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Supersymmetry Breaking Introduction to Supersymmetry Lecture IV : Supersymmetry Breaking Athanasios Dedes Department of Physics University of Ioannina, Greece Pre-SUSY 2014 July 15-19, Univ. of Manchester, UK Outline Bibliography Spontaneous SUSY breaking General considerations Fayet-Iliopoulos (D-term) SUSY breaking O' Raifeartaigh (F-term) SUSY breaking Goldstino and Gravitino Mediation of SUSY breaking Hidden SUSY breaking sector Gravity Mediated SUSY breaking Gauge Mediated SUSY breaking Bibliography J. Wess and J. Bagger, Supersymmetry and supergravity, [Ch. VIII], Princeton, Univ. Press (1992) 259 p. S. P. Martin, A Supersymmetry primer, [Ch. 7], hep-ph/9709356, v6 2011. I use g µν = (1; −1; −1; −1), but otherwise Martin's notation. Outline Bibliography Spontaneous SUSY breaking General considerations Fayet-Iliopoulos (D-term) SUSY breaking O' Raifeartaigh (F-term) SUSY breaking Goldstino and Gravitino Mediation of SUSY breaking Hidden SUSY breaking sector Gravity Mediated SUSY breaking Gauge Mediated SUSY breaking Spontaneous SUSY breaking y µ Taking the trace of fQα; Qα_ g = 2σαα_ Pµ (see Lec.1) we obtain: 1 H = (Q Qy + Q Qy + QyQ + QyQ ) 4 1 1 2 2 1 1 2 2 y If SUSY is unbroken in the vacuum state, Qαj0i = 0 ; Qα_ j0i = 0 then Hj0i = 0 which means that the vacuum has zero energy. Conversely, if SUSY is spontaneously broken then y Qαj0i 6= 0 ; Qα_ j0i 6= 0 and therefore 1 h0jHj0i = jjQyj0ijj2 + jjQ j0ijj2 + jjQyj0ijj2 + jjQ j0ijj2 > 0 4 1 1 2 2 )h 0jVj0i > 0 Recall the relations: ∗ ∗i i (α) ∗ (a) Fi = −Wi ; F = −W ; D = −g(φ T φ) with ∗ ∗ @W i @W Wi = ∗i ; W = @Φ Φ=φ @Φi Φ=φ Spontaneous SUSY breaking But remember: X 1 X V(φ, φ∗) = jF j2 + D(a)D(a) i 2 i a so if the system of equations (a) Fi = 0 and D = 0 (1) are not satisfied simultaneously by all values of the fields then SUSY is spontaneously broken. Spontaneous SUSY breaking But remember: X 1 X V(φ, φ∗) = jF j2 + D(a)D(a) i 2 i a so if the system of equations (a) Fi = 0 and D = 0 (1) are not satisfied simultaneously by all values of the fields then SUSY is spontaneously broken. Recall the relations: ∗ ∗i i (α) ∗ (a) Fi = −Wi ; F = −W ; D = −g(φ T φ) with ∗ ∗ @W i @W Wi = ∗i ; W = @Φ Φ=φ @Φi Φ=φ Outline Bibliography Spontaneous SUSY breaking General considerations Fayet-Iliopoulos (D-term) SUSY breaking O' Raifeartaigh (F-term) SUSY breaking Goldstino and Gravitino Mediation of SUSY breaking Hidden SUSY breaking sector Gravity Mediated SUSY breaking Gauge Mediated SUSY breaking Consider two chiral superfields, Φ1 and Φ2, charged under an abelian group with charges +1 and -1, respectively. Then from eq.(Lec. II - 37) we write the SuperQED+FI theory 1 α ∗ 2eV ∗ −2eV LFI = [W Wα]F + c:c + Φ1e Φ1 + Φ2e Φ2 4 D D ∗ ∗ + (mΦ1Φ2 + mΦ1Φ2)F − 2 κ [V ]D (2) From eq.(Lec. II - 35, 33,14) we write the LFI in components form. The scalar potential is: 1 V(φ, φ∗) = − D2 − e (jφ j2 − jφ j2) D + k D 2 1 2 2 2 + jF1j + jF2j + (m φ2 F1 + m φ1 F2 + c:c) (3) Fayet-Iliopoulos (D-term) SUSY breaking F-I (1974) noticed that a [V ]D = D is both Supersymmetric and gauge invariant and therefore is allowed for an abelian gauge group. They showed that this term breaks SUSY spontaneously. From eq.