m f f unchanged under the field redefinition (1), the complex mass matrix m should transform into

Transcription

1 PHY 396 T: SUSY Solutions or problem set #8. Problem (a): To keep the net quark mass term L QCD L mass = ψ α c m ψ c α + hermitian conjugate (S.) unchanged under the ield redeinition (), the complex mass matrix m should transorm into m ( Ũ ) m ( U ) (S.2) Or in matrix notations using N c matrix Ψ α or the LH quarks and N c matrix Ψ α or the LH antiquarks, so that the ield redeinitions () become Ψ α Ψ α U, Ψα Ũ Ψ α, (S.3) the mass matrix m should transorm to m U m Ũ (S.4) to keep the net mass term ( Ψα ) L mass = tr Ψ α m + H. c. (S.5) invariant. Now consider the chiral anomaly o the chiral ield redeinition (). For an abelian chiral redeinition with U = exp(ia), Ũ = exp(iã), the chiral anomaly may be compensated by changing the Θ angle by Θ = (a + ã). (S.6) Likewise, or the non-abelian U = exp(iat ), Ũ = exp(iã T ) or some hermitian matrices T and

2 T, the Θ angle should be changed by Θ = a tr(t ) + ã tr( T ). (S.7) To rewrite this ormula in terms o the U and Ũ matrices themselves, note that log(det(u)) = tr(log(u)) = ia tr(t ) (S.8) and likewise log(det(ũ)) = iã tr( T ), (S.9) so eq. (S.7) becomes Θ = phase(det(u)) + phase(det(ũ)). (S.0) At the same time, eq. (S.4) implies det(m) det(m) det(u) det(ũ) (S.) and hence phase(det(m)) = phase(det(u)) phase(det(ũ)). (S.2) Thereore, the combination Θ = Θ + phase(det(m)) (2) changes by Θ = 0; in other words, it remains invariant under the ield redeinitions (). 2

3 Problem (b): In matrix orm, the transormations (3) act as Q Q U, Q Ũ Q, Q U Q, Q Q Ũ. (S.3) To keep the net quark Lagrangian L quark = ( d 4 θ tr ( ) Z Qe 2V Q ( ) d 2 θ tr m QQ ( )) + tr Z 2V Qe Q d 2 θ tr (m Q Q ) (S.4) the matrices m, Z, and Z should transorm into m U m Ũ, Z U Z ( U ), Z (Ũ ) Z Ũ. (S.5) Note that or a non-unitary matrix U, U U and ( U ) U, and likewise or the Ũ. Also, a chiral redeinition o quark and antiquark superields aects the measure o the path integral, which leads to the Konishi anomaly. To compensate or this anomaly, we should adjust both the Θ angle and the Wilsonian gauge couplings. In terms o the holomorphic gauge coupling = g 2 W iθ 8π 2 the adjustment is simply the complexiied orm o eq. (S.0): (S.6) 8π 2 ( tr log(u) + tr log(ũ) ) = 8π 2 ( log det(u) + log det(ũ) ). (S.7) Moreover, the adjustment here is related to the change o log det(m). Indeed, eq. (S.5) or the mass matrix leads to and hence det(m) det(m) det(u) det(ũ) (S.8) log det(m) log det(m) log det(u) log det(ũ). (S.9) 3

4 Comparing this ormula to eq. (S.7), we immediately see that the combination 8π 2 = 8π 2 log det(m) (4) remains invariant under the ield redeinition. Problem (c): In terms o the holomorphic dimensional transmutant Λ SQCD o the Wilsonian gauge coupling, Λ 3Nc SQCD = exp ( 8π 2 w ) Λ UV a numeric constant, (S.20) eq. (S.7) means that i we redeine the quark and antiquark ields while keeping the Wilsonian UV cuto Λ UV ixed, then the holomorphic Λ SQCD must change according to Λ 3Nc SQCD Λ 3Nc SQCD det(u) det(ũ). (S.2) However, in light o eq. (4), the product Λ 3Nc SQCD det(m) (S.22) remains invariant under the redeinition. Now suppose all the quarks are heavy and let s integrate them out rom the low-energy eective QFT which is thereore the SU(N c ) SYM theory without any quarks. Since the low-energy theory does not have any quarks, its couplings cannot be aected by any redeinitions o the quark ields we had integrated out. Consequently, the low-energy gauge coupling or its dimensional transmutant Λ SYM may depend only on the reparametrization-invariant combinations o the -energy SQCD couplings. Moreover, unbroken supersymmetry requires Λ SYM to be a holomorphic unction o the SQCD s holomorphic couplings namely or Λ SQCD, and the quark mass matrix m while it cannot depend on the non-holomorphic couplings such as Z or Z matrices. This, the only SQCD couplings that may aect the low-energy Λ SYM are holomorphic combinations o Λ SQCD and m which are invariant under any chiral redeinitions (S.3) o the quark superields. 4

