Quantum Theory of Matter

Transcription

1 Imperial College London Department of Physics Professor Ortwin Hess Quantum Theory of Matter Spring Periodic Structures 1.1 Direct and Reciprocal Lattice (a) Show that the reciprocal lattice of a reciprocal lattice is the real lattice. (b) Show that the fcc and bcc lattices are reciprocal to each other. 1. Face-centred rectangular lattice (a) Determine the reciprocal lattice of a face-centred rectangular lattice. Which Bravais-lattice does one obtain? (b) Sketch the first Billouin zone of a face-centred rectangular lattice. 1

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15 Quantum Theory of Matter Problem Sheet Answers Ortwin Hess and Marc Coury April 4, 014 Consider the following 1D Hamiltonian for two coupled harmonic oscillators H = p 1 M + p m + K x 1 + k (x 1 x ), (1) where particles 1 and, of masses, M and m respectively, have momentum squared operators p 1 = h x 1 p = h x and x 1, x are the displacements from equilibrium of particle 1 and.,, ().1 Part (i) Calculate the eigenvalues of H To find the eigenvalues of the Hamiltonian, Equation (1), we need to express the problem as decoupled simple harmonic oscillators. The energy of simple harmonic oscillators is just ε n = hω(n + 1 ). To do this we do something very similar to the process described in the notes, section 3.1. This simplified Hamiltonian with only two particles is already in the harmonic form. This means that we can express it as H = p 1 M + p m + 1 V x i x j, (3) x ij i x j which is the kinetic energy operator plus the harmonic potential, where 1 ij V x i x j x i x j = 1 ( ) ( K + k k x1 x k k ) ( x1 x ). (4) You can prove to yourself that expanding out Equation (4) and inserting it into Equation (3) gives you the original Equation (1). We now change the weighting of the displacements from equilibrium to be x 1 = Mx 1, x = mx, (5) p 1 = p 1 M, p = p m. (6) Note that to find Equation (6) we have substituted Equation (5) into Equation (). We can now express the 1 Hamiltonian using the dynamic matrix, D ij = V mim j x i x j, H = p 1 + p + 1 D = ( K+k M k Mm D ij x i x j, (7) ij k Mm k m ). (8) 1

16 A physical explanation the dynamic matrix Recall that the final term of the Hamiltonian, Equation (1), is the effective potential that describes the energy landscape that the particles move within. In Equation (3) we are assuming that we can adequately describe this landscape by the second order Taylor expansion around the equilibrium position of the particles (in this case it is exact but when you re describing atoms in materials it is approximate) the first order term is zero because the expansion is around the particles equilibrium positions. The idea is that each of the particles are at the bottom of valleys in the energy landscape and they oscillate around their equilibrium as simple harmonic oscillators (because the second order Taylor expansion describes the valleys as if they were quadratic). The curvature of these valleys determines the frequency of oscillation. The curvature is found from the second derivative of the potential energy, at the equilibrium position of the particles, with respect to position, and is therefore coded into the dynamic matrix. In general, the diagonalised dynamic matrix has a simple form with the frequencies (the second derivatives of the potential energy) along the diagonal, i.e. ω CDC 0 ω... 0 = Ω =....., ωnd where C is a unitary matrix (each column is one of the eigenvectors of the dynamic matrix D), Nd is the product of the number of particlces, N, and the number of dimensions, d. We now diagonalise the dynamic matrix to find the frequencies of oscillation and thus the eigenvalues. Diagonalising a matrix can be done by hand K+k M ω k Mm k k Mm m ω = 0 (9) ω )( k ( ) k m ω ) = 0 (10) Mm ( K + k M ω 4 ω ( K + k M ω = K+k M The two frequencies are therefore ( 1 K + k ω + = M + k m ( 1 K + k ω = M + k m and the corresponding eigenvalues are + k m ( + k m ± K+k M ) k(k + k) + Mm k Mm = 0 (11) ) 4 Kk + k m Mm. (1) ) ( 1 K + k + 4 M + k ) Kk m Mm, (13) ) ( 1 K + k 4 M + k ) Kk m Mm, (14) E n+,n = hω + (n ) + hω (n + 1 ). (15) In summary, we have decomposed the Hamiltonian, Equation (1), into two independent harmonic oscillators and added their energies together to find the eigenvalues, Equation (15), that depend upon the energy levels n + and n.

