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peskin量子场论课后答案(芝加哥大学版)

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Physics 443 Homework #1 Due Thursday, October 7, 20101.) Peskin the constant T must still be integrated over, however. a) Path integrate over x(t), subject to the boundary conditions xµ(0) = xµ, xµ(1) = yµ, yielding a deltafunction δ( ˙ p) along the path. Solve this constraint (find the set of functions that solve it) and path integrateover those p(t) to find the quantum mechanical propagation amplitudehy|xi = DF(x − y) =Z∞0dT(2πiT)−d/2exp?−i2? (m2− iǫ)T +(x − y)2 T?? ,where d is the number of spacetime dimensions. b) Use this integral representation to show that DFsatisfies(∂2+ m2)DF= iδ(d)(x − y) .c) Evaluate the T integral in terms of Bessel functions.2.) Peskin and Schroeder 9.2a-cHints: For 9.2a, it is sufficient to formulate the partition function in terms of a path integral; you are goingto evaluate it in part (b). For 9.2c, first show that the partition function can be formulated as a path integralover fields in Euclidean 4-space that are periodic in the imaginary time direction. The spatial field modes are harmonic oscillators; take the log of the partition function to get the free energy as a sum over modes of the free energy of each oscillator, and use your result for (b) to evaluate it. A second approach to (c) uses the methods of problem (1). Use the representationlog[Z] = log[det(−∂2E+ m2)] = trlog(−∂2 E+ m2)=Z d4xZ∞0dT Thx|e−T(−∂2 E+m2)|xi ,together with the representation of the matrix element derived in problem (1), to evaluate the functional determinant and hence the partition function. You may want to take ∂/∂m of the above expression toremove an m-independent divergence and render the integral finite.3.) Write a field theory action describing nonrelativistic scalar particles interacting via a potential U(~ x−~ y) (this action-at-a-distance form of interaction is permissible in a nonrelativistic setting, but not in relativisticfield theory, where it would break Lorentz invariance by selecting a preferred surface of simultaneity). Findthe corresponding Hamiltonian for the field. Use your expression for the energy in terms of fields and evaluate it to show that the expectation value of the Hamiltonian in the noninteracting ground state of a system ofN particles in a volume V is, to first order in perturbation theory,E(1) N=N − 1 2V˜U(0) ,where ˜U(~ q) =Z d3xU(~ x)exp[−i~ q · ~ x] .Use of ‘first quantized’ methods to derive this answer is not acceptable (the point of the exercise is to gainfamiliarity with quantized fields; you may find it useful, however, to compare the two approaches).QFT 1 :Problem Set 21.)We begin by attempting to motivate the action for a relativistic point particle appearing in the homework set. Perhaps the more familiar action is that given by the invariant length of the worldline: S1[x] = −mZ ds (ηµν˙ xµ˙ xν)1/2Where ˙ xµ≡ ∂sxµ≡ ∂xµ/∂s. In addition to being Poincare invariant, this action is invariant under