(Lec. II - 35, 33,14) we write the LFI in components form. The scalar potential is: 1 V(φ, φ∗) = − D2 − e (jφ j2 − jφ j2) D + k D 2 1 2 2 2 + jF1j + jF2j + (m φ2 F1 + m φ1 F2 + c:c) (3) Fayet-Iliopoulos (D-term) SUSY breaking F-I (1974) noticed that a [V ]D = D is both Supersymmetric and gauge invariant and therefore is allowed for an abelian gauge group. They showed that this term breaks SUSY spontaneously. Consider two chiral superfields, Φ1 and Φ2, charged under an abelian group with charges +1 and -1, respectively. Then from eq.(Lec. II - 37) we write the SuperQED+FI theory 1 α ∗ 2eV ∗ −2eV LFI = [W Wα]F + c:c + Φ1e Φ1 + Φ2e Φ2 4 D D ∗ ∗ + (mΦ1Φ2 + mΦ1Φ2)F − 2 κ [V ]D (2) Fayet-Iliopoulos (D-term) SUSY breaking F-I (1974) noticed that a [V ]D = D is both Supersymmetric and gauge invariant and therefore is allowed for an abelian gauge group. They showed that this term breaks SUSY spontaneously. Consider two chiral superfields, Φ1 and Φ2, charged under an abelian group with charges +1 and -1, respectively. Then from eq.(Lec. II - 37) we write the SuperQED+FI theory 1 α ∗ 2eV ∗ −2eV LFI = [W Wα]F + c:c + Φ1e Φ1 + Φ2e Φ2 4 D D ∗ ∗ + (mΦ1Φ2 + mΦ1Φ2)F − 2 κ [V ]D (2) From eq.(Lec. II - 35, 33,14) we write the LFI in components form. The scalar potential is: 1 V(φ, φ∗) = − D2 − e (jφ j2 − jφ j2) D + k D 2 1 2 2 2 + jF1j + jF2j + (m φ2 F1 + m φ1 F2 + c:c) (3) The scalar potential after integrating out F − and D− terms becomes 1 2 V(φ, φ∗) = m2 (jφ j2 + jφ j2) + κ − e (jφ j2 − jφ j2) (7) 1 2 2 1 2 2 Case I: hφ1i = hφ2i = 0 for m > e k 2 Case II: hφ1i = 0; hφ2i = v2 for m < e k Fayet-Iliopoulos (D-term) SUSY breaking Then F1;2 and D obey the e.o.m 2 2 D − κ + e (jφ1j − jφ2j ) = 0 (4) ∗ F1 + m φ2 = 0 (5) ∗ F2 + m φ1 = 0 : (6) F1 = F2 = 0 and D = 0 cannot be satisfied simultaneously (because of κ 6= 0) and therefore SUSY is spontaneously broken. Fayet-Iliopoulos (D-term) SUSY breaking Then F1;2 and D obey the e.o.m 2 2 D − κ + e (jφ1j − jφ2j ) = 0 (4) ∗ F1 + m φ2 = 0 (5) ∗ F2 + m φ1 = 0 : (6) F1 = F2 = 0 and D = 0 cannot be satisfied simultaneously (because of κ 6= 0) and therefore SUSY is spontaneously broken. The scalar potential after integrating out F − and D− terms becomes 1 2 V(φ, φ∗) = m2 (jφ j2 + jφ j2) + κ − e (jφ j2 − jφ j2) (7) 1 2 2 1 2 2 Case I: hφ1i = hφ2i = 0 for m > e k 2 Case II: hφ1i = 0; hφ2i = v2 for m < e k Fayet-Iliopoulos (D-term) SUSY breaking κ = 0 Vmin = 0 ) SUSY unbroken 2 Case I: hφ1i = hφ2i = 0 for m > e k ) Gauge Symmetry unbroken Fayet-Iliopoulos (D-term) SUSY breaking κ 6= 0 1 2 Vmin = 2 κ ) SUSY broken 2 Case I: hφ1i = hφ2i = 0 for m > e k ) Gauge Symmetry unbroken Fayet-Iliopoulos (D-term) SUSY breaking κ 6= 0 Vmin > 0 ) SUSY broken 2 Case II: hφ1i = 0; hφ2i = v2 for m < e k ) Gauge Symmetry broken Fayet-Iliopoulos (D-term) SUSY breaking Masses (case I): Boson masses: 2 complex scalar fields with m2 − e κ, m2 + e κ and a massless gauge boson Aµ Fermion masses: 2 Weyl fermions 1; 2 with equal mass m and a massless gaugino λ (the Goldstino) A Sum Rule The SuperTrace of the squared mass eigenvalues vanish 2 X 2sj 2 STr(m ) ≡ (−1) (2sj + 1)Tr(mj ) = j 2 y 2 Tr(mB) − 2 Tr(mFmF) + 3 Tr(mV) = 0 (8) Application eq.