17 .1 Part (ii) Solve H using the Born-Oppenheimer Approximation We are asked to consider particle 1 to be very slow moving and to solve for particle as if particle 1 were stationary, just like in the Born-Oppenheimer (BO) approximation. The Born-Oppenheimer Approximation To put this into context, the Born-Oppenheimer approximation is one of the main approximations made when solving the Schrödinger equation for real systems of atoms and electrons. As atoms are so much heavier than electrons they move much more slowly than electrons. It is sometimes called the adiabatic approximation because one chooses a configuration for the atoms and then solves for the ground state of the electrons, then the atoms are moved infinitesimally and then the ground state of the electrons is solved for again. Repeating this adiabatic proceedure until one finds the minimum in energy one can find the equilibrium spacings of atoms in crystals. It is not always correct but it is a very good approximation and is widely used (luckily for us because it means that we can actually make the Schrödinger equation much more tractable this way). We therefore separate the two particles wavefunctions as Ψ(x 1, x ) = α χ α(x 1 )ψ α (x ; x 1 ) and rephrase the problem, Equation (1), as HΨ(x 1, x ) = H α χ α (x 1 )ψ α (x ; x 1 ), (16) H = H 0 (x 1 ) + p 1 M, (17) H 0 (x 1 ) = p m + K x 1 + k (x 1 x ), (18) where the χ α (x 1 ) are the wavefunctions of particle 1 (or the nuclei in the BO approximation) and ψ α (x ; x 1 ) are the wavefunctions of particle (or the electrons in the BO approximation), given that particle 1 is at position x 1. We now look more carefully at the term H 0 (x 1 ) H 0 (x 1 )Ψ(x 1, x ) = α = α ε α (x 1 )χ α,β (x 1 )ψ α (x ; x 1 ), (19) ( ( p χ α (x 1 ) m + k ) (x 1 x ) ψ α (x ; x 1 ) + K ) x 1χ α (x 1 )ψ α (x ; x 1 ). (0) The next step in the BO approximation is to say that just one term in the expansion of the wavefunctions dominates, i.e. Ψ(x 1, x ) χ 0 (x 1 )ψ 0 (x ; x 1 ). This enables us to completely separate the two particles wavefunctions and solve Equation (0) just for particle H 0 (x 1 )χ 0 (x 1 )ψ 0 (x ; x 1 ) =χ 0 (x 1 ) ( p m + k (x 1 x ) ( ε 0 (x 1 ) K ) ( p x 1 ψ 0 (x ; x 1 ) = m + k (x 1 x ) ) ψ 0 (x ; x 1 ) + K x 1χ 0 (x 1 )ψ 0 (x ; x 1 ), ) (1) ψ 0 (x ; x 1 ) () We can see that this is just the quantum harmonic oscillator equation (remember that x 1 is fixed), see http: //en.wikipedia.org/wiki/quantum_harmonic_oscillator if you need a reminder on how to solve the quantum harmonic oscillator. We need to add another index to the energy and wavefunction, i.e. ψ 0,n (x ; x 1 ) for the quantum harmonic oscillator solutions ε 0,n (x 1 ) K ( x 1 = hω n + 1 ) (3) ( ε 0,n (x 1 ) = hω n + 1 ) + K x 1, (4) k ω = m. (5) 3

18 We now expand out Equation (16) using what we have learned for particle ( ) Hχ 0 (x 1 )ψ 0,n (x ; x 1 ) = H 0 (x 1 ) + p 1 χ 0 (x 1 )ψ 0,n (x ; x 1 ), (6) M =ε 0,n χ α,β (x 1 )ψ α (x ; x 1 ) + ψ 0,n (x ; x 1 ) p 1 M χ 0(x 1 ) + χ 0 (x 1 ) p 1 M ψ 0,n(x ; x 1 ) + 1 M (p 1χ 0 (x 1 ))(p 1 ψ 0,n (x ; x 1 )), (7) where the notation in the final term is not completely clear, the final term means the differential with respect x 1 of both χ α,β and ψ α separately, as would be obtained from the chain rule. The next important step in the BO approximation is to ignore the final two terms of Equation (7). 4

