Tom Ouldridge Keble College Lattice Normal Modes and via

The Hamiltonian for the 1-D monatomic chain is a sum of coupled oscillators:

2 2 X pj mω  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 mω  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 mω     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 mω     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 mω     =⇒ H = + 0 e−ika − 1 eika − 1 X X , (6) 2m 2 k −k k

2 ! X PkP−k mω =⇒ H = + 0 (2 − 2 cos(ka)) X X , (7) 2m 2 k −k k

2 ! X PkP−k mω =⇒ 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 ). In the quantum case, we need to explicitly consider the observables a operators.

1 † † 1. Show that [Xk,Pq] = i¯hδk,q, and that Xk = X−k (and similarly for Pk ). To do this you will † need to use the hermitian property of xj: xj = xj.

2. Define the operator ak by: 1/2 mωk  i † ak = (Xk + Pk ), (9) 2¯h mωk † where ωk = 2ω0| sin(ka/2)|. Show that [ak, aq] = δk,q, and hence that the Hamiltonian simplifies to the form: X † H = ¯hωk(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.   −1/2 X ikja −1/2 X −iqla [Xk,Pq] = N e xj,N e pl , (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:

 † † −1/2 X ikja −1/2 X −ikja † −1/2 X −ikja Xk = N e xj = 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 , a†] = −(ω /ω )1/2[X ,P ] + (ω /ω )1/2[P †,X†] = δ . (15) k q 2¯h k q k q q k k q k,q

† To prove that we can rewrite the Hamiltonian, it is easiest to expandhω ¯ k(akak + 1/2):  2        † ¯hmwk † −i i † † 1 † ¯hωk ¯hωk(akak + 1/2) = XkXk + XkPk + Pk Xk + 2 Pk Pk + . 2¯h mωk mωk (mωk) 2 (16) † † We now commute Pk and Xk, and use the conjugation relation derived earlier:  2        † mwk 1 −i i ¯h ¯hωk ¯hωk(akak+1/2) = XkX−k + 2 PkP−k + XkPk + X−kP−k − + . 2 (mωk) mωk mωk mωk 2 (17)  2        † mwk 1 −i i =⇒ ¯hωk(akak +1/2) = XkX−k + 2 PkP−k + XkPk + X−kP−k) . 2 (mωk) mωk mωk (18) If we sum over all k, the third and fourth terms will cancel each other.

 2    X † X mwk 1 =⇒ ¯hωk(akak + 1/2) = XkX−k + 2 PkP−k . (19) 2 (mωk) k k Which is the Hamiltonian!

2 We have now decoupled the system into independent quantum oscillators. What is the spec- trum? Explain what each of the new operators means physically. Referring to your answer, or otherwise, explain what a is. P E = k ¯hωk(nk + 1/2) † ak increases the energy of the kth normal mode byhw ¯ k, and ak decreases it by the same amount. These quantized excitations of the normal modes can be interpreted as bosons, which we call phonons.

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