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Heavy quark symmetry: ideas and applications

arXiv:hep-ph/9207223v1 8 Jul 1992SHEP–91/92–22hep-ph/9207223Heavy quark symmetry: ideas and applicationsJ M Flynn† and N Isgur‡† Physics Department, University of Southampton, Highfield, Southampton SO9 5NH‡ CEBAF, 12000 Jefferson Avenue, Newport News, Virginia 23606Abstract. This report is a combined version of two talks presented by the authors atthe Edinburgh b-physics Workshop, December 1991.

It presents the ideas of heavyquark symmetry and gives an introduction to some applications.The referencesindicate where to go for more information: they are not intended to be complete,nor do they necessarily refer to the original work on any particular subject.PACS number(s): 11.30.Hv, 13.20.-v, 13.30.CeSubmitted to: J. Phys. G: Nucl.

Part. Phys.Date: May 1992

1. The basic idea: no subtletiesConsider two hadrons each containing a single heavy quark.

To make our point, letthe heavy quark in the first hadron, Qi, have a mass of 10 kg and let the other quarkQj, have a mass of 1 kg.ւidenticalbrown muck ց····································································································•Qi(10 kg)····································································································•Qj(1 kg)As seen in their rest frames, the hadronic systems that can be built on each heavyquark out of the light degrees of freedom of QCD will be identical: we are just lookingat how QCD distributes the “brown muck” of light quarks and glue around a staticcolour charge. Since the scale of the interactions of the brown muck is set by ΛQCDwe can say that whenmQi, mQj ≫ΛQCDthe heavy quark will in fact be effectively static so that the light degrees of freedomwill be independent of the heavy flavour.

For Nh flavours of heavy quark there willbe an SU(Nh) symmetry [1] [2]. This is not a model: the solution of the QCD fieldequations will be independent of mQ as mQ →∞.

The symmetry is analogous tothe isotope effect in atomic physics — the brown muck is independent of the heavyquark mass just as the electronic structure of an atom is independent of the numberof neutrons in the nucleus.We can compare this new symmetry to the ordinary SU(3) flavour symmetryof the light quarks u, d, and s which arises since the light quark masses are smallcompared to ΛQCD. In each case one must be careful to apply the symmetry to theappropriate observables.

For example, since the light flavour symmetry arises fromthe fact that the light quark masses are near the chiral limit, and not because theyare nearly degenerate, it does not imply that pions and kaons have the same mass.In contrast, the light flavour symmetry does imply that the baryons built out of lightquarks are approximately degenerate. The pion and kaon are pseudogoldstone bosonswhose masses vanish as the quark masses vanish, whereas the baryon masses have afinite limiting value.

Similarly, the new symmetry among the heavy quarks, c, b and tarises because they are much heavier than ΛQCD. If the b and c quarks weighed 10 kgand 1 kg, the brown muck distributed around them would obviously look the same.The heavy quark symmetry doesn’t say that hadrons containing a single b or c quarkhave the same mass (these masses don’t approach a finite value in the heavy quarklimit), but it does say that if you line up the lowest energy states (at around 10 kgand 1 kg respectively), the spectra will then match.1

The strange quark mass is actually not very small compared to ΛQCD, so there aresizeable corrections to the predictions of light quark symmetry (about 30%). Likewise,the deviations from heavy quark symmetry will be biggest for the charm quark whosemass is not extremely large compared to ΛQCD: one would expect deviations of aboutΛQCD/mQ, or 10% for charm quarks and 3% for b quarks.

Heavy quark symmetrywould be best for the top quark, but it decays so fast via weak interactions that wedo not expect to see top hadrons.The new symmetry is of an unfamiliar kind. In the two flavour example abovewe saw the symmetry when both heavy quarks were at rest.

By boosting we can seethat the heavy flavour symmetry will apply between any heavy quarks of the samevelocity, not the same momentum.····································································································•bSU(2) flavour−→····································································································•c→⃗v→⃗v⇓pµb ̸= pµcThat is, the SU(Nh) maps |HQi(⃗v, λ)⟩↔HQj(⃗v, λ).As a result, heavy quarksymmetry is a symmetry of certain matrix elements, not a symmetry of the S-matrix.For example, the symmetry can relate form factors in spacelike regions to those intimelike regions.In fact there is much more symmetry than we have mentioned so far. Since theheavy quark spin decouples like 1/mQ, just as in atomic physics, in the heavy quarklimit the brown muck doesn’t care about the spin:ւidenticalbrown muck ց····································································································•Qi ↓····································································································•Qj ↑The SU(Nh) flavour symmetry becomes an SU(2Nh) spin-flavour symmetry.

