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Representations of the Lorentz and Poincare Groups and Transformation Properties of Physical Quantities

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Representations of the Lorentz and Poincare Groups and Transformation Properties of Physical Quantities CHAPTER 2 Representations of the Lorentz and Poincare Groups and Transformation Properties of Physical Quantities At the very outset, it is pertinent to mention that finite dimensional irreducible representations (IRRs) of both these groups are not unitary and all unitary representations are infinite dimensional since the groups are non-compact. 2.1 UNITARY REPRESENTATIONS OF THE LORENTZ GROUP (WIGNER, 1939; GEL'FAND, 1963; NAIMARK, 1964; OHNUKI, 1976; TUNG, 1984) 2.1.1 Reduction of Representations and Relabelling of Basis The Lie algebra for the proper Lorentz group is given by JJμλσv , = iejJ σvvg μλ J λg μσ J μλg v σ+ J μσg v λ (1) or equivalently by mnl JJmn, = ie J l (2) mnl KJmn, = ie K l (3) mnl KKmn, = ie J l (4) Representations of the Lorentz and Poincare Groups and Transformation Properties... 49 related to the eigenvalue of the Casimir operator M 2 and write eq. (12) as M 3 |ukÒ = k|ukÒ (13) and, further 2 M |ukÒ = u(u + 1)|ukÒ (14) Of these, the index u corresponding to the eigenvalue of the Casimir operator M 2 labels out the IRR, since it is left invariant by group operations and hence, is constant across a given IRR, whereas the other index k, which is the eigenvalue of M 3 labels out the basis vectors within an IRR. We have M M |uk = M M |uk + MM , uk 3 ± Ò ± 3 Ò 3 ± = M ± M 3|ukÒ ± M ± |ukÒ = (k ± 1) M ± |ukÒ (15) Thus, M ± |ukÒ is an eigenstate of M 3 with eigenvalue k ± 1. This means that, in line with our ± labelling scheme for eigenstates, we must have M ± |ukÒ•|u, k ± 1Ò or M ± |ukÒ = Am|u, k ± 1Ò. Similar calculations apply to SU(2)N. Hence, we can label the IRRs of the direct product algebra SU(2)M ƒ SU(2)N that provides the IRRs for the Lorentz group by two numbers u, v corresponding to the two Casimir operators M 2 2 and N of the subalgebras SU(2)M and SU(2)N respectively. The basis vectors in the representation space are, then, chosen as {|uv; klÒ} ∫ {|klÒ} that consist of the direct products of the basis vectors from the canonical basis {|ukÒ} and {|vlÒ} of the two subalgebras SU(2)M and SU(2)N respectively. On this basis, the action of the generators of the Lorentz group is given by the following equations: |kl = MNkl = (k + l)|kl (16) J 3 Ò ej33+ Ò J kl = MNkl = A±(u, k)|k ± 1, l + B± (v, l)|k, l ± 1 (17) ± ej±±+ Ò Ò K 3 kl = iejNM33 kl = i(l k|klÒ (18) K kl = i NM kl = iA± (u, k)|k ± 1, l + iB± (v, l)|k, l ± 1 (19) ± ej±± Ò Ò It is seen from Eqs. (16) and (17) that the restriction of the representation space to the subgroup of three-dimensional rotations is equivalent to that of a direct product of representations of the SO(3) group labelled by the indices u and v respectively. We can reduce this direct product of representations to a direct sum of invariant subspaces with respect to rotations. For this purpose, we identify these constituents of the direct product representations in terms of u, v corresponding the eigenvalues of the respective Casimir operators and denote them by Du [R] and Dv[R] respectively. We denote the direct product of these two 50 Relativistic Quantum Mechanics representations by Du × v[R]. We designate, as above, the basis vectors in the product u × v representations by {|uv; klÒ} ∫ {|klÒ} so that |klÒ = |ukÒ ƒ |vlÒ. The action of D [R] on the direct product space W = VM ƒ VN is obtained by its action on the direct product basis |klÒ = |ukÒ ƒ |vlÒ, e.g., u × v u m v n D [R]|klÒ = D [R] k D [R] l |mnÒ (20) We show that Du × v [R] defines a representation of SO(3). This can be easily done by u + v u × v computing the character of the direct product representation. We have c = TrD [R] u × v k u i v j u v = D [R]k = D [R]i D [R] j = c c establishing our proposition. We examine the reducibility of Du × v[R]. For this purpose, we write u × v λ D [R]~ ∑ aDλ [] R (21) λ⊕ i.e., we attempt to reduce the direct