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1 2 2 1 4 where W (φ) = 2 µ φ + 4 λφ , and the s means summing over the three directions x, y, z with ~n + s meaning the vector (nj + δjs), that is, one of the nearest neighbors in each direction. P The same form is a standard Hamiltonian in solid state physics, although W need not be given by φ2 and φ4 terms there. (This limitation comes from requiring renormalizability for a relativistic continuum theory.) Chapter 15 Let us first ask what translational invariance imposes on the states of this system. Under a translation ~n → ~n +~t, where ~t has integer components, U Phonons, Bloch Waves; implemented by an operator ~t, U −1φ U = φ , ~t ~n ~t ~n+~t Spontaneous Symmetry the Hamiltonian is invariant because it is a sum over all sites equally. We expect the states to transform as a sum of irreducible representations of the Breaking group k U~t |ψji =Γij(U~t) |ψii . The translation group is Abelian, so all of its irreducible representations are one dimensional, because all of the Γ(U)’s commute and can be simul- 15.1 Translations on a Lattice taneously diagonalized. The group is generated by the three translations by iakj tj We have seen that symmetry groups may have different implications on the one lattice spacing, which must have unitary representations Γ(U~t)= e . ~ physics. The first we have seen is an “internal symmetry”, global, manifested The k is a momentum. For an infinite lattice it may take on any value. But 2π by multiplets of states which form the space acted upon by a representation kj → kj + a has no effect on any of the fields. So both k’s are the same of the group. Quite different is a local symmetry, which, although it is a representation of lattice translations. much larger group, being a product of groups at each space-time point, does It is possible to reformulate the Hamiltonian in terms of operators which not have a corresponding map of different physical states into each other. transform more simply under translations. Let a Rather, through gauge invariance, it makes some of the apparent fields Aµ i~k·(a~n) a q~k = e φ~n unphysical, so that the 4n fields Aµ (n is the dimension of the group) only describe 2n degrees of freedom, transversely polarized “photons”. X~n Another form of symmetry’s effect is given by the translation group. Con- ~ π π where we restrict k to the interval − a , a . Then sider a spatial cubic lattice with sites ~x = (nx, ny, nz)a, with ni ∈ Z, a the π/a π/a  lattice spacing, and with a field φ on each site. If we think of this as an 3 3 ia~k·(~n−~n ′) d k q˙ q˙ = d k e φ˙ φ˙ ′ approximation to a relativistic field theory with Lagrangian density ~k −~k ~n ~n −π/a −π/a ′ Z Z X~n,~n 1 µ 1 2 2 1 4 3 3 L = (∂ φ)(∂µφ) − µ φ − λφ , 2π 2π ˙ ˙ ′ ′ ˙2 2 2 4 = φ~n φ~n δ~n,~n = φ~n ′ a a the latticized Hamiltonian density is X~n X~n     X~n 1 1 H = φ˙2 + (φ − φ )2 + W (φ ), so the momentum term can be easily reexpressed. Without the time deriva- 2 ~n 2a2 ~n+s ~n ~n 1 2 2 s tives the same thing holds, so the 2 µ φ is no problem either. X 163 618: Last Latexed: April 25, 2017 at 9:45 165 166. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro

