Physics 582, Problem Set 1 Solutions TAs: Hart Goldman and Ramanjit Sohal Fall 2018 1. THE DIRAC EQUATION [20 PTS] Consider a four-component fermion Ψ(x) in 3+1D, L[ Ψ, Ψ] = Ψ(i/ m)ψ, (1.1) where we use the notation /A A µ γ µ, γ µ are the Dirac gamma matrices, and Ψ Ψ γ 0. The Dirac equation is the equation of motion for Ψ, 0 = δl δ Ψ δl µ = (i/ m)ψ. δ( µ Ψ) (1.2) Equivalently, we can consider the equation of motion for Ψ, 0 = i( µ Ψ)γ µ + m Ψ, (1.3) which is the complex conjugate of Eq. (1.2). 1. [4 pts] Notice that the Lagrangian is invariant under a global U(1) charge symmetry Ψ e iθ Ψ. By Noether s theorem, this means that there should be a conserved current j µ, where µ j µ = 0. We can see that this current should be Ψγ µ Ψ using the Dirac equation, 0 = µ j µ = ( µ Ψ)γ µ Ψ + Ψ/ Ψ (1.4) = m ΨΨ + m ΨΨ (1.5) = 0. (1.6) Why is j µ associated with the aforementioned U(1) symmetry? You can see this explicitly by varying L with respect to the infinitesimal U(1) transformation Ψ Ψ + iδθ Ψ and plugging in the Dirac equation. 2. [4 pts] Now we will see that Ψ satisfies the Klein-Gordon equation. We can act on Eq. (1.2) with i/ m to obtain 0 = / / Ψ + im(/ / )Ψ + m 2 Ψ (1.7) = ( µ µ + m 2) Ψ. (1.8)
2 where we have used the fact that the Clifford algebra {γ µ, γ ν } = 2η µν implies / / = 2 µ ν η µν / / / / = µ µ. 3. We now proceed to derive some more gamma matrix identities. These will prove to be extremely useful when we start calculating Feynman diagrams. (a) [4 pts] We start by proving a more general version of the identity we used in the last problem. Consider the product /A /B, /A /B = A µ γ µ B ν γ ν (1.9) = A µ B ν (2η µν γ ν γ µ ) (1.10) = A µ B ν (2η µν (γ µ γ ν [γ µ, γ ν ])) (1.11) /A /B = A µ B ν (η µν + 1 ) 2 [γµ, γ ν ] (1.12) = A B(id 4 4 ) iσ µν A µ B µ, (1.13) where we recall the definition σ µν = i 2 [γµ, γ ν ]. Notice that when A = B we get back the identity derived in the previous part of this problem by the antisymmetry of σ µν. (b) [4 pts] We now use this to compute Tr[ /A /B], Tr[ /A /B] = A B Tr[id 4 4 ] i Tr[σ µν ]A µ B ν = 4A B, (1.14) since it is easy to see that Tr[σ µν ] Tr[γ µ γ ν γ ν γ µ ] = Tr[γ µ γ ν ] Tr[γ µ γ ν ] = 0. (c) [4 pts] Another useful identity involves the product γ µ γ ν γ µ, γ µ γ ν γ µ = 2γ µ η µν γ µ γ µ γ ν = (2 η µ µ)γ ν = 2γ ν, (1.15) since η µ µ = D = 4 (D is the number of spacetime dimensions). 2. TRANSFORMATION PROPERTIES OF DIRAC FERMION BILINEARS [20 PTS] In this problem, we investigate the behavior of Dirac fermion bilinears under Lorentz transformations. Recall that a particular Lorentz transformation Λ induces the following transformation of a Dirac spinor Ψ, Ψ α(x = Λx) = S αβ [Λ]Ψ β (x), (2.1)
3 where α, β = 1,..., 4 are spinor indices and S[Λ] = exp ( i 2 ω µνσ µν) for some real constants ω µν. We will not raise or lower spinor indices to emphasize that they are not the same as 4-vector indices. It is important to recognize that S[Λ] does not constitute a unitary representation of the Lorentz group, i.e. in general. (S[Λ]) (S[Λ]) 1 (2.2) Consequently, Ψ Ψ is not Lorentz invariant (notice that this is the density operator, so it is actually one component of the current 4-vector!), but we will see that ΨΨ is very soon. What is the relationship