Symmetries and Feynman Rules for Ramond Sector in WZW-type Superstring Field Theories

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Symmetries and Feynman Rules for Ramond Sector in WZW-type Superstring Field Theories Hiroshi Kunitomo (YITP) 215/5/11 SFT215@Sichuan U. arxiv:1412.5281, 15x.xxxxx Y TP YUKAWA INSTITUTE FOR THEORETICAL PHYSICS 1

1. Introduction WZW-type Superstring field theory A formulation with no explicit picture changing operator and working well for NS sector: Open [Berkovits (1995)] Hetetrotic [Berkovits, Okawa and Zwiebach (24)] Type II [Matsunaga (214)] Difficult to construct an action for R sector EOM [Berkovits (21)] [HK (214)] A pseudo-action [Michishita (25)] [HK (214)] We can construct the (self-dual) Feynman rules reproducing the on-shell four point amplitudes. However, these rules do not reproduce the five-point amplitudes [Michishita (27)], or have an ambiguity [HK (213)]. Q: Can we construct Feynman rules for R sector reproducing correct on-shell physical amplitudes? 2

Plan of the Talk 1. Introduction 2. New Feynman rules in the WZW-type open Super SFT 2.1 EOM and the pseudo-action 2.2 Gauge fixing and the self-dual Feynman rules 2.3 Gauge symmetries and the new Feynman rules 3. amplitude with the external fermions 4. Summary and discussion 3

2. New Feynman rules in the WZW-type Open Super SFT 2.1 EOM [Berkovits] and the pseudo-action [Michishita] Large Hilbertspace : (η η ) H large A = α + ξ a, with α, a H small (i.e. ηα = ηa = ). String fields : NS : Φ H (NS) large, (G, P ) = (, ), Φ =, R : Ψ H (R) large EOM including the R sector :, (G, P ) = (, 1/2), Ψ =, η(g 1 Qg) + (ηψ) 2 =, Q ηψ =, (1) where g = e Φ and Q A = QA + A(g 1 Qg) ( 1) A (g 1 Qg)A. Gauge symmetry : g 1 δg = Q Λ + ηλ 1 (ηψ)λ 12 Λ 12 (ηψ), δψ = Q Λ 12 + ηλ 32 + Ψ(ηΛ 1 ) (ηλ 1 )Ψ. (2a) (2b) 4

NS action : S NS = 1 2 (g 1 Qg)(g 1 ηg) where ĝ = g(t) = e tφ. Auxiliary field : Ξ The pseudo-action : Z 1 dt(ĝ 1 t ĝ){(ĝ 1 Qĝ), (ĝ 1 ηĝ)}, H (R) large, (G, P ) = (, 1/2), Ξ =. The variation of S = S NS + S R yields S R = 1 2 (QΞ)g(ηΨ)g 1. (3) η(g 1 Qg) + 1 2 {ηψ, Q Ξ } =, Q ηψ =, ηq Ξ =. (4) where Ξ = g 1 Ξg. If we eliminate Ξ by imposing a constraint EOMs (4) become EOMs (1). Q Ξ = ηψ, 5

2.2 Gauge fixing and the self-dual Feynman rules Since the pseudo-action (3) is not the true action we cannot derive the Feynman rules, but we can construct them. Quadratic part of the action has gauge symmetries S = 1 2 (QΦ)(ηΦ) 1 (QΞ)(ηΨ), (5) 2 δφ = QΛ + ηλ 1, δψ = QΛ 12 + ηλ 32, δξ = QΛ 1 2 + η Λ 12. (6) Fix them by (the simplest) gauge conditions b Φ = ξ Φ =, b Ψ = ξ Ψ =, b Ξ = ξ Ξ =. Then (5) can be inverted as ΦΦ = ξ b L = dτ(ξ b )e τl, ΨΞ = ΞΨ = 2 ξ b L = 2 ΦΦ. 6

The self-dual Feynman rules : [Michishita] An essential prescription Taking into account the constraint by replacing ηψ and QΞ in vertices by their (linearized) self-dual part ω = (ηψ + QΞ)/2. ( A part vanishing under the constraint should be decoupled.) Propagator : NS : ΦΦ = ξ b L, R : ωω = 1 2 Q ««ξ b ξ b η + η L L «Q (diagonal) Vertices :... 7

On-shell external fermions : ω = ηψ by imposing the linearized constraint QΞ = ηψ. (It automatically implies the on-shell condition QηΨ =.) The self-dual Feynman rules give the well-known (on-shell) 4-point amplitudes but do not reproduce the 5-point amplitudes. [Michishita] 8

