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

~† µ ~ ~† 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 ≡ √1 (ϕ + iϕ ). Then, the Lagrangian becomes j 2 j N+j

1 m2 1 m2 L = ∂ ϕ ∂µϕ − ϕ ϕ ≡ ∂ ~ϕT ∂µ ~ϕ − ~ϕ2 . (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 = iψjγ ∂µψj − ψjMjkψk , (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 ! ψ ψ = L ψR j† µ j j† µ j L = iψL σ ∂µψL + iψR σ ∂µψR . (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 ∗ σ2ψL transforms as

σ∗ h σ i σ ψ∗ −→ σ Λ∗ ψ∗ = σ exp (−iθ − φ ) i ψ∗ = exp (iθ + φ ) j σ ψ∗ = Λ σ ψ∗ . (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 j† N+j χR ≡ ψR , χR ≡ iσ2ψL =⇒ ψL = −iσ2 χR , ψL = i χR σ2 . (1.6)

2 2 CONSERVED U(1) CHARGE

The Lagrangian is now

T ∗ N+j µ N+j j† µ j L = i χR σ2σ σ2∂µ χR + iχR σ ∂µχR (1.7) T ∗ N+j µ N+j j† µ j (integration by parts) = −i ∂µχR σ2σ σ2 χR + iχR σ ∂µχR (1.8) † N+j µ T N+j j† µ j (transpose) = i χR σ2 (σ ) σ2∂µχR + iχR σ ∂µχR (1.9) † µ T µ ∗ N+j µ N+j j† µ j (σ ) = (σ ) = i χR σ ∂µχR + iχR σ ∂µχR (1.10)

n† 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 = eiωaX , the ﬁeld transforms as

0 −1 φ(x) −→ φ (x) = DR(U) φ DR0 U x (2.1) 1 a 2 = ( + iωaTR) φ(x − δx) + O ωa (2.2) µ 2 = φ(x) + [iωaTaφ(x) − δx ∂µφ(x)] + O ωa , (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 = iωaTRφ − δx ∂µφ , ∆aφ = δφ . (2.4) ∂ωa Under a general symmetry transformation, the action is invariant. Therefore the Lagrangian must be invariant up to a total derivative

µ δL = ωa∂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 = ωa∂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 nˆiJa = 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 = iψγ ∂µψ − mψψ . (2.10)

The internal symmetry of the ﬁelds is a global U(1), under which

ψ −→ eiωψ , ψ −→ e−iωψ . (2.11)

If we compare to the general case discussed above, we have only one generator and one ω parameter. Under this transformation, F = 0. The variation of the ﬁelds are

δψ = iωψ , ∆ψ = iψ , (2.12) δψ = −iωψ , ∆ψ = −iψ . (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 = iψ + −iψ = −ψγµψ , Q = − d3x ψ†ψ . (2.14) ∂ (∂ ψ) µ ∂ ∂µψ

We know that after quantizing the Dirac field, the fields become operators. Specifically, we define the equal-time anti-commutation relations

n † o (3) ψa(~x) , ψb (~y) = δ (~x − ~y) δab , (2.15) with all other anti-commutators zero. We can transform the ﬁeld operators under the transformation induced by the conserved charge

iωQ −iωQ 2 ψa(~x) −→ e ψa(~x) e = ψa(~x) + iω [Q, ψa(~x)] + O ω . (2.16)

In order to get δψ right, as in Eq. (2.12), we must have [Q, ψa(~x)] = ψa(~x). Let us check it Z 3 h † i [Q, ψa(~x)] = − d y ψb (~y) ψb(~y) , ψa(~x) (2.17) Z 3 † † = d y ψa(~x) ψb (~y) ψb(~y) − ψb (~y) ψb(~y) ψa(~x) (2.18) Z 3 n † o = d y ψa(~x) , ψb (~y) ψb(~y) = ψa(~x) , (2.19)

In terms of the creation and annihilation operators

X Z d3p Q = as†as − bs†bs ∼ Nˆ − Nˆ . (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),

ψ −→ eiωγ5 ψ , (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 † 5 JA = iγ ψ = −ψγ γ ψ , QA = − d x ψ γ ψ . (2.22) ∂ (∂µψ)

5 3 THE NON-RELATIVISTIC LIMIT

Similarly to the computation before, we get

5 [QA, ψa(~x)] = γabψb(~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 = ψ (iγ ∂µ − ieγ Aµ − m) ψ , (3.1) where Aµ = ϕ, A~ is the electromagnetic potential. We want to write Dirac equation, so we switch to the Hamiltonian formalism. The conjugate ﬁeld is

∂L π = = iψ† . (3.2) ∂ψ˙

The Hamiltonian density is thus

˙ † 0 k µ H = πψ − L = ψ γ −iγ ∂k + eγ Aµ + m ψ . (3.3) | {z } H

One can thus write the Lagrangian as

† L = ψ (i∂t − H) ψ . (3.4)

The time independent Dirac equation is given by

H |ψi = E |ψi . (3.5)

First, let us move to the momentum operators using the known exchange

~ ~ Pk = i∂k ⇐⇒ P = −i∇ . (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 + eϕ . (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. But, there is another useful basis - the Dirac basis. In this basis, the Dirac spinor decomposes to the fermion component and anti-fermion component. The γ matrices in the Dirac basis are ! ! 1 0 0 σk γ0 = , γk = . (3.8) 0 −1 −σk 0

We would like to take the non-relativistic limit soon, namely assume that the fermion mass, m, is the largest scale. We thus further write E = m + E0. The Dirac equation is now

 0  ! ! eϕ − E ~σ · P~ − eA~ |χi 0   = . (3.9) −~σ · P~ − eA~ eϕ − 2m − E0 |ηi 0

The second equation gives us

|ηi = (2m + E0 − eϕ)−1 ~σ · P~ − eA~ |χi . (3.10)

Substituting this relation into the ﬁrst equation results in

h i ~σ · P~ − eA~ (2m + E0 − eϕ)−1 ~σ · P~ − eA~ + eϕ |χi = E0 |χi . (3.11)

0 −1 The non-relativistic limit states m E , eϕ, |~pχ|, hence, to leading order in m

1 2 ~σ · P~ − eA~ + eϕ |χi = E0 |χi . (3.12) 2m

Using σjσk = δjk + ijklσl , (3.13) and

j k j k j j k j A (t, ~x) ,P = i∂kA (t, ~x) ⇐⇒ P A (t, ~x) = A (t, ~x) P − i∂kA (t, ~x) , (3.14)

7 3 THE NON-RELATIVISTIC LIMIT the ﬁrst term is simpliﬁed to

h i2 ~σ · P~ − eA~ = σjσk P j − eAj P k − eAk (3.15)

= δjk + ijklσl P j − eAj P k − eAk (3.16) 2 = P~ − eA~ − iejklσl P jAk + AjP k (3.17) 2 = P~ − eA~ − e~σ · ∇~ × A~ . (3.18)

We have found the known non-relativistic Hamiltonian

1 2 e H = P~ − eA~ − ~σ · ∇~ × A~ +eϕ . (3.19) NR 2m 2m | {z } e ~ ~ g 2m S·B

In order to collect extra terms (such as the one which comes from the relativistic kinetic energy, the spin-orbit term, the Darwin term, and so on), one needs to follow the method of Tani-Foldy-Wouthuysen Transformation, but this is out of our scope.