8.324 Relativistic Quantum Field Theory II

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1 Lecture Relativistic Quantum Field Theory II Fall Relativistic Quantum Field Theory II MIT OpenCourseWare Lecture Notes Hong Liu, Fall 00 Lecture 6.5: BRST SYMMETRY, PHYSICAL STATES AND UNITARITY.5.: Becchi-Rouet-Stora-Tyutin (BRST) Symmetry From the last lecture, we have Z = is ef f [A a,c, C ] DA a µdc a DC a e µ, () with d 4 x f δf a (A Λ (x) S A, C, C a (A) + d 4 xd 4 eff = S 0 [A] y C a (x) ξ C b (y), () δλ b (y) Λ=0 where f (A) is the gauge-fixing function and S 0 [A] = a 4 under the gauge transformation F a µν F µνa is the pure Yang-Mills action. S 0 [A] is invariant A a µ A a + D µ Λ a (3) µ where D µ Λ a = µ Λ a + f abc b A µ Λ c. We note that in () we integrate over all A a µ(x), including the unphysical configurations, but by construction Z should only receive contributions from the physical A a µ(x). We also note that Z = 0, + 0,. S eff A, C, C no longer has gauge symmetries, but it has a hidden global fermionic symmetry, the BRST symmetry, which is, in fact, a remnant of the gauge symmetry. To see this, it is convenient to introduce an auxillary field h a (x) : is ef f [A a,c, Z = DA a µdh a DC a DC a e µ C,h] (4) with ξ S eff A, C, C, h = S 0 [A] + Now, consider the following (BRST) transformations: d 4 x h a + d 4 x h a (x)f a (x) + L gh. (5) δ B A a µ = η(d µ C) a ηs(a a µ) δ B Ca = ηh a ηs( C a ) δ B C a = gηf abc C b C c ηs(c a ) δ B h a = 0 ηs(h a ) (6) (7) (8) (9) with η an anticommuting constant parameter. Then, in general, δ B ϕ ηs(ϕ), ϕ = A a µ, C a, C a, h a. (0) s(ϕ) takes ϕ to a field of opposite fermionic parity. We note some of the important properties of s : i. ii. s(ϕ ϕ ) = s(ϕ )ϕ ± ϕ s(ϕ ), where the + sign is for ϕ bosonic, and the sign is for ϕ fermionic. s (ϕ) = 0. For example, s ( C a ) = 0 and s (C a ) = 0, which follows from the Jacobi identity. () ()

2 Lecture Relativistic Quantum Field Theory II Fall 00 iii. From (i) and (ii), we have that s (F (ϕ)) = 0. (3) iv. s(a a µ) is the same as the infinitesimal gauge transformation of A a µ with Λ a Based on the above properties, we will now prove that δ B S = 0. replaced by C a. We first show that S = S 0 + d 4 x s(f (x)) (4) with F (x) = C a f a ξ C a h a, so that This can be established by showing that d 4 x C a(x)s(f a (A µ (x))) = s(f (x)) = h a f a + C a s(f a (A µ )) + ξ h a. [ d 4 yd 4 δfa (A Λ (x) x C a(x) δλ b (y) Λ=0 (5) ] C b (y), (6) which is left as an exercise to the reader. Then, we have that δ B S = δ B S 0 + η d 4 x s (F (x)), (7) and these terms are separately zero by the properties (iii) and (iv) shown above. The BRST symmetry implies the existence of a conserved fermionic charge Q B. or, equivalently, δ B ϕ = i [ηq B, ϕ] = ηs(ϕ) (8) s(ϕ) =i [Q B, ϕ] ± { i [Q B, ϕ], = i {Q B, ϕ}, ϕ bosonic, ϕ fermionic. Since s (ϕ) = 0, we have that That is, [ Q B, ϕ] = 0 for any ϕ, and hence, [ QB, [Q B, ϕ] ± ] We can also define a ghost number, which is conserved: for any other field ϕ..5.: Physical States and Unitarity = 0. (9) Q B = 0. (0) gh [C] =, gh[c ] =, gh [QB ] =, gh[ϕ ] = 0 () Physical states should be independent of the gauge choice. Z = 0, + 0, is so by construction, as it should be independent of f a (A). We now consider more general observables. More generally, we should have that 0 = δ g f i = i f δ g S i ()

3 Lecture Relativistic Quantum Field Theory II Fall 00 where δ g represents the change under the variation of the gauge-fixing condition f a (A). Note from (4) we have that δ g S = d 4 x s(δ g F (x)) = d 4 x s(c aδf a ) = i d 4 x { Q B, C a δf a (A) }, and so it must be true that { } d 4 x f Q B, C aδf a (A) i = 0 (3) for arbitrary δf a (A) for a physical observable, and so That is, a physical state ψ should satisfy Similarly, by considering we find that Q B i = Q B f = 0. (4) Q B ψ = 0. (5) δ g f O... O n i = 0, (6) [Q B, O] = 0 (7) and so O should be gauge invariant (if it does not contain ghost fields). Note that any state of the form ψ = Q B... (8) satisfies Q B ψ = 0, but that in this case, χ ψ = 0 for any physical state χ. Such a state ψ is called a null state. All physical observables involving a null state vanish. If ψ, ψ satisfying (5) are related by ψ = ψ + Q B..., (9) they will have the same inner product with all physical states, and thus are equivalent. We introduce H closed = { ψ : Q B ψ = 0}, H exact = { ψ : ψ = Q B... }, H closed H phys =. H exact That is, H phys is the cohomology of Q B. In summary:. Defining H big to be the Fock space composed from A µ, C, C, we have that H phys H closed H big. (30). By restricting to H phys and gauge invariant O, f O... O n i does not depend on the gauge choice. 3. Our path integral construction guarantees that only physical states contribute in the intermediate state. Example : Quantum electrodynamics in Feynman gauge (ξ = ) L = 4 Fµν F µν ξ ( µa µ ) + µ C µ C = ( µ A ν )( µ A ν ) + µ C µ C 3

4 Lecture Relativistic Quantum Field Theory II Fall 00 Under the BRST transformation: and so δ B A µ =η µ C δ B C = η µ A µ δ B C =0, [Q B, A µ ] = i µ C Q B, C =i µ A µ [Q B, C] =0. It is left as an exercise for the reader to find the explicit form for Q B. Now, we set H big to be the set of states formed by acting with creation operators for A µ, C, C on the ground state 0. Imposing Q B ψ = 0 gives us H closed and H phys. For illustration, consider the one-particle state: H big = { e µ, p, c, p, c, p }. (3) Then, Q B e, p = Q B e A 0 = e p c, p from [Q B, A µ ] = i µ C, and we obtain the physical state condition: e.p = 0 (3) For e.p 0, we get c, p null states. Q A µ B c, p p µ (p) 0 = p µ A µ (p) 0 = e = p, p is a null state. So, we have that 0, and so the c, p are non-physical states, and H phys = { e, p : e.p = 0, e µ e µ + p µ }. (33) Take p µ = ( p 0, 0, 0, p 3), p = 0. Then, e.p = 0 implies that e µ = ( p 0, e, e, p 3), and e e µ = (0, e, e, 0), and so, only transverse components of A µ generate physical states. Remarks: e + p implies that. While A 0 0 creates negative-norm states, they do not lie in the physical state space (giving a positivedefinite norm on the physical state space).. Ghosts C, C make sure that these negative norm states do not contribute in intermediate steps. 4

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