Teoria quântica de campos - equação de Dirac

Prévia do material em texto

<p>Dirac equation: are some matrices = = = 28" + = - Tr (ri) = -Tr in red box. All other relations written above follow. Dirac form Clifford algebra Hermiticity: for any</p><p>(det = eigen values are +1, eigen values are Ii is sum of eigen Tr=0 # of tve eigen values = # of -ve eigen are even-dimensional In 4 spacetime dimensions we need 4 anticommuting cannot be realized as 2x2 matrices, canbe realized as 4x4 matrices. In lower space-time dimensions, 2x2 are realizable. Substitute It at - my =0. = / = multicomponent Next, construct a classical field theory whose field give Dinac</p><p>Dirac field : = Vary - = = O. (a) = results of w.r.t. Grassman variables - why Indentify, =0. -i - = Reality of action: complex of eqn conjugation (a). nothing but - + - - L - Grassman variables derivatives, not any</p><p>Classical description of Fermi/Dirac fields can be given in terms of anticommuting variables Grassmann variables Suitable for imposing quantization rules on anticommutators of canonical conjugate quantities leads Pauli exclusion principle Grasmann 1. = =0 2. F(Q, a function is defined by the polynourial (2) + (n) in Oin C (k) totally antisymmetric, ordinary complex numbers E.g. n=1, F(0) Cot / n=2, = + C2O2 + 3. Differentiation: the variable of differentiation must be brought to the left, take derivative. E.g. a O2) =</p><p>Forbids existence of terms quadratic in derivatives in the L leads to EOM, first order in derivatives etc. Lorentz Spin What transformation property should we assign to Dirac field the Dirac equation is Lorentz = = (ir dr -m) = (ir m) = - = M Y'(x') = Condition on L(A) that = (a) (a) L</p><p>Infinitesimal transformations, = around = - = = Ex.9. above condition is solved by 4 SMY matrices also satisfy Lorentz (as it should, for to be in Lorentz = group) Recall, - rit_ = = i,j=1,2,3 sist = Since are</p><p>- =0, as n diagonal matrix = = = commutes with and anticommutes with Using which one can show, Arguments: = = ~ = WMV SMV) L(1) YY (not is Lorentz invariant Prove Lorentz inv. of the Dirac Lagrangian L</p><p>Solutions of = = f(k) ) k= = For non-trivial k-m) =0 (a) (K+m) =0 = (1/2 = AB+BA =0 number We have squared by are both the sol's satisfy egn (a) ? YES go to rest - m)f(k) = = However, r° 4x4 matrix eigen values of each eigen value two independent eigen vectors (as = 1) (doubly degenerate) -ik.x ) k Wk</p><p>Dirac (K = us(k) = =0, Normalizations: = = (k) Spin N = N = for above S normalizations check 2 = +mr = r = 2w- = + k-m) = 0 Both sides of the spin sums act similary on all the basis spinors.</p><p>Fourier decomposition 2 + S=1,2 4(x) = 2 S=1,2 = 24 = Hamiltonian 4-L + = (ik) + = -ik.x Similarly, = =</p><p>Using the normalizations, I - Now promote as,bs, ast, as S=1,2 In order to avoid -ve energy eigen values, usual normal ordering will not work impose anti commutation (k), = br t all other anticommutators Normal ordered Hamiltonian b,b}; {a, vanish. N(H) I + bs t (k) ) S=1,2 = ast N = - etc Using above anticommutation relations show that (x°, ), B which is same as = is - Vacuum & excited states = 10> One particle as Pauli exclusion principle = O= 2 particles or 2 antiparticles with same spin and same momentum cannot be</p><p>U(1) charge, antiparticle a fixed real number - L has symmetry under e 4(x) continuous real parameter Infinitesimal O So = Noether current, = q Noether Q = 3 x 2 (ast S=1,2 have opposite = =-q Normal ordered that vacuum has zero charge and zero</p><p>Green's Feynman propagator factor is taken out from Green - = x2) 4x4 will be identified matrix as Feynman propagator SF = i.k. Fourier expansion Fourier a matrix (i - I SF(k)e -ik = - ik. 11 4 (ir - SF(k)e (x1-x2) K -m (K = i II SF(k) = 2 (k = i (K + m) SF(k) = i(K+m) which has poles take Feynman prescription along the real at end E integration contour = i L ) the last expression is written for simplicity</p><p>this choice of SF (x, = (x2) Ex.12 where time ordered product for fields is defined as need = S B xi > x2 spin sums - x2 > xi (xn) T 0> correlators vanish unless Bm consequence of charge unless there are equal number conservation of Y's and unless n=m (x1) O= Wicks theorem applies to free fermions, with the difference that a sign must be included infront of each term, according to the number of anticommutations required to bring the contracted field's next to each other = - S +</p><p>Weyl representation of Dirac matrices are the 2x2 ( ) = Pauli matrices ) 1,2,3 12x2 O2x2 O2x2 = become block diagonal they form a reducible rep. of Lorentz algebra = = 1/2 ) 523 1/2 = 1/2 02) Angular momenta Eijk 1/2 = 1/2 J3 eigen values are each doubly degenerate two spin 1/2 particles General comments on Lorentz not restricted to Dirac spinor rep. [MMV = i = i Eijk angular momentum Eijk Kk algebra i Eijk Jk</p><p>1/2 Yi 1/2 whose physical significance is obscure Eijk = two different SU(2) addition of "angular momenta".</p>