Lie Theory in Particle Physics

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1 Lie Theory in Particle Physics Tim Roethlisberger May 5, 8 Abstract In this report we look at the representation theory of the Lie algebra of SU(). We construct the general finite dimensional irreducible representation of the complexified Lie algebra. Using tensor product representations we find the structure behind the baryons and mesons observed in experiments in the early 96s, which leads us to the idea of the quark model. Introduction In the 95s and 96s the increase in the energies of particle accelerators led to the discovery of many unknown particles. There were so many different particles this became known as the particle zoo. The big question was whether there was an underlying structure to this particle zoo. To answer this question people tried to order the particles depending on their properties. For example the spin / particles could be ordered by plotting their hypercharge Y against their isospin I. The hypercharge of a particle is the sum of its strangeness S and baryon number B, so Y = S + B. The strangeness S, baryon number B and isospin I are all quantum numbers that can be measured by experiment. In the early 96s nine spin / particles were known. In figure we have plotted their hypercharge against their isospin. This plot already exhibits some structure, as did other such plots for particles with different spin. There was however no known particle with I =, Y =, which would complete the triangle in figure. In this report we will discuss how to find a structure behind this pattern. For this we will consider representations of the Lie algebra of SU() which will lead us to the quark model. This model describes the experimental results and led people to predict the missing particle with I =, Y =. All figures are taken from Jan B. Gutowski s lecture notes Symmetry and Particle Physics [] SU() Lie algebra representations We consider the Lie Group SU() = {U Mat UU =, det(u) = } of the special unitary matrices. Its Lie algebra, which is the tangent space at the identity, is then

2 Figure : The spin / baryons known in the early 96s. The missing particle at I =, Y = had not been discovered yet. the set of all antihermitian and traceless matrices denoted by L (SU()) = su() = {A Mat A = A, tr(a) = }. We now allow for complex linear combinations of the elements of su(), that is the complexification of su(). Since we allow complex linear combinations the new elements will not be necessarily antihermitian, but they will however still be traceless, which is exactly the defining property of the Lie algebra of SL() = {U Mat det(u) = }. We choose a basis of this complexified Lie algebra as follows: h = h = e + = e + = e + = () e = e = e = It is important to note that the elements h and h are both diagonal and they commute. The maximal commuting and diagonalisable subalgebra of a Lie algebra is called the Cartan subalgebra. In our case of sl() the Cartan subalgebra is spanned by h and h. This Cartan subalgebra is important because if d is a representation of the Lie algebra then it follows that d(h ) = H and d(h ) = H commute and thus they are simultaneously diagonalisable. This

3 allows us to find a common eigenbasis of the vector space V on which the representation acts and in turn we can label the basis states of the vector space V with the eigenvalues with respect to H and H. If v is such a basis state and H v = λ v and H v = λ v then we call (λ, λ ) the weight of the state. We are only interested in finite-dimensional representations. The first representation we look at is the adjoint representation, which will play an important role for all the other representations.. The adjoint representation The adjoint representation of an element x in the Lie algebra g acts on the Lie algebra itself and is defined as: { g g ad(x) : y [x, y] The representation property is satisfied as a consequence of the Jacobi identity of the commutator. So for su() we have an eight dimensional representation. The common eigenstates of H and H are just the basis states of the complexified Lie algebra we defined earlier. We can calculate their weights by evaluating the commutator. Doing so we find: State v [h, v] [h, v] Weight (λ, λ ) h (, ) h (, ) e + e + (, ) e e (-,) e + e + e + (, e e + e e + e (, e + (, ) e e e (, ) We can plot these states on the so called weight-plane as shown in figure. The non-zero weights of the adjoint representation are called roots of the Lie algebra and will play an important role for general representations. We will call the adjoint representation the 8 representation.. The su() subalgebras In preparation for considering the general irreducible finite dimensional representation, let us consider the following commutation relations: [h, e ±] = ±e ± [e, e ] = h [ h h, e ±] = ±e ± [e +, e ] = h h [ h + h, e ±] = ±e ± [e +, e ] = h + h