between (S[Λ]) and (S[Λ]) 1? We can write (S[Λ]) = exp ( i2 ) ω µν(σ µν ). (2.3) But (σ µν ) = i 2 [(γν ), (γ µ ) ] = i 2 [(γµ ), (γ ν ) ], (2.4) If we choose the chiral representation of the gamma matrices, γ µ = 0 σµ σ µ 0 where σ = (σ x, σ y, σ z ), then it is easy to see that Thus, since (γ 0 ) 2 = 1,, σ µ = (1, σ), σ µ = (1, σ), (2.5) (γ µ ) = γ 0 γ µ γ 0. (2.6) (σ µν ) = i 2 γ0 [γ µ, γ ν ]γ 0 = γ 0 σ µν γ 0 (S[Λ]) = γ 0 (S[Λ]) 1 γ 0 (2.7) With this identity in hand, we can now solve the problem. 1. [6 pts] (4 pts for establishing Eq. (2.7) and 2 pts for the remainder of the problem) We start with ΨΨ. On its own, Ψ transforms as Ψ (x ) = Ψ (S[Λ]) γ 0 = Ψ γ 0 (S[Λ]) 1 (γ 0 ) 2 = Ψ(S[Λ]) 1. (2.8) Thus, Ψ Ψ (x ) = Ψ(S[Λ]) 1 S[Λ]Ψ (x) = ΨΨ(x). (2.9)
4 2. [5 pts] Now we consider the pseudoscalar operator, Ψ γ 5 Ψ (x ) = Ψ(S[Λ]) 1 γ 5 S[Λ]Ψ (x) (2.10) where γ 5 = iγ 0 γ 1 γ 2 γ 3 = i ε 4! µνλσγ µ γ ν γ λ γ σ. Since S[Λ] induces a 4-vector Lorentz transformation on γ µ, (S[Λ]) 1 γ µ S[Λ] = Λ µ νγ ν, (2.11) then (S[Λ]) 1 γ 5 S[Λ] = i 4! ε µ ν λ σ Λ µ µλ ν νλ λ Λ σ λ σγ µ γ ν γ λ γ σ (2.12) = det Λ i 4! ε µνλσγ µ γ ν γ λ γ σ (2.13) = det Λ γ 5. (2.14) Thus, Ψ γ 5 Ψ (x ) = det Λ Ψγ 5 Ψ(x) (2.15) 3. [3 pts] Now we move on to the current, Ψ γ µ Ψ (x ) = Ψ(S[Λ]) 1 γ µ S[Λ]Ψ(x) = Λ µ Ψγ ν µ Ψ(x). (2.16) 4. [3 pts] And the axial current, Ψ γ 5 γ µ Ψ (x ) = Ψ(S[Λ]) 1 γ 5 S[Λ]Λ µ νγ ν Ψ (x) = det Λ Λ µ Ψγ ν 5 γ µ Ψ(x). (2.17) This follows immediately from Eqs. (2.11) and (2.14). 5. [3 pts] Finally, we consider the matrix bilinear, Ψ σ µν Ψ (x ) = i 2 Ψ(S[Λ]) 1 [γ µ, γ ν ]S[Λ]Ψ (x) = Λ µ λ Λν Ψ σ σ µν Ψ(x) (2.18) where we have again used Eq. (2.11). 3. CHIRAL SYMMETRY [20 PTS] Let us continue to work in the chiral representation (2.5) (note the sign difference between the definition used here and the definition in the problem set). We can write the fourcomponent spinor Ψ in terms of two two-component (Weyl) spinors ψ and χ, Ψ = ψ. (3.1) χ
5 1. [2 pts] In this language, the Dirac equation becomes 0 = iσ µ µ χ mψ = i t χ + iσ i i χ mψ (3.2) 0 = i σ µ µ ψ mχ = i t ψ iσ i i ψ mχ (3.3) Notice how the direction of motion of ψ (the left-handed Weyl fermion) is opposite that of χ (the right-handed Weyl fermion). 2. [8 pts] 3pts for the dispersion, 4pts for the eigenvectors, 1pt for the chirality eigenvalues Let m = 0. Then these equations decouple, and we just get i t χ = iσ i i χ (3.4) i t ψ = iσ i i ψ (3.5) Solutions to these equations are plane waves with energy-momentum 4-vector p µ = (ɛ p, p), ψ(x) = d 3 p 1 (2π) 3 ɛp ψ p e i(ɛpt p x), χ(x) = where ψ p and χ p both satisfy the equation d 3 p 1 (2π) 3 ɛp χ p e i(ɛpt p x), (3.6) ɛ p ψ p = σ i p i ψ p, ɛ p χ p = σ i p i χ p. (3.7) Or, in matrix form, ɛ p χ p = p z p x ip y p x + ip y p z χ p. (3.8) From here on, we suppress the equations for ψ p since they just have an extra minus sign. The dispersion relation is obtained simply by calculating the eigenvalues of the matrix on the right hand side, which are ɛ ± p = ± p i p i = ± p, (3.9) What are the eigenvectors ψ p corresponding to each eigenvalue? For this, it is convenient to notice that the Clifford algebra implies that for massless fermions (p µ σ µ )(p µ σ µ ) = (ɛ p ) 2 p 2 = 0, (3.10)