This isreminiscent of the old SU(6) quark model, but this time it is exact in the mQ →∞limit. The SU(2Nh) is also like Wigner’s SU(4) in nuclear physics.1.1.

SpectroscopySince the spin of the heavy quark decouples, we should find states occurring in doubletscorresponding to the two possible orientations of the heavy quark spin.We can2

classify states using the heavy quark spin ⃗SQ and the remaining “spin” (combined spinand orbital angular momentum) of the brown muck, which are both good quantumnumbers, and which combine to make the total spin of the state,⃗S = ⃗SQ + ⃗Sℓ.Hence we should find degenerate doublets characterised by the spin sℓof the brownmuck, with total spin sℓ± 12 (unless, of course, sℓ= 0).Of course, heavy quarksymmetry can’t tell us which sℓquantum numbers will be associated with whichstates in the spectrum: this is a dynamical issue. In nature we observe (as predictedby the naive quark model) that the lowest-lying mesons with Q¯q quantum numbershave sπℓℓ= 12−(i.e., the spin and parity of an antiquark).

Therefore the degenerateground state mesons have JP = 0−and 1−and are the B and B∗or D and D∗mesons. Similarly, for baryons we observe, as expected in the quark model, that thelightest states with Qqq quantum numbers have zero light spin (sπℓℓ= 0+) giving theΛb or Λc baryons, whilst light spin of one (sπℓℓ= 1+) gives the degenerate ΣQ andΣ∗Q baryons with JP = 12+ or 32+.

These predictions don’t depend on a valence quarkapproximation, or the assumption that sℓis dominated by the light quark spins, butthey do depend on the identification of the sℓmultiplets with the physical states.As mentioned above, the heavy flavour symmetry tells us that if we line up theground states, corresponding to subtracting the mass of the heavy quark, then thespectra built on different flavours of heavy quark should look the same. The splittingsare flavour independent, although the overall scale is not.

This is illustrated in figure 1.The four strong transitions between any two pairs of doubly degenerate states,occurring via the emission of light hadrons, will be related just by Clebsch-Gordancoefficients. For example, the following factors relate transitions from D1 and D∗2 toDπ and D∗π states:relative coefficientD∗2→Dπp2/5[D∗π]S0[D∗π]Dp3/5relative coefficientD1→Dπ0[D∗π]S0[D∗π]D1The double degeneracy of the states is lifted at order 1/mQ where the first spindependence operates.

The prediction is that the splitting is 1/mQ times a function atmost logarithmic in mQ. For the vector and pseudoscalar mesons, if you approximatemQ by the average, (mV + mP )/2, you predict m2V −m2P should be roughly constant.Experimentally, this is very well satisfied for B, D and even K mesons.m2B∗−m2B=0.55 GeV2m2D∗−m2D=0.56 GeV2m2K∗−m2K=0.53 GeV23

Initial lattice calculations have been done [3] to find the B–B∗splitting using heavyquark methods. However, even including a perturbative matching correction [4] to getfrom the lattice result to the continuum value, the result, (0.19 ± 0.04 −0.07) GeV2,is still about one third of the experimental value.1.2.

Current matrix elementsConsider the matrix element of the b-number current between a B meson of velocity⃗v and one of velocity ⃗v ′. Compare this to the matrix element of the current cγµbbetween a B of velocity ⃗v and a D meson of velocity ⃗v ′.B(⃗v )B(⃗v ′)····································································································•bbγµb−→····································································································•b····································································································•bcγµb−→····································································································•cB(⃗v )D(⃗v ′)The heavy flavour symmetry says the brown muck doesn’t know the difference, sinceit cares only about the velocity of the colour sources carried by the heavy quarks.

Inequations:⟨B(⃗p ′B)| bγµb |B(⃗p )⟩= FB(tBB)(p + p′B)µ⟨D(⃗p ′D)| cγµb |B(⃗p )⟩= f+(tDB)(p + p′D)µ + f−(tDB)(p −p′D)µwhere p′Xµ = mXv′µ and tXB = (p −p′X )2. The symmetry says that f± are relatedto FB.

We find, equating coefficients of v and v′ and removing the trivial effects ofthe heavy quark masses from the normalisation of states,f±(tDB) = mD ± mB2√mDmBFB(tBB),with tDB = (mB −mD)2 + tBB mD/mB. Furthermore, since bγµb is a symmetrycurrent, counting b-number, the absolute normalisation of FB is known at v = v′ ortBB = 0.

