Quantum Field Theory Part II: Spin One Half Mark Srednicki Department of Physics University of California Santa Barbara, CA 93106 [email protected] This is a draft version of Part II of a three-part textbook on quantum field theory. 1 Part II: Spin One Half 33) Representations of the Lorentz Group (2) 34) Left- and Right-Handed Spinor Fields (3, 33) 35) Manipulating Spinor Indices (34) 36) Lagrangians for Spinor Fields (4, 22, 35) 37) Canonical Quantization of Spinor Fields I (36) 38) Spinor Technology (37) 39) Canonical Quantization of Spinor Fields II (38) 40) Parity, Time Reversal, and Charge Conjugation (39) 41) LSZ Reduction for Spin-One-Half Particles (39) 42) The Free Fermion Propagator (39) 43) The Path Integral for Fermion Fields (9, 42) 44) Formal Development of Fermionic Path Integrals (43) 45) The Feynman Rules for Dirac Fields and Yukawa Theory (10, 13, 41, 43) 46) Spin Sums (45) 47) Gamma Matrix Technology (36) 48) Spin-Averaged Cross Sections in Yukawa Theory (46, 47) 49) The Feynman Rules for Majorana Fields (45) 50) Massless Spin-One-Half Particles and Spinor Helicity (48) 51) Loop Corrections in Yukawa Theory (19, 40, 48) 52) Beta Functions in Yukawa Theory (27, 51) 53) Functional Determinants (44, 45) 2 Quantum Field Theory Mark Srednicki 33: Representations of the Lorentz Group Prerequisite: 2 In section 2, we saw that we could define a unitary operator U(Λ) that implemented a Lorentz transformation on a scalar field ϕ(x) via U(Λ) 1ϕ(x)U(Λ) = ϕ(Λ 1x) . (1) − − As shown in section 2, this implies that the derivative of the field transforms as U(Λ) 1∂µϕ(x)U(Λ) = Λµ ∂¯ρϕ(Λ 1x) , (2) − ρ − where the bar on he derivative means that it is with respect to the argument x¯ = Λ 1x. − Eq.(2) suggests that we could define a vector field Aµ(x) which would transform as U(Λ) 1Aρ(x)U(Λ) = Λµ Aρ(Λ 1x) , (3) − ρ − or a tensor field Bµν(x) which would transform as U(Λ) 1Bµν(x)U(Λ) = Λµ Λν Bρσ(Λ 1x) . (4) − ρ σ − Note that if Bµν is either symmetric, Bµν(x) = Bνµ(x), or antisymmetric, Bµν(x) = Bνµ(x), then this symmetry property is preserved by the Lorentz − transformation. Also, if we take the trace to get T(x) g Bµν(x), then, µν ≡ using g Λµ Λν = g , we find that T(x) transforms like a scalar field, µν ρ σ ρσ U(Λ) 1T(x)U(Λ) = T(Λ 1x) . (5) − − Thus, given a tensor field Bµν(x) with no particular symmetry, we can write Bµν(x) = Aµν(x)+Sµν(x)+ 1gµνT(x) , (6) 4 3 where Aµν is antisymmetric (Aµν = Aνµ) andSµν is symmetric (Sµν = Sνµ) − and traceless (g Sµν = 0). The key point is that the fields Aµν, Sµν, and T µν do not mix with each other under Lorentz transformations. Is it possible to further break apart these fields into still smaller sets that do not mix under Lorentz transformations? How do we make this decomposi- tion into irreducible representations of the Lorentz group for a field carrying n vector indices? Are there any other kinds of indices we could consistently assign to a field? If so, how do these behave under a Lorentz transformation? The answers to these questions are to be found in the theory of group representations. Let us see how this works for the Lorentz group in four spacetime dimensions. For an infinitesimal transformation Λµ = δµ +δωµ , we can write ν ν ν U(1+δω) = I + iδω Mµν , (7) 2 µν where Mµν = Mνµ is a set of hermitian operators, the generators of the − Lorentz group. As shown in section 2, these obey the commutation relations [Mµν,Mρσ] = i gµρMνσ (µ ν) (ρ σ) . (8) − ↔ − ↔ (cid:16) (cid:17) ~ We can identify the components of the angular momentum operator J as J 1ε Mjk and the components of the boost operator K~ as K Mi0. i ≡ 2 ijk i ≡ We then find from eq.