6 so we conventionally write 1 χ p = σ µ p µ ξ (3.11) for any constant two-component spinor ξ. Here the square root of a matrix denotes the matrix one obtains by taking the square root of the eigenvalues. The solution for ψ p is ψ p = σ µ p µ ξ. (3.12) We can assign chirality quantum numbers to ψ and χ using γ 5, which in the chiral basis is so γ 5 = 1 0 0 1 (3.13) γ 5 χ = +1 (right-handed), γ 5 ψ = 1 (left-handed). (3.14) 3. Consider the chiral (also called axial) rotation Ψ e iθγ 5 Ψ (3.15) (a) [1 pt] This transformation clearly maps the two-component spinors as ψ e iθ ψ, χ e iθ χ (3.16) (b) [4 pts] Ψ transforms as follows Ψ Ψ (e iθγ 5 ) γ 0 = Ψ (1 iθγ 5 1 ) 2 θ2 (γ 5 ) 2 + γ 0 (3.17) ( = Ψ 1 + iθγ 5 1 ) 2 θ2 (γ 5 ) 2 + (3.18) = Ψe iθγ 5 (3.19) since γ 5 = γ 5 and γ 5 anticommutes with γ µ. (c) [4 pts] Thus, the mass operator transforms as ΨΨ Ψe i 2θγ 5 Ψ = cos(2θ) ΨΨ + i sin(2θ) Ψγ 5 Ψ (3.20) since (γ 5 ) 2 = 1. In other words, the existence of a mass breaks chiral rotations! 1 There are obviously many equivalent ways of writing this answer, and I will accept any of them.
7 On the other hand, the current operator is invariant, Ψγ µ Ψ Ψe iθγ 5 γ µ e iθγ 5 Ψ = Ψγ µ e iθγ 5 e iθγ 5 Ψ = Ψγ µ Ψ, (3.21) again because γ 5 anticommutes with γ µ. (d) [1 pt] Clearly, then, the Dirac equation is not invariant under chiral rotations. It becomes 0 = (i/ me i 2θγ 5 )Ψ = (i/ m cos(2θ) im sin(2θ)γ 5 )Ψ (3.22) Notice that there is a new pseudoscalar mass term. 4. THE LANDAU THEORY OF PHASE TRANSITIONS [20 PTS] In this problem, we consider a Landau-Ginzburg free energy density of the form E = 1 2 ( φ)2 + U[φ], (4.1) where U[φ] = r 2 φ2 + λ 4 4! φ4 + λ 6 6! φ6. (4.2) Here r = a(t T 0 ), λ 4 < 0, and λ 6 > 0. Notice that λ 6 is required for the stability of the system. 1. [1 pt] Let us restrict ourselves to the case where the local magnetization φ is uniform. Then we can neglect the gradient term and see that the free energy is minimized when 0 = δe δφ φ= φ = r φ + λ 4 3! ( φ )3 + λ 6 5! ( φ )5. (4.3) 2. [8 pts] 3pts for the plots of U[φ], 3pts for the qualitative behaviour of φ, 2pts for solving for φ ± and plugging into U[φ] Consider the case where λ 4 < 0. Then we can plot U[φ] at various temperatures,
8 Notice that there is a temperature T c > T 0 at which the ordered phase where φ 0 is degenerate with the disordered phase, where φ = 0. Below this temperature, the ordered phase is energetically favored, but the value of the order parameter jumps from zero to a nonzero value, i.e. the (qualitative) behavior of φ near T c looks like In other words, this is a first order transition! We can obtain the energy in the ordered phase by solving Eq. (4.3) for φ = 0 and plugging the result into U[φ]. We can divide Eq. (4.3) out by φ to get (defining c 4 λ 4 /4!, c 6 λ 6 /6!) 0 = r + 4 c 4 ( φ ) 2 + 6 c 6 ( φ ) 4. (4.4) This has solutions ( φ ) 2 = 1 ( ) 4c 4 ± 16c 2 4 24c 6 r. (4.5) 12c 6
