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# Nuclear & Particle Physics PHYS 771

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This 9 page Class Notes was uploaded by Hayley Jenkins on Thursday October 29, 2015. The Class Notes belongs to PHYS 771 at College of William and Mary taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/231184/phys-771-college-of-william-and-mary in Physics 2 at College of William and Mary.

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Date Created: 10/29/15

Young Tableaux Combining representations in SU2 is very straightforward the way we have done it by merely using the rules of angular momentum addition and Clebsch Gordan coefficients to determine the irreducible represen tationsi Of course this is possible for other groups but even for SU2 it becomes cumbersomei Remember we need to know the symmetry properties of these representations so that we can put together completely symmetric representations to create mesons or baryons since they then will combine with color in just the right way so as to make a state with the right symmetry under quark interchange way to make this simpler and generalize to any SUN group is to use Young Tableauxi This is a schematic technique to write down representations and fully understand the symmetry properties just from a diagram of boxes I suggest you read through Lie Algebras in Particle Physics by Howard Georgi to understand the basis behind this These notes will basically guide you through the rules Lets start with some group SUNi The rules are as follows for a given representation we write down a set of boxes which are leftjusti ed and ordered such that a given row cannot have more boxes that the row above it although they can have the same number Additionally for SUN we can have at most N rows but as many columns as one would want Lets be concrete for now and set N 3 Some possible representations can be visualized as DHEFEEE Okay this is not useful unless we give rules for how these correspond to given irreducible representations For now we care only about the symmetry properties of a given representation and about the dimension So rst let s explain how to write down the dimension of a given representation given its Young Tableauxi The dimension of a given tableaux A is determined by a simplelooking formu a where F is the factor determined as follows for SUN Put N in the upperleft hand box and increase by l as you go to the right and decrease by one as you go down F is just the product of all of these numbers For SU3 then the last diagram above would look like and the corresponding F would be F lt3gtlt4gtlt5gtlt6gtlt7gtlt8gtlt2gtlt3gtlt4gtlt1gt 483 840 The factor H is the number of hooks77 which we calculate as follows A hook is a line that starts at the bottom of the diagram moving verticaly upward to some point then turning right and going out of the diagrami If we call the number of boxes the hook goes through h then H is the product of all possible h7s we can make with a diagrami There are ten hooks for this diagram that we re talking about for SU3 so E Just bottom box E Bottom box and middle row E 2 right boxes middle row E 1 right box middle row E Left column top row E 2nd column top row E 3rd column top row E 4th column top row E 5th column top row 1 6th column top row So H 1421865321 11 520 and 483 840 D 11520 42 This corresponds to a representation Which has dimension 42 We can see that the fundamental representation is given by a single box DN What about the antifundamental representation Which is the complex conjugate of the fundamental It is given by N 7 1 boxes in the rst column so for SU3 E4 At this point you can see Why 5112 is special because both the 2 and the Q are represented by a single box so they are equivalent representations The singlet representation is always represented by N boxes in the rst column so in SU3 Ea Now what else can we gain from the Young Tableaux We need the symmetry properties from them The rule is simple Each box in the Young diagram corresponds to an index on a tensor which transforms in a speci c way under the group The indices which corresponds to the rows are always symmetrized and those in the columns are antisymmetrized Thus if we have a tensor a with components aijk then the YT would give us the combination azjk ajlk 7 aka 7 am If we were going to call this an SU3 representation then we would have dimension 3427 D131 8 This is actually the adjoint representation of SU3 ijk run from 1 to 3 since this is SU3 and we can see there is a mixed symmetry here it s symmetric in the rst two indices and antisymmetric in the second two Later we will see another 8dimensional representation that is also mixed and it will be the combination of these two that creates physical particles What we are doing is saying that in any group we can build a general representation by putting together more and more of the fundamental representations That is why this is single box Think of SU2 The fundamental representation is a spin12 and every other spin can be formed with this So a spin1 oject would have two spinl2 indices just put together in a very speci c way to give a spin1 object A spin 52 object would have ve spinl2 indices again in just the right arrangement to give spin52 other arrangements would give lower spin states This is what the Clebsch Gordon coef cients do for us in SU2 and we can do it diagrammatically with Young Tableaux Why now is the N the YT with N 7 1 boxes in the rst column We know why the dimension is what it is but what makes it the complex conjugate of the N Well if we lled the 3 YT with indices and said that