4 Figure : The eigenstates of the adjoint representation plotted on the weight-plane. The states h and h lie at the origin since they commute with each other and themselves. These are exactly the commutation relations of the complexified su() Lie algebra. The generators e m ± play the role of raising and lowering operators and h, h h and h + h are three linearly dependent matrices that play the role of the Cartan generator of su(). The eigenstates of the su() representation are thus also eigenstates of the three su() subalgebras. This puts constraints on the possible weights of the representation since we know that the eigenvalues of a finite-dimensional su() representation are half-integers. Thus for a weight (λ, λ ) we find λ Z, λ λ Z, λ + λ Z. Adding the latter two equations we find λ Z. Using the fact that in our normalization the eigenvalues of an su() representation are spaced by all the weights of an irreducible su() representation are constrained to lie on a lattice of equilateral triangles with length as depicted in figure. Now let v be an eigenstate of H and H with weight (λ, λ ). We can then look at the effect of E m ± = d(e m ±) acting on v: ( H H ) ( E± m [H, E± v = m ] [H, E± m ] ) v + E± m = (root(e m ±)) t E m ± v + ( λ ( H λ ) v H ) E± m v () So either E m ± v is zero, that is E m ± annihilates v, or it is again an eigenstate of H and H with its weight shifted by the root of e m ±. This can be depicted in the weight-plane as shown in figure. The blue lines in figure represent all the points on the weight plane which have zero eigenvalue with respect to one of the three su() subalgebras. Since representations of su() are symmetric about the zero eigenvalue the su() representations have to be symmetric 4

5 Figure : Effect of the operators E m ± acting on the state in the center of the arrows about the three blue axes. Therefore it is enough to know the structure of the representation in one of the sectors confined by the blue lines.. The general irreducible finite-dimensional representation We define the highest weight state of an irreducible finite-dimensional representation as the state w that is an eigenstate of H and H and is annihilated by all the raising operators E m +. Such a state has to exist, since otherwise one could construct infinitely many new states by repeated application of the raising operators. The highest weight state has to be unique as well, since otherwise one can find two subspaces V and V of V, by applying the different raising and lowering operators, on each of which the representation would be irreducible and thus on V the representation would not be irreducible. Since the highest weight state is a highest weight state of all three su() subalgebras we know it has non-negative eigenvalues with respect to all three of the subalgebras. Geometrically this means the highest weight has to lie in the red region of figure 4 or on its boundary. We will consider three different cases, (i) the highest weight state lies on the vertical edge of Figure 4: The highest weight state of an irreducible finite-dimensional su() representation has to lie in the red region of the weight-plane or on its boundary. the red region, (ii) it lies on the diagonal edge of the red region or (iii) it lies inbetween. 5

6 If the highest weight state lies on the vertical axis it is in the j = representation with respect to the first su() subalgebra, thus E v = and there cannot be another state to its left. We can repeatedly act on the highest weight state with one of the other lowering operators (namely m =, ) until we reach the state that we get by mirroring the highest weight state in the diagonal blue lines. In this manner we can find all the states on two edges of a triangle (see figure 5), which all have multiplicity one. These states are all highest weight Figure 5: A 5-dimensional representation with the highest weight state on the vertical axis. All states have multiplicity one. states of representations of one (or two) of the su() subalgebras. We can then proceed in the same way as before and construct the remaining states in the triangle. These states also have multiplicity one which can be shown using the commutation relations following from the matrices in equation (), and the fact that E v =. The case with the highest weight state lying on the diagonal blue line is very similar. This time the highest weight state is the j = representation with respect to the second su() subalgebra. Again we can find the states on two edges of the triangle by repeatedly applying E and E until we reach the mirrored states. The remaining states can be reached by applying E and E in combination. Again all the states have multiplicity. This case is depicted in figure 6. We now consider the case with the highest weight state lying inbetween the blue lines, that is in the red region of figure 4. This time we have E m w, so none of the lowering operators annihilate the highest weight state w. This will result in higher multiplicities for the states in the inner layers. For the outer layers we can proceed with the same analysis as before. The outer layer of the resulting hexagon (as in figure 7) can be produced by applying the different lowering operators and using the symmetry of the su() representations. Thus all the states of 6

7 Figure 6: A -dimensional representation with the highest weight state on the diagonal axis. All states have multiplicity one. the outer layer have multiplicity one. If we proceed to the next layer of the hexagon the multiplicity of the states increases by one. This happens for every layer until we reach the first triangular layer. Then all the following triangular layers will have the same multiplicity as the first triangular layer. This is illustrated in figure 7. The outer layer is colored in black and has multiplicity one. The next layer, colored in red, has multiplicity two. This continues until the first triangular layer, colored in bright blue, is reached. This layer has multiplicity 5. The next triangular layer then also has multiplicity 5. Once again we will not give the full proof for the multiplicities of states. The idea however is again to use the commutation relations between the different raising and lowering operators and derive upper and lower bounds for the multiplicities. Interested readers can find the proof in the lecture notes Symmetry and Particle Physics by Jan B. Gutowski on pages 74 to 76. To find the dimension of the general representation we can now just count up the number of states. We will call the length of the upper edge of the hexagon m and the length of the other edge n, as depicted in figure 7. Without loss of generality we assume m > n. The length of the edge of the first triangular layer will thus be (m n + ). So there are (m n + ) = (m n + )(m n + ) states in the interior triangle. Each of these states has multiplicity (n + ). There are n hexagonal layers labelled from k = to k = n. Then in the k-th layer there are ((m + (k )) + (n + (k ) )) = (m + n k + ) states, each of them 7

