PHY 396 L. Solutions for homework set #21.

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1 PHY 396 L. Solutions for homework set #21. Problem 1a): The difference between a circle and a straight line is that on a circle the path of a particle going from point x 0 to point x does not need to be straight but may wrap around the whole circle one or more times. Indeed, let us compare a particle moving on a circle according to xt) modulo 2πR) with a particle moving on an infinite line according to yt). If the two particles have exactly the same velocities at all times, dx dt dy dt S.1) and similar initial positions x 0 = y 0 according to some coordinate systems) at time t = 0, then after time T one generally has yt) = xt) + 2πR n S.2) for some integer n = 0,±1,±2,±3,... because the xy) path may wrap around the circle n times while the yt) path may not wrap. For example, the two paths depicted below have same constant) velocities and begin at y 0 = x 0 but end at yt) = xt)+2πr 2: yt) yt) 2πR xt) modulo 2πR) xt) x 0 0 t y 0 t It is easy to see that the paths xt) modulo 2πR) and yt) modulo nothing) are in one-toone correspondence with each other, provided we restrict the initial point y 0 of the particle on 1

2 the infinite line to a particular interval of length L = 2πR, say 0 y 0 < 2πR. Consequently, in the path integral for the particle on the circle xt=t)=x modl) xt=0)=x 0modL) D [xt)modl)] = n= yt=t)=x +nl yt=0)=x 0 D [yt)]. S.3) Furthermore, in the absence of potential energy, the circle path xt)mod L) and the corresponding line path yt) have equal actions S[xt)modL)] = S[yt)] = dt [ M 2 ẋ 2 = M 2 ẏ2], T 0 S.4) and therefore U circle x ;x 0 ) = xt=t)=x modl) D [xt)modl)]e is[xt)modl)]/ h = = xt=0)=x 0modL) n= n= yt=t)=x +nl yt=0)=x 0 D [yt)]e is[yt)]/ h U line y = x +nl;y 0 = x 0 ). 1) Q.E.D. Problem 1b): For a free particle living on an infinite line the evolution kernel is given by U line y ;y 0 ) = M i 2πi ht exp h S classical = ī h hence according to eq. 1), a particle on a circle has kernel Mx x 0 ) 2 2T ), 3) U circle x ;x 0 ) = M 2πi ht + n= ) im exp 2 ht x x 0 +nl) 2. S.5) 2

3 To evaluate this sum, we use Poisson re-summation formula 2), which gives n= ) im exp 2 ht x x 0 +nl) 2 = + l= dν exp ) im 2 ht x x 0 +νl) 2 e 2πilν. S.6) Rearranging the exponential, we have im 2 ht x x 0 +νl) 2 + 2πilν = iml2 2 ht and therefore ν + x x 0 L + 2πl ht ML 2 ) 2πil x x 0 L i ht2πl)2 ML 2, S.7) + dν exp Consequently, ) im 2 ht x x 0 +νl) 2 e 2πilν = 2πi ht ML 2 exp 2πil x x 0 L 2πl)2 i ht ML 2 ). S.8) U circle x ;x 0 ) = = 1 L M 2πi ht + 2πi ht ML 2 exp 2πil x x 0 L l= l= e ipx x 0)/ h e ite/ h 2πl)2 i ht ML 2 ) S.9) where p = 2π hl L = hl R and E = p2 2M. S.10) Problem 1c): This is obvious from eqs. S.9) and S.10). 3

4 Problem 3 the textbook problem 9.2, part c): In class, we have learned the path-integral formula for the partition function of a quantum particle, ZT) exp itĥ) = xt)=x0) D[xt)] exp is[xt)] ). S.11) In statistical mechanics, the partition function at temperature T is defined as ZT) Trexp βĥ), β = 1 T. S.12) The two partition functions are related by analytic continuation of the real time T to the imaginary time iβ = it 1, thus Z SM T) = Z QM T = it 1 ). S.13) In terms of the path integrals, this relation corresponds to going from the Minkowski path integral to the Euclidean path integral ZT) = xβ)=x0) D[xt E )] exp S E [xt E )] ) S.14) where the Euclidean action is S E [xt E )] = β 0 dt E m 2 ) 2 dx + Vx)). S.15) dt E Note the boundary conditions for the Euclidean path integral S.14): after the Euclidean time interval β = 1/T, the particle must come back to its starting point. In other words, at finite temperatures, the motion in Euclidean time is periodic with period β = 1/T. 4

