QFT 3 : Problem Set 1

Transcription

1 QFT 3 : Problem Set.) Peskin & Schroeder 5. (a.) The basis for the fundamental representation of SU(N) is formed by N N traceless Hermitian matrices. The number of such matrices is = N. For SU(3) this number evaluates to = 3 = 8. (b.) For this problem one needs to do the tedious algebra of computing matrix commutators. This is most easily done using a numerical package.the structure constants are given by the algebra [t a, t b ] = if abc t c. Using the above relation, one finds for instance f 3 = f 3 = f 3 = f 3 = f 3 = f 3 =, while f c = 0 for c 3. Similarly some of the other structure functions come out to be f 47 = /, f 367 = /, f 56 = /, f 678 = 3/ and so on. In general one finds f abc = f bac = f bca for all values of a, b, and c from through 8. Therefore f abc is completely antisymmetric. (c.) The Orthogonality condition is tr(t a r t b r) = C(r)δ ab. It is easy to check that tr(t a rt a r) = C(r) = a, and that the trace reduces to zero unless a = b. for any (d.) The definition of the quadratic Cassimir operator is t a rt a r = C (r). Thus multiplying the matrices together and summing up we find C (r) = 4 3. d(r) = 3 and d(g) = 8. Also we found in part (c.) that C(r) =. Thus we verify the relationship between C(r) and C (r): d(r)c (r) = d(g)c(r)..) Peskin & Schroeder 5. The basis matrices in the adjoint representation are given by the structure constants of the algebra. (t b G ) ac = if abc, where a and c denote the row and column indices of the matrix t b G. The structure constants of SU() are given by the Levi-Civita tensor f abc = ǫ abc where a, b, and c can take values,, nd 3. Thus the basis matrices are as follows: t G = i, t G = 0 0 +i 0 0 0, t 3 G = 0 i 0 +i i 0 i To compute C(G) we need to use tr(t a G tb G ) = C(G)δab. For any value of a, tr(t a G ta G ) = C(G) =. To compute C (G) we need to use t a G ta G = C (G). Thus multiplying the matrices together and summing up we find C (G) =. As expected we find C(G) = C (G) = for SU() as expected from the general relation C(G) = C (G) = N for SU(N). 3.) Peskin & Schroeder 5.5 (a.) In the spin j representation of SU() the Casimir operator is J and the number of independent J z s represent the dimension of the algebra. It is known to us that J has as its eigenvalues C (j) = j(j + )

2 and also that J z has eigenvalues from j to j so that d(j) = j +. The adjoint representation of SU() has dimension d(g) = 3. Thus we find for a spin j rep of SU() d(j)c (j) = d(g)c(j) or 3C(j) = j(j + )(j + ). Now we consider the general group G, an irrep r of which decomposes into a sum of reps of SU(): r j i. The contribution to C(r) for the Group G may then be obtained by summing over C(j i )s which are the contributions from individual SU() reps. We have already seen that for a spin j i rep of SU() the contribution is 3C(j i ) = j i (j i + )(j i + ). Hence summing over i (note that t a r are traceless) we find: 3C(r) = 3 i C(j i) = i j i(j i + )(j i + ). (b.) In this part we use the relation from part (a.): 3C(r) = i j i(j i + )(j i + ). The fundamental rep of SU(N) transforms as one j = (-component spinor) and (N - ) j = 0 (singlets.) Thus we have: 3C(N) = = 3. Hence we find C(N) =. We know that the fundamental rep of SU(N) decomposes as: N = + (N ), where is the spin part while represents the singlets. Similarly the anti-fundamental rep of SU(N) decomposes as: N = + (N ). Hence the adjoint rep of SU(N) may be decomposed as: N N = ( + (N )) since the reps and are the same. Now we know that = 3 + where 3 is a spin rep of SU(). Hence we find that the adjoint rep of SU(N) decomposes as: N N = 3 + (N ) where we have dropped the singlets. We compute C(G) as follows: 3C(G) = 3 + (N ) 3 = 3N. Hence C(G) = N. (c.) If we ignore the singlets the symmetric and anti-symmetric reps of SU(N) have the following decompositions under SU() (This follows from observations we made in the previous part ): S = 3 + (N ) () A = (N ) () The dimensions of these reps are respectively (See Cheng. & Li., pp. 06 for the general case of a symmetric tensor with k indices): Therefore we obtain: d(s) = d(a) = N(N + ) N(N ) 3C(S) = 6 + (N ) 3 = 3N + 3 (5) (3) (4) 3C(A) = (N ) 3 (6) We may now use the relation d(r)c (r) = d(g)c(r) to obtain: (N )(N + ) C (S) = N (N + )(N ) C (A) = N Now since N N = S + A we have: tr(t a N N ) = (C (N) + C (N))d(N)d(N) = N N N = N(N ) (9) = C (S)d(S) + C (A)d(A) = N(N ) (0) Thus the identity for product representations is satisfied. (7) (8)

