0 – Lifetime Difference in D D 0 – D D 0 0 Lifetime Difference in Mixing within R- -parity parity Mixing within R Violating Supersymmetry Violating Supersymmetry Gagik Yeghiyan Gagik Yeghiyan Wayne State University, Detroit, MI, USA Wayne State University, Detroit, MI, USA Based on Based on A. A. Petrov, G. K. Yeghiyan, Phys. Rev. D 77, 034018 (2008) A. A. Petrov, G. K. Yeghiyan, Phys. Rev. D 77, 034018 (2008)
Outline Outline • We show that the contribution from RPV SUSY RPV SUSY models with models with • We show that the contribution from ∆Γ Γ D to ∆ the leptonic number violation leptonic number violation to is negative negative, i.e. opposite , i.e. opposite the D is in sign to what is implied by the recent experimental in sign to what is implied by the recent experimental evidence. evidence. • • It is possibly quite large It is possibly quite large in absolute value (may exceed the in absolute value (may exceed the ∆Γ Γ D experimentally allowed values for ∆ ). experimentally allowed values for D ). • We derive new constraints on the relevant RPV coupling pair RPV coupling pair • We derive new constraints on the relevant ∆ m . Unlike those coming from the study of ∆ products. Unlike those coming from the study of m D , our products D , our bounds are insensitive or weakly sensitive to assumptions bounds are insensitive or weakly sensitive to assumptions on R R- -parity conserving (pure MSSM) parity conserving (pure MSSM) sector. sector. on • • We emphasize the necessity of taking into account of the We emphasize the necessity of taking into account of the transformation of RPV couplings RPV couplings from the from the weak eigenbasis weak eigenbasis transformation of to the quark mass eigenbasis to the quark mass eigenbasis. .
∆Γ D ∆ C=1 It is known that ∆Γ by the physics of ∆ is driven driven by the physics of C=1 sector sector It is known that D is both in the SM and beyond. both in the SM and beyond.
∆Γ D ∆ C=1 It is known that ∆Γ by the physics of ∆ is driven driven by the physics of C=1 sector sector It is known that D is both in the SM and beyond. both in the SM and beyond. Let Let (NP) , (SM) + A , |n> |n> is a charmless state A[D 0 0 → → n] = A n] = A n + A n is a charmless state A[D n(SM) n(NP)
∆Γ D ∆ C=1 It is known that ∆Γ by the physics of ∆ is driven driven by the physics of C=1 sector sector It is known that D is both in the SM and beyond. both in the SM and beyond. Let Let (NP) , (SM) + A , |n> |n> is a charmless state A[D 0 0 → → n] = A n] = A n + A n is a charmless state A[D n(SM) n(NP)
∆Γ D ∆ C=1 It is known that ∆Γ by the physics of ∆ is driven driven by the physics of C=1 sector sector It is known that D is both in the SM and beyond. both in the SM and beyond. Let Let (NP) , (SM) + A , |n> |n> is a charmless state A[D 0 0 → → n] = A n] = A n + A n is a charmless state A[D n(SM) n(NP)
∆Γ D ∆ C=1 It is known that ∆Γ by the physics of ∆ is driven driven by the physics of C=1 sector sector It is known that D is both in the SM and beyond. both in the SM and beyond. Let Let (NP) , (SM) + A , |n> |n> is a charmless state A[D 0 0 → → n] = A n] = A n + A n is a charmless state A[D n(SM) n(NP) ∆Γ D Γ D = ∆Γ /(2 Γ Then for y y D ) Then for D = D /(2 D )
The SM contribution is rather uncertain (due to long- -distance effects): distance effects): y y SM The SM contribution is rather uncertain (due to long SM ÷ 10 4 ÷ ~ 10 - ~ 10 -4 10 - -2 2 3 – ± 2.1 ) = (6.6 ± y D 2.1 )• • 10 10 - – is this due to the is this due to the SM SM or or NP NP or both of them? or both of them? y exp = (6.6 exp -3 D
The SM contribution is rather uncertain (due to long- -distance effects): distance effects): y y SM The SM contribution is rather uncertain (due to long SM ÷ 10 4 ÷ ~ 10 - ~ 10 -4 10 - -2 2 3 – ± 2.1 ) = (6.6 ± y D 2.1 )• • 10 10 - – is this due to the is this due to the SM SM or or NP NP or both of them? or both of them? y exp = (6.6 exp -3 D NP ∆ ∆ C=1 The second term (interference of the (interference of the SM SM and and NP C=1 transitions) transitions) – – The second term ≤ 10 D ≤ - yields - yields y y D 10 - -4 4
Consider the last term in this equation: Consider the last term in this equation:
