Tom Ouldridge Keble College Lattice Normal Modes and Phonons via Quantum Mechanics The Hamiltonian for the 1-D monatomic chain is a sum of coupled harmonic oscillators: 2 2 X pj m! H = + 0 (x − x )2 (1) 2m 2 j+1 j j Show that the following Fourier decomposition decouples them; −1=2 X −ikja −1=2 X ikja xj = N e Xk pj = N e Pk (2) k k into the from: 2 XP−kPk m! H = + 0 (4 sin2(ka=2))X X (3) 2m 2 −k k k where N is the number of atoms in the chain and a is the equilibrium atomic spacing. P 0 Hint: use the identity j exp(i(k + k)ja) = Nδk;−k0 . What are the allowed values of k? To do this part we simply substitute in the new operators as given to us. Be careful with your indeces! i(k+l)ja 2 ! X PkPle m! H = N −1 + 0 e−ika − 1 e−ikjaX e−ila − 1 e−iklaX : (4) 2m 2 k l j;k;l If we perform the j summation first, we can use the identity provided in the hint: 2 ! X PkPl m! H = δ + 0 e−ika − 1 e−ila − 1 X X δ ; (5) 2m k;−l 2 k l k;−l k;l 2 ! X PkP−k m! =) H = + 0 e−ika − 1 eika − 1 X X ; (6) 2m 2 k −k k 2 ! X PkP−k m! =) H = + 0 (2 − 2 cos(ka)) X X ; (7) 2m 2 k −k k 2 ! X PkP−k m! =) H = + 0 4 sin2(ka=2) X X : (8) 2m 2 k −k k We are perfectly at liberty to change the dummy index k to −k, obtaining exactly the form in Eqn. 3. In defining our Fourier transforms (Eqn. 2), our k values are exactly the the values of k within 2πm the first Brillouin zone which are allowed by periodic boundary conditions: i.e., k = Na . This is to provide the correct periodicity of xj. So far we have made no assumptions about whether we are dealing with a quantum or a classical system. In the classical case, we could use this form of H to derive the same dispersion as in question 2 (don't worry about the details of how to do this - the easiest way is with Hamiltonian mechanics). In the quantum case, we need to explicitly consider the observables a operators. 1 y y 1. Show that [Xk;Pq] = i¯hδk;q, and that Xk = X−k (and similarly for Pk ). To do this you will y need to use the hermitian property of xj: xj = xj. 2. Define the operator ak by: 1=2 m!k i y ak = (Xk + Pk ); (9) 2¯h m!k y where !k = 2!0j sin(ka=2)j. Show that [ak; aq] = δk;q, and hence that the Hamiltonian simplifies to the form: X y H = ¯h!k(akak + 1=2) (10) k 1. To do the commutator, we again just substitute in. To do this, first invert Eqn. 2, which is reasonably simple. 2 3 −1=2 X ikja −1=2 X −iqla [Xk;Pq] = 4N e xj;N e pl5 ; (11) j l −1 X ikja −iqla =) [Xk;Pq] = N e e [xj; pl] : (12) j;l Now we can use the good old commutation relations of second year, [xj; pl] = i¯hδj;l, i¯h X [X ;P ] = ei(k−q)ja = i¯hδ : (13) k q N i;j j The conjugation property is not too hard to show: 0 1y y −1=2 X ikja −1=2 X −ikja y −1=2 X −ikja Xk = @N e xjA = N e xj = N e xj = X−k (14) j j j 2. We now have everything we need to evaluate the second commutator: i [a ; ay] = −(! =! )1=2[X ;P ] + (! =! )1=2[P y;Xy] = δ : (15) k q 2¯h k q k q q k k q k;q y To prove that we can rewrite the Hamiltonian, it is easiest to expandh! ¯ k(akak + 1=2): 2 y ¯hmwk y −i i y y 1 y ¯h!k ¯h!k(akak + 1=2) = XkXk + XkPk + Pk Xk + 2 Pk Pk + : 2¯h m!k m!k (m!k) 2 (16) y y We now commute Pk and Xk, and use the conjugation relation derived earlier: 2 y mwk 1 −i i ¯h ¯h!k ¯h!k(akak+1=2) = XkX−k + 2 PkP−k + XkPk + X−kP−k − + : 2 (m!k) m!k m!k m!k 2 (17) 2 y mwk 1 −i i =) ¯h!k(akak +1=2) = XkX−k + 2 PkP−k + XkPk + X−kP−k) : 2 (m!k) m!k m!k (18) If we sum over all k, the third and fourth terms will cancel each other. 2 X y X mwk 1 =) ¯h!k(akak + 1=2) = XkX−k + 2 PkP−k : (19) 2 (m!k) k k Which is the Hamiltonian! 2 We have now decoupled the system into independent quantum oscillators. What is the energy spec- trum? Explain what each of the new operators means physically. Referring to your answer, or otherwise, explain what a phonon is. P E = k ¯h!k(nk + 1=2) y ak increases the energy of the kth normal mode by ¯hwk, and ak decreases it by the same amount. These quantized excitations of the normal modes can be interpreted as bosons, which we call phonons. 3.