Hence we know f± at the “zero-recoil” point tDB = (mB −mD)2 where aB at rest decays to a D at rest.4

The spin symmetry of the heavy quark theory lets us say even more. We canconsider the matrix element of a current cΓb, where Γ is any Dirac matrix, betweenstates where the b and c quarks have any spin, and relate it to FB.B(⃗v )B(⃗v ′)····································································································•b ↑bγµb−→····································································································•b ↑····································································································•b ↑cΓb−→····································································································•c ↓B(⃗v )D(⃗v ′) or D∗(⃗v ′)FB contains the non-perturbative information on the response of the brown muck toa change in the velocity of the colour source from v to v′.

The spin-flavour symmetrytells us that we can use any current to kick the heavy quark and change its velocity(and spin and flavour).1.3. One subtletySo far we have been economical with the truth.

The preceding arguments apply in alow energy effective theory with a cutoffµ, withΛQCD ≪µ ≪mQj ≤mQi.Momenta above mQ, however, probe non-static heavy quarks Q. The results discussedabove applied to the current J in the low energy theory in which momenta larger thanµ are cut off.

This current is related to the full current j according tojjiν = CjiJjiν + O(1/mQ) + O(αs/π).That is, since the effective theory and the full theory differ at high energy there is acalculable perturbative QCD matching between the full and effective currents. Thismatching gives a correction factor Cji between the two currents, as well as generatingextra operators in the effective theory which match to the full theory current (theαs/π terms).

There are also additional corrections of order 1/mQ for heavy quarkswhich are not infinitely massive.5

If you think of scaling down from very high energy to low energy the picture lookslike this:both “massless”Qi↕different evolutionQjboth staticµΛQCDAt scales above the mass of both quarks Qi and Qj, the full vector and axial vectorcurrents are partially conserved and so have zero anomalous dimension. The quarksare roughly “massless” and there is no contribution to Cji in this region.

Once wecome below mQi, however, the i quark is regarded as heavy whilst the j quark isnot, so in this region the current is not conserved, and Cji is different from unity [5][6] [7]. Once we move below mQj, both quarks are heavy.

Again the current is notconserved unless the quarks have the same velocity, in which case they are relatedby the heavy flavour symmetry. Hence in the low energy region there is a velocitydependent contribution to Cji which reduces to 1 if vj = vi.

We will see this in moredetail below.2. Heavy Quark Effective Field TheoryIn this section we describe how the ideas of heavy quark symmetry [1] [2] can beembodied in a low energy heavy quark effective field theory (HQET) [8] [9].

Thiswill allow us to derive Feynman rules for the heavy quarks and give a recipe fordoing calculations.See the TASI Summer School lecture notes by Georgi [10] formore details. (Incidentally, several other reviews of heavy quark symmetry and itsapplications are available: see [11] [12] [13] for example).First consider a bound state with velocity vµ, and mass MQ, which contains asingle heavy quark (or antiquark) together with some brown muck.

The momentumof the bound state isP µ = MQvµ.For a heavy quark we expect the quark mass mQ to be nearly equal to the boundstate mass, mQ ≈MQ, with the difference independent of mQ. We also expect theheavy quark to carry nearly all of the momentum of the bound state, although thebrown muck will carry a small momentum qµ.

We can write an equation for the quarkmomentum, pµ,pµ = P µ −qµ = mQvµ + kµ,where we define the residual momentum, kµ,kµ = (MQ −mQ)vµ −qµ.6

The four velocity of the heavy quark is,vµQ = pµmQ= vµ + kµmQso that the velocity of the bound state and the quark are the same in the heavy quarklimit. As mQ →∞the heavy quark is nearly on-shell and carries nearly all of thebound state’s momentum.The QCD interactions do not change the heavy quark’s velocity at all.

Any kinksin the trajectory of the heavy quark must be caused by external, non-QCD, agencieslike weak or electromagnetic interactions.2.1. mQ →∞in strong interaction diagramsFirst look at the spinor u for a heavy quark.Since the final momentum in anystrong interaction diagram in the low energy effective theory will differ from the initialmomentum by an amount much less than mQ, the heavy quark spinor satisfies /vu ≃u.Now consider the usual fermion propagator,i/p −mQ.Again, let pµ = mQvµ + kµ and look at the limit of large mQ to see that,i/p −mQ= i(/p + mQ)p2 −m2Q= i(mQ/v + /k + mQ)2mQv·k + k2≈iv·k1 + /v2.The (1+ /v)/2 projection operator can always be moved to a spinor u satisfying /vu = u(since, as we will see below, /v commutes with the heavy quark-gluon vertex), so wereplace the projector by 1.