product space W into a direct sum of invariant subspaces W l, a where labels out the IRR and =1, 2,... , distinguishes the spaces that correspond to the l a al same l. We can, then, choose a basis in W corresponding to any IRR, i.e., any l, such that the l l l l first n1 basis vectors are in W 1, the next n1 in W 2, and so on. Therefore, the representation matrices are in block diagonal form in the new basis. To examine the reducibility of our direct product representation, we consider an infinitesimal rotation about an arbitrary axis n . Then, we must have u v u × v D [Rn(dq)] D [Rn(dq)] = D [Rn(dq)] (22) whence, to first order in dq u u v v u v uvu v uv×× uv ejI iJdn θ . ejI iJdn θ = I ƒ I iejJJn ⊗+⊗II n dq = ejI iJdn θ (23) uv× u v u v which gives J n = J n ƒ I ¥ I ƒ J n . Thus, we find that the generators of a direct product representation are the sums of the corresponding generators of its constituent representations. uv× u v u v The generators defined by J n = J n ƒ I + I ƒ J n satisfy the same Lie algebra as the uv×× uv m uv¥ original one, i.e., JJk , l = iekl Jm . We have uv×× uv uvu v uvu v JJk , l = ejejJJJJk ƒ+ƒIIk , l ƒ+ƒ II l = JJuvu⊗+⊗II v JJuvu⊗+⊗II v ejk k ejl l uvu v uvu v ejJJl ⊗+⊗II l ejJJk ⊗+⊗II k uvuv u v u v = ejJ k ⊗ I ejJ l ⊗ I + ejI ⊗ J k ejI ⊗ J l uvuv u v u v ejJ l ⊗ I ejJ k ⊗ I ejI ⊗ J l ejI ⊗ J k Representations of the Lorentz and Poincare Groups and Transformation Properties... 51 u u v u v v u u v u v v = ejJJk l ƒ I + I ƒ ejJJk , l ejJJl , k ƒ I I ƒ ejJJl , k u u v u v v m uvu v m uv× = JJk , l ƒ I + I ƒ JJk , l = iekl ejJJm ⊗+⊗II m = iekl J m Our next step is to regroup the basis vectors of the product representation {|klÒ} to form invariant subspaces thereby leading to the reduction of the direct product representation. We also have uv¥ uvu v J3 kl = ejJJ33×+II × (|ukÒ × |vlÒ) = (k + l) (|ukÒ × |vlÒ) = (k + l)|klÒ (24) uv× so that |klÒ is an eigenvector of J 3 with eigenvalue (k + l). 2 uv× 2 uv× uv× uv× uv× Further, (J ) = ejJ 3 + J 3 + J J + whence 2 u × v uvuvu v u v ejJ uk ƒ|vl Ò = ejJJ33⊗+⊗II ejJJ33ƒ+ƒII uk ƒ|vlÒ + JJu ⊗+⊗IIv u v uk |vl + JJuvuv⊗+⊗II JJuvuvII uk |vl ej33ƒ Ò ejej++ƒ+ƒ ƒ Ò 2 uvuv = (k + l) |ukÒ ƒ|vlÒ + (k + l)|ukÒ ƒ |vlÒ + ejJJ⊗+⊗II + + [A (u, k)|u, k + 1Ò ƒ|v, lÒ + B (v, l)|uk ƒ|v, l + 1Ò] 2 = (k + l) |ukÒ ƒ|vlÒ + (k + l)|ukÒ ƒ |vlÒ + A (u, k + 1) + + A (u, k)|u, kÒ ƒ|v, lÒ + B (v, l + 1)B (v, l)|u, kÒ ƒ|v, lÒ 2 so that |ukÒ ƒ |vlÒ is also an eigenvector of ejJ uv× . We also know from the elementary theory of angular momentum in three dimensions, of which the group SO(3) provides a representation, that the range of the eigenvalues of the generator u 2 M ∫ J 3 extends from u £ k £ + u, where u relates to the eigenvalue of the Casimir operator M . uv× Similarly, the range of l extends from v £ l £ + v. Therefore, the highest eigenvalue of J 3 is u + v and there is only one eigenvector corresponding to this eigenvalue given by |uvÒ. There will be uv× two eigenvectors corresponding to the eigenvalue u + v 1 of J 3 given by |u 1, vÒ and |u, v 1Ò uv¥ and so on. The lowest eigenvalue of J3 is u v and there is only one eigenvector corresponding uv¥ to this eigenvalue given by | u, vÒ. Since the eigenvalues of J3 change in integral steps, the total uv¥ number of eigenvalues of J3 will be 2(u + v) + 1 and the corresponding eigenvectors can be uv× obtained by applying the lowering operator J to the highest state |uvÒ. 52 Relativistic Quantum Mechanics Since there is only one state with the eigenvalue u + v and there exists no state with a higher u ¥ v eigenvalue of J 3 , this state i.e. |uvÒ is the highest member of the basis of the IRR labelled by u + v. The remaining basis vectors can be generated by successive application of the lowering operator u¥v J , as mentioned above. The total number of basis vectors in this representation will be u ¥ v 2(u + v) + 1 (because the range of eigenvalues of J 3 is from u + v through 0 to (u + v)) and they constitute the basis of an invariant subspace corresponding to the IRR labelled by u + v relating it to the eigenvalue of ( J u ¥ v)2.
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