PHY{396 K. Solutions for homework set #6. Problem 1(a): h ^i ^ji 1 ik` jmn h ^k` ^mni J ; J ≡ 4 J ; J = hh by eq. (1) ii 1 ik` jmn km ^`n kn ^`m `m ^kn `n ^km = 4 −ig J + ig J + ig J − g J hh by antisymmetry of the 's ii = ik`jmn × −igkmJ^`n = iJ^`n × −gkmik`jmn = +δkmik`jmn = δijδ`n − δinδ`j = 0 − iJ^ji = +iJ^ij ≡ +iijkJ^k: (S:1) h ^i ^ ji 1 ik` h ^k` ^0ji J ; K ≡ 2 J ; J = hh by eq. (1) ii 1 ik` k0 ^`j kj ^`0 `0 ^kj `j ^k0 = 2 −ig J + ig J + ig J − ig J 1 ik` kj ^`0 `j ^k0 = 2 0 − iδ J + 0 + iδ J 1 ik` kj ^ ` `j ^ k ≡ 2 +iδ K − iδ K 1 ij` ^ ` 1 ikj ^ j = 2 K − 2 K = ijkK^ k; (S:2) h i h i K^ i; K^ j ≡ J^0i; J^0j = hh by eq. (1) ii = −ig00J^ij + ig0jJ^i0 + igi0J^0j − igijJ^00 = −iJ^ij + 0 + 0 + 0; ≡ −iijkJ^k: (S:3) Problem 1(b): h ^ i ^ji 1 jk` h ^ i ^k`i 1 jk` ik ^ ` i` ^ k V ; J ≡ 2 V ; J = 2 ig V − ig V 1 jk` ik ^ ` i` ^ k i ji` ^ ` i jki ^ k = 2 −iδ V + δ V = − 2 V + 2 V = iijkV^ k; (S:4) 1 h ^ 0 ^ji 1 jk` h ^ 0 ^k`i 1 jk` 0k ^ ` 0` ^ k V ; J = 2 V ; J = 2 i\g V − i\g V = 0; (S:5) h i h i V^ i; K^ j = V^ i; J^0j = i\gi0V^ j − igijV^ 0 = +iδijV^ 0; (S:6) h i h i V^ 0; K^ j = V^ 0; J^0j = ig00V^ j − i\g0jV^ 0 = +iV^ j: (S:7) Note that the Hamiltonian of a relativistic theory is a member of a 4-vector multiplet P^µ = (H;^ P^) where P^ is the net momentum operator. Applying the above equations to the P^µ vectors, we obtain h i P^i; J^j = iijkP^k; h i H;^ J^j = 0; h i (S:8) P^i; K^ j = +iδijH;^ h i H;^ K^ j = +iP^j: In particular, the Hamiltonian H^ commutes with the three angular momenta J^j but it does not commute with the three generators K^ k of the Lorentz boosts. Problem 1(c): In the ordinary quantum mechanics, it is often said that generators of continuous symmetries must commute with the Hamiltonian operator. However, this is true only for the symmetries that act in a time independent manner | for example, rotating the 3D space by the same angle at all times t. But when the transformation rules of a symmetry depend on time, the Hamiltonian must change to account for this time dependence. In a Lorentz boost, the transform x ! x0 obviously depends on time, which changes the way the transformed quantum fields such as Φ^ 0(x; t) depend on t. Consequently, the Hamiltonian H^ of the theory must change so that the new Heisenberg equations would match the new time dependence. In terms of the generators, this means that the boost generators K^ i should not commute with the Hamiltonian. 2 Note that this non-commutativity is not caused by the Lorentz boosts affecting the time 0 0 µ itself, t = L µx 6= t. Even in non-relativistic theories | where the time is absolute | the generators of symmetries which affect the other variables in a time-dependent matter do not commute with the Hamiltonian H^ . Indeed, consider a Galilean transform from one non-relativistic moving frame into an- other, x0 = x+vt but t0 = t. This is a good symmetry of non-relativistic particles interacting with each other but not subject to any external potential, X 1 X H^ = ^p2 + 1 V (x^ − x^ ): (S:9) 2M a 2 a b a a6=b A unitary operator G^ realizing a Galilean symmetry acts on coordinate and momentum operators as y y G^x^aG^ = x^a + vt; G^^paG^ = ^pa + Mv; (S:10) and it also transforms the Hamiltonian into ^ ^ ^y ^ ^ 1 2 GHG = H + v · Ptot + 2 Mtotv : (S:11) In terms of the Galilean boost generators K^ G, G^ = exp −iv · K^ G ; (S:12) hence under an infinitesimal boost v = ~, various operators transform according to i h ^ i i O!b Ob + δOb; δOb = −i KG; Ob : (S:13) Hence, the commutation relations with the boost generators follow from the infinitesimal boosts, for example h i ^ j i ij δx^a = ~t =) x^a; KG = iδ × t; h i ^ j i ij δp^a = ~M =) p^a; KG = iδ × M; (S:14) ^ ^ h ^ ^ j i ^i δH = ~ · Ptot =) H; KG = iPtot : In particular, the Hamiltonian does NOT commute with the Galilean boost generators. 