9 There are four saddle points (as visible from the above figure). The ones corresponding to global minima in the ordered phase are φ ± = ± 4c 4 + 16c 2 4 24c 6 r 1/2. (4.6) One way of seeing this is to note that deep in the ordered phase (r < 0), the other solutions are imaginary for c 4 < 0. Of course, one can also just calculate directly whether each saddle point corresponds to a maximum or a minimum. The energy of the ordered state is obtained by plugging this result into U[φ]. 3. [6 pts] 3pts for the plots of U[φ] and 3pts for the qualitative behaviour of φ Now let us briefly consider the case where λ 4 > 0. Now U[φ] has the following behavior as a function of temperature The phase transition is now continuous, or second order, occurring at T 0. Notice that the two phases are no longer degenerate at the transition. 4. [5 pts] 2pts for realizing that λ 4 (T c ) is determined by 0 = U[ φ ± ] (note that students might calculate T c (λ 4 ) earlier in the problem), 1pt for the value of λ 4 (T c ), and 2pts for the phase diagram We now return to the case of the firstorder transition λ 4 < 0. In contrast to the second order transition discussed in the previous sub-problem, which always occurs at T = T 0, the transition temperature T c depends on the values of the coupling constants λ 4 and λ 6. We thus wish to determine the value of λ 4 at the phase transition as a function of r c = T c T 0. We can do this by noticing that the phase transition occurs when the ordered and disordered phases
10 each have zero energy. Thus, we need to solve 0 = U[ φ ± ] (4.7) for λ 4. After some algebra or Mathematica usage, one finds a solution 8 λ 4 (T c ) = 5 λ 6(T c T 0 ). (4.8) Thus, the phase diagram for fixed λ 6 is where we have placed the origin at λ 4 = r = 0. The black line is the second order transition, while the blue curve is the first order phase transition. 5. SCALAR ELECTRODYNAMICS [20 PTS] In this problem, we will study scalar electrodynamics, which has the Lagrangian L = D µ φ 2 r 2 φ 2 λ 4! φ 4 1 4 F µνf µν, (5.1) where φ is a complex scalar field, A µ is a U(1) gauge field, and D µ is the covariant derivative D µ = µ + iea µ. (5.2) 1. [3 pts] 2pts for transformation of D µ φ and 2pts for transformation of F µν and potential terms. Consider the gauge transformation, φ(x) e ieλ(x) φ(x), A µ A µ (x) + µ Λ(x). (5.3)
11 Then clearly D µ φ µ ( e ieλ φ ) + ie(a µ + µ Λ)e ieλ φ = e ieλ D µ φ, (5.4) (D µ φ) e +ieλ (D µ φ), (5.5) D µ φ 2 D µ φ 2. (5.6) and φ 2 φ 2 (5.7) φ 4 φ 4 (5.8) F µν µ A ν ν A µ + µ ν Λ ν µ Λ = F µν. (5.9) Thus, we immediately see that L is invariant under gauge transformations. 2. [4 pts] 2pts for each EOM The equation of motion for φ is which can be written as 0 = δl δφ = δl µ δ( µ φ ), (5.10) ((D µ ) 2 + r2 + 2 λ4! φ 2 ) φ. (5.11) The equation of motion for φ is simply the complex conjugate of this. Similarly, the equation of motion for A µ is Gauss /Ampere s Law, ν F νµ = ie (φ D µ φ φ(d µ φ) ). (5.12) 3. [4 pts] 2pts for the conjugate momenta and 2pts for the Hamiltonian The conjugate momenta to φ, φ, and A µ are π φ = δl δ(( t φ) ) = Dt φ, π φ = δl δ( t φ) = (Dt φ), π Aµ = Letting U[φ] = r 2 φ 2 + λ 4! φ 4, the Hamiltonian density is δl δ( t A µ ) = F µt = (0, E i ). (5.13) H = π φ t φ + π φ ( t φ) + E i t A i L (5.14) = π φ (π φ iea t φ) + π φ (π φ + iea t φ ) π φ π φ D i φ 2 + U[φ] + 1 2 ( E 2 + B 2 ) A t i E i = π φ π φ D i φ 2 + 1 2 ( E 2 + B 2 ) A t [ i E i ie(π φ φ π φ φ) ] + U[φ], (5.15) where the i E i term is obtained by integrating by parts. Notice that A t is now acting as a Lagrange multiplier which imposes Gauss Law, Eq. (5.12).