(8) to FI-model (case I) gives 2 (m2 − e κ) + 2 (m2 + e κ) − 2 m2 − 2 m2 − 2 · 02 + 3 · 02 = 0 Fayet-Iliopoulos (D-term) SUSY breaking Question 1 For the FI-model above, can you break the U(1) gauge symmetry without breaking supersymmetry (at tree level)? Question 2 Can you construct an MSSM+FI ? Discuss possible problems you might encounter. Outline Bibliography Spontaneous SUSY breaking General considerations Fayet-Iliopoulos (D-term) SUSY breaking O' Raifeartaigh (F-term) SUSY breaking Goldstino and Gravitino Mediation of SUSY breaking Hidden SUSY breaking sector Gravity Mediated SUSY breaking Gauge Mediated SUSY breaking O' Raifeartaigh (F-term) SUSY breaking O'R's (1975) idea is the following: pick up some chiral superfields such that the Fi = 0 equations are not simultaneously satisfied. Consider the (minimal) superpotential: y W = −k Φ + m Φ Φ + Φ Φ2 (9) 1 2 3 2 1 3 Then the F-terms for the three complex scalar fields are ∗ @W y 2 F1 = − = k − φ3 @Φ1 2 ∗ @W F2 = − = −m φ3 @Φ2 ∗ @W F3 = − = −m φ2 − y φ1φ3 (10) @Φ3 Fi = 0 cannot be satisfied simultaneously (because of k 6= 0) and therefore SUSY is spontaneously broken O' Raifeartaigh (F-term) SUSY breaking ∗ P3 2 The scalar potential is VTREE (φ, φ ) = i=1 jFi j The minimum happens for hφ2i = hφ3i = 0 but hφ1i is undetermined (called moduli field): it follows a “flat direction" in a 2 scalar potential )VTREE min = k However this flat direction is only accidental: hφ1i can be determined by loop corrections (Coleman-Weinberg potential) Veff = VTREE + V1−LOOP Now the global minimum of Veff corresponds to hφ1i = hφ2i = hφ3i = 0 Real scalar field masses : 0; 0; m2; m2; m2 − yk; m2 + yk Weyl fermion masses : 0; m2; m2 R-symmetry of the model: rΦ1 = rΦ2 = 2 ; rΦ3 = 0 Theorem (Nelson-Seiberg (1993)) If a SUSY theory is broken spontaneously by a non-zero F-term (and the superpotential is generic) then this theory must have an exact R-symmetry This theorem places phenomenological obstacles : If R is exact then gaugino masses are forbidden If R is spontaneously broken then an R-(pseudo) Goldstone boson exists in the spectrum O' Raifeartaigh (F-term) SUSY breaking The zero scalar mass can be modified at 1-loop: y 4k2 m2 ' ; φ1 48π2 m2 but the massless fermion ( 1)- the Goldstino - remains massless to all orders in perturbation theory! Theorem (Nelson-Seiberg (1993)) If a SUSY theory is broken spontaneously by a non-zero F-term (and the superpotential is generic) then this theory must have an exact R-symmetry This theorem places phenomenological obstacles : If R is exact then gaugino masses are forbidden If R is spontaneously broken then an R-(pseudo) Goldstone boson exists in the spectrum O' Raifeartaigh (F-term) SUSY breaking The zero scalar mass can be modified at 1-loop: y 4k2 m2 ' ; φ1 48π2 m2 but the massless fermion ( 1)- the Goldstino - remains massless to all orders in perturbation theory! R-symmetry of the model: rΦ1 = rΦ2 = 2 ; rΦ3 = 0 This theorem places phenomenological obstacles : If R is exact then gaugino masses are forbidden If R is spontaneously broken then an R-(pseudo) Goldstone boson exists in the spectrum O' Raifeartaigh (F-term) SUSY breaking The zero scalar mass can be modified at 1-loop: y 4k2 m2 ' ; φ1 48π2 m2 but the massless
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