8 Figure 7: A 5-dimensional representation representing the general case. The different colors stand for different multiplicities of states. The length m of the upper edge of the hexagon would be 8 in this case. The length n of the other edge would be 4 in this case. with multiplicity k. Summing up we find: n (n + )(m n + )(m n + ) + k(m + n k + ) = (m + )(n + )(m + n + ) k= Taking m = or n =, this formula works for the triangular representations as well. The lowest dimensional irreducible representations have dimension d =,, 6, 8, and so on. Here the 8-dimensional representation is the adjoint representation we saw earlier..4 Complex conjugate representation If we look at a finite dimensional representation d then d(v) is a matrix and we can define d(v) = d(v) as the complex conjugate of d(v). Since d is a representation it satisfies d([v, w]) = d(v)d(w) d(w)d(v). If we take the complex conjugate of this equation we get d([v, w]) = d(v)d(w) d(w)d(v) which shows that d is again a representation. For our case of su() we can look at ih : d(ih ) = (d(ih )) = (ih ) = ih. 8

9 Using the linearity of d we can also write this as d(ih ) = id(h ) = ih. This means H = H and in the same way we find H = H. This shows that if v is an eigenstate of H and H with weight (λ, λ ) then v is an eigenstate of H and H with weight ( λ, λ ). Since the weights of our representations are real this means that the weight diagram of the complex conjugate representation is just the weight diagram of the original representation inverted in the origin..5 The, the and the The lowest dimensional irreducible representation of su() is the trivial representation denoted by. It is called the singlet and is one-dimensional; m = n = in figure 7. For all elements v su() d(v) = and hence there is just one eigenstate with weight (, ). The next representation we look at is the. It is the fundamental representation with d(v) = v and it is -dimensional; m = and n = in figure 7. We can choose a basis of eigenstates: state weight u = (, ) d = (, ) s = (, ) already anticipating the quark model we will see later. The weight diagram of the is shown in figure 8. The highest weight state is u. The next representation we consider is the complex conjugate of the which is the. Since the states in the were real we can just take the same states and, as we saw in the previous section, the weights are just inverted: state weight u = (, d = (, s = (, ) We can again plot the weight diagram which is just the inverted weight diagram of the as shown in figure 9. This time s is the highest weight state. 9 ) )

10 Figure 8: The weight diagram of the representation. The states are labeled as in the quark model..6 Tensor product representations We will now consider tensor product representations, which will in general not be irreducible. Consider the irreducible representations d and d on V and V respectively. Then d = d + d is the corresponding representation on the tensor product V = V V. We saw above that we can choose an eigenbasis of V and an eigenbasis of V. The tensor product of these basis states is then a basis of V. Let v V and v V be states with weights (p, q ) and (p, q ) respectively. Then We can do the same with d(h ) to find d(h )v v = (d (h ) + d (h ))v v = p v v + p v v = (p + p )v v. d(h )v v = (q + q )v v. So the weight of a state in the tensor product is the sum of the individual weights, that is v v has weight (p + p, q + q ). Starting with a highest weight state, which is not necessarily unique, we can find the associated set of states on which the representation is irreducible using the analysis of section.. We can then choose a new highest weight state from the remaining states and do the same again. We end up with a decomposition of the tensor product V into invariant subspaces V = W W W n, where each W i is an irreducible representation. The same can be done for a triple tensor product V = V V V where the weights are just the sum of the three individual weights. We now consider the tensor product representations that will bring us to the quark model.

11 Figure 9: The weight diagram of the representation. The states are labeled as in the quark model..6. The representation We will use the basis for the and introduced in section.5. With the analysis above we find the weights of the to be: state weight u s (, u d (, ) d s (, u u, d d, s s (, ) d u (, ) s u (, s d (, In figure the weight diagram is plotted. The shape of the 8-dimensional adjoint representation is already clearly visible. However the weight (, ) has multiplicity three and not two. If we act on the highest weight state u s with all the lowering operators we find an 8 and we are left with a single state with weight (, ), a singlet. Explicitly this state is (u u + d d + s s). This means is reducible and decomposes as = 8.

12 Figure : The weight diagram of the 9-dimensional reducible representation. The states are labeled as in the quark model..6. The representation In the representation we have = 7 states. As discussed above, the weights of these states are the sum of the individual weights of their constituents: state weight u u u (, s s s (, ) d d d (, u u s, u s u, s u u (, ) u u d, u d u, d u u (, s s u, s u s, u s s (, s s d, s d s, d s s (, d d s, d s d, s d d (, ) d d u, d u d, u d d (, u d s, u s d, d u s, d s u, s u d, s d u (, ) The highest weight state of the representation is u u u and lies on the diagonal blue line as seen in figure. We saw in section. that acting on this state with the different lowering operators gives us an irreducible -dimensional representation. We will call it the. The remaining states are plotted in figure.