12 4. [5 pts] 1pt for the Lagrangian, 2pts for the ρ and θ EOM, 1pt for the gauge-fixed Lagrangian, and 1pt for the gauge-fixed ρ EOM If we write φ = 1 2 ρ(x) e iθ(x), (5.16) where ρ, θ are real (the reason for the factor of 1/ 2 will become apparent in a minute), then so the Lagrangian becomes D µ φ = 1 2 e iθ [ µ ρ + iρ ( µ θ + ea µ )], (5.17) L = 1 2 ( µρ) 2 + 1 2 ρ2 ( µ θ + ea µ ) 2 U[ρ] 1 4 F µνf µν. (5.18) The equations of motion for ρ and θ are respectively ρ : 0 = 2 ρ ρ( µ θ + ea µ ) 2 + rρ + λ 3! ρ3, (5.19) θ : 0 = µ [ρ 2 ( µ θ + ea µ )]. (5.20) Notice that the equation of motion for θ is just the statement that the gauge current is conserved. We can eliminate θ by choosing unitary gauge, i.e. φ = ρ = e iθ φ, A µ = A µ + 1 e µθ. This leads to a new, gauge fixed Lagrangian L g.f. = 1 2 ( µρ) 2 + 1 2 ρ2 e 2 A µa µ U[ρ] 1 4 F µνf µν. (5.21) It looks like the gauge field has acquired a mass! This is the famous Higgs mechanism. Note that this does not run afoul of gauge invariance since the presence of the mass term is a consequence of gauge fixing. The equation of motion for ρ is now 0 = 2 ρ + [ e 2 A µa µ + r ] ρ + λ 3! ρ3. (5.22) 5. [4 pts] 1pt for v, 1pt for L eff, 2pts for deriving the Proca equation If r < 0 and we neglect the term involving A, we can solve this equation for ρ(x) = v 0, where v is a constant, v = ± 6r λ. (5.23) Plugging this back into the Lagrangian (5.21), we obtain the effective action for the gauge field A µ L eff = 1 4 F µν F µν + 1 2 v2 e 2 A µa µ. (5.24)
13 The equation of motion for A µ is now 0 = ν F νµ + v 2 e 2 A µ. (5.25) If we contract both sides of this equation with µ, we can see immediately that the antisymmetry of F µν implies µ A µ = 0. (5.26) Plugging this back into Eq. (5.25) implies 0 = ( 2 + v 2 e 2) A µ, (5.27) which looks like the Klein-Gordon equation for a field with mass m 2 = v 2 e 2! It turns out that Eq. (5.27) encodes the Meissner effect, or the expulsion of magnetic fields from within superconductors. To see this, let us consider magnetostatic solutions so that Eq. (5.27) can be written in terms of the magnetic field B = A as 2 B = v 2 e 2 B 1 λ B, (5.28) 2 where λ = 1/ve is the London penetration depth of the superconductor. Now consider a planar interface between our superconductor and vacuum (where the gauge field is massless) in the x, y plane at z = 0, with the superconductor (vacuum) filling the space z > 0 (z < 0). If we set up a uniform magnetic field B 0ˆx in the vacuum (say choosing A = B 0 yẑ), then the field inside the superconductor will be given by zb 2 x (z) = 1 λ B x(z). (5.29) 2 Since we don t want B to blow up at ±, only exponentially decaying solutions are allowed B = B 0 e z/λˆx. (5.30) Thus, the magnetic field can penetrate only to a length scale of order λ before being expelled: the photon is too heavy to go any further!