5 Generalizing eq. S.14) from particle mechanics to field theory is quite straightforward. For a real scalar field field φx) with a Euclidean Lagrangian L E = 1 2 φ)2 + Vφ) S.16) we have finite-temperature Partition function Zβ) = φx,β)=φx,0) D[φx,x 4 )] exp d 3 x β 0 dx 4 12 φ) 2 + Vφ)). S.17) Again, the finite temperature translates into the geometry of the Euclidean 4D spacetime: The Euclidean time x 4 = it is of finite extent β = 1/T and the scalar field is subject to the periodic boundary condition; the other 3 dimensions x 1,x 2,x 3 are infinite as usual. For the free scalar field, the Euclidean action is a quadratic functional S E [φx E )] = 1 2 d 4 x E φm 2 2 )φ, S.18) so the path integral S.17) is a Gaussian integral that can be formally evaluated as the determinant Zβ) = const Det[m 2 2 E ] ) 1/2 S.19) Or in terms of the Helmholtz s free energy, F T logz = const + T 2 logdet[m2 E 2 ] = const+ T 2 Trlog[m2 E 2 ]. S.20) To actually evaluate the trace here, we diagonalize the m 2 E 2 operator via Fourier transform to the momentum space. However, because of the periodicity of the Euclidean time coordinate, 5

6 the Euclidean energies k 4 have discrete rather than continuous spectrum, k 4 = 2π β integer. S.21) Consequently, the trace evaluates to Trlog[m 2 2 E ] = d 3 k 2π) 3 k 4=2πT n logm 2 +k 2 +k 2 4) S.22) and the free energy of the quantum field becomes F = const + T 2 d 3 k 2π) 3 k 4=2πT n logm 2 +k 2 E ) S.23) Now let s use the Poisson s resummation formula 2) to re-arrange the sum over discrete k 4 momenta as k 4 Fk 4 ) = β Consequently, the free energy S.23) becomes l= dk4 2π Fk 4)e iβlk4. S.24) F = const l= d 4 k E 2π) 4 eiβlk4 logk 2 E +m2 ). S.25) In the zero-temperature limit β, the sum l reduces to the l = 0 term while all the other terms are suppressed by the rapidly changing phase e iβlk4. In the general spirit of subtracting the zero-point energy contribution, we should therefore get rid of the l = 0 term. Since all the other terms come in symmetric pairs ±l 0, we arrive at F = l=1 d 4 k E 2π) 4 eiβlk4 logk 2 E +m2 ). S.26) Formula S.26) has a nice 4D form, but for the purpose of comparison with the ordinary statistical mechanics, let us integrate over the k 4 before we integrate over the 3 momentum k. 6

7 For fixed k and l, we need to calculate I = dk4 2π eiβlk4 logk 2 4 +E2 ) S.27) where E 2 = m 2 + k 2, and the best way to evaluate this integral is to deform the integration contour in the complex k 4 plane away from the real axis. In terms of k 4, the logarithm logk4 2+ E 2 ) has one branch cut running from +ie to +i and another cut running from ie to i. Lets move the integration contour up towards +i ) until it wraps around the upper branch cut: k 4 Inotherwords, k 4 = ie1+x+iǫ)onitswaydownfromx = + tox = 0andk 4 = ie1+x iǫ) on its way up from x = 0 back to k = +. Consequently, the integral becomes I = ie 2π = E and hence dxe βle1+x) [ loge 2 2x x 2 +iǫ)) loge 2 2x x 2 iǫ)) = 2πi ] dxe βle1+x) = e βle βl F = l=1. S.28) d 3 k e βlek) 2π) 3. S.29) βl At this point, it s convenient to sum over l before integrating over the 3D momentum k, thus 1 β e βe ) l l=1 l = T log 1 e βe), S.30) 7

8 and therefore FT,m) = d 3 k ) 2π) 3 T log 1 e βe k. S.31) Finally, let us compare our result S.31) for the free energy of the free scalar quantum field with the conventional statistical mechanics of identical spinless relativistic bosons. In the SM of identical bosons, FT,m) = where each oscillator mode contributes d 3 k 2π) 3 Fharmonic oscillator T,E k) S.32) F harmonic oscillator T,E k) = T logz harmonic oscillator = T log2sinheβ/2)) = 1 2 E + T log 1 e βe). S.33) Subtracting the zero-point energy 1 2E and substituting into eq. S.32), we arrive at precisely eq. S.31). Thus, functional quantization of the field theory correctly reproduces the free energy of the field s quanta. Problem 11.1a): The easiest way to compute the correlation function of exp+iφx 1 )) and exp iφx 2 ) is in terms of functional integrals: Te +iˆφx 1) e iˆφx 2) = = D[Φx)]e is[φx)] e +iφx1) e iφx2) D[Φx)]e is[φx)] ) D[Φx)] exp i L+JΦ)d d x ) Z[J] D[Φx)] exp i Ld d Z[0] x S.34) where Jx) = δ d) x x 1 ) δ d) x x 2 ). S.35) 8