13 Figure : The weight diagram of the 7-dimensional reducible representation. The states are labeled as in the quark model. Figure : The weight diagram of the representation after removing the. There are two linearly independent highest weight states. There are now two linearly independent highest weight states. We can choose one of them and act on it with the lowering operators. This gives us an irreducible 8 representation. Removing this invariant subspace as well we can take the other highest weight state and do the same thing. Again we get an 8. Finally we are left with a single state with weight (, ), the singlet. This state is 6 (s d u s u d + d u s d s u + u s d u d s). Indeed, it is the only state that is annihilated by all the raising and lowering operators. We

14 have now found that the representation is reducible and decomposes to = 8 8. The quark model Having constructed representations of su() and studied their tensor products, let us now use this to understand the structure behind the particle zoo. The spin and the spin particles can be plotted with their hypercharge Y on the y-axis and the isospin component I on the x-axis as in figures and 4. There are two more particles with hypercharge and isospin, one with spin (denoted η ) and one with spin (denoted ϕ). The spin and spin particles each can be identified with the basis states of the 8 and the in the decomposition if we set (λ, λ ) = (I, Y ). Figure : The spin particles plotted with respect to their hypercharge and isospin component. These particles are known as pseudoscalar mesons. 4

15 Figure 4: The spin particles plotted with respect to their hypercharge and isospin component. These particles are known as vector mesons. We can do the same thing with the spin and spin and 6. particles as depicted in figures 5 Figure 5: The spin particles plotted with respect to their hypercharge and isospin component. These particles are known as the baryon decuplet (including the missing Ω particle). 5

16 Figure 6: The spin particles plotted with respect to their hypercharge and isospin component. These particles are known as the baryon octet. The spin particles match the basis states of the and the spin particles match an 8 of the decomposition, again making the identification (λ, λ ) = (I, Y ). This motivates the idea that the baryons and mesons consist of smaller particles, called the quarks. Here we have encountered three different quarks, the up, the down and the strange quarks which are the basis states in the representation and have spin. The anti-quarks are the basis states in the complex conjugate representation. Baryons are particles that consist of three quarks, so mathematically they are multiplets in the representation. Mesons consist of one quark and one antiquark, mathematically they are multiplets in the representation. Were this SU() flavour -symmetry exact, it would imply that all the particles in the same multiplet have the same mass. Experimentally this is not the case and hence the SU() flavour - symmetry is only approximate. While the up (m u = MeV) and down (m d = MeV) quarks have similar masses, the mass of the strange quark (m s = MeV) is considerably higher []. Therefore the SU()-symmetry of the up and down quarks is more accurate. This is called the SU() isospin symmetry, of which I is the third component. With the emergence of the quark model there were new questions to answer. Why do two quark states not exist? Why does the ++ particle exist? In the quark model the ++ paricle should consist of three up quarks with aligned spin (since ++ is a spin particle). That is all the quarks should be in the same state, which is not allowed by the Pauli exclusion principle. To circumvent this problem a new color label for the quarks was introduced. There are three different color charges, red, blue and green. Each quark carries one of the three colors, while the anti-quark carries an anti-color. Now the three quarks in the ++ particle can each carry a different color and the problem is solved. It is observed experimentally that color charged particles do not exist. It appears that particles have to be color-neutral. An isolated quark and also a two-quark particle cannot be color-neutral thus they should not exist. This gives rise to a new symmetry, the SU() color -symmetry, which is an exact symmetry. The requirement of color-neutrality implies that particles 6

17 should transform in the singlet representation of this SU() color -symmetry. This SU() color - symmetry is the gauge-symmetry of quantum chromodynamics. 4 The discovery of the missing particle Murray Gell-Mann and Yuval Ne eman were the first to organize the baryons and mesons in this way in 96. This led to the prediction of the missing particle in figure 5 with isospin, hypercharge - and spin. The mass differences between the different isospin multiplets in the baryon decuplet also allowed them to predict the mass of the missing particle to be around 68 MeV. In 964 such a particle was discovered with almost exactly the properties predicted. This particle is the Ω particle and has mass 67.45±.9 MeV []. In 969 Murray Gell-Mann received the Nobel price in Physics for this prediction and the classification of elementary particles. There are three more quark flavours known today, the charm, top and bottom quark. They are however much heavier than the three light quarks and thus the SU() flavour - symmetry cannot be extended to an SU(6) flavour -symmetry in a meaningful way. References [] Jan B. Gutowski. Symmetry and Particle Physics [] C. Patrignani et al. (Particle Data Group), Chin. Phys. C, 4, (6) and 7 update 7

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