + 4 ( b1 4 eiφ b 1 S 2 + a 1 e iφa 1 S + c.c. We study the cases A1 S = 0; a 1 = b 1 = 0. 2 H H S 2 + b 2 2 S 2 + d 2 4 S 4 (Unbroken U(1 A2 S = 0; a 1 = 0, b 1 0. (explicitly broken U(1 B1 S 0; a 1 = b 1 = 0. (spontaneously broken U(1 B2 S 0; a 1 0, b 1 0. (explicitly broken U(1 and Z 2 14 / 22

20 Constraints on the potential V = m2 ( H H λ ( 2 H δ 2 H + 4 ( b1 4 eiφ b 1 S 2 + a 1 e iφa 1 S + c.c. 2 H H S 2 + b 2 2 S 2 + d 2 4 S 4 Existence of a vacuum (v, v S : We take λ > 0, d 2 > 0 if δ 2 < 0 then λd 2 > δ 2 2. For simplicity, we take φ b1 = φ a1 = π A = 0. The vacuum must be a local minimum, so the mass squared matrix must be positive definite. 15 / 22

21 Case A: S = 0. The mass matrix in (v, 0 is M 2 = diag ( M 2 h, M2 S, M2 A,where M 2 h = 1 2 λv 2, M 2 S = 1 2 b b 2 + δ 2v 2 4, M 2 A = 1 2 b b 2 + δ 2v 2 4. For case A1, that is b 1 = 0, we obtain two phenomenologically aquivalent particles. xsm. 16 / 22

22 Case A2: S = 0; a 1 = 0, b 1 0 V = m2 ( H H λ ( 4 ( b 1 4 S2 + c.c. H H M 2 S/A = 1 2 b b 2 + δ 2v 2 4. No mixing of the scalars. Stable two-component dark matter scenario. 2 + δ 2 2 H H S 2 + b 2 2 S 2 + d 2 4 S 4 17 / 22

23 Figure: taken from V. Barger et al., [arxiv: ] Contribution to the relic density over the mass of the light scalar MS 2 = 1 2 b b 2 + δ2v 2 4. M H = 120 GeV, b 2 = GeV 2, d 2 = / 22

24 Case B1: S 0; a 1 = b 1 = 0. V = m2 ( H H + λ ( 2 H δ 2 H H H S 2 + b 2 2 S 2 + d 2 4 S 4. λv 2 /2 δ 2 vv S /2 0 M 2 = δ 2 vv s /2 d 2 vs 2/ Two unstable mixed scalars. A is stable but massles. no dark matter candidate. 19 / 22

25 Case B2: S 0; a 1 0, b 1 0. V = m2 ( H H 2 + λ ( 2 H δ 2 H H H S 2 + b 2 2 S 2 + d 2 4 S 4 ( + b 1 4 S2 a 1 S + c.c.. λv 2 /2 δ 2 vv S /2 0 M 2 = δ 2 vv s /2 d 2 vs 2/2 + 2 a 1 /v S b a 1 /v S Two unstable mixed scalars. A remains stable (no mixing and MA 2 = b 1 + A candidate for dark matter! 2 a1 v S > / 22

26 Figure: Contribution to the relic density over the mass M A. v S = 100 GeV, = 120 GeV, M h1 M h2 = 250 GeV. (V. Barger et al., [arxiv: ] Figure: Contribution to the relic density over the mass M A. v S = 10 GeV, = 120 GeV, M h1 M h2 = 140 GeV. (V. Barger et al., [arxiv: ] 21 / 22

27 cxsm - Conclusive remarks The cxsm model yields a simple two-component DM scenario, if the U(1 symmetry is explicitly but not spontaneously broken. yields a single-component DM scenario and allows for first order EWPT, as required for electroweak baryogenesis, if the U(1 symmetry is both explicitly and spontaneously broken. 22 / 22