2.3 Gauge symmetries and the new Feynman rules Gauge symmetries : The (pseudo-) action S = S NS + S R is invariant under g 1 δg = Q Λ + ηλ 1, δψ = ηλ 32 + [Ψ, ηλ 1 ], δξ = QΛ 1 2 + [Q, Λ ]. These are compatible with the self-dual/anti-self-dual decomposition. δ(q Ξ ± ηψ) = [(Q Ξ ± ηψ), Λ 1 ], and so respected by the self-dual rules. However, these symmetries are not enough to gauge away all the un-physical states. They do not contain all the symmetries of the quadratic action (6). The missing symmetries generated by Λ1 2 and Λ1 2 can be extended to g 1 δg = 1 2 {Q Ξ, Λ 12 } + 1 2 {ηψ, Λ 12 }, (7a) δψ = Q Λ 12, δξ = g(η Λ 12 )g 1. (7b) 9

The variation of the action under these transformations becomes δs = 1 4 Λ 1 2 [(Q Ξ ) 2, (Q Ξ ηψ)] + 1 4 Λ 12 [(ηψ) 2, (Q Ξ ηψ)], which vanishes under the constraint. These symmetries (7) are not respected by the self-dual rules because they are not compatible with the self-dual/anti-self-dual decomposition. Ansatz The correct Feynman rules should respect these symmetries too. 1

The new Feynman rules : An essential prescription Do not impose any restriction on the off-shell (propagating) states. Propagator : NS : ΦΦ = ξ b L, R : ΨΞ = ΞΨ = 2 ξ b L Π R, (off-diagonal) Vertices :... 11

Notes : Fermion number, F (Ψ) = 1, F (Ξ) = 1, is conserved. Then R-propagator is directed. On-shell external fermions : ηψ by imposing the linearized constraint QΞ = ηψ, after adding (averaging) two possibilities of the external fermions, ηψ and QΞ. We can show that the new Feynman rules correctly reproduce all the on-shell physical 4- and 5-point amplitudes. 12

3. Amplitudes with the external fermions Four-point amplitudes are correctly reproduced as in the self-dual rules. Difference between two rules comes from the diagram with either (i) at least two fermion propagators. The amplitude is schematically A (QΠ R η + ηπ R Q) (QΠ R η + ηπ R Q) W, in the self-dual rules and A QΠ R η QΠ R η W + ηπ R Q ηπ R Q W, in the new rules, where we can assign a direction to fermion lines, + + + 13

or, (ii) two-fermion-even-boson interactions, e.g. S (4) R = 1 Φ 2 (QΞ)(ηΨ)) Φ 2 (ηψ)(qξ)) 4 + 1 Φ(QΞ)Φ(ηΨ) Φ(ηΨ)Φ(QΞ), 4 with at least one off-shell fermion, which gives non-trivial contribution to the amplitude. These differences improve the discrepancy in the five-point amplitudes! 14

3.2 Five-point amplitudes Example : (ABCDE) = (F F BBB) Five two-propagator (2P) diagrams A E B A C B B D C E D A C (a) D (b) E (c) D C E D E B A C A (d) B (e) 15

A E B D C (a) A (2A)(a) F F BBB = 1 dτ 1 dτ 2 8 (QΞ A ηψ B + ηψ A QΞ B )(ξ c1 b c1 Q)Φ C (ηξ c2 b c2 )(QΦ D ηφ E + ηφ D QΦ E ) W + (QΞ A ηψ B + ηψ A QΞ B )(ξ c1 b c1 η)φ C (Qξ c2 b c2 )(QΦ D ηφ E + ηφ D QΦ E ) W!. (3) (4) (2) (5) Witten diagram (1) 16

Using dτ{q, b }e τl = dτ τ e τl, we can align three external bosons as (QΦ C, QΦ D, ηφ E ): A (2A)(a) F F BBB = 1 2 + 1 8 d 2 τ QΞ A ηψ B + ηψ A QΞ B dτ QΞ A ηψ B + ηψ A QΞ B (ξ c1 b c1 ) QΦ C b c2 QΦ D ηφ E W (ξ c b c ) Φ C QΦ D ηφ E + ηφ D QΦ E W 2 QΞ A ηψ B + ηψ A QΞ B (ξ c b c ) QΦ C η(φ D Φ E ) W QΦ D ηφ E + ηφ D QΦ E (ξ c b c ) QΞ A ηψ B + ηψ A QΞ B Φ C W 2 η(φ D Φ E ) (ξ c b c ) QΞ A ηψ B + ηψ A QΞ B QΦ C W!. 17

