Quntum Field Theory Exmple Sheet 1, Michælms 25 Jcob Lewis Bourjily 1. A string of length with mss per unit length σ under tension T with fixed endpoints is described by the Lgrngin ( 2 1 y L = dx 2 σ 1 ( ] 2 y t 2 T, x where y(x, t is the trnsverse displcement of the string t (x, t. We re to express this in terms of the Fourier modes of y(x, y, compute the field equtions nd show tht the system is equivlent to n infinite set of hrmonic oscilltors. Let us first note tht we cn write the displcement y(x, t in terms of Fourier sine eries expnsion, 2 ( nπx y(x, t = q n sin. From this we my directly observe tht y 2 ( nπx t = q n sin nd from whence it is cler tht ( σ L = dx q n sin y 2 x = ( nπx 2 ( π2 T 3 q n n cos nπ ( nπx q n cos, ( nπx 2. By the orthogonlity of the sine nd cosine functions for different n, we see tht upon integrting with respect to x we obtin { σ ( L = dx q n 2 sin 2 nπx T ( π2 3 qnn 2 2 cos 2 nπx }, { = dx q n 2 σ ( nπx ( cos2 q 2 σ n + T } π2 3 n2, L = { σ 2 q2 n T 2 ( πn } 2 q 2 n. By direct ppliction of the vritionl principle, hence the Euler-Lgrnge equtions, we obtin the field eqution ( πn 2 σ q n + T qn =, for ech n Z. This is obviously the eqution for mode of simple hrmonic oscilltor with frequency T ( nπ ω n =. σ 1
2 JACOB LEWIS BOURJAILY 2. A string of length with unit mss density nd tension cn vibrte in two trnsverse directions, y 1 nd y 2. We re to determine the Lgrngin describing this system, find the equtions of motion, nd show tht y 1,2 (x, t ech stisfy the wve eqution with unit velocity. We re then to show tht the system hs n (infinitesiml SO(2 symmetry, determine the ssocited Noether current nd show tht the chrge is independent of time. Let us consider prticulrly nice coordinte system loclized in some smll neighborhood U bout x (, in which the displcement is entirely long the direction ỹ which is liner superposition of y 1 nd y 2. Notice tht our work in exercise one bove implies tht the Lgrngin density for this neighborhood is L x U = 1 2 ( tỹ 2 1 2 ( xỹ 2. Becuse we re working in Eucliden system it is cler tht dỹ 2 = dy1 2 + dy2, 2 nd so therefore L = dx 1 2 µy i µ y i, where µ =, 1 nd i = 1, 2 nd summtion is implied 1. For ech i, we see tht the field equtions re L L t + x ẏ i ( x y i = ÿ i 2 y i x 2 =, which is the wve eqution with unit velocity. Notice tht the system hs n obvious SO(2 symmetry given by ( ( ( y1 cos θ sin θ y1. sin θ cos θ y 2 This cn be seen to( be symmetry of the Lgrngin (leving L invrint s follows: y1 if we consider y = to be the dynmicl vrible, the Lgrngin only involves y 2 the (Eucliden inner product y i y i = y1 2 + y2 2 nd its derivtives. Becuse the inner product is preserved under SO(2 we see tht SO(2 is globl symmetry of the system. The Lie lgebr ssocited with this rottionl symmetry, so 2 is generted by the infinitesiml trnsformtions y 2 y 1 y 1 θy 2 nd y 2 y 2 + θy 1, s cn be seen from tking the Tylor series of sin θ nd cos θ for smll θ in the mtrix for n SO(2 rottion bove. Becuse the Lgrngin is itself invrint under these trnsformtions, we hve tht the following conserved current j µ = L ( µ y i δy i; which, by direct computtion, is j µ = (j, j 1 = (y 1 t y 2 y 2 t y 1, y 2 x y 1 y 1 x y 2. Let us now show tht the chrge ssocited with this current is itself constnt in time. By direct computtion we see tht dh dt = = dx (ẏ 1 ẏ 2 + y 1 ÿ 2 ẏ 2 ẏ 1 y 2 ÿ 1, ( 2 y 2 dx y 1 x 2 y 2 y 1 2 x 2. Notice tht this expression is odd under the interchnge of y 1 y 2, but tht the physicl sitution is of course independent of this choice of xes so tht dh dt =. 1 We re using the two-dimensionl Minkowski metric dig(hµν = (1, 1.