it corresponded to a tensor a then it would correspond to something that looks like an 7 ajl So it is an antisymmetric combination of the indices ij Well in the context of color we could see where this comes from If we are talking about R GB then RR BB CC is colorless which means that R must correspond to some combination of GB and G N RB B N GR If you look at the completely antisymmetric combination of color for baryons this would give that R GB 7 BC So the complex conjugation of the fundamental rep is an antisymmetric combination like this You really need the full power of Group Theory or abstract algebra to prove this but the rule is as follows For SUN two representations are complex conjugates if when put together they form a rectangle that has a height of N the number of rows is irrelevant so long as it forms a rectangle Any Young Tableau which is of this form height of N and any number of columns is a singlet in the group and it can be formed by a speci c combination of a representation and its complex conjugate representation Now we can enumerate all sorts of representations this way but what we really would like to do is combine representations and so how do we do this with YTls Let7s look at a concrete example and that is combining the reps 3 and 3 of SU3 This would be 3 3D B The easiest way to do this is by putting the YT with the most boxes on the left Bw In the second diagram we put letters in the boxes as follows In the rst row we put as second row we put bls then c d Here its a little trivial but welll do it anyway B E Then we take the boxes with the as and put them into the other diagram to make valid SU3 YT s E EBEE The rule is that if you had more as then you never put more than one in a given column If we had bls we would then enlarge the current diagrams and continue onward For this example we are done and we have Em j And when calculating the dimensions we would read this as 3 381 So we have an 8dimension irreducible representation irrep and a singlet Lets do something less trivial We7ll combine two 87s together 3 First we put the as everywhere So from these we can get rid of the last one since there are two as in the same column and there are 4 rows which is a nono for 5113 Additionally at this stage any nondistinct diagrams we combine B3 m Now we put the bls 69 HI I I E E E 69 3 Let7s drop the YTls that are illegal F EB 69 In 69 Now we use the restriction of admissable sequences to avoid doublecountingi Admissable sequences made up of letters a 25 are those where at any point in the sequence7 Na 2 Nb 2 Nc So abcd aabcaaabcb are all admissable7 but abbc acb are not The restriction is that we read these boxes from right to left top to bottom and cut out YTls that have inadmissable sequences The sequences above would correspond to BE ME baaEBaabEBaab EB baa EB aba EB aab EB baa EB aba EB baa EB aba The following must be dropped baa and all the others are ne aaa aa ia j j Working through these dimensions we get 8 82710 881 Note that we have two 87s and two lOdimensional representations and the second 10 is the conjugate of the rst Our use of Young Tableaux will end with this We could be more systematic to be sure which reps are which but we really just want to know the symmetry properties and the dimensions of the representations We won t need the representations shown by combining two 87s but this was just a good exercise to use t emf SU3 Flavor symmetry Now lets look at the quark model in more detail We will assume that the three quarks are u d s and that there is an SU3 avor symmetry among theml This assumes that they all have the same mass or are all masslessl They don7t and arenlt and in fact many quantitative predictions just do not hold with SU3 nowhere near as well as SU2l Nonetheless it allows for a good way to organize everything Additionally predictions come out right at the 30 level which isn7t too bad for a rst pass So we have a quark triplet of SU3 u Q d s which transforms in the 3 of SU3 This is denote with the YT as Let7s create some mesons since those are easiestl We have also already done this We saw that we had 3 381 So there is an octet of states all with the same mass and a singlet state The YT for the 8 is and for the 1 Now how do these correspond to states Well this part for mesons is easy We combine the states with appropriate names as follows For spin0 we have 7T ud 7F d1 K4r u K 311 K0 d3 F0 33 There are three other qt states that we can create and these are the ones that we must determine The six above and these three form the nonet that is the 3 3 representation and one of these is a singlet This must contain the three quarks on equal footing so we write this as 1 7 7 mi dd s n This is the l or the singlet Then the other two states are given by realizing that they must be orthogonal to the 77 and we choose one by assuming that isospin is still a good symmetry so that 7r dd 0 1 7 wu The last is then making sure its orthogonal to these other avorneutral note the terminology this is not a avorsinglet but it is avorless states 77 u ddii 23E Notice this is why I said that the equivalent of the singlet now the 77 didn t exist It has to have an S term How do we really determine what the combinations are Mesons are simple The avor charged77 elds are all isolated particles like above Then we just have to determine the neutral elds and in every meson nonet one is a singlet The other two are linear 39 quot that come pond to 39 t of the diagonal generators of the group here A33 The spinzero mesons are as above but the vector spinl mesons happen to form different combinations and these things can be studied experimentally