20201021 Quantum Mechanics II Tutorial 11 - Dirac Teaching Assistant: Oz Davidi January 16, 2020 Notations and Conventions 1. We use τ as a short for 2π.1 1 Enhanced Symmetries In this section, we will see that sometime the Lagrangian has a larger symmetry than naively thought. It is usually the case in non-interacting theories (as interaction terms give more constraints). We state here the symmetry group, and it is left as an exercise to deduce the transformation rules. 1.1 Free Complex Fields Consider a free theory of N > 1 complex fields ~y µ ~ ~y 2 ~ L = @µφ @ φ − φ M φ ; (1.1) where M 2 is, without loss of generality, a diagonal positive-definite mass matrix. The symmetry in the case of no degeneracy is U(1)N , as the fields are distinct. If we assume all masses are degenerate, one can naively think that the symmetry is U(N), but in fact, the symmetry is larger. Define φ ≡ p1 (' + i' ). Then, the Lagrangian becomes j 2 j N+j 1 m2 1 m2 L = @ ' @µ' − ' ' ≡ @ ~'T @µ ~' − ~'2 : (1.2) 2 µ j j 2 j j 2 µ 2 1See https://tauday.com/tau-manifesto for further reading. 1 1.2 Weyl Spinors 1 ENHANCED SYMMETRIES This Lagrangian describes 2N real scalar fields, which can not be distinguished. The symmetry 2N(2N−1) 2 2 group is O(2N), which has more generators than U(N)( 2 = 2N − N against N ). Exercise 1.1. Repeat this exercise for the case of no mass degeneracy. Is the symmetry of the scalar fields larger? 1.2 Weyl Spinors Consider a free theory of N left-handed and N right-handed Weyl spinors (j; k = 1;:::;N) µ L = i jγ @µ j − jMjk k ; (1.3) where M is, without loss of generality, a diagonal positive-definite mass matrix. The symmetry in the case of no degeneracy is U(1)N , as the fields are distinct. If we assume all masses are degenerate, it means that we can not distinct between the different copies (generations) of the fields, so the symmetry is now U(N). Let us assume that the matrix M vanishes identically. Fine, this is a special case of a diagonal mass matrix. But it is more than that. Now, we can not even define what is left and right. We rewrite the Lagrangian in terms of the Weyl spinors ! L = R jy µ j jy µ j L = i L σ @µ L + i R σ @µ R : (1.4) From this expression we can clearly see a U(N)2 symmetry that mixes the left-handed and ∗ right-handed fields separately. However, we saw that using σ2σj = −σjσ2, the combination ∗ σ2 L transforms as σ∗ h σ i σ ∗ −! σ Λ∗ ∗ = σ exp (−iθ − φ ) i ∗ = exp (iθ + φ ) j σ ∗ = Λ σ ∗ : (1.5) 2 L 2 L L 2 j j 2 L j j 2 2 L R 2 L We can now rename ∗ T j j N+j j∗ j N+j jy N+j χR ≡ R ; χR ≡ iσ2 L =) L = −iσ2 χR ; L = i χR σ2 : (1.6) 2 2 CONSERVED U(1) CHARGE The Lagrangian is now T ∗ N+j µ N+j jy µ j L = i χR σ2σ σ2@µ χR + iχR σ @µχR (1.7) T ∗ N+j µ N+j jy µ j (integration by parts) = −i @µχR σ2σ σ2 χR + iχR σ @µχR (1.8) y N+j µ T N+j jy µ j (transpose) = i χR σ2 (σ ) σ2@µχR + iχR σ @µχR (1.9) y µ T µ ∗ N+j µ N+j jy µ j (σ ) = (σ ) = i χR σ @µχR + iχR σ @µχR (1.10) ny µ n (n = 1;:::; 2N) = iχR σ @µχR : (1.11) The symmetry is enhanced to a U(2N) symmetry. 2 Conserved U(1) Charge In this section, we are going to go over the Noether procedure for extracting the conserved currents and charges, and apply it for the free massive fermion theory. 2.1 Noether Procedure Consider a system with a symmetry group G, a field φ in a representation R, and the coordinates a xµ in a representation R0. For a group element U = ei!aX , the field transforms as 0 −1 φ(x) −! φ (x) = DR(U) φ DR0 U x (2.1) 1 a 2 = ( + i!aTR) φ(x − δx) + O !a (2.2) µ 2 = φ(x) + [i!aTaφ(x) − δx @µφ(x)] + O !a ; (2.3) a a where TR = πR(X ) is the matrix form of the generators in the R representation. We will denote a µ @ δφ = i!aTRφ − δx @µφ ; ∆aφ = δφ : (2.4) @!a Under a general symmetry transformation, the action is invariant. Therefore the Lagrangian must be invariant up to a total derivative µ δL = !a@µFa : (2.5) 3 2.2 Charge