3 Problem 2(a): 1 ^µν Consider a linear combination 2 NµνJ of Lorentz generators with some generic coefficients iφ ^µν Nµν = −Nνµ. The Lorentz symmetries L(N; φ) = exp( 2 NµνJ ) generated by this combi- nation preserve the momentum pµ of the particle state jpi if and only if iφ ^µν ^α iφ ^µν ^µ exp 2 NµνJ P exp − 2 NµνJ jpi = P jpi : (S:15) For an infinitesimal φ, this condition becomes h iφ ^µν ^αi 2 NµνJ ; P jpi = 0: (S:16) Applying eqs. (13) to the left hand side of this formula gives us h iφ ^µν ^αi iφ αµ ^ν αν ^µ 2 NµνJ ; P jpi = 2 Nµν −ig P + g P jpi αν = φN P^ν jpi (S:17) αν = φN pν jpi ; 1 ^µν µ so the condition for the generator 2 NµνJ to preserve particle's momentum p is simply αν N × pν = 0: (S:18) In 3D terms, N ij = ijkak and N 0k = −N k0 = bk for some 3-vectors a and b, the generator in question is 1 ^µν ^ ^ 2 NµνJ = a · J + b · K; (S:19) and the condition (S.18) becomes a × p − bE = 0 and b · p = 0: (S:20) Actually, the second condition here is redundant, so the general solution is p any a; b = a × : (S:21) E In terms of eq. (S.19), these solutions mean (a × p) · K^ a 1 N J^µν = a · J^ + = · EJ^ + p × K^ for any a: (S:22) 2 µν E E In other words, the Lorentz symmetries preserving the momentum pµ have 3 generators, 4 namely the components of the 3-vector R^ = E J^ + p × K^ : (4) Quod erat demonstrandum. Problem 2(b): Consider a massive particle moving at a slower-than-light speed β in z direction, so the energy is E = γm (where γ = 1=p1 − β2) while the 3-momentum is p = (0; 0; βγm). Consequently, the three components of the R^ vector this energy-momentum are R^x = γmJ^x − βγmK^ y; R^y = γmJ^y + βγmK^ x; R^z = γmJ^z: (S:23) These 3 operators generate the little group G(p) of the particle's 4{momentum pµ = (E; p). To see that this little group happens to be isomorphic to the 3D rotation group SO(3), we need to find 3 linear combinations J~x;y;z of the operators (S.23) which obey the angular- momentum commutation relations h i J~i; J~j = iijkJ~k: (S:24) My choice of the (properly normalized) generators J~x;y;z is spelled out in eqs. (5). Let's see that they indeed obey the commutation relations (S.24): h i h i h i J^z; Jex = γ J^z; J^x − βγ J^z; K^ y = γ × iJ^y − βγ × (−iK^ x) = iJey; h i h i h i J^z; Jey = γ J^z; J^y + βγ J^z; K^ x = γ × (−iJ^x) − βγ × (+iK^ y) (S:25) = −iJex; h x yi 2 h ^x ^yi 2 h ^ y ^yi 2 h ^ ^ x i 2 2 h ^ y ^ xi Je ; Je = γ J ; J − βγ K ; J + βγ Jx; K ; − β γ K ; K = γ2 × iJ^z − 0 + 0 − β2γ2 × iJ^z = iJ^z × γ2(1 − β2) = 1 = iJ^z: Quod erat demonstrandum. 5 Problem 2(c): For a massless particle, we cannot rescale the little group generators as in eqs. (5) since the 1=m factor for the J~x;y generators would be infinite. Instead, the best we can do is to use the 1=E factor for all three rescaled generators, hence eqs. (6) and (7). Consequently, instead of eqs. (S.25) for the commutators of the rescaled generators, we get h i h i h i J^z; I^x = J^z; J^x − J^z; K^ y = iJ^y − (−iK^ x) = iI^y; h i h i h i J^z; I^y = J^z; J^y + J^z; K^ x = (−iJ^x) − (+iK^ y) = −iI^x; (S:26) h i h i h i h i h i I^x; I^y = J^x; J^y − K^ y; J^y + J^x; K^ x − K^ y; K^ x = iJ^z − 0 + 0 − iJ^z = 0; precisely as in eq. (8). The commutation relations (8) are different from the angular-momentum commutation relations, and they cannot be brought to the form (S.24) by any finite rescaling of the generators. Consequently, the little group of a light-like momentum pµ = (E; 0; 0;E) is NOT isomorphic to the SO(3). Instead, the commutation relations (8) are similar to the commutation relations between the z component of the angular momentum and the x and y components of the linear momentum, h i h i h i J^z; P^x = +iP^y; J^z; P^y = −iP^x; P^x; P^y = 0: (S:27) The J^z operator generates rotations around the z axis, i.e., within the xy plane, while the P^x and P^i operators generate translations in that plane.