3 4.) Heavy Quark Lagrangian. The first part of this problem is about rewriting the Lagrangian in a convenient way, as a functional of h(x) and H(x) which are appropriate projections on the light cone. Let us start from the definitions of the new fields: and hence we also have: ψ(x) = e imv x (h + H) () h(x) = imv x + v e ψ(x) () H(x) = imv x v e ψ(x) (3) Hence the redefined Lagrangian becomes: ψ(x) = e imv x (h + H) (4) h(x) = e imv x ψ(x) + v (5) H(x) = e imv x ψ(x) v (6) L = ψ(i D m)ψ (7) = (h + H)(i D m( v))(h + H) (8) where we obtain the last term from the derivative acting on the exponential. Furthermore we know the action of v on h and H: Using Eqn.s () and () we may rewrite our Lagrangian as: + v vh = e imv x v ψ (9) imv x v + v = e ψ (0) = h () vh = H () L = h(i D)h + H(i D m)h + H(i D)h + h(i D)H (3) where we have used hh = 0 since ( + v)( v) = v = 0. At this point we notice the following, (these follow from the properties of gamma matrices): ( ± v ± v ± v γµ ) = ± v = ± v ( ±v µ + γ µ v ) ± v = ± ± v ± v vµ (4) (5) Using these we may once more rewrite the Lagrangian as follows: L = h(iv D)h H(iv D + m)h + H(i D)h + h(i D)H (6) Let us now project out the component of D µ that is perpendicular to v: D µ Dµ v D v µ (7) 3

4 Now since h vh = 0 we may replace the i D factors in Eqn. (6) by i D so as to obtain: L = h(iv D)h H(iv D + m)h + H(i D )h + h(i D )H (8) We may now write down the equation of motion for the fields as follows: (iv D + m)h = i D h (9) ( iv D)h = i D H (30) From Eqn. (9) we conclude that as m H(x) is O ( m) suppressed compared to h(x). Hence in the limit m we can neglect H(x). We may substitute H(x) from Eqn. (9) into Eqn. (8) to obtain: H = iv D + m i D h (3) ( ) L = h iv D + i D iv D + m i D h (3) = h(iv D)h h D D m h + O( m ) (33) Hence the leading term and the leading order correction to the Lagrangian are: A simplification of this form can be made using the following relations: L 0 = h(iv D)h (34) L = h D D m h (35) γ µ γ ν = g µν + [γ µ, γ ν ] (36) [D µ, D ν ] = igf a µν ta (37) D D = γ µ γ ν D µ Dν = D + [γ µ, γ ν ] D µ Dν (38) = D + 4 [γ µ, γ ν ] [D µ, D ν ] (39) where in the last line we have used the fact that the commutators are antisymmetric under interchange of µ and ν. Hence we may re-write the leading order correction to the Lagrangian (using σ µν = i [γµ, γ ν ]) as follows: (b.) L = h D m h + ghσµν F a µνt a 4m h (40) In this section we note that the transformation of a Wilson line under a gauge transformation of the field ψ given a boundary condition on the gauge parameter is as follows: ψ(x) U(x)ψ(x) (4) h(x) U(x)h(x) (4) V (x, ) U(x)V (x, )U ( ) (43) { [ where V (x, ) = V (x) = P exp ig 0 dsvµ A a µ(x + sv)t ]}. a Now, U(x) = e iαa (x)t a and α a (x) vanishes outside a finite region, so that U ( ) =. Thus we see V (x) U(x)V (x) and hence the field redefinition h(x) = V (x) h(x) implies that under the gauge transformation h(x) doesn t undergo any change. 4

5 Let us now introduce the field redefinition h(x) = V (x) h(x) into the leading Lagrangian L 0 : L 0 = hv (x)(iv D)V (x) h (44) = h(iv ) h (45) where we have used: (v D)V (x) = V (x)v. 5

6 5. a) We use a variation of the Baker-Campbell-Hausdorff expansion e A Be A = B + [A, B] +! [A, [A, B]] + [A, [A, [A, B]]] +..., () 3! with A = iπj, B = J i. For i =, this reduces to J. For i, we use the identity [J, [J, J i ]] = J i. Then e iπj J i ɛ iπj (iπ) n = (n)! J i (iπ) n+ + (n + )! [J, J i ] = cos(π)j i + i sin(π)[j, J i ] = J i. () n=0 n=0 In the spin-/ case, j i = σ i /, and we use the identity (σ i ) =. Then e iπj = cos(π/) + i sin(π/)σ = iσ, (3) and equivalently, e iπj = iσ. Using commutation relations for the Pauli matrices, for i, we have (iσ ) σi ( iσ ) = σi = J i (4) 5. b) As they are in the same representation, the field strength transforms like the gauge field (one can check this by explicitly writing F µν ), The transformation of the Yang-Mills term is F a µνt a F a µν( t a ). (5) Tr(F µνf µν ) F a µνf aµν Tr(t a t b ), (6) and Tr(t a t b ) = Tr(t at t bt ) = Tr(t b t a ) = Tr(t a t b ), where we have used hermiticity of the group generators, Tr(A T ) = Tr(A), and cyclicity of the trace. It is obvious that the (W i ) term is invariant. For the third term, i ψγ µ µ ψ G iψ T (γ ) γ 0 γ µ γ (S S) µ ψ T iψ (γ (γ µ ) T γ 0 γ ) µ ψ, (7) where we have applied the G-parity transformation in the first arrow and the transpose in the second arrow. We used S S = (S acts only on flavour indices), (γ ) T = γ = (γ ), and ignored the total derivative. For any µ, we apply the gamma matrix anticommutation relations and (γ 0 ) T = γ 0, (γ k ) T = ( ) k γ k to verify that this term is invariant. Similarly for the fourth and last term, ψγ µ A a µt a ψ G ψ T (γ ) γ 0 γ µ γ A a µ( t a )ψ T ψ (γ (γ µ ) T γ 0 γ )A a µt a ψ, (8) ψγ µ W i µj i ψ G ψ T (γ ) γ 0 γ µ γ W i µ(s J i S)ψ T ψ ( γ (γ µ ) T γ 0 γ )W i µ( ) i (J i ) T ψ. (9) Using the hermiticity of t a and (J i ) T = ( ) i+ J i, we see that these terms are invariant for any µ. 6