9 Moreover, for a free scalar field Φx) ) Z[J] = Z[0] exp 2 1 d d x d d yjx)g F x y)jy). S.36) For the source as in eq. S.35), the double integral inside the exponential is simply 2G F 0) 2G F x 1 x 2 ), hence Te +iˆφx 1) e iˆφx 2) = Z[J] Z[0] ) exp G F x 1 x 2 ) G F 0). S.37) Problem 11.1b): Under the axionic symmetry Φx) Φx) α, the derivatives µ Φ, µ ν Φ, etc. are invariant but the field Φ is not. Consequently, the most general effective Lagrangian for the quantum Φx) field must be a function of its derivatives only, L = ρ 2 Φ)2 + A 2 2 Φ) 2 + B 4 Φ)4 +. S.38) Now consider the renormalizability. In d spacetime dimensions, the scalar field Φ has dimensionality d 2 1, hence a Lagrangian term involving n fields and m derivatives has dimensionality = nd 2 + m n). S.39) Renormalizability requires d and hence d n 2) + m n) 0. S.40) 2 On the other hand, axionic symmetry requires m n no field without a derivative) while any term with n < 2 can be disregarded as a total derivative. All these conditions leave just one possibility assuming d > 0), namely m = n = 2 and hence L renorm. = ρ 2 Φ)2 + nothing else. S.41) In other words, in the infrared regime where the non-renormalizable interactions become irrelevant, Φx) is a free massless field. 9

10 Problem 11.1c): Suppose a theory with a global phase symmetry U1) appears to have a symmetry-breaking VEV of a complex field Sx) = Ax) e iφx), A > 0. S.42) The radial field A is massive, so its fluctuations decouple from the low-energy effective theory. All we have at low energies is the VEV A > 0 andthe massless Goldstonefield Φx) and we saw in part b) that Φx) is effectively a free field with Lagrangian S.41). It s non-canonically normalized, so TΦx)Φy) = 1 ρ G 0x y) S.43) where G 0 x y) is the usual Feynman propagator for m 2 = 0. Consequently, according to eq. S.37), TSx)S y) = A 2 Te +iˆφx 1) e iˆφx 2) ) = A 2 G0 x y) G 0 0) exp S.44) ρ C 2 expg 0 x y)/ρ) where C = A e G00)/2ρ absorbs the UV corrections to the bare A parameter. Similarly, in the Euclidean d-dimensional space of the statistical mechanics, the correlation function becomes Sx)S y) = C 2 expg E 0 x y)/ρ) S.45) where G E 0 x y) = d d p E 2π) E e ipx y) p 2 E S.46) is the Euclidean propagator of a massless scalar. By the SOd) symmetry, it depends only on 10

11 the Euclidean distance r = x y, and for m 2 = 0 this dependence is a pure power law: G E 0 x) = Γd 2 1) 4π d/2 r 2 d. S.47) Specifically, in d = 3 Euclidean dimensions D E 0 = 1/4πr and hence 1 Sx)S y) d=3 = C 2 exp 4πρ 1 ). S.48) r Likewise, in d = 4 dimensions, 1 Sx)S y) d=4 = C 2 exp 2π 2 ρ 1 ) r 2. S.49) In both cases, we see extremely strong self-correlations of the Sx) field at very short distances. On the other hand, in the long-distance limit the correlated expectation valuess.48) ands.49) remain finite. Indeed, in any dimension greater then two Sx)S y) d>2 r C2 > 0. S.50) Physically, such asymptotic behavior is characteristic of non-trivial vacuum expectation values: By cluster expansion, Sx)S y) d>2 r Sx) S y) S 2, S.51) thus the physical meaning of the limit S.50) is Sx) = C > 0. S.52) In other words, the U1) symmetry of the complex Sx) field is spontaneously broken. 11

12 One the other hand, in one Euclidean dimension, G E 0 = 1 2r and hence Sx)S y) d=1 = C 2 exp r ). S.53) 2ρ This correlated expectation value remains finite at short distance and decreases exponentially at largedistances. Indeed, foranydimension dless thantwo G E 0 x y) forr = x y and hence Sx)S y) d<2 r 0. S.54) In terms of the cluster expansion S.51), this means S = 0 despite the classical formula Sx) A > 0, the quantum theory has a zero VEV in d < 2 and the U1) phase symmetry remains unbroken. In the borderline case of exactly two dimensions, D0 E = 1 2π logr + const and hence Sx)S y) d=2 r 1/2πρ. S.55) Such scaling behavior correspond to the absence of dimensionful parameters in the theory the spinwave modulusρisdimensionless ford = 2. Amongotherthings, thescaling behaviors.55) provides for vanishing of the correlated expectation value in the infinite distance limit. Thus, similarly to the d < 2 case, we again have S = 0 and the unbroken O2) symmetry. The bottom line of this exercise is to illustrate the general rule: In d > 2 spacetime dimensions, exact symmetries of the action may become spontaneously broken by vacuum expectation values such as S. But in d 2 dimensions, quantum effects destroy any VEV that would break a continuous symmetry. However, the discrete symmetries may be spontaneously broken in two dimensions or in fractional dimensions d > 1. 12

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