Conservation 2 CONSERVED U(1) CHARGE On the other hand, we can the variation of L as @L @L @L @L δL = δφi + δ@µφi = EOM + @µ δφi = !a@µ ∆aφi : (2.6) @φi @ (@µφi) @ (@µφi) @ (@µφi) Equating the two terms (δL = δL) we get µ @L µ @µJa = @µ ∆aφi − Fa = 0 : (2.7) @ (@µφi) µ Ja are the Noether currents. There is one conserved current for every symmetry generator. Every current leads to a conserved charge which is given by Z 3 0 Qa = d x Ja : (2.8) It is conserved due to Z Z Z 3 0 3 i 2 i @tQa = d x @tJa = − d x @iJa = − d x n^iJa = 0 ; (2.9) V V @V where we assumed that the boundary integral is zero (no flux). 2.2 Charge Conservation Let us work a concrete example for the system of one massive and free Dirac fermion µ L = i γ @µ − m : (2.10) The internal symmetry of the fields is a global U(1), under which −! ei! ; −! e−i! : (2.11) If we compare to the general case discussed above, we have only one generator and one ! parameter. Under this transformation, F = 0. The variation of the fields are δ = i! ; ∆ = i ; (2.12) δ = −i! ; ∆ = −i : (2.13) 4 2.2 Charge Conservation 2 CONSERVED U(1) CHARGE The conserved current and the charge associated with this symmetry are 0 @L @L> Z J µ = i + −i = − γµ ; Q = − d3x y : (2.14) @ (@ ) µ @ @µ We know that after quantizing the Dirac field, the fields become operators. Specifically, we define the equal-time anti-commutation relations n y o (3) a(~x) ; b (~y) = δ (~x − ~y) δab ; (2.15) with all other anti-commutators zero. We can transform the field operators under the transfor- mation induced by the conserved charge i!Q −i!Q 2 a(~x) −! e a(~x) e = a(~x) + i! [Q; a(~x)] + O ! : (2.16) In order to get δ right, as in Eq. (2.12), we must have [Q; a(~x)] = a(~x). Let us check it Z 3 h y i [Q; a(~x)] = − d y b (~y) b(~y) ; a(~x) (2.17) Z 3 y y = d y a(~x) b (~y) b(~y) − b (~y) b(~y) a(~x) (2.18) Z 3 n y o = d y a(~x) ; b (~y) b(~y) = a(~x) ; (2.19) In terms of the creation and annihilation operators X Z d3p Q = asyas − bsybs ∼ N^ − N^ : (2.20) τ 3 p p p p a b s=1;2 The conservation of Q implies that the number of particles minus the number of anti-particles is a constant, and therefore pairs of particle-anti-particle can annihilate each other or appear together such that Q is conserved. In the massless limit of the Dirac action there is another U(1), −! ei!γ5 ; (2.21) which rotates the left and right spinors with an opposite phase. The conserved current and charge are Z µ @L 5 µ 5 3 y 5 JA = iγ = − γ γ ; QA = − d x γ : (2.22) @ (@µ ) 5 3 THE NON-RELATIVISTIC LIMIT Similarly to the computation before, we get 5 [QA; a(~x)] = γab b(~x) : (2.23) 3 The Non-Relativistic Limit How can one see the Schr¨odingerequation arising from the Dirac one? Here we will derive the known result from Quantum Electro-Dynamics (QED), and get the chance to work in a different spinor basis. We start by writing the Lagrangian of QED2 µ µ L = (iγ @µ − ieγ Aµ − m) ; (3.1) where Aµ = '; A~ is the electromagnetic potential. We want to write Dirac equation, so we switch to the Hamiltonian formalism. The conjugate field is @L π = = i y : (3.2) @ _ The Hamiltonian density is thus _ y 0 k µ H = π − L = γ −iγ @k + eγ Aµ + m : (3.3) | {z } H One can thus write the Lagrangian as y L = (i@t − H) : (3.4) The time independent Dirac equation is given by H j i = E j i : (3.5) First, let us move to the momentum operators using the known exchange ~ ~ Pk = i@k () P = −ir : (3.6) 2The student is encouraged to take the courses on Quantum Field Theory and Elementary Particles, to understand the logic and beauty behind this Lagrangian. 6 3 THE NON-RELATIVISTIC LIMIT This results in h i H = γ0~γ P~ − eA~(t; ~x) + γ0m + e' : (3.7) As you may have heard (and know from your experience with matrices), the γ matrices are basis dependent. It is convenient to work in the chiral (Weyl) basis, where the Dirac spinor decomposes to two Weyl spinors.