(8) that [J ,J ] = +iε J , (9) i j ijk k [J ,K ] = +iε K , (10) i j ijk k [K ,K ] = iε J . (11) i j ijk k − We would now like to find all the representations of eqs.(9–11). A repre- sentation is a set of finite-dimensional matrices with the same commutation relations. For example, if we restrict our attention to eq.(9) alone, we know (from standard results in the quantum mechanics of angular momentum) that we can find three (2j+1) (2j+1) hermitian matrices , , and 1 2 3 × J J J that obey eq.(9), and that the eigenvalues of (say) are j, j+1,...,+j, 3 J − − where j has the possible values 0, 1,1,.... We further know that these ma- 2 trices constitute all of the inequivalent, irreducible representations of SO(3), 4 the rotation group in three dimensions. (Inequivalent means not related by a unitary transformation; irreducible means cannot be made block-diagonal by a unitary transformation.) We would like to extend these conclusions to encompass the full set of eqs.(9–11). In order to do so, it is helpful to define some nonhermitian operators whose physical significance is obscure, but which simplify the commutation relations. These are N 1(J iK ) , (12) i ≡ 2 i − i Ni† ≡ 21(Ji +iKi) . (13) In terms of Ni and Ni†, eqs.(9–11) become [N ,N ] = iε N , (14) i j ijk k [Ni†,Nj†] = iεijkNk† , (15) [Ni,Nj†] = 0 . (16) We recognize these as the commutation relations of two independent SO(3) groups [or, equivalently, SU(2); see section 32]. Thus the Lorentz group in four dimensions is equivalent to SO(3) SO(3). And, as just discussed, we × are already familiar with the representation theory of SO(3). We therefore conclude that the representations of the Lorentz group in four spacetime dimensions are specified by two numbers n andn, each a nonnegative integer ′ or half-integer. This turns out to be correct, but there is a complication. To derive the usual representation theory of SO(3), as is done in any text on quantum mechanics, we need to use the fact that the components J of the angular i momentum operator are hermitian. The components N of eq.(13), on the i other hand, are not. This means that we have to redo the usual derivation of the representations of SO(3), and see what changes. As we have already noted, the final result is the naive one, that the representations of the Lorentz group in four dimensions are the same as the representationsofSO(3) SO(3). Thoseuninterestedinthe(annoyinglycom- × plicated) details can skip ahead all the way ahead to the last four paragraphs of this section. 5 We begin by noting that N~2 commutes with N ; this is easily derived i from eq.(14). Similarly, N~†2 commutes with Ni†. Eq.