5 Eq. (S.22)gives us one such invariant, and it is easy to see that there are no other holomorphic invariants. Indeed, the mass matrix m can be transormed into any other mass matrix m (as long as both m and m have det 0) by using U = µ m, Ũ = µ m or some mass scale µ (S.23) U m Ũ = m, c. eq. (S.5). Thus, we cannot make any invariants at all rom just the mass matrix. Consequently, by involving one more input parameter the Λ SQCD we might be able to make at most one independent invariant. (I we could make two invariants, we would be able to eliminate the Λ SQCD rom their combination and construct an invariant just rom the mass matrix m, but we just saw that that s impossible.) Note: this argument does not tell us whether an invariant exists at all, only that i it does exist, it is unique. The bottom line is, Λ SYM must be a holomorphic unction o the invariant (S.22), and by dimensional analysis the only suitable unction is Λ 3Nc SYM = Λ3Nc SQCD det(m) a numeric constant. (5) Quod erat demonstrandum. Problem (d): Let s start by integrating out the heaviest quark lavor. To do that, we need to compare two energy regimes: the -energy regime E M all other quark masses, and the lowerenergy regime E M but E M 2, etc. In the -energy regime we use the original SQCD with N c colors and lavors; in the lower-energy regime, we use the eective theory rom which the heaviest lavor has been integrated out, thus N c colors but only lavors. 5

6 have Integrating the RG equations or the -energy regime using the NSVZ beta-unction, we 8π 2 g 2 (E) N c log g 2 (E) = 8π2 Re SQCD = (3N c ) log (3N c ) log Λ UV E + E Λ SQCD + a numeric constant + = = log Z (E) Z (E) log Z (E) Z (E) (S.24) where the constant depends on the details o the renormalization scheme we use and on the precise deinition o the Λ SQCD but not on any couplings o the theory. Likewise, in the lowerenergy regime the eective theory gives us 8π 2 g 2 (E) N c log g 2 (E) = (3N c + ) log + a numeric constant. E Λ lower + =2 log Z (E) Z (E) (S.25) The two regimes meet at the threshold energy or the heaviest quark, so or E = M (the physical mass o that quark) eqs. (S.24) and (S.25) should produce the same running gauge coupling g(e). Comparing the two equations, we see that this matching condition requires (3N c + ) log = (3N c ) log log Z = M ) + const M Λ lower M Λ SQCD (S.26) since all the other terms in eqs. (S.24) and (S.25) cancel each other. Exponentiating both sides o eq. (S.26) and some simple algebra give us Λ lower 3Nc + = Λ SQCD + M Z = M ) const. (S.27) Finally, thanks to eq. (6), M Z = M ) = m (S.28) where m is the holomorphic mass o the heaviest quark. Consequently, Λ lower 3Nc + = ΛSQCD m a numeric constant. (S.29) In eq. (S.29), the label SQCD reers to the original -energy SQCD while lower reers 6

7 to the eective theory or lower energies rom which the heaviest quark has been integrated out. But this eective theory is itsel an SQCD, so when we go to energies below M 2 we may integrate out the second-heaviest quark in exactly the same ashion. By repeating the above argument almost verbatim, we obtain Λ second lower 3Nc +2 = Λ irst lower 3Nc + m 2 a numeric constant (S.30) and hence Λ second lower 3Nc +2 = ΛSQCD m m 2 a numeric constant. (S.3) Similarly, integrating out the third-heaviest lavor gives the eective theory with Λ third lower 3Nc +2 = ΛSQCD m m 2 m 3 a numeric constant, (S.32) etc., etc. Eventually, ater integrating out all the lavors, we inally arrive at the eective low-energy SYM without any quarks and with Λ SYM 3Nc = ΛSQCD N = m a numeric constant, (S.33) in perect agreement with eq. (5). Problem (e): Suppose there are N h heavy lavors with masses Λ and N l = N h light lavors. We may integrate out the heavy lavors exactly as we did in part (d), except now we stop ater N h steps. Consequently, the eective low energy theory is an SQCD with N c colors and N l lavors, and it has Λ 3Nc N l low = Λ 3Nc N l N h N h = m a numeric constant. (S.34) For simplicity, this ormula assumes a diagonal mass matrix or the heavy lavors. I it is not diagonal, replace the product o masses with the determinant, thus Λ 3Nc N l low = Λ 3Nc N l N h det ( m heavy ) a numeric constant. (S.35) 7

