Chapter 15 Phonons, Bloch Waves; Spontaneous Symmetry Breaking

Chapter 15 Phonons, Bloch Waves; Spontaneous Symmetry Breaking

164. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro 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 4 618: Last Latexed: April 25, 2017 at 9:45 167 168. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro 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.

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