Similarly, the contribution of the diagram (b) is given by B A C E D (b) A (2P)(b) F F BBB = 1 2 dτ 1 dτ 2 QΞ B Φ C (ηξ c1 b c1 Q) Φ D (ηξ c2 b c2 Q) Φ E ηψ A W + ηψ B Φ C (Qξ c1 b c1 η) Φ D (Qξ c2 b c2 η) Φ E QΞ A W! 18

= 1 2 Z d 2 τ QΞ B QΦ C (ξ c1 b c1 ) QΦ D b c2 ηφ E ηψ A W + ηψ B QΦ C (ξ c1 b c1 ) QΦ D b c2 ηφ E QΞ A W! 1 2 Z dτ QΞ B Φ C (ξ c b c ) ηφ D QΦ E ηψ A W QΞ B QΦ C (ξ c b c ) η(φ D Φ E ) ηψ A W QΦ E ηψ A (ξ c b c ) QΞ B η(φ C Φ D ) W + ηφ E QΞ A (ξ c b c ) ηψ B QΦ C Φ D W + ηφ E ηψ A (ξ c b c ) QΞ B QΦ C Φ D W Φ E ηψ A (ξ c b c ) QΞ B QΦ C ηφ D W!, 19

A (2P)(c) F F BBB =1 2 d 2 τ QΦ C QΦ D (ξ c1 b c1 ) ηφ E b c2 QΞ A ηψ B + ηψ A QΞ B W + 1 dτ 2 Φ C QΦ D (ξ c b c ) ηφ E QΞ A ηψ B + ηψ A QΞ B W 8 + QΦ C ηφ D ηφ C QΦ D (ξ c b c ) Φ E QΞ A ηψ B + ηψ A QΞ B W 2 QΦ C ηφ D (ξ c b c ) Φ E QΞ A ηψ B + ηψ A QΞ B W 2 QΦ C Φ D (ξ c b c ) ηφ E QΞ A ηψ B + ηψ A QΞ B W 2 QΞ A ηψ B + ηψ A QΞ B (ξ c b c ) Φ C QΦ D ηφ E W QΞ A ηψ B + ηψ A QΞ B (ξ c b c ) QΦ C ηφ D ηφ C QΦ D Φ E W + 2 QΞ A ηψ B + ηψ A QΞ B (ξ c b c ) QΦ C η(φ D Φ E ) W!, 2

A (2P)(d) F F BBB =1 2 + 1 4 d 2 τ QΦ D ηφ E (ξ c1 b c1 ) dτ QΦ D ηφ E + ηφ D QΦ E QΞ A b c2 ηψ B + ηψ A b c2 QΞ B (ξ c b c ) ηψ A QΞ B Φ C W + η(φ D Φ E ) (ξ c b c ) QΞ A ηψ B + ηψ A QΞ B QΦ C W + QΞ B Φ C (ξ c b c ) QΦ D ηφ E + ηφ D QΦ E ηψ A W QΞ B QΦ C (ξ c b c ) η(φ D Φ E ) ηψ A W QΦ C W ηψ B QΦ C (ξ c b c ) η(φ D Φ E ) QΞ A W!, 21

A (2P)(e) F F BBB =1 2 d 2 τ ηφ E QΞ A (ξ c1 b c1 ) ηψ B b c2 QΦ C QΦ D W + ηφ E ηψ A (ξ c1 b c1 ) QΞ B b c2 QΦ C QΦ D W! 1 4 dτ Φ E ηψ A (ξ c b c ) QΞ B QΦ C ηφ D + ηφ C QΦ D W ηφ E QΞ A (ξ c b c ) ηψ B QΦ C Φ D Φ C QΦ D W ηφ E ηψ A (ξ c b c ) QΞ B QΦ C Φ D Φ C QΦ D W QΦ C ηφ D + ηφ C QΦ D (ξ c b c ) Φ E ηψ A QΞ B W QΦ C Φ D Φ C QΦ D (ξ c b c ) ηφ E (QΞ A ηψ B + ηψ A QΞ B ) W!. 22

Five one-propagator diagrams A E B A C B D E A B (a) C C (b) D D (c) E D C E D B C E (d) A A (e) B 23