8 Problem 2(a): In matrix orm, M is a matrix product o the quark and antiquark chiral superields, M = Q Q, so under chiral linear redeinition (S.3) it transorms according to M Ũ M U. (S.36) Consequently, log det(m) log det(m) + log det(u) + log det(ũ). (S.37) At the same time, the Konishi anomaly o the chiral ield redeinition should be compensated by adjusting the Wilsonian gauge coupling o the theory according to 8π 2 ( log det(u) + log det(ũ) ). (S.7) Combining the last two equations, we immediately obtain a holomorphic reparametrizationinvariant combination o and the moduli matrix, namely 8π 2 + log det(m). (0) Or in terms o the dimensional transmutant o the holomorphic gauge coupling Λ 3Nc SQCD = exp ( 8π 2 ) Λ 3Nc UV a numeric constant, (S.38) the ratio Λ 3Nc SQCD det(m) (S.39) is invariant under all linear redeinitions o the quark and antiquark superields. Moreover, similar to what we had in problem (c), (S.39) is the only independent holomorphic invariant combination o Λ SQCD and the moduli matrix M. Indeed, were there two such invariants, we would be able to eliminate Λ SQCD rom their combination and construct a holomorphic invariant rom just the M matrix. But that s impossible since any non-degenerate 8

9 matrix M can be turned into any other non-degenerate matrix M bu suitable transorm (S.36), or example Ũ = µ 2 M, U = µ 2 M = Ũ M U = M. (S.40) As in problem (c), the Λ low o the eective low-energy theory whose gauge group is the SU(N c ) which is let unbroken by the Higgs mechanism, and which does not have any quarks (since the Higgs mechanism ate them all) must be a holomorphic unction o the -energy SQCD s Λ SQCD and the holomorphic moduli M o the squark VEVs, and it cannot depend on any non-holomorphic couplings o SQCD such as the Z and Z matrices. At the same time, Λ low cannot be aected by any redeinitions o the ields which have been integrated out rom the low-energy EFT. Consequently, Λlow must be a holomorphic unction o the invariant (S.39), and it cannot depend on anything else. In particular, Λlow cannot depend on any dimensionul parameter besides the (S.39), so by dimensional analysis we must have Λlow 3(Nc ) = Λ3Nc SQCD det(m) a numeric constant. (9) Problem 2(b): The hierarchy φ φ 2 φ N vector ields physical masses 2, o squark VEVs leads to similar hierarchy o massive M 2 v () = 2 g2 (E = M v ) M v () M 2 v (2) M 2 v ( ). Z Z (E = M v ) φ 2, (S.4) (S.42) Consequently, the thresholds associated with massive vectors are ar rom each other on the energy scale, so we may work out the RG low one threshold at a time. As in problem (d), let s start with the est-energy threshold at E = M v () and compare the NSVZ ormulae or the renormalized gauge couplings g(e) in two regimes: the er energy regime E > M v () governed by the original SQCD, and the lower-energy regime E < M v () 9

10 (but E M v (2), etc) governed by the eective theory rom which the heaviest massive vector multiplets have been integrated out. As we had learned in early September, this eective theory is itsel an SQCD but with N c = N c and N =. For the er-energy regime, the NSVZ ormula gives us 8π 2 g 2 (E) N c log g 2 (E) = (3N E c ) log Λ + a numeric constant, = 2 log Z Z (E) (S.43) c. eq. (S.24). Likewise, or the lower-energy regime we have a similar ormula, except or a dierent Λ o the eective theory and dierent N c N c and 3N c 3N c 2), thus 8π 2 g 2 (E) (N c ) log g 2 (E) = (3N c 2) log + a numeric constant. E Λ lower =2 (hence 2 log Z Z (E) (S.44) The two regimes connect at the threshold point E = M v (), so at that point both eqs. (S.43) and (S.44) should produce the same gauge coupling g(e). = M v (), LHS (S.44) LHS (S.43) = RHS (S.44) RHS (S.43), (S.45) which ater cancellation o similar terms in both ormulae becomes log g 2 (E = M v ()) = (3N c 2) log M v() Λ lower (3N c ) log M v() Λ + 2 log Z Z ((E = M v ()) + a numeric constant. (S.46) Exponentiation this ormula gives us (ater some algebra) Λ lower 3Nc 2 = Λ g2 Z = M v ()) M 2 v () a numeric constant (S.47) where the second actor on the RHS looks like a painul mess but thanks to eq. (S.4)it is simply 2/ φ 2. Thereore, Λ lower 3Nc 2 = Λ φ 2 a numeric constant. (S.48) And this ormula is the bottom line o integrating out the massive ields at the irst threshold. 0