Hence we have a rule for replacing propagators of heavyquarks according to:−→i/p −mQiv·kFor the vertex between a heavy quark and a gluon, observe that it will alwaysoccur between propagators or on-shell spinors, so we can sandwich it between (1+/v)/2projectors, and use1 + /v2γµ 1 + /v2= vµ 1 + /v27

to obtain the replacement rule:·······································································−→········································································−igγµ λa2−igvµ λa2where we have again moved the projection operator to act on a spinor where it gives 1.The new Feynman rules contain no reference to the heavy quark mass so they showexplicitly the symmetry under change of heavy quark flavour. Similarly, there are noγ-matrices in the Feynman rules, so the heavy quark spin symmetry is also apparent.In the static theory, where we expand around v = (1, 0, 0, 0), the propagator andvertex become the nonrelativistic propagator and charge density coupling respectively,iv·k→ik0 =iE −mQ,−igvµ λa2→−igδµ0 λa2 .2.2.

A systematic expansionWe would like to have a low energy effective theory which will allow us to incorporateαs and 1/mQ corrections systematically [9]. For each velocity vµ, the Feynman rulesabove are those arising from a LagrangianLv = iQvvµDµQvwhere Dµ = ∂µ −igGµaλa/2 is the colour covariant derivative, and the field Qv isrelated to the ordinary heavy quark field Q byQ = exp(−imQv·x)Qv,/vQv = Qv.We have done this for the heavy quark only.

There is a similar procedure to incorporatethe heavy antiquark. The antiquark is different only in being in a conjugate colourrepresentation.

There is not enough energy to create heavy quark-antiquark pairs,so the quark and antiquark fields are independent in the HQET (see [10] for moredetails).The above applies at lowest order in 1/mQ. Corrections suppressed by inversepowers of the heavy quark mass can be straightforwardly incorporated.

For example,at dimension five there are two new operatorsQv(iD)2 −(iv·D)22mQQv,QviσµνDµDν2mQQv,8

which are the heavy quark kinetic energy and a colour magnetic moment term,respectively. The appearance of the σµν in the magnetic moment term is the firsttime anything has distinguished the heavy quark spins.

Hence, the magnetic momentoperator will be responsible for splitting the vector and pseudoscalar heavy-quarkmeson masses at order 1/mQ.The underlying spin and flavour symmetries are easy to see in the effectivelagrangian,L =X⃗vNhXj=1iQjvv·DQjv.For each ⃗v we can rotate any spin component of any flavour of heavy quark intoanother, giving the SU(2Nh) spin-flavour symmetry. Lorentz transformations mix upthe different velocities of heavy quark.

The overall symmetry has been christened“SU(2Nh)∞⊗Lorentz” by Georgi.2.3. Relation to QCDBy construction, the heavy quark effective theory (HQET) described in the twosteps above reproduces the low energy behaviour of QCD: we build the heavy theorydemanding that it give the same S-matrix elements as QCD.

This means that the naiveheavy quark limit must be corrected for the effects of high energy virtual processes.For example, the weak flavour changing b to c current is corrected at order αs inboth QCD and the HQET. The difference between these corrections tells us how wemust modify the coefficient of the HQET current, as well as possibly introducing newstructures in the HQET current, so that the HQET reproduces the physics.We illustrate for the case of the current bγµc.The γµ in the QCD current isreplaced in the HQET current by [14]γµ −→IΓµ =1 + C0αsπγµ + αsπXiCiΓµiso there is a new strength for the naively matched γµ together with new structures Γµi(such as vµ and v′µ, where v and v′ are the heavy quark velocities).

We say that we“match the low energy approximation to the full theory”. Diagrammatically, the twosets of diagrams shown in figure 2 are matched to determine IΓ to order αs (the figureillustrates the case where both quarks are treated as heavy in the effective theory).If mb ≫mc ≫µ, the new coefficient of the naive γµ turns out to have the formshown above withC0 = ln mbmc−43[w r(w) −1] ln mcµwhere w = v·v′ andr(w) = ln(w +√w2 −1 )√w2 −1.9

The leading logs can be summed by the renormalisation group with the result [15][16],IΓµ = Ccbγµ + αsπXiCiΓµiwithCcb =αs(mb)αs(mc)−6/25 αs(mc)αs(µ)8[w r(w)−1]/27.The µ-dependence in Ccb is cancelled by the µ-dependence of the non-perturbativefunction describing the brown muck transition from v to v′. The two factors in Ccbcorrespond to the renormalisation of the current between mb and mc and then betweenmc and µ, respectively.

As advertised earlier, we see explicitly that the second factoris 1 when w = 1, in which case the heavy flavour symmetry relates the heavy b andc quarks, so that the current is conserved and has no anomalous dimension. Strictlyspeaking, if we perform the matching at order αs we should use the two loop anomalousdimension for the renormalisation group scaling.