as to what the quark content of a given meson actually is Now lets move onto baryons Lets do something more concrete and combine 3 3 rst This is easy D EDIE This gives us a 3 36 The 6 is the completely symmetric combination and g is an antisymmetric combination of the quarks When adding in the third quark let s label for the YT so we know which quarks are which D D DIB D D EB Sowehave 3 3 310 B8 B81 The 10 decuplet is completely symmetric in the interchanges of all the quarks The two 87s have different symmetries as we can see The rst one is symmetric in the rst two quarks and the other is antisymmetric Finally there is the singlet which is 39 ic under the 39 t of the quarks We can do the same thing for combining the spins except that singlet can t exist because its SU2 so we have Cl Cl Sowehave 2 2 24EB2692 Now remember that we want our baryons to have completely antisymmetric wavefunctions and the antisymmetry is provided by the color piece so we want everything else to be symmetric For now we are talking about ground states so the spatial wavefunctions we assume to be symmetric This means that under the combination of spin and avor it must be symmetric One option is the 10 of avor with the 4 of spin as these are both symmetric The 10 cannot be combined with the 27s of spin because these all have some form of antisymmetry This implies that the baryon decuplet must all have spin 32 which they do Worrying from now on about the flavor combinations we will always add in the spin later and it ll be implied this gives us four states with zero strangeness the 3 1Agt 1uuugt lAgt i3 lduugt l 1A gt V 1441 ldud ludd 1M 7 W You can see now how I came up with this earlier in class The states must be symmetric under the interchange of any two quarksi The coef cients come from the Clebsch Gordon coef cients when combining a spin1 and spinl2 particle to get spin321 Here we just used the fact that isospin is a subgroup of SU3 flavori Now what about those with a single strange quark 7 1 l2 gt 7 73 lsuugt l2mgt udsgt lusdgt ldusgt ldsugt lsudgt lsdu 1 ll mm gwda dsdgt sdd The and states get the coef cients they get because if s 7gt ud then we must get a A due to the symmetry There is also the Eu dss7s and the Qsss in the decupletw The octet is more dif cult due to the mixed symmetriesi We need to take the rst of the SU3 avor group and put it with the second of the spin group then add that to the other order For the spin group we have the two mixed symmetry combinations 1 X1 7 HT HT 72 TN 1 X2 EMT 71W Of course these are for spin up77 combinationsi We would have to construct the appropriate mixed symmetry states for the two octets and combine then appropriately with X131 Therels an easier way to do this There are two options for the spinup A N udsi We can take the antisymmetric 3 of SU3 with the 3 and connect that with the antisymmetric 2 of SU2 spin so that we have ud7 duMlW 71TH Now this is symmetric under u lt gt d as long as we keep the spins coupled to the flavor To make this completely symmetric we just cyclicly permute the indices so that we have lA712gt ludsgt 7 WSW 7gt 7 l 7 gtgt WW 7 lusd l 7 gt 7 l 4 M 7 WW lt1 7gt71 7m The overall factor is for normalization Lets say uud so two u quarks and a d quark the proton This is a little trickier than the Al The easier way to create an overall symmetric state is as follows First we just take a avor state symmetric in two indices say uud Then we multiply this by something which is symmetric in spin in those same indices but has overall spin 12 This is XL The result is 1 m In this notation the spin and quark states are couple under the cyclic permutationsi The overall factor is for normalization lp 12 2 7gt 7 l 7 7 l 7 gt cyclic permutations Constructing all of the possible states is possible and has been done of course Lets see what this really can do for us thoug i One thing we notice is that SU3 is not a great symmetry because if it were the octet should all have the same mass and they do not For the decuplet for example we have A 1232 2 1385 3 1530 9 1672 So clearly these are not degenerate in mass However we can determine relationships among masses based upon the quark contenti That is if we assume that m ma so isospin is a good symmetry and ms is different we can make comments about the masses How does this work Well let s forget the rest of the interactions for now and say that there is a part of the QCD Hamiltonian that pertains to the masses Well if we sandwich this between the particle states we should get that states mass ltlemmlpgt mp Hmass is independent of spin and has the form Hmass qu q q where qu is essentially a number operator for the quark 4 like ala for the simple harmonic oscillator Now we know that there is a representation of baryons which transforms in the 8 of SU3 so we can put them in the following matrix 2 A TW 202 A p B 2 777 n 7 V520 6 7 H H x3 Clearly the differences in the masses for the decuplet are small compared to the actual masses and the same is true for the octeti Now the idea is that we could write the part of the Hamiltonian that is responsible for the strong interaction contribution to the masses can be written as Hstrong Hm H6m The rst term gives a common term to all the masses in a given representation and the second gives rise to the splittingsi We would like to assume that the second term transforms like the T8 component of the octet representation of SU3 This is because we assume that the strong interactions are independent of electric charge and baryon number so the only possibility is if Hm X I rBTBT3 Y TrBTT8B

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