11 Since the eective theory below the irst threshold is itsel an SQCD, we may integrate out the ields that become massive at the second threshold in exactly the same way. Repeating eq. (S.48) almost verbatim, we obtain Λ second lower 3Nc 4 = Λ irst lower 3Nc 2 φ 2 2 a numeric constant = Λ φ 2 φ 2 2 a numeric constant. (S.49) Likewise, ater three thresholds the eective theory has Λ third lower 3Nc 6 Λ φ 2 φ 2 2 φ 3 2 a numeric constant, (S.50) etc., etc., until ater thresholds we inally arrive at the lowest-energy theory a quarkless SYM with SU(N c ) unbroken gauge group and Λ lowest 3Nc 3 = Λ = φ 2 a numeric constant. (S.5) Or in terms o the moduli M, = N φ 2 = = M = det(m) (S.52) since or diagonal squark VEVs () the moduli matrix is also diagonal, hence Λ lowest 3Nc 3 = Λ det(m) a numeric constant, (S.53) in perect agreement with eq. (9)

12 Problem 3(a): For N c 2, the ields that survive the Higgs mechanism and remain massless are the vector superields o the unbroken gauge group SU(N c ) and the N 2 chiral moduli superields M. The moduli are neutral with respect to the SU(N c ), but the low-energy gauge coupling depends on the moduli according to eq. (9). At low energies E Λ low, the SU(N c ) gauge coupling becomes strong, the eective SYM theory becomes conining, the chiral symmetry o the gauginos is spontaneously broken, so 2-gaugino bound states orm a Bose Einstein condensate. In terms o the gaugino ield themselves, S tr(λ α λ α ) = 0. (S.54) As I explained in class, the magnitude and the phase o this gaugino-bilinear condensate (which everybody calls simply gaugino condensate ) are controlled by the Λ SYM, S = N Λ 3N SYM = Λ3 SYM N. (S.55) For the present case, N and Λ SYM belong to the eective low-energy theory, thus N = N c while Λ SYM should be identiied as Λ low in eq. (9), hence S = Nc N Λ3Nc a numeric constant. det(m) (S.56) I have also explained in class that when the SYM s gauge coupling depend on some moduli ields whatever the origin o such dependence, gaugino condensation creates sources or the auxiliary components o the moduli superields, and that such sources can be accounted by a non-perturbative eective superpotential or the moduli W NP (M) = N S (M). 6π2 (S.57) In light o eq. (S.56), or the problem at hand this eective superpotential has orm W NP (M) = N c 6π 2 Λ 3 Nc N low (M) = Λ3Nc a numeric constant. (3) det(m) Quod erat demonstrandum. 2

13 Problem 3(b): In any supersymmetric vacuum chiral SF φ i, Wnet φ i = 0 (S.58) where W net comprises both the tree-level and the non-perturbative superpotentials. For the problem at hand the relevant chiral superields are the matrix elements o the moduli matrix M, while the net superpotential is spelled out in eq. (4), so all we need to do is to take the derivatives. The derivative o the tree-level term is obviously tr(mm) M while or the non-perturbative term we use = m (S.59) det(m) M = [ minor(m) ] = det(m) ( M ) (S.60) and hence M Λ3Nc det(m) /(N c ) = Λ3Nc N c det(m) /(N c ) ( M ). (S.6) In the context o eq. (4), this means W NP (M) M without any additional numeric actors. Altogether, in matrix orm = Λ3 low = S 6π 2 ( M ) (S.62) W net (M) M S (M) = m + 6π 2 M. (S.63) To make this tadpole vanish, we need M m = m M = N S (M) 6π 2 (S.64) 3