arbitrary reparameter- izations of the worldline coordinate s. The coordinates xµof course transform like scalars under this transformation. The equation of motion for x(s) is the familiar:∂ ∂s? ˙ xν(ηµν˙ xµ˙ xν)−1/2? = 0We now introduce a new action which incorporates a field which transforms like a metric under reparameterizations of the worldline coordinate s. We will see that it leads to the same equations of motion for x(s).S2[x, g] = −m 2Z ds√g ss(gss∂sx · ∂sx + 1)Where ∂sx · ∂sx ≡ ηµν˙ xµ˙ xνand gssgss= 1. The equation of motion for g is found to be:gss= ∂sx · ∂sxThus, if the equations of motion are satisfied, g coincides with the metric along the worldlineinherited (through its imbedding in spacetime) from η . From this we find:S2[x, ∂sx · ∂sx] = S1[x]Thus the action S2leads to the same classical equations of motion for x as does S1. Wenow make a change of field variables by taking advantage of the fact that, in one dimension, we may replace the metric by a one-form. Thus we choose N =√g ss/m. This leads to the action : S3[x, N] = −1 2Z ds?N−1˙ x · ˙ x + N m2?From this action we derive the Hamiltonian:H = pµ˙ xµ− L = −1 2N?p · p − m2?Here pµ= ∂L/∂ ˙ xµ= −N−1˙ xµand the canonical momentum of the N variable vanishes.We may define the path integral associated with S3as follows :DF(y − x) =Z DNZx(s1)=yx(s0)=xDx eiS3[x,N]Perhaps more fundamentally, we may consider the path integral to be defined through the use of the following action :S4[x, p, N] =Z ds (pµ˙ xµ− H(p,x)) =Z ds?pµ˙ xµ+1 2N?p · p − m2??1We write the path integral as :DF(y − x) =Z DNZ DpZx(s1)=yx(s0)=xDx eiS4[x,p ,N]This may be seen to give the same result as the path integral involving S3since the action S4 is quadratic in p. Note that we do not integrate over the canonical momentum for N since it is identically zero. To ensure that this path integral converges we substitute m2→?m2− iǫ? and constrain the path integral to positive values of N .In the discussion of gauge fixing we are going to diverge a bit from the statement of the problem. The problem presents a symmetry under canonical transformations induced bythe Hamiltonian where N is treated as a Lagrange Multiplier for a first-class constraint. The symmetry is δx = αp, δp = 0, δN = −∂sα for arbitrary α(s). I am more comfortablediscussing fixing the symmetry of the action under diffeomorphisms; that is reparameteriza- tions of the time parameter s. This symmetry treats x and p as scalars and N as a one-form so that δx = −β∂sx, δp = −β∂sp, δN = −∂s(βN) for arbitrary β(s). I strongly suspect that the symmetries are equivalent and certainly lead to the same result. This symmetry allows us to transform N subject to the condition thatRdsN(s) is preserved as it mustbe under diffeomorphisms. The finite transformation of N is just the tensor transformation law: ˜N(˜ s)∂˜ s ∂s= N(s)We may use this freedom to transform any N(s) to˜N(˜ s) = 1. Then we have: Zs1s0dsN(s) =Z˜ s1˜ s0d˜ s = (˜ s1− ˜ s0) ≡ TWe may now do a further transformation to set the limits of the integral to s0= 0 and s1= 1 with N(s) = T. We cannot gauge away N entirely and T must be integrated over in the path integral. The principal reason for avoiding the tranformation in the homework is that I am not sure what the analog of theRdsN(s) constraint is. We are thus lead to thefollowing gauge fixed path integral:DF(y − x) =Z∞0dTZ DpZx(1)=yx(0)=xDx exp? iZ10ds?p · ˙ x +1 2T?p2− m2+ iǫ???