3 π/a 3 −ia~k·~n Inverting, φ~n =(a/2π) −π/a d k e q~k, so we see that So we have interactions of phonons with the total incoming momentum not conserved, but conserved modulo 2π/a. If we write a momentum operator R3 π/a † a 3 −ia~k·~n −ia~k·~s P~ = ~ ~ka a~ we can show that k ~k k φ~n+~s − φ~n = 3 d k e e − 1 q~k, (2π) −π/a Z   P ia~t·P~ so U~t = e . 6 1 1 1 ~ ~ ′ P~ is not a conserved quantity, because only U are symmetries, not other (φ − φ )2 = d3kd3k′ e−ia(k+k )·~n ~t 2a2 ~n+~s ~n 2π 2a2 translations by non-integer numbers of lattice spaces. But P~ is conserved ~n,s   Z ~n X X modulo 2π/a, so U~t is conserved. −ia~k·~s −ia~k ′·~s e − 1 e − 1 q~kq~k ′ . The nuclei of a crystal lattice carry degrees of freedom which are displace-     ments from the lattice points, but they are not fields defined for all ~x, so the But 3 translational modes only involve integer numbers of lattice spaces. The elec- ~ ~ ′ 2π e−ia(k+k )·~n = δ3(~k + ~k ′), trons, however, at least the ones not tightly bound, need to be treated as fields a ~n and therefore dependent on ~x as a continuous variable. We can ask how the X   ~ ~ ′ ~ lattice translational invariance affects the wave functions ψ(~x) for an electron so ~k ′ = −~k, e−iak·~s − 1 e−iak ·~s − 1 = 4 sin2 ak·~s , so the gradient term 2 which experiences a periodic potential V (~x) = V (~x + a~t). Again the possi- is       ble states can be decomposed into irreducible representations, which as the a 3 2 a~k · ~s ~ ~ 3 2 group is Abelian are one-dimensional and unitary, so ψ(~x + a~t)= eiak·tψ(~x). d k sin q~kq−~k. 2π a2 2 ~ s ! Then, if we write ψ(~x) = eik·~xu (~x), we see that u (~x + a~t) = u (~x), so the   Z X ~k ~k ~k † We note that hermiticity of φ implies q = q . So we see that the Hamilto- Bloch function u~k(~x) is periodic on the lattice. But we must keep in mind n −~k ~k nian, except for the terms in W higher order than 2nd order in φ, are diagonal that the electron wave function is not periodic. ~ ~ quadratic operators in q , of a harmonic oscillator type. The quartic term As before, the irreducible representation k is defined only for k modulo ~k Bravais gives a coupling interaction between different q~ ’s. the reciprocal (or ) lattice, so is essentially defined only within the k π π Brillouin zone (kj ∈ [− , )). But there may be several Bloch functions, in The q~k operators transform very simply under the translations: a a bands, for the same ~k. For example, a free electron (V = 0) is trivially in a −1 ia~k·~n ia~k·(~n ′−~t) −ia~k·~t U q U = e φ = e φ ′ = e q , i~k·~x ~t ~k ~t ~n+~t ~n k periodic potential, and has eigenfunctions e for all ~k, so when considered ~n ~n ′ ~ ′ π π X X as Bloch functions with k j ∈ [− a , a ), the higher k values will be shown as ′ so U only changes the phase of q~k. higher energy states with the same ~k . ~ The q~k andq ˙~k, for each k separately, can be combined into harmonic oscillator raising and lowering operators a† and a , which correspond to the ~k ~k 15.2 Spontaneous Symmetry Breaking creation and annihilation of quanta of momentum ~k. The quadratic terms † in the Hamiltonian become ~ ω~ a a~ , which simply counts the number of k k ~k k We will return to momenta and the translation group in a relativistic contin- ~ uum theory later, only for the continuum. But first let us consider another phonons of momentum k, multipliesP by the energy ω~k of each, and sums. ~ −ia(P ki)·~t symmetry. Invariance of a term in the Hamiltonian ∝ i q~k requires e = 1, or i The Hamiltonian ~ 2π Q ki = × integer. 1 ˙2 1 2 1 2 2 1 2 2 a H = φ + φ ~ − φ~n + µ φ + λ φ i ~n a2 ~n+t ~n ~n X 2 2 2 4

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is invariant under a global transformation φ~n → −φ~n. In fact, if one considers mechanical states then have wave functions ψ(φ1,φ2,...) centered around 2 φ~n to be a vector in some N dimensional real vector space, with φ~n := φ1 = φ2 = ··· = φV = ±φ0 if there are V points in our lattice. The overlap 2 j(φ~n)j , H is invariant under SO(N) global transformations on the φ’s. of these states, however, is approximately the V ’th power of the single site Consider the classical ground state of the system. Any time derivative overlap, V is likely to be very large, and for the infinite lattice limit, there P or gradient only increases the energy, so the ground state is φ~n = constant is no overlap at all! And so the quantum system has a true degeneracy of =: φ0, with eigenstates, including the ground state. 1 1 V (φ)= µ2φ2 + λφ4 Consider a ferromagnet below the Curie point. The magnetic dipoles 2 4 want to line up, but which way is immaterial. They must all line up in the taking its minimum value at φ0. same way, however, to get the lowest energy state. Once the whole material is lined up, the possibility of a transition to one of 2 φ If µ and λ are positive, the minimum is clearly at V( ) the other degenerate vacuum states is an exponentially decreasing function of φ0 = 0, and this classical ground state is invariant the volume of the system. So the statement that the states of the system form 2 under sign reversal or rotation of φ. If µ < 0, multiplets of the group becomes irrelevant — the relevant physical states are maximum however, the potential has a local at the ones close enough to the particular original ground state for transitions φ = 0, with two degenerate minima at φ = ±φ0 =6 to be possible. The Hamiltonian still has a symmetry group, but the relevant 0. If φ is a vector, this diagram is rotated into φ states no longer transform as an irreducible representation of it. This is called the bottom of a wine bottle, and the minimum spontaneous symmetry breaking φ . becomes ring or sphere. 0 A particularly interesting feature emerges when the symmetry group is a If we were talking about a one-dimensional continuous one. Consider φ as an N-vector, and let φ0 be the “vacuum” value. 2 2 b p |µ | 2 Then the Hamiltonian is invariant under a global symmetry eiω Lb with L the quantum mechanics problem H = 2m − 2 x + ψ1ψ 2 b λ 4 generators of the Lie algebra of SO(N). If we consider a local transformation 4 x , the two classical ground states at x = ±φ0 = iωb(~x)L 2 e b , the Hamiltonian is invariant except for the gradient term, which ± |µ |/λ would become two approximate eigen- b ~ ~ b iω (~x)Lb states ψ and ψ concentrated around the two φ gets a piece ∼ ∇φ · ∇ω Lbφ. Then the energy of the state e |Ωi is p 1 2 minima. If there were no overlap, ψ1 and ±ψ2 b b 1 would be eigenstates of the same energy. hΩ| e−iω (~x)Lb Heiω (~x)Lb |Ωi = hΩ| H + i[H,ωbL ] − [[H,ωbL ],ωcL ]+ ···|Ωi . b 2 b c ψ+ But as there is some overlap, the true eigen- The first term is just the ground state energy E0. The second term is states would be linear combinations 3 b d x∇~ ω (~x) hΩ| ∇~ φ(~x)Γ(Lb)φ(~x) + h.c. |Ωi. As Γ(Lb) is hermitean, the op- φ erator is ∇~ (φΓ(L )φ), which therefore vanishes as hΩ| φ(~x)Γ(L )φ(~x) |Ωi is ~x ψ1(x)+ ψ2(x) ground state R b b independent by translational invariance. Therefore the leading term in ∆E ψ1(x) − ψ2(x) first excited state ψ− is proportional to the integral over space of the square of the gradient of ω. with an energy separation proportional to the If the variation of ω takes place over a region of length L, for a total fixed + variation of ∆ω, ∇ω ∝ ∆ω/L, and the region over which this energy density overlap between ψ1 and ψ2. ψ has slightly lower energy than ψ−, which bends a bit more. φ is increased is ∝ L, so