Then we match once at the b quarkscale, where the b quark becomes heavy, scale between mb and mc and match againat mc where the c quark becomes heavy. The anomalous dimension calculations havenow been taken to two loops [17] [18], so the matching and scaling can be done.3.

Some ApplicationsWhenever a symmetry can be identified it gives us calculational power. By relatingvarious matrix elements, and fixing the absolute normalisation of some, heavy quarksymmetry enhances our predictive ability, allowing us in some cases to finesse thedifficulties of understanding hadronic structure.

This is analogous to pion, kaon, andeta physics where the effective chiral lagrangian can be used to extract systematicallythe consequences of the pattern of chiral symmetry breaking. For heavy quarks wehave identified a new symmetry and have developed the heavy quark effective theoryto calculate its predictions.Heavy quark ideas also help us to model b quarks in lattice calculations, givingaccess to the actual values of further matrix elements.The problem with puttingb quarks on the lattice by conventional methods is that their Compton wavelengthis smaller than the lattice spacing, and the b quarks “fall through”.

Heavy quarksymmetry allows us to extract the b mass dependence, leaving an effective theorywithout such a large mass scale, which can be modelled on present day lattices. Theseideas were discussed elsewhere at this workshop, and we refer readers to the latticegroup’s contributions.

Here we will concentrate on one area of great potential forheavy quark ideas: constraining the CKM matrix using b-physics.10

3.1. Determining VcbOne of the first applications of the ideas of heavy quark symmetry was to semileptonicB meson decays and the extraction of the b to c mixing angle [19] [1] [2].Vcb can be determined from semileptonic decays of B mesons to D and D∗mesons.There are altogether six form factors in the two decays,⟨D(p′)| V µ |B(p)⟩= f+(tDB)(p + p′)µ + f−(tDB)(p −p′)µ,⟨D∗(p′, ǫ)| Aµ |B(p)⟩= f(tDB)ǫ∗µ + a+(tDB)ǫ∗·p(p + p′)µ + a−(tDB)ǫ∗·p(p −p′)µ,⟨D∗(p′, ǫ)| V µ |B(p)⟩= ig(tDB)ǫµνλσǫ∗ν(p + p′)λ(p −p′)σ.In these equations, V µ = cγµb, Aµ = cγµγ5b and tDB = (p −p′)2.

In section 1.2 wesaw how the heavy flavour symmetry related the vector current matrix element to thatof the b-number current. Now the spin symmetry relates all the B →D and B →D∗matrix elements, so they can each be expressed in terms of one universal functionξ(w), where w = v ·v′.

This function is the same for any heavy quark transition,Qi →Qj with the same brown muck. It describes the response of the brown muck tothe change in velocity of the colour source from v to v′.

Furthermore, when v = v′there is a flavour symmetry between the two heavy quarks. Then the current causingthe transition is a symmetry current so the normalisation of ξ(w) is fixed at the pointw = 1 which is maximum tDB.

We can take the normalisation to be ξ(1) = 1. For Bto D(∗) decays this point is the “zero recoil” point where a B at rest decays to a D orD∗at rest.The relations of the form factors to ξ come out as follows:f± = CcbmD ± mB2√mBmDξ(w),g = a+ = −a−= Ccb12√mBmDξ(w),f = Ccb(w + 1)√mBmD ξ(w).The factor Ccb is the perturbatively calculable correction to the strength of the currentin the HQET that we described above.

For details see [16].If experimental measurements can be reliably extrapolated to the zero recoil pointthen we can determine the CKM matrix element Vcb. Alternatively, we can actuallycalculate ξ(w) on the lattice, by looking at the B to D(∗) matrix element as a functionof the velocity transfer between the heavy quarks.3.2.

Determining VubThe idea behind using heavy quark symmetry to extract the b →u mixing angle is11

illustrated by the following picture [20]:B−ρ0····································································································•buγµb Vub−→····································································································•uSU(2) = HQS↕↕SU(2) = isospin····································································································•cdγµc Vcd−→····································································································•dD0ρ−The heavy quark symmetry relates B →ρ0 to the unphysical D →ρ0 matrix elements.However, light quark isospin relates ρ0 and ρ−, and the decay D0 →ρ−is determinedby the known CKM matrix element Vcd. This means that we can close the loop inthe diagram above to fix Vub from the B−→ρ0 decay.