(16) then implies that N~2, N3, N~†2, and N3† are all mutually commuting. Therefore, we can define a set of simultaneous eigenkets n,m;n,m , where the eigenvalues of N~2, ′ ′ | i N3, N~†2, and N3† are f(n), m, f(n′), and m, respectively. [Later we will see that n and n must be nonnegative integers or half-integers, and that ′ f(n) = n(n+1), as expected.] We also define a set of bra states n,m;n,m ′ ′ h | that, by definition, obey n ,m ;n ,m n ,m ;n ,m = δ δ δ δ ∆ (17) h 2 2 ′2 ′2| 1 1 ′1 ′1i n2n1 m2m1 n′2n′1 m′2m′1 ≡ 21 and n,m;n,m n,m;n,m = 1 . (18) ′ ′ ′ ′ | ih | X In eq.(18), the sum is over all allowed values of n, m, n, and m; our goal is ′ ′ to determine these allowed values. From the discussion so far, we can conclude that n ,m ;n ,m N~2 n ,m ;n ,m = f(n )∆ , (19) h 2 2 ′2 ′2| | 1 1 ′1 ′1i 1 21 n ,m ;n ,m N n ,m ;n ,m = m ∆ , (20) h 2 2 ′2 ′2| 3| 1 1 ′1 ′1i 1 21 n ,m ;n ,m N~ 2 n ,m ;n ,m = f(n )∆ , (21) h 2 2 ′2 ′2| † | 1 1 ′1 ′1i ′1 21 n ,m ;n ,m N n ,m ;n ,m = m ∆ . (22) h 2 2 ′2 ′2| 3| 1 1 ′1 ′1i ′1 21 Note that we have not yet made any assumptions about the properties of the states under hermitian conjugation. From eqs.(14) and (15), we see that hermitian conjugation exchanges the two SO(3) groups. Therefore, we must have n,m;n,m = n,m;n,m , (23) ′ ′ † ′ ′ | i h | n,m;n,m = n,m;n,m , (24) ′ ′ † ′ ′ h | | i up to a possible phase factor that turns out to be irrelevant. Compare the ordering of the labels in eqs.(23) and (24) with those in eqs.(17) and (18); a state n,m;n,m has zero inner product with its own hermitian conjugate ′ ′ | i if n = n or m = m. ′ ′ 6 6 6 Next, take the hermitian conjugates of eqs.(19) and (20), using eqs.(23) and (24). We get n ,m ;n ,m N~ 2 n ,m ;n ,m = [f(n )] ∆ , (25) h ′1 ′1 1 1| † | ′2 ′2 2 2i 1 ∗ 21 hn′1,m′1;n1,m1|N3†|n′2,m′2;n2,m2i = m∗1∆21 , (26) Comparingeqs.(25)and(26)witheqs.(21)and(22),wefindthattheallowed values of f(n) and m are real. We now define the raising and lowering operators N N iN , (27) 1 2 ± ≡ ± (N†)± ≡ N1† ±iN2† ; (28) note that (N ) = (N ) . (29) † † ± ∓ The commutation relations (14) become [N ,N ] = N , (30) 3 ± ± ± [N ,N ] = 2N , (31) + 3 − plus the equivalent with N N . By inserting a complete set of states † → into eq.(30), and mimicking the usual procedure in quantum mechanics, it is possible to show that n ,m +1;n ,m N n ,m ;n ,m = λ (n ,m )∆ , (32) h 2 2 ′2 ′2| +| 1 1 ′1 ′1i + 1 1 21 n ,m ;n ,m N n ,m +1;n ,m = λ (n ,m )∆ , (33) h 1 1 ′1 ′1| −| 2 2 ′2 ′2i − 1 1 21 where λ (n,m) and λ (n,m) are functions to be determined. By inserting + − a complete set of states into eq.(31), and using eqs.(32) and (33), we can show that λ (n,m 1)λ (n,m 1) λ (n,m)λ (n,m) = 2m . (34) + + − − − − − The solution of this recursion relation is λ (n,m)λ (n,m) = C(n) m(m+1) , (35) + − − 7 where C(n) is an arbitrary function of n. Next we need the parity operator P, introduced in section 23. From the discussion there, we can conclude that P 1MµνP = µ ν Mρσ , (36) − ρ σ P P where +1 1 µ =  − . (37) P ν 1  −   1  −    Eq.(36) implies P 1J P = +J , (38) − i i P 1K P = K , (39) − i i − or, equivalently, P−1NiP = Ni† , (40) P−1Ni†P = Ni . (41) Since P exchanges Ni and Ni†, it must be that P n,m;n,m = n,m;n,m , (42) ′ ′ ′ ′ | i | i P 1 n,m;n,m = n,m;n,m , (43) − ′ ′ ′ ′ | i | i up to a possible phase factor that turns out to be irrelevant. Taking the hermitian conjugate of eqs.(42) and (43), we get n,m;n,m P 1 = n,m;n,m , (44) ′ ′ − ′ ′ h | h | n,m;n,m P = n,m;n,m , (45) ′ ′ ′ ′ h | h | where we have used the fact that P is unitarity: P = P 1. † − Now we can take eq.