14 in accordance with the irst equation (5), or equivalently M = m S 6π 2. (S.65) At the same time, the gaugino condensate S itsel depends on the moduli ields according to eq. (S.56), hence the second eq. (5). Problem 3(c): The Veneziano Yankielowicz Taylor eective superpotential (6) obtains rom the superpotential (4) or the moduli via integrating in the gaugino condensate S. Given the VYT superpotential, the VEVs o S and M in SUSY vacua obtain by solving the equations W VYT S = 0 and W VYT M = 0. (S.66) Let s start with the derivative WRT to S, W VYT S = log SNc det(m) 6π 2 Λ 3Nc = 6π 2 log SNc det(m) Λ 3Nc + const + S 6π 2 N c S + (N c ) + const. (S.67) To make this derivative vanish, we need log S Nc det( M ) Λ 3Nc = a numeric constant (S.68) and hence S = Nc Λ3Nc a numeric constant, det( M ) (S.69) exactly as in the second eq. (5). 4

15 Now let s take the derivative o the W VYT WRT the moduli matrix M. Using eq. (S.60) and hence log det(m) M = det(m) det(m) M = M, (S.70) we ind W VYT M = m + S 6π 2 M. (S.7) To make this derivative vanish as a matrix, we need which is equivalent to the irst eq. (5). M = S 6π 2 m, (S.65) Problem 3(d): Taking the determinant o both sides o the irst equation (5) gives us det( M ) det(m) = ( ) N S 6π 2. (S.72) At the same time, taking both sides o the second eq. (5) to the power N c gives us S Nc = Λ3Nc a numeric constant. det( M ) (S.73) Combining these two equations, we arrive at S Nc ( ) N S 6π 2 = det( M ) det(m) Λ3Nc a numeric constant det( M ) (S.74) and hence S Nc = Λ 3Nc det(m) a numeric constant. (S.75) Clearly, this equation has N c distinct solutions S = Nc Λ 3Nc det(m) a numeric constant any Nc. (S.76) 5

16 Once we have solved or the gaugino condensate, the moduli VEVs ollow rom the irst eq. (5): M = S 6π 2 m (S.77) = m Nc Λ 3Nc det(m) a numeric constant Nc. (S.78) To make the irst line here work, on the second line we must take the same branches o the power N c roots as in eq. (S.76). Consequently, the net number o solutions to eqs. (5) or both S and M is N c rather than N 2 c. Physically, this means that SQCD with non-zero masses or all quark lavors has precisely N c supersymmetric vacua. This agrees with the Witten index or SQCD: N c, same as or the SU(N c ) SYM without the quarks. Problem 3(e): Let s simpliy eq. (S.78) or a diagonal mass matrix m. In this case det(m) = m, while the inverse mass matrix m is also diagonal. Consequently, eq. (S.78) yields a diagonal matrix M o the moduli VEVs, and or each diagonal element we have M diag = Λ (3Nc )/N c In terms o the semiclassical squark VEVs φ = φ const (m ) /Nc (m ) (/Nc). (S.79) m negative M diag, this means that m positive. (S.80) Thus, when we make any particular lavor lighter, the corresponding squark VEV becomes larger while the other squark VEVs become smaller. What i we make several lavor s masses smaller at the same time? In that case, the speciic powers in eq. (S.79) tell us that φ m Nc m /2N c. (S.8) Consequently, as long as all the lighter lavors become lighter at the same rate, and the #lighter lavors < N c which is automatic or < N c then all the corresponding squark 6

17 VEVs become larger. In particularly, i several lavors become very light, then all the corresponding squark VEVs become very large. Problem 3(): Without the tree-level quark masses, the eective superpotential or the moduli comes only rom the gaugino condensation, so eq. (3) gives the complete superpotential. In terms o the diagonal squark VEVs, where W net = C = φ ν (S.82) C = Λ (3Nc )/(N c ) a numeric constant and ν = 2 N c. (S.83) The speciic values o C and ν will not be important or the ollowing argument, all we need is C 0 and ν < 0, so the superpotential is proportional to the negative powers o the squark VEVs. Plugging the superpotential (S.82) into eq. (8) or the scalar potential, we obtain W net φ = νc φ ν φ ν (S.84) and hence V s = ν2 C 2 2 = φ 2ν = φ 2. (S.85) Clearly, this potential decreases monotonically when any o the φ. Consequently, all o the squark VEVs want to runaway to ininity, and the theory has no stable vacua, supersymmetric or otherwise. 7