Note that the two ρ states onthe right hand side above are “pure brown muck”: the heavy quark symmetry relatesmatrix elements for b and c quarks surrounded by the same brown muck to go to thesame pure brown muck states.The matrix elements of the vector and axial vector weak currents between B or Dand π or ρ involve six form factors (defined like those in B →D(∗) decay). Since theheavy quark symmetry relates states of the same velocity, one difficulty is that theallowed range of momentum transfer in the D decays does not allow us to cover thewhole range for the B decays of interest.

In its purest form, the above extraction musttherefore be done in the region of the B →ρ Dalitz plot where the ρ momenta do notexceed those available in the D →ρ Dalitz plot. Another potential difficulty may beillustrated by the example of the f+ form factor.

The value of this form factor in Bdecay is determined by both f+ and f−for D decay, but the f−contribution to the Ddecay rate is suppressed by a factor of (mℓ/mD)2, where mℓis the mass of a chargedlepton. Since the decay to the τ lepton is not kinematically allowed, f−will be verydifficult to obtain from D decay.

Fortunately, one can show that in the heavy quarklimit there are enough relationships amongst heavy-to-light form factors to overcomethis lack of data [20]. In this case, f−= −f+ up to corrections of order ΛQCD/mc.3.3.

Rare B decaysRare B decays are expected to be a good probe of new physics, but if we are to seenew physics we had better know the standard model expectation first. HQET canhelp us [20] by relating the matrix elements of interest, such as in B →Ke+e−, to12

more easily measured processes like D →K−e+νe. In fact for these examples we needthe matrix element of the vector current between the heavy and light pseudoscalarmeson states.

Using light quark flavour symmetry, this is the same problem as lookingat B →π or D →π matrix elements considered in section 3.2.The rare decays B →K∗γ and B →K∗e+e−have contributions from a transitionmagnetic moment operator,sLσµνbR.We can use the heavy quark spin symmetry to relate the matrix elements of thisoperator to those of a current. Set µ = 0 and ν = i and observe that,σ0i ∝[γ0, γi] = −2γiγ0.In the rest frame of the heavy b quark, γ0b = b, so, in the rest frame we find,sLσµνbrelated to−→sLγibL.Furthermore, the heavy flavour symmetry allows us to relate the current to sLγicL, sothat B →K∗γ is related to D0 →K−e+νe.

One problem here is that in B →K∗γ,the K∗has a fixed momentum outside the kinematic range of the corresponding kaonin the semileptonic D decay (this was discussed at the workshop by A. Ali).3.4. Heavy baryon weak decaysStates in which the brown muck has spin sℓ= 0 form spin-1/2 baryons.

The heavyquark spin symmetry relates up and down spin states of the baryon, so it will relatebaryon form factors among themselves.As mentioned in section 1.1, the lowest-lying Qqq state containing a single heavyquark is expected to be a ΛQ with sπℓℓ= 0+. The simplest example is the Λ wherea strange quark is bound to an isospin zero sπℓℓ= 0+ light quark state.The Λchas been observed with a mass of 2285 MeV, but the corresponding Λb is yet to beconfirmed [21] [22].

Heavy quark symmetry tells us that the mass splitting of thepseudoscalar meson and the baryon is independent of the heavy quark mass in leadingorder in the HQET, so we expect mΛb = mB + mΛc −mD.The rate for the semileptonic decay Λb →Λceν is given in terms of six form factors:ΛQj(p′, s′) V µ |ΛQi(p, s)⟩= us′(p′)F ji1 γµ + F ji2 vµ + F ji3 v′µus(p)ΛQj(p′, s′) Aµ |ΛQi(p, s)⟩= us′(p′)Gji1 γµγ5 + Gji2 vµγ5 + Gji3 v′µγ5us(p)where p = mQiv and p′ = mQjv′. Heavy quark symmetry implies that,F ji1 = Gji1 = Cji η(w)13

where Cji is the same renormalisation factor we discussed earlier, arising frommatching the HQET to QCD: it depends only on the heavy quarks. The function η(w)is a universal (brown muck dependent) function of the velocity transfer, w = v·v′.

Theremaining form factors are zero in the heavy quark limit.Heavy quark symmetry makes the same prediction for decays of Qsu and Qsdbaryons, Ξb →Ξceν, and similar predictions for the decay of Qss baryons,Ωb →Ωceν, Ω∗ceν.One can also prove that the following decays are forbidden in the heavy quark limit(ΣQ is a Quu or Qdd baryon):Λb →ΣceνΛb →Σ∗ceνandΞb →Ξ′ceνΞb →Ξ∗ceνIn each case we find a common QCD correction to the matrix element, determined bythe renormalisation of the current which changes b into c, together with a function ofw = v·v′ which contains the response of the brown muck [23] [24].Heavy quark symmetry can be applied to some purely hadronic decays.Forexample, Λb →ΛcDs can be related to Λb →ΛcD∗s, since the underlying process,b →ccs involves three heavy quarks (the c and c are independent in the HQET) [25].3.5. FactorisationAttempts have long been made to justify factorisation in two body decays ofpseudoscalar mesons.Factorisation means that the matrix element for B →Dπ,for example, can be separated as,⟨Dπ| dγµucγµb |B⟩≈⟨D| cγµb |B⟩⟨π| dγµu |0⟩.This cannot really be true since the two sides have different renormalisation pointdependence.