Consider the last term in this equation: Consider the last term in this equation: It may seem unreasonable to consider this term, as it is suppressed as It may seem unreasonable to consider this term, as it is suppres sed as (M W 2 /M /M NP 2 ) ) 2 2 . On the other hand . On the other hand… … (M 2 2 W NP The SM contribution vanishes in the limit of the exact flavor SU(3) The SM contribution vanishes in the limit of the exact flavor SU(3) symmetry. symmetry. In many popular SM extensions and in particular within the RPV SUSY RPV SUSY, , In many popular SM extensions and in particular within the NP ∆ ∆ C=1 the second term (interference of the SM SM and and NP C=1 transitions) transitions) the second term (interference of the also vanishes in the limit of the exact vanishes in the limit of the exact flavor SU(3) flavor SU(3) symmetry. symmetry. also ∆ C=1 contribution to ∆ Then the last term (pure Then the last term (pure NP NP contribution to C=1 transitions) transitions) if non if non- - vanishing, dominates in the dominates in the exact exact flavor SU(3) flavor SU(3) limit! limit! vanishing, In the real world flavor SU(3) flavor SU(3) is of course broken, contributions of first is of course broken, contributions of first In the real world two term are suppressed in powers m two term are suppressed in powers m s s /m /m c c but they are not zero. but they are not zero. The last term may give numerically large contribution if The last term may give numerically large contribution if 2 /m M W M 2 /M /M NP 2 > m > m s /m c 2 . . 2 2 2 2 W NP s c ∆ C=1 generated ∆ Consider the diagrams with two NP NP generated C=1 transitions as well! transitions as well! Consider the diagrams with two
The purpose of our work was to revisit the problem of the The purpose of our work was to revisit the problem of the NP contribution to contribution to y y D and provide constraints on RPV RPV NP D and provide constraints on SUSY models as a primary example. models as a primary example. SUSY We considered a general low- -energy SUSY scenario with no energy SUSY scenario with no We considered a general low assumptions made on a SUSY breaking mechanism at unification assumptions made on a SUSY breaking mechanism at unification scales. scales. Superpotential: Superpotential: λ ’’ To avoid rapid proton decay, we put λ ’’ = 0. = 0. In the quark mass In the quark mass To avoid rapid proton decay, we put eigenbasis eigenbasis where where
Very often in the literature one neglects the difference between λ λ ’ ’ and and Very often in the literature one neglects the difference between based on based on there is no hierarchy in couplings λ λ ’ Subtlety: this is true if only Subtlety: this is true if only there is no hierarchy in couplings ’ ! ! More generally, one can show that More generally, one can show that The above approximation is valid when studying or using constraints on nts on The above approximation is valid when studying or using constrai individual couplings λ λ ’ individual couplings ’ However, when considering bounds on RPV coupling However, when considering bounds on RPV coupling pair products, one must specify if these bounds are for pair products, one must specify if these bounds are for λ ’ λ ’ λ x λ ’ x ’ or or x x pair products. pair products. Otherwise: S. L. Chen, X. G. He, A. Hovhannissyan, H.S. Tsai Otherwise: S. L. Chen, X. G. He, A. Hovhannissyan, H.S. Tsai - - RPV RPV SUSY contribution to y D SUSY contribution to y D is rather small is rather small – – but this is not true! but this is not true!
The dominant diagrams. The dominant diagrams. Neglecting numerically subdominant terms, Neglecting numerically subdominant terms,
The dominant diagrams. The dominant diagrams. Neglecting numerically subdominant terms, Neglecting numerically subdominant terms, It is non- -vanishing in the exact vanishing in the exact flavor SU(3) flavor SU(3) It is non limit. Else, | | λ λ ss | ≤ ≤ 0.29, | 0.29, | λ λ dd | ≤ ≤ 0.29 0.29 or or limit. Else, ss | dd | 2 ≤ λ ss ≤ 0.0841 0.0841 and and λ λ dd ≤ 0.0841 0.0841, thus , thus λ ss2 dd2 2 ≤ contribution of this type of diagrams is large. contribution of this type of diagrams is large.
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