QUANTUM FIELD THEORY EXAMPLE SHEET 1 3 3. Let us consider the dynmics of complex sclr field ϕ(x governed by the Lgrngin density L = µ ϕ µ ϕ m 2 ϕ ϕ λ (ϕ ϕ 2. We re to compute the system s field equtions, demonstrte the system s globl U(1 symmetry, compute its ssocited Noether current, nd show tht this current is conserved. By direct nlogy with the Klein-Gordon field, we see tht the field equtions for ϕ nd ϕ re ( + m 2 + 2λ ϕ 2 ( ϕ = nd + m 2 + 2λ ϕ 2 ϕ =. Notice tht the Lgrngin is invrint under the U(1 symmetry ϕ e iα ϕ nd ϕ e iα ϕ. The Lie lgebr u 1 ssocited with this U(1 symmetry is described by infinitesiml trnsformtions of the form ϕ (1 iαϕ nd ϕ (1 + iαϕ. This cn be seen s Tylor expnsion of the U(1 symmetry for smll vlues of α. Noting tht the Lgrngin density is itself invrint under this trnsformtion, the Noether current ssocited with this symmetry is given by j µ = i {ϕ µ ϕ ϕ µ ϕ }. By direct computtion nd using the equtions of motion clculted bove, we see tht µ j µ ( ϕϕ + µ ϕ µ ϕ ( ϕ ϕ ϕ µ ϕ, ( m 2 + 2λ ϕ 2 ϕ 2 ( m 2 + 2λ ϕ 2 ϕ 2, µ j µ =. 4. We re to verify tht the Lgrngin density for n isovector field {ϕ i } for i = 1, 2, 3, L = 1 2 µϕ i µ ϕ i 1 2 m2 ϕ i ϕ i, is invrint under globl SO(3 symmetry which is generted by the infinitesiml iso-rottion of ngle θ bout the unit vector n j given by ϕ i ϕ i + θɛ ijk n j ϕ k. We re to compute the Noether current ssocited with this symmetry nd show tht the components of this isospin current re seprtely conserved. Let us begin by noting tht this is the theory of three (independent sclr fields tht ech stisfy the Klein-Gordon field equtions ( + m 2 ϕ i =. The fct tht L is invrint under SO(3 is cler becuse it only depends on the (Eucliden norm of the vector ϕ (ϕ 1, ϕ 2, ϕ 3 which is invrint under SO(3 rottions 2. The Lie lgebr so 3 describes the infinitesiml trnsformtions close to the identity of SO(3 nd these re generted by the ntisymmetric mtrices. To see this nother wy, notice tht the infinitesiml trnsformtion ϕ i ϕ i + θɛ ijk n j ϕ k cn be exponentited to form globl trnsformtions. We clim tht these trnsformtions re in SO(3 nd hence leve L invrint. Let σ j ik (θ θɛ ijk where e j is the j th unit vector. Notice tht σ is ntisymmetric in the indices ik so tht ( exp(σ j ik exp(σ j ik = exp(s j ik exp( sj ik = id. Therefore we see tht these infinitesiml trnsformtions generte SO(3 s climed. 2 Actully, L is invrint under ll of O(3 but we re only interested in the component connected to the identity for this problem.
4 JACOB LEWIS BOURJAILY Let us describe the rottion bout the j th unit vector. Using our definition of σ j ik bove, we see tht σik 1 = 1 σik 2 = 1 σik 3 = 1 1. 1 1 From this we clculte the iso-current ssocited with rottions bout the e th j j µ j to be j µ 1 = µ ϕ 1 ( + µ ϕ 2 ( ϕ 3 + µ ϕ 3 (ϕ 2 = ϕ 2 µ ϕ 3 ϕ 3 µ ϕ 2 ; j µ 2 = µ ϕ 1 (ϕ 3 + µ ϕ 2 ( + µ ϕ 3 ( ϕ 1 = ϕ 3 µ ϕ 1 ϕ 1 µ ϕ 3 ; The divergence of j µ i j µ 3 = µ ϕ 1 ( ϕ 2 + µ ϕ 2 (ϕ 1 + µ ϕ 3 ( = ϕ 1 µ ϕ 2 ϕ 2 µ ϕ 1 ; is directly seen to be j µ i = ɛ ijk ϕ j µ ϕ k. µ j µ i = ɛ ijk µ ϕ j µ ϕ k ɛ ijk ϕ j (m 2 ϕ k direction, where we hve mde use of the field equtions. Notice tht both terms re symmetric in the indices jk so tht ech must vnish under contrction with the totlly ntisymmetric symbol ɛ ijk. Therefore, µ j µ =. 5. Let us consider ϕ 4 -theory described by the Lgrngin L = 1 2 ( µϕ 2 1 2 m2 ϕ 2 λ 4! ϕ4. We re to compute the equtions of motion for the field ϕ, determine its stress-energy tensor T µ ν nd show tht it is conserved. By direct ppliction of the Euler-Lgrnge equtions, we see tht the equtions of motion for ϕ re given by ( + m 2 + λ 3 ϕ2 ϕ =. We cn compute the stress-energy tensor s the Noether currents ssocited with infinitesiml spce-time trnsltions s follows T µ ν = L ( µ ϕ nuϕ δ µ νl, ( 1 = µ ϕ ν ϕ δ µ ν 2 ( µϕ 2 1 2 m2 ϕ 2 λ 4! ϕ4, = 1 2 µ ϕ ν ϕ + δ µ ν Agin by direct computtion, we see tht ( 1 2 ϕ2 + λ 4! ϕ4 µ T µ ν = ϕ ν ϕ + ( µ ϕ ( ν µ ϕ 1 2 ν ( µ ϕ 2 + m 2 ϕ ν ϕ + λ 3! ϕ3 ν ϕ, = ϕ( ν ϕ + m 2 ϕ ν ϕ + λ 3! ϕ3 ν ϕ, nd so by the equtions of motion we hve µ T µ ν =..