(32) and insert PP 1 on either side of N to get − + λ (n ,m )∆ = n ,m +1;n ,m PP 1N PP 1 n ,m ;n ,m + 1 1 21 h 2 2 ′2 ′2| − + − | 1 1 ′1 ′1i = n ,m ;n ,m +1 P 1N P n ,m ;n ,m h ′2 ′2 2 2 | − + | ′1 ′1 1 1i = n ,m ;n ,m +1 (N ) n ,m ;n ,m . (46) h ′2 ′2 2 2 | † +| ′1 ′1 1 1i 8 In the second line, we used eqs.(43) and (45). In the third, we used eq.(40). Now taking the hermitian conjugate of eq.(46), and using eqs.(23), (24), and (29), we find n ,m ;n ,m N n ,m +1;n ,m = [λ (n ,m )] ∆ . (47) h 1 1 ′1 ′1| −| 2 2 ′2 ′2i + 1 1 ∗ 21 Comparing eq.(47) with eq.(33), we see that λ (n,m) = [λ (n,m)] . (48) + ∗ − This is the final ingredient. Putting eq.(48) into eq.(35), we get λ (n,m) 2 = C(n) m(m+1) . (49) + | | − From here, everything can be done by mimicking the usual procedure in the quantum mechanics of angular momentum. We see that the left-hand side of eq.(49) is real and nonnegative, while the right-hand side becomes negative for sufficiently large m . This is not a problem if there are two values of | | m, differing by an integer, for which λ (n,m) is zero. From this we can + deduce that the allowed values of m are real integers or half-integers, and that if we choose C(n) = n(n+1), then n is an integer or half-integer such that the allowed values of m are n, n+1, ..., +n. We can also show that − − f(n) = C(n) = n(n+1). Thus the representations of the Lorentz group in four dimensions are just the same as those of SO(3) SO(3). × We will label these representations as (2n+1,2n+1); the number of com- ′ ponents of a representation is then (2n+1)(2n+1). Different components ′ within a representation can also be labeled by their angular momentum rep- resentations. To do this, we first note that, from eqs.(12) and (13), we have J~ = N~ +N~ . Thus, deducing the allowed values of j given n and n becomes † ′ a standard problem in the addition of angular momenta. The general result is that the allowed values of j are n n , n n +1,...,n+n, and each of ′ ′ ′ | − | | − | these values appears exactly once. The four simplest and most often encountered representations are (1,1), (2,1), (1,2), and (2,2). These are given special names: (1,1) = Scalar or singlet 9 (2,1) = Left-handed spinor (1,2) = Right-handed spinor (2,2) = Vector (50) It may seem a little surprising that (2,2) is to be identified as the vector rep- resentation. To see that this must be the case, we first note that the vector representation is irreducible: all the components of a four-vector mix with each other under a general Lorentz transformation. Secondly, the vector rep- resentation has four components. Therefore, the only candidate irreducible representations are (4,1), (1,4), and (2,2). The first two of these contain angular momenta j = 3 only, whereas (2,2) contains j = 0 and j = 1. This 2 is just right for a four-vector, whose time component is a scalar under spatial rotations, and whose space components are a three-vector. In order to gain a better understanding of what it means for (2,2) to be the vector representation, we must first investigate thespinor representations (1,2) and (2,1), which contain angular momenta j = 1 only. 2 Problems 33.1) Express Aµν(x), Sµν(x), and T(x) in terms of Bµν(x). 33.2) Verify that eqs.(14–16) follow from eqs.(9–11). 10