19 The Born-Oppenheimer Approximation Continued The Born-Oppenheimer approximation cannot actually be derived, it is just seen to be a good approximation. It cannot be derived because it is not always correct, one cannot always ignore the last two terms of Equation (7). The intuitive reason why the final two terms of Equation (7) can be ignored in a system of atoms and electrons is that the atomic wavepackets χ α,β are far more localised than the electronic wavepackets ψ α. The more localised the wavepacket then the greater the value of the derivative, i.e. in the limit of χ α,β being much more localised than ψ α then p 1χ α,β (x 1 ) >> (p 1 χ α,β (x 1 ))(p 1 ψ α (x ; x 1 )) >> p 1ψ α (x ; x 1 ). I have included a very schematic diagram in Figure 1. In fact, here we can find an approximate wavepacket size for fast moving particle, by noting that the ground state solution for the harmonic oscillator is a Gaussian, i.e. ψ 0 (x ) e ax, where a = h 1 mk. The spread of the wavepacket is e x ax x e ax = = 1 e ax e ax 4a = h mk. (8) The standard deviation of the Gaussian is the square root of the variance, so 68% of the wavepacket ( ) h is within 1 4 4mk of the equilibrium position. Similarly for the slower particle 1, which will be shown very shortly to also satisfy the quantum harmonic oscillator equation. Therefore one would expect ( ) h that 68% of the wavepacket to be within 1 4 4MK of the equilibrium position. Let s put in some numbers to get a reasonable comparison. Consider h, k and K to be of the order of unity, so we just compare m 1 4 with M 1 4. To continue the metaphor with electrons and nuclei, the difference in magnitude between an electron and a proton is 1836, so for the electron mass of 1 and proton mass of 1836, the standard deviation is 1 and 0.15 respectively. Usually nuclei contain many protons and neutrons. Hence for Aluminium with atomic mass of 7, the standard deviation of the wavepacket is M 1 4 = (7 1836) 1 4 = compared with 1 for an electron. Doing a similar comparison for atoms on a lattice, one finds that electrons have a standard deviation of the order of a lattice spacing whereas the nuclei are very localised. One can carry out a similar process for momentum as a post-hoc justification of the Born- Oppenheimer approximation for these two particles. Starting from the fast particle p = e ax ( i h x ) e ax e ax e ax = h a = h mk. (9) The standard deviation is the square root of the variance, so 68% of the momentum wavepacket is ( ) h σ p = 1 ( ) 4 mk h 4 (for particle 1: σ p = 1 4 MK 4 )and so a typical speed would be σp m (for particle 1: 3 ). Considering h, k and K to be of the order of unity again we just compare m σ p M 4 for particle and M 3 4 for particle 1. If we insert the same numbers we used before, the typical speed for particle 1 (aluminium atom) is compared with 1 for particle (electron). By substituting Equation (3) into Equation (7) and ignoring the final two terms we obtain another quantum harmonic oscillator equation, this time for the nuclear wavefunctions ( Hχ 0 (x 1 )ψ 0,n (x ; x 1 ) = ψ 0,n (x ; x 1 ) ( hω α + 1 ) + K ) x 1 + p 1 χ 0 (x 1 ). (30) M We now define Hχ 0,m (x 1 )ψ 0,n (x ; x 1 ) = E 0 (n, m)χ 0,m (x 1 )ψ 0,n (x ; x 1 ), where we have added another index, this time to the χ, as it also turns out to be a quantum harmonic oscillator. We then divide through Equation 5

20 χ ψ Figure 1: Very simplistic diagram of the localised nuclear wavepackets χ and the extended electronic wavepackets ψ. (30) by ψ 0,n and we find ( (E 0 (n, m) hω n + 1 )) ( K χ 0,m (x 1 ) = ω 1 = = ) x 1 + p 1 χ 0,m (x 1 ) (31) M ( ( hω 1 m 1 )) χ 0,m (x 1 ), (3) K M. (33) Hence, the solution is ( E 0 (n, m) = hω n + 1 ) ( + hω 1 m 1 ). (34).1 Part (iii) compare the answers to Part (i) and Part (ii) We see that Equation (15) is very similar to Equation (34), but with different frequencies. What we have obtained with our approximate approach (Part (ii)) appears to be a rather good representation of the exact solution, (Part(i)). Let s see to what order in m M our approximation is correct to by expanding out Equations (13) and (14) in the limit of m M << 1, ω ± = 1 (K K + k M + k + k m ± M + k ) 4Kk m Mm. (35) We then go through the terms, using the binomial theorem, starting with ( K + k M + k ) = k m m ( 1 + ) ( m(k + k) = k m(k + k) Mk m 1 + Mk = k k(k + k) + m Mm + 1 ( m Mm O M ( )) m + O M, (36) ), (37) which we substitute back into Equation (35) ω ± = 1 K + k M + k k m ± k(k + k) + m Mm + 1 ( m ) Mm O 4Kk M Mm, (38) = 1 K + k M + k m ± k 1 + m((k + k) 4K + O ( ) m M ). (39) m Mk 6

21 We now expand 1 + m((k + k) 4K + O ( ) m M ) = Mk and substitute back into Equation (39) 1 + m(k K + O ( ) m M ) = ( ) K) m m(k + O Mk Mk M +..., (40) ω ± = 1 K + k M + k m ± k ( 1 + (k K) m ( )) m m Mk + O M. (41) We now expand these for the two terms separately ω + = 1 K + k M + k m + k ( 1 + (k K) m ( )) m m Mk + O M, (4) = 1 k M + k m + k ( ) m m O M, (43) = 1 k 1 + m ( 1 + O ( )) m M, (44) m M k ( ( m )) = 1 + O, (45) m M hence ω is equal to ω + up to zeroth order, i.e. it contains an error of first order in m M. We now continue the process ω = 1 K + k M + k m k ( 1 + (k K) m ( m m Mk + O M = 1 K M k ( m m O M = 1 K 1 Mk ( m M Km O M = )), (46) ), (47) ), (48) K ( ( m )) 1 O. (49) M M We have obtained a similar result that ω 1 is equal to ω up to zeroth order, i.e. it contains an error of first order in m M. It is worth noting that the two errors are of the opposite sign, so a more careful annalysis may have uncovered some potential for cancellation of error for E 0 (m, n) with m = n. 7

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