QUANTUM FIELD THEORY EXAMPLE SHEET 1 5 6. Recll tht Lorentz trnsformtion x µ Λ µ νx ν preserves the Minkowski metric g µν so tht g µν x µ x ν = g µν x µ x ν. We re to show tht this implies g µν = g στ Λ σ µλ τ ν nd use this to show tht n infinitesiml Lorentz trnsformtion hs the form Λ µ ν = δ µ ν + ω µ ν where o µν = ω µν]. Lstly, we re to write the mtrix form of ω µ ν nd use this to determine the finite mtrices for boost long nd rottion bout the 3-direction. Using the fct tht nd hence g µν x µ x ν = g µν x µ x ν g µν x µ x ν = gµν Λ µ ρλ ν σ }{{} g ρσ by definition x ρ x σ, we see immeditely tht g ρσ = g µν Λ µ ρλ ν σ. An infinitesiml Lorentz trnsformtion is represented by n element of so(3, 1 which is quite generlly given by x µ (δ µ ν + ω µ νx ν, where ω µ ν is some infinitesiml mtrix. Using the identity bove nd ignoring terms qudrtic in ω µ ν we see tht g ρσ = g µν ( δ µ ρ + ω µ ρ ( δ ν σ + ω ν σ, = (g ρν + ω νρ ( δ ν σ + ω ν σ, = g ρσ + (ω ρσ + ω σρ + O(ω 2, = g ρσ + 2ω (ρσ, which implies tht ω (ρσ =, or equivlently, ω ρσ = ω σρ. An infinitesiml rottion of θ degrees bout the x 3 -direction then hs the form id + ω where ω µ ν = θ θ. Exponentiting this, we see tht the ssocited Lorentz trnsformtion (for rbitrry θ is given by Λ µ ν = 1 cos θ sin θ sin θ cos θ 1 An infinitesiml boost of rpidity η in the x 3 -direction hs the form id + ω where ω is then given by η ω µ ν =. η Exponentiting this, we see tht the ssociting Lorentz trnsformtion (for rbitrry η is given by Λ µ ν =. cosh(η sinh(η sinh(η cosh(η.
6 JACOB LEWIS BOURJAILY 7. We re to show tht the infinitesiml Lorentz genertor ω µ ν cn be expressed s liner combintion of bsis mtrices (M στ µ ν = gµσ δ τ ν g µτ δ σ ν. We re to verify tht the genertors M στ stisfy the Lie lgebr brcket M στ, M αβ] = g τα M σβ + g σβ M τα g σα M τβ g τβ M σα, nd tht M 1, M 23] =, for ll permuttions of {, 1, 2, 3}. Defining J 1 = M 23 J2 = M 31 J3 = im 12, we re to show tht Ji, J ] j = i J k for even permuttions of {1, 2, 3}. To show tht the mtrices (M στ µ ν generte the Lorentz lgebr so(3, 1 it is sufficient to show tht 3 ω ρν = g ρµ (M στ ω στ µ ν is ntisymmetric this being the only criterion on the mtrix ω ρν. This is reltively esy to demonstrte: by direct clcultion, we see tht g ρµ (M στ ω στ µ ν = ( δρ σ δν τ δρδ τ ν σ ωστ = ω ρν ω νρ, which is clerly ntisymmetric. To check the commuttion reltion M στ, M αβ] µ ν = (M στ µ ( ρ M αβ ρν ( M αβ µ ρ (M στ ρ ν, we will lower the index µ nd proceed using the definition of M στ 4. We see tht M στ, M αβ] ην = (M στ ηρ ( M αβ ρ ν ( M αβ ηρ (M στ ρ ν, = ( δ σ ηδ τ ρ δ τ ηδ σ ( ρ g ρα δ β ν g ρβ δ α ( ν δ α η δ β ρ δ β η δ α ( ρ g ρσ δ τ ν g ρτ δ σ ν, = δ σ ( η g τα δ βν g τβ δ α ( ν δ τ η g σα δ β ν g σβ δ α ( ν δ α η g βσ δ τ ν g βτ δ σ ( ν + δ β η g ασ δ τ ν g ατ δ σ ν, = g τα ( δ σ ηδ β ν δ σ νδ β η g τβ ( δ σ ηδ α ν δ α ηδ σ ν g σα ( δ τ ηδ β ν δ β ηδ τ ν + g σβ ( δ τ ηδ α ν δ α ηδ τ ν, = { g τα M σβ g τβ M σα g σα M τβ + g σβ M τα} ην, M στ, M αβ] = g τα M σβ g τβ M σα g σα M τβ + g σβ M τα. Notice tht becuse g µν is digonl mtrix, M 1, M 23] = ; nd likewise for ny permuttion of {, 1, 2, 3} t lest two indices must gree to hve contribution on the right-hnd-side of the boxed eqution bove. Given the opertors J i defined bove, let us evlute Ji, J ] k. Notice tht by the ntisymmetry, it is sufficient to determine this for Ji, J ] i+1, where i {1, 2, 3}. In this cse, by the definition of the mtrices J i nd using the Lorentz brcket condition we see tht Ji, J ] i+1 = M (i+1(i+2, M (i+2i] = g (i+2(i+2 M (i+1i = M (i+1i = M i(i+1 = i J i+2. 3 We re neglecting n irrelevnt fctor of ½ in the definition of ω µ ν reltive to our previous discussion. 4 Notice tht we clculted (M στ µν in exercise 6 bove.
QUANTUM FIELD THEORY EXAMPLE SHEET 1 7 8. We re to show tht the representtion of the Lorentz lgebr generted by the differentil opertors M στ = x σ τ x τ σ is in fct representtion i.e. tht it stisfies the Lorentz lgebr brcket condition. We will proceed by direct computtion, using identities of differentil opertors fmilir to most young children 5. M στ, M αβ] = (x σ τ x τ σ ( x α β x β α ( x α β x β α (x σ τ x τ σ, = x σ ( τ x α β + x σ x α τ β x σ ( τ x β α x σ x β τ α + x τ ( σ x α β x τ x α σ β + x τ ( σ x β α + x τ x β σ α + x α ( β x σ τ x α x σ β τ + x α ( β x τ σ + x α x τ β σ + x β ( α x σ τ + x β x σ α τ x β ( α x τ σ x β x τ α σ, =g τα x σ β g τβ x σ α g σα x τ β + g σβ x τ α g βσ x α τ + g βτ x α σ + g ασ x β τ g ατ x β σ, =g τα ( x σ β x β σ g τβ (x σ α x α σ g σα ( x τ β x β τ + g σβ (x τ α x α τ, M στ, M αβ] = g τα M σβ g τβ M σα g σα M τβ + g σβ M τα. 9. Recll tht the (free electromgnetic field is described by the Lgrngin L = 1 4 F µνf µν, where F µν = µ A ν ν A µ where A µ is the vector potentil. We re to show tht this theory is invrint under U(1 guge trnsformtions with A µ A µ + µ ξ(x, where ξ(x is n rbitrry differentible mp ξ : M 4 C. We re to compute the field equtions. Under the guge trnsformtion A µ A µ + µ ξ(x, we hve tht F µν µ A ν + µ ν ξ(x ν A µ ν µ ξ(x = µ A ν ν A µ = F µν, so tht F µν is itself invrint. Therefore, we see tht L is invrint under guge trnsformtions. Using the Euler-Lgrnge equtions, we see tht µ ( µ A ν ν A µ = A ν ν µ A µ = µ F µν =. Therefore, if one mkes the guge choice µ A µ =, then the field equtions demnd tht A ν =. 1. The Lgrngin for mssive vector field B is L = 1 4 F µνf µν + 1 2 B µb µ, where, s usul, F µν = µ B ν ν B µ. We re to derive the field equtions nd show tht they imply tht the field B µ is divergenceless. One cn compute the Euler-Lgrnge equtions directly nd even guess the nswer by nlogy to other, similr problems in this exercise sheet. One obtins µ F µν + B ν =. Becuse this eqution holds independently of x, we should find tht the expression on the left hnd side is lso divergenceless. Therefore, we see tht ν µ F µν + ν B ν = ν ( µ µ B ν µ ν B µ + ν B ν, = ( ν B ν ( µ B µ + ν B ν, = µ B µ, µ B µ =. 5 The one less-trivil identity tht we will use is x µ x ν = g ρν xµ x ρ = g ρν δ µ ρ = g µν.
8 JACOB LEWIS BOURJAILY