(a)We will respect the apparent time-honored tradition in theoretical physics of treating thesolution of the path integral somewhat loosely. But first we present an expression that may be worked with to provide a perhaps more careful solution (here ∆ = 1/n) :DF(y − x) =limn→∞Z∞0dTnYj=0Z ddxjnYk=1Zddpk (2π)dδd(x0− x)δd(xn− y)×expi∆nXk=1?pk· (xk− xk−1)/∆ − T/2?p k· pk− m2+ iǫ??!We treat the x path integral, following an integration by parts in the action, as a functional Fourier transform : Zx(1)=yx(0)=xDx exp? iZ10ds (p · ˙ x)? = δ [ ˙ p] exp(i [p(1) · y − p(0) · x])2Inserting this into the path integral we blithely convert the p functional integral into an ordinary integral since ˙ p = 0:DF(y − x) = λdZ∞0dTZddp (2π)dexp?i?p · (y − x) +1 2T?p2− m2+ iǫ???Note that the integral over s in the action produces its integrand since ˙ p = 0. Also note that a factor λdhas been inserted to provide the normalization to be determined below. We now use the following formula for the integral of a gaussian : Z ddp e−ip·αe−ap2= e−α2/(4a)?πa?d/2Setting a = −iT/2 and α = (x − y) we find :DF(y − x) = λd?i 2π?d/2Z∞0dT T−d/2exp? −i/2h T?m2− iǫ?+ T−1(y − x)2i?(b)We now show that DFis a Green function for the Klein-Gordon equation. We find that : ?∂2+ m2− iǫ?DF(x)=λd?i 2π?d/2Z∞0dT T−d/2?m2− iǫ − x2/T2− id/T?× exp?−i/2?T?m2− iǫ?+ T−1x2??=2iλd?i 2π?d/2Z∞0dTd dT? T−d/2exp?−i/2?T?m2− iǫ?+ T−1x2???=−2iλd?i 2π?d/2 lim T→0h T−d/2exp?−ix2/(2T )?i= −2iλdδd(x)To see that the distribution given here is a delta function we may integrate it against a testfunction. We will find that the integral oscillates wildly in the T → 0 limit except near x = 0. We remove the test function (evaluated at x = 0) and the distribution integrates to 1 since it is a normalized gaussian for all T . This is of course pretty loose language but is essentially correct. To verify this result we return to the (normalized) expression for DFprior to performing the momentum integral. To conform with the definition in Peskin and Schroeder we choose the normalization λd= 1/2 and take p → −p in the integral :DF(x) =1 2Z∞0dTZddp (2π)dexp?−i?p · x −1 2T?p2− m2+ iǫ???We perform the T integral to find ;DF(x) =Zddp (2π)di e−ip·x (p2− m2+ iǫ)Thus,?∂2+ m2− iǫ?DF(x) = −iδd(x)3(c)The most straightforward way to approach this problem is to use a table of integrals or plug the expression for DFas an integral over T into a program like mathematica. The result is:DF(x) =1 2π?im2π√x2?