However, with some extensions of HQET ideas it has been possible toprove [26] that,Γ(B →Dπ) = 6π2f 2πA2 dΓ(B →Deν)dm2eνm2eν=m2πwith corrections of order ΛQCD/mc.Here A ≈1.15 is a renormalisation groupmatching factor. This relation agrees well with experiment.The same method suggests that factorisation need not hold in B →ππ orB →DD, but it does suggest a test: factorisation should hold for B →Dππ whenthe two final state pions are collinear.14

Factorisation and heavy quark symmetry give absolute predictions for ratios ofdecay rates, for example, Γ(B →Dπ)/Γ(B →D∗π). The order αs(mb)/π corrections(both factorisable and non factorisable) to this result are small [27], although theirprecise value is not well determined because of cancellations.3.6.

A Bjorken sum-rule for ξ(w)Consider the following picture in which a b quark with velocity v surrounded by brownmuck is kicked by a cγµb current, giving a c quark moving at velocity v′. The brownmuck must rearrange itself and reform with some surrounding the moving c quark,and some possibly left behind.

Clearly we have a set of possible outcomes where the cquark has all possible brown muck configurations reachable from the initial one, eachoutcome having an associated probability. This will give us a sum rule [28] [29].····································································································•bcγµb−→······················································································································································→•cSince the brown muck cannot change the heavy quark velocity in the heavy quarklimit (the “velocity superselection rule”) one can obtain the following sum rule:Rate[b(⃗v ) →c(⃗v ′)] =XXcRate[B(⃗v ) →Xc(⃗v ′)].The inclusive rate for a b quark of velocity ⃗v to go to a c quark of velocity ⃗v ′ isobtained by summing over all possible states containing a c quark of velocity ⃗v ′ whichcan be reached from a B containing a b quark of velocity ⃗v.

That is, the heavy quarkis undisturbed by the “splash” of the brown muck. Explicitly:1 = w + 12|ξ(w)|2+ 12(w −1)2(w + 1)nmax(µ)Xn=1|ξ(n)(w)|2+ 2(w −1)mmax(µ)Xm=1|τ (m)1/2 (w)|2+ (w −1)(w + 1)2pmax(µ)Xp=1|τ (p)3/2(w)|2+ · · ·The successive lines on the right hand side refer first to the sum over D and D∗followed by the sums over other states with light spin sπℓℓ= 12−, then sπℓℓ= 12+ and15

sπℓℓ= 32+ and so on. The heavy quark symmetry makes the RHS doable.

For example,each sπℓℓ= 32+ multiplet has eight form factors all related to a single function τ3/2.The interpretation is that as rate disappears from the “elastic” channel (D(∗)) asyou move away from w = 1, it appears in the excited states. This is analogous tothe Cabibbo–Radicati sum rule for the proton form factor.

If you write the universalfunction ξ asξ(w) = 1 −ρ2(w −1),where ρ is the slope or “charge radius”, you can prove thatρ2 = 14 +X τ (m)1/2 (1)2+ 2X τ (p)3/2(1)2,i.e., the compensation is only by the sπℓℓ=12+, 32+ states, and ρ2 ≥14.In theharmonic oscillator quark model, the compensating states are the lowest excitationsof the brown muck, obtained by combining a light quark of spin- 12 with an orbitalangular momentum of 1. The sπℓℓ= 32+ states are known and ρ2 is now measured.This sum rule should soon give an interesting constraint on theory/experiment.There is an analogous sum rule for the Λb [30]:η(w) = 1 −ρ2Λ(w −1) + · · ·whereρ2Λ = 0 +X σ(n)(1)2.This time, the one-quarter is replaced by a zero and the sum is over the states withsπℓℓ= 1−, which are again the lowest excitations in the quark model.3.7.

ΛQCD/mQ corrections vanish at w = 1Heavy quark symmetry makes many predictions in the mQ →∞limit. However, sincethe quarks to which we wish to apply these ideas are not really infinitely massive, wemust ask about ΛQCD/mQ corrections.