For our lattice, there are also a set of degenerate classical ground states, ∆ω 2 1 ∆E ∝ L × ∝ ∝ k with all φ~n = ±φ0, but all the φ~n’s must be the same. The quantum L L   618: Last Latexed: April 25, 2017 at 9:45 169 170. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro if ω(x) varies roughly as eikx. Thus if the variation is slow enough, the energy vanishes, so that this corresponds to a massless excitation. Thus there is one of this state differs from the ground state by arbitrarily little, proportional to massless Goldstone boson for each dimension in the coset space G/K. the momentum of the “spin-wave”, which is rotating the φ field with a gentle We have not exhausted the ways symmetries can work their magic, how- long wavelength. The added energy can be attributed to this spin-wave or ever. We see that a broken symmetry can give rise to massless scalar particle Goldstone boson, which has an energy proportional to its momentum as for directions in the Lie algebra. We previously saw that gauge theories would be expected for a zero-mass particle. gave massless vector particles for each direction in a Lie algebra. The Higgs Let’s consider this more generally. Suppose the theory is invariant under mechanism combines these two — and the massless vectors eat the Goldstone a Lie group G with Lie algebra G, but the vacuum state |Ωi is invariant bosons, gaining the third degree of freedom which allows them to become fat, only under a subgroup K with Lie Algebra K. If the fields transform under a i.e. massive. This is the basis of the electroweak interactions, with the weak j iω Γ(Lj ) ± 0 constant transformation φa → e φb ≈ φa + α∆a(φ) acting on |Ωi vector particles W and Z getting mass from the Higgs field, while the pho- ab ton retains its masslessness in the direction of the unbroken Maxwell gauge the energy of the transformed vacuum state 1 becomes invariance. Before we do this, however, let us consider what happens in ∂V (φ) more detail in SO(N) scalar field theory. V (φa + α∆a(φ)) ≈ V (φa)+ α∆a(φ) = V (φa) ∂φa because the theory, and V , must be invariant under G. So

∂V (φ) ∆a(φ) =0 ∂φa for all fields φ. Differentiate with respect to φb and set φ to its value in the vacuum state φ0:

∂∆ ∂V ∂2V a + ∆ (φ ) =0. ∂φ ∂φ a 0 ∂φ ∂φ b φ0 a φ0 a b φ0

∂V But the first term vanishes as = 0 because φ minimizes V . So ∂φ 0 a φ0

∂2V ∆ (φ ) =0. (15.1) a 0 ∂φ ∂φ a b φ0

2 2 ∂ V Note that mab = is a mass term in the lagrangian. Now if our ∂φa ∂φb j ω Lj lies within K and leaves the vacuum state φ0 invariant, ∆a(φ0) = 0 and j we learn nothing from (15.1). But if ω Lj 6∈ K, ∆a(φ0) is a nonzero vector in the space of φ, and then (15.1) tells us that the mass matrix acting on ∆a(φ0)

1V can represent the energy of a full quantum state |φi for a φ(x) field configuration.