(d/2−1) K(d/2−1)(−im√ x2)The following are graphs of x(1−d/2)Kd/2−1(x) (spacelike) in green and the real and imagi- nary parts of (−ix)(1−d/2)Kd/2−1(−ix) (timelike) in blue and red respectively. For d = 2 :0246810121416For d = 3 :0246810121416For d = 4 :024681012141642.) Peskin find the corresponding Noether current.Show that the corresponding Noether charge measures the total helicity of a collection of massless Dirac particles, and that the additionof a mass term to the Lagrangian violates the symmetry; find an equation that expresses the violation of current conservation by the mass. c) Find the Noether current related to charge conservation by considering a phaserotation of a Dirac field (of arbitrary mass)ψ′= exp[iα]ψ .2.) Peskin and Schroeder 3.13.) Peskin and Schroeder 3.4; you may omit part (d) – you did it in problem 1.4.) Peskin and Schroeder 3.71QFT 1 :Problem Set 31.)(a)We consider the Dirac Lagrangian: L[ψ] =¯ψ (i/ ∂ − m)ψWhere we have: {γµ, γν} = 2ηµνand¯ψ = ψ†γ0And we use the conventions:/ ∂ψ ≡ γµ∂µψand(γµ)†= γ0γµγ0Lorentz invariance means that:L[ψ](x) = L[ψ′](Λx)whereψ′(Λx) = MΛψ(x)Where Λ ∈ SO(3,1) such that: Λµν=?exp?−12iωαβLαβ??µν Where,?Lαβ?µν= i?ηαµδβν− ηβµδαν?And where MΛ∈ Spin(3,1) such that:MΛ= exp?−12iωαβSαβ?Where,Sαβ=i 4?γα, γβ?Here we have suppressed spinor indices. We now considerL[ψ′](Λx) =¯ψ′(Λx)(i/ ∂ψ′)(Λx) − m¯ψ′(Λx)ψ′(Λx)Now, M†Λγ0= γ0M−1Λsinceγ0?Sαβ?†γ0= −SαβThus, ¯ψ′(Λx) = ψ′†(Λx)γ0= ψ†(x)M†Λγ0=¯ψ(x)M−1Λ Also,(i/ ∂ψ′)(Λx) = iγµ∂ ∂yµψ′(y)???? y=Λx= iγµMΛ∂ ∂yµψ(Λ−1y)???? y=ΛxDefining a = Λ−1y and using the chain rule,∂ ∂yµψ(Λ−1y)???? y=Λx=∂ψ ∂aν(a)∂aν ∂yµ???? y=Λx=∂ψ ∂xν(x)?Λ−1?νµThus, (i/ ∂ψ′)(Λx) = i?Λ−1?νµγµMΛ∂νψ(x)1Thus we have: L[ψ′](Λx) =?Λ−1?νµ¯ψ(x)M−1ΛγµMΛi∂νψ(x) − m¯ψ(x)ψ(x)Now, using the γ matrix algebra, for infinitesimal ωαβ, one may show:?1 + i/2ω αβSαβ?γµ?1 − i/2ω αβSαβ?=?1 − i/2ω αβLαβ?µνγνThus, M−1ΛγµMΛ= ΛµνγνAnd, finally, L[ψ′](Λx) = L[ψ](x) =¯ψ(x)(i/ ∂ − m)ψ(x)(b)We consider the Lagrangian of a massless Dirac field:L0[ψ] =¯ψ (i/ ∂)ψA chiral rotation is given by:ψ′= eiαγ5ψwhereγ5= iγ0γ1γ2γ3This transformation can be seen to be a symmetry of the Lagrangian and thus of the action:L0[ψ′] =¯ψ′(i/ ∂)ψ′= ψ†e−iα(γ5)†γ0(i/ ∂) eiαγ5ψNow, it may be shown, (γ5)†= γ5and?γµ, γ5?= 0Thus, L0[ψ′] = ψ†e−iαγ5γ0e−iαγ5(i/ ∂) ψ =¯ψ(i/ ∂)ψ = L0[ψ]We now derive the Noether current. For an infinitesimal variation δψ , after integrating by parts andapplying the equations of motion, we find:S [ψ + δψ] − S [ψ] ≃Z d4x ∂µ?i¯ψγµδψ?Using δψ = iǫγ5ψ we find the conserved axial current:JµA=¯ψγµγ5ψwhere∂µJµA= 0This leads to the Noether charge:QA=Z d3xJ0A=Z d3xψ†γ0γ0γ5ψ =Z d3xψ†γ5ψUsing,ψ =?ψL ψR? andγ5=?−10 01?We find,QA=Z d3x? ψR†ψR− ψL†ψL?2To show that this operator measures the helicity of a collection of massless particles we introduce the matrices: σµ≡ (1,σ)and¯ σµ≡ (1,−σ)The γ matrices take the form:γµ=?0σµ ¯ σµ0?This allows us to decouple the Dirac equation for massless particles:i¯ σ · ∂ψL= 0andiσ · ∂ψR= 0We write the field operators as:ψL=Zd3p (2π)31p2E p?aL(p)αL(p)e−ip·x+ b†L(p)βL(p)eip·x?ψR=Zd3p (2π)31p2E p?aR(p)αR(p)e−ip·x+ b†R(p)βR(p)eip·x?Imposing the equations of motion we find:p · ¯ σαL(p) = p · ¯ σβL(p) = 0wherep · ¯ σ = Ep+ p · σAnd, p · σαR(p) = p · σβR(p) = 0wherep · σ = Ep− p · σThus, defining the helicity operator, h = p · σ/EpWe find: hαL(p) = −αL(p)andhβL(p) = −βL(p)And, hαR(p) = αR(p)andhβR(p) = βR(p)Using the orthogonality relations (see P+S pg 48):αL,R†(p)αL,R(p) = βL,R†(p)βL,R(p) = 2EpαL,R†(p)βL,R(−p) = βL,R†(−p)αL,R(p) = 0We find:: QA: =Zd3p (2π)3?a†RaR+ b†RbR− a†LaL− b†LbL?Thus we see that the Noether charge measures the helicity of a collection of particles. We now add a mass term to the Lagrangian and verify that the chiral symmetry is violated.L[ψ] =¯ψ (i/ ∂ − m)ψ = L0[ψ] − m¯ψψThus, L[ψ′] = L0[ψ′] − mψ†e−iαγ5γ0eiαγ5ψ = L0[ψ] − m¯ψ e2iαγ5ψThus for an infinitesimal variation δψ = iǫγ5ψ we find:S [ψ + δψ] − S [ψ] ≃ −ǫZ d4x?2im¯ψγ5ψ?3Thus the chiral symmetry is violated by the mass term. From above:S [ψ + δψ] − S [ψ] ≃ −ǫZ d4x ∂µ?¯ψγµγ5ψ?Thus we may express the violation of current conservation as:∂µ?¯ψγµγ5ψ?= 2im¯ψγ5ψIt is simpler to derive this by computing ∂µJµAand making use of the Dirac equation.(c)We now consider the symmetry under phase rotations:ψ′= eiαψClearly, L[ψ′] = L[ψ]For an infinitesimal variation δψ = iǫψ we find:S [ψ + δψ] − S [ψ] ≃ −ǫZ d4x ∂µ?¯ψγµψ?Thus we find the conserved current:Jµ=¯ψγµψwhere∂µJµ= 02.) Peskin translated in Sov. J. Nucl. Phys.1QFT 1 :Problem Set 41.) Peskin for instance the speed of sound in different directions, or for longitudinally versustransversely polarized phonons, typically differ.However these features are orientational in nature; theeffective action is still independent of the lattice spacing at leading order.)1QFT 1 :Problem Set 51.) Peskin however the last three terms take the form:2?=−2iλδijδkl2γ2µ2 (p1+ p2)2− m2σ3?=−2iλδikδjl2γ2µ2 (p1− p3)2− m2σ4?=−2iλδilδjk2γ2µ2 (p1− p4)2− m2σAt threshold (pa= 0):(p1+ p2)2= 4m2πand(p1− p3)2= (p1− p4)2= 0After a little algebra we find:iM = i(λ)3/2a µ3(3δijδkl− (δikδjl+ δilδjk))62.) Peskin & Schroeder 4.4We consider the quantized Dirac field interacting with a classical potential Aµ(x):HI=Z d3xe¯ψγµψ Aµ(a)We consider the (O(e)) matrix element:hp0s0|iT |psi = −ieZ d4xAµ(x) hp0s0|¯ψ(x)γµψ(x)|psiHere S = 1 + iT . Now, ignoring vacuum bubbles:hp0s0|¯ψ(x)γµψ(x)|psi = hp0s0|¯ψ(x)γµψ(x)|psiWhere,ψ(x)|psi = e−ip·xus(p)|0iandhp0s0|¯ψ(x) = h0| ¯ us0(p0)eip0·xThus, hp0s0|iT |psi = −ie ¯ us0(p0)γµus(p)˜Aµ(p0− p)Where, ˜Aµ(q) =Z d4xAµ(x)eiq·x(b)If Aµ(x) is time independent:˜Aµ(q) =Z dteiq0tZ d3xAµ(x)e−iq·x= (2π)δ(q0)˜Aµ(q)Where it is to be understood that˜Aµ(q) is the spatial Fourier transform of Aµ(x).With the following definition of M:hp0s0|iT |psi = iM(p0s0| ps) (2π)δ(Ep0− Ep)We find: iM(p0s0| ps) = −ie ¯ us0(p0)γµus(p)˜Aµ(p0− p)Thus, with the usual momentum space Feynman rules for fermions, we find the following vertex rule:pp0q⇒ieγµ˜Aµ(q)(q0= 0)7We consider the following expression for dσ :dσ =d3p0 (2π)31 2Ep0Z d2b??hp0s0|iT eib·P|φsi??2Here eib·Pis the translation operator perpendicular to the beam. Also,|φsi =Zd3k (2π)31√2E kφ(k)|ksiwherehφs| φsi =Zd3k (2π)3|φ(k)|2= 1Note that
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