Fortunately, the effective field theory organisesthese correction effects, so there is some predictive power from the symmetry evenwhen it is broken [31]. This is reminiscent of the Gell-Mann–Okubo and Coleman–Glashow formulas.The most surprising case so far is for the form factors in the matrix elements forthe heavy baryon decay Λb →Λc [32].

Just one new constant, ∆m is required, andthe universal function η(w) is still normalised, η(1) = 1. The situation is summarised16

as follows:leading orderwith Λ/mQ correctionsF1η(w)η(w)(1 + ∆)F20−η(w)∆F300G1η(w)η(w)G20−η(w)∆G300The 1/mQ corrections are given in terms of ∆= ∆m/[mc(1 +w)]. Perturbative QCDcorrections will add a multiplicative correction factor Ccb to all terms, as we discussedabove.

Then the result is good up to corrections of orders, (∆m/mc)2, ∆m/mb (onlythe c-quark was given a finite mass) and αs(∆m/mc).The fact that the form factor G1 retains a known normalisation at zero recoilw = 1 offers the possibility of extracting Vcb with reasonably small uncertainties. Infact, at zero recoil there are no 1/mQ corrections to the matrix elements of the vectorand axial vector currents [33].

To understand why, recall that the charges associatedwith V µ and Aµ are symmetry generators when v = v′, and b quarks can be rotatedinto c quarks of the same velocity. More explicitly, write the ground state baryon forfinite mQ as a perturbative sum over states in the mQ →∞limit:|ψ0mQ⟩= |ψ0∞⟩+Xn̸=0|ψn∞⟩⟨ψn∞| O(∆m/mQ) |ψ0∞⟩(En −E0).Since the mQ →∞states are eigenstates of the charges, we have ⟨ψ0∞| Q |ψn∞⟩= δn0,so that,ψ0m2 Qψ0m1= 1 + O(∆m/mQ)2.For semileptonic B →D(∗) decays things are not so simple.

The new constant∆m enters and only two of the six form factors are unaffected by 1/mQ corrections.In particular, the form factor proportional to vµ + v′µ in the vector current matrixelement is unaffected. However, experiments measure a form factor proportional topµ + p′µ = mbvµ + mcv′µ (the form factor accompanying pµ −p′µ picks up a factor ofthe lepton mass when contracted with the leptonic part of the matrix element for thedecay), which contains an admixture of a corrected form factor.

Fortunately, at zerorecoil, the B →D∗matrix element depends solely on an uncorrected form factor, sothis may be the best way to extract Vcb from semileptonic B decays. Initial analysisby Neubert gives [34],|Vcb|τB1.18 ps= 0.045 ± 0.007which is slightly more precise than values extracted using model dependent analyses,and is model independent.17

4. Status and prospectsThe number of papers on heavy quark symmetry produced in the last two years is inthe hundreds.

New papers on this subject appear nearly every day. Corrections forfinite heavy quark masses and for perturbative QCD matching have been classifiedand calculated, and some phenomenology done.

The absence of 1/mQ corrections atv = v′ is possibly the most significant recent development.It now appears that it may be useful to think of hadrons containing a single heavyquark as the “hydrogen atoms of QCD”. There are many advantages to this limit:relativistic effects are simplified and the heavy quark acts as a pointlike probe of thelight “constituent quarks”.

The symmetries and rigorous results of the heavy quarklimit can be used for consistency checks (in the form of “boundary conditions”) onmodels.In a more practical vein, new data from beauty and charm factories should allowus both to test heavy quark symmetry and obtain tighter limits on standard modelparameters. Heavy quark ideas applied to lattice calculations may allow the theoreticalprediction of strong interaction matrix elements which were unavailable before.Ironically, heavy quarks may prove to be an essential tool in finally helping us tounderstand the nature of the brown muck of QCD.18

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......←levels characterised by sℓtotal spin: s = sℓ± 1/2←strong transitions related byClebsch–Gordan coefficientsD∗2(2460)D1(2420)D∗(2010)D(1870)B∗(5330)B(5280)←line up groundstatesD statesB statesFigure 1. Spectrum of states predicted by heavy quark symmetry∨∨∨∨∨∨∨∨+ ∨∨∨∨∨∨∨∨······································································+ ∨∨∨∨∨∨∨∨·······································································+ ∨∨∨∨∨∨∨∨·············································································································································Full QCD to order αs∨∨∨∨∨∨∨∨IΓ + ∨∨∨∨∨∨∨∨·····································································+ ∨∨∨∨∨∨∨∨·······································································+ ∨∨∨∨∨∨∨∨··········································································································································HQET to order αsFigure 2.

Matching a current between QCD and the heavy quark effective theory20

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