From the straightforward calculation, we obtain: A (1P)(a) F F BBB = 1 24 dτ QΞ A ηψ B + ηψ A QΞ B (ξ c b c ) Φ C QΦ D ηφ E ηφ D QΦ E W 2 QΞ A ηψ B + ηψ A QΞ B (ξ c b c ) QΦ C Φ D ηφ E ηφ C Φ D QΦ E + QΞ A ηψ B + ηψ A QΞ B W (ξ c b c ) QΦ C ηφ D ηφ C QΦ D Φ E W!, 24

A (1P)(b) F F BBB =1 4 dτ ηψ B ηφ C (ξ c b c ) Q(Φ D Φ E ) QΞ A W QΞ B ηφ C (ξ c b c ) Q(Φ D Φ E ) ηψ A W ηψ B Φ C (ξ c b c ) QΦ D ηφ E ηφ D QΦ E QΞ A W! A (1P)(c) F F BBB =1 8 1 4 QΞ A ηψ B Φ C Φ D Φ E W, dτ QΦ C ηφ D + ηφ C QΦ D (ξ c b c ) Φ E QΞ A ηψ B ηψ A QΞ B W, A (1P)(d) F F BBB =1 8 dτ QΦ D ηφ E + ηφ D QΦ E (ξ c b c ) QΞ A ηψ B ηψ A QΞ B Φ C W, 25

A (1P)(e) F F BBB =1 4 dτ Φ E ηψ A (ξ c b c ) QΞ B QΦ C ηφ D ηφ C QΦ D W + ηφ E QΞ A (ξ c b c ) ηψ B ηψ A (ξ c b c ) QΞ B Q(Φ C Φ D ) W! 1 4 ηψ A QΞ B Φ C Φ D Φ E W. One no-propagator diagram, A E 5 B 3 C D gives A (NP) F F BBB = 1 12 QΞ A ηψ B + ηψ A QΞ B Φ C Φ D Φ E W. 26

The boundary contributions are cancelled in the total amplitude: A F F BBB = Z = ex i=a A (2P)(i) F F BBB + ex i=a A (1P)(i) F F BBB + A(NP) F F BBB d 2 τ QΞ A ηψ B + ηψ A QΞ B (ξ c1 b c1 ) QΦ C b c2 QΦ D ηφ E W + ηψ B QΦ C (ξ c1 b c1 ) QΦ D b c2 ηφ E QΞ A W + QΞ B QΦ C (ξ c1 b c1 ) QΦ D b c2 ηφ E ηψ A W! + QΦ C QΦ D (ξ c1 b c1 ) ηφ E b c2 QΞ A ηψ B + ηψ A QΞ B W + QΦ D ηφ E (ξ c1 b c1 ) QΞ A b c2 ηψ B + ηψ A b c2 QΞ B QΦ C W + ηφ E QΞ A (ξ c1 b c1 ) ηψ B b c2 QΦ C QΦ D W + ηφ E ηψ A (ξ c1 b c1 ) QΞ B b c2 QΦ C QΦ D W! 27

After eliminating Ξ from the external states by imposing QΞ = ηψ, Z A F F BBB = d 2 τ ξ(ηψ A ηψ B b c1 QΦ C b c2 QΦ D ηφ E ) W + ξ(ηψ B QΦ C b c1 QΦ D b c2 ηφ E ηψ A ) W + ξ(qφ C QΦ D b c1 ηφ E b c2 ηψ A ηψ B ) W + ξ(qφ D ηφ E b c1 ηψ A b c2 ηψ B QΦ C ) W + ξ(ηφ E ηψ A b c1 ηψ B b c2 QΦ C QΦ D ) W!. Since this final expression has the same form (up to ξ) as the bosonic SFT (with ηψ, QΦ, ηφ A), we can conclude that this is nothing but the well-known amplitude. Similarly, we can calculate the amplitudes with FFFFB and FBFBB, and show that the results agree with the well-known amplitues. 28

4. Summary and discussion We have found that the missing gauge-symmetries are realized as those under which the variation of the pseudo-action is proportional to the constraint. We have proposed the new Feynman rules for the open super SFT respecting all the gauge- symmetries. We have shown that the correct on-shell 4- and 5-point amplitudes are reproduced by the new rules. We can apply the similar argument to the heterotic SFT, and obtain the new Feynman rules which reproduce the correct 4- and 5-point on-shell physical amplitudes. 29

We have to clarify whether the new Feynman rules reproduce all the on-shell physical amplitudes at the tree level. general theory of the pseudo-action and its symmetries We have to extend the Feynman rules to those applicable beyond the tree level. gauge fixing by means of the BV method extra factor 1/2 for each fermion loop? It is interesting to calculate the off-shell amplitudes and to study their properties. 3