Chapter 5 Phonons II Density of States Chapter 5 Phonons II Density of States Density of States � Last time, we learned that you have to calculate the heat capacity by first calculating the total ener[.]
Chapter 5: Phonons II Density of States Density of States • Last time, we learned that you have to calculate the heat capacity by first calculating the total energy with: U = ∑∑ κ wavevectors p hωκ , p exp(hω / kT ) − U = ∑ ∫ dω ( D (ω )) p p hω exp(hω / kT ) − polarization • So, what we first have to calculate is the density of states (no of modes at a particular frequency) dN D(ω ) = dω No of modes Frequency Density of States : 1D • Our strategy is to start in 1D and work our way up to 3D • For 1D, consider a chain of N + atoms, separated by a (N = 10) s=0 a • Each normal vibrational mode of polarization p has the form of a standing wave: determined by dispersion relation displacement of atom s us = u(0) sin(sKa – ωκ,Pt) ~ u(0) exp (-i ωκ,Pt) sin(sKa) s = 10 Density of States: 1D • What wavelengths, or modes, are allowed? (keeping the ends fixed) Wavelength = 2L → Wavevector κ = 2π/λ = 2π/(2L) = π/L Wavelength = L → Wavevector κ = 2π/λ = 2π/L (Minimum wavelength = 2L/(N-1) = 2a) Therefore, the allowed wavenumbers are: π/L, 2π/L, 3π/L, … (N-1) π/L Maximum value Density of States: 1D • • • • You can show that this is the maximum number by substituting the next K value = Nπ/L into us ~ sin (sKa) which is us ~ sin (sNπa/L) = sin(πa/a) = sin π = This means that the atoms are not moving for this mode (and any smaller wavelengths) This also proves what we did before – the maximum wavevector corresponds to a minimum wavelength of 2a, which is the minimum you can have for a wave in a discrete system So, each value of K corresponds to a different standing wave, and there is one mode for each interval ∆K = π/L The allowed wavenumbers are: π/L, 2π/L, 3π/L, … (N-1) π/L ∆K modes in total (or, 1/mobile atom) Density of States in 1D • To calculate the density of states, we will use our definition: D(ω) dω = (No of states/unit frequency) (freq unit) = No of states • We know that we have one state per unit range of K of π/a The allowed wavenumbers are: π/L, 2π/L, 3π/L, … (N-1) π/L ∆K • So, no of states = (L/π) ∆K = (L/π) dK = (L/π) dω (dK/dω) • And therefore: D(ω) dω = (L/π) dω/ (dω/dK) Group velocity – we can get this from dispersion curve Density of States: 1D • So, in 1D we have: ω dω D(ω )dω = • π dω / dK L • • Knowing the dispersion curve, we can calculate dω/dK (which is the slope, or the group velocity) Whenever the group velocity is zero (like near the zone boundary), the density of states goes to infinity! (called a singularity) Slope = here, so the density of states is undefined At this point, there is no group velocity – no net movement of atoms Density of States : 3D • In 3D, this becomes a bit more complicated, but we can still calculate D(ω) • Let’s take a cubic solid, with a side of length L There are N3 primitive cells within this solid (so N per length L) How many modes are there? For N3 atoms, there are N3 modes (one for the x, y and z direction) L How many modes per unit range of ∆K? L (N atoms on a side) L What happens at the edges of the solid? Periodic Boundary Conditions • • Another common way of approaching this problem as N becomes large is by using periodic boundary conditions In 1D, we take our string and make it into a ring so we don’t have to worry about the fixed points at the ends s=2 s=1 s=N or The number of possible wavelengths is still equal to the number of mobile atoms (N in this case instead of N-1, but as N gets large, this doesn’t matter) However, in this case, you now have ways of making waves (clockwise or ccw – the ends aren’t fixed) This changes ∆K from π/L to 2π/L Density of States: 3D • • Let’s use periodic boundary conditions, and assume that waves can travel in the +/- x, y or z direction Now you have a ∆K = 2π/L instead of ∆K = π/L for each wave travelling in the x, y or z direction (N = 10) s=0 Real space values s = 10 Before we had: (in K-space) π/L 2π/L (N-1)π/L K values Now, allowing for +/- direction: -Nπ/L -4π/L -2π/L Separated by 2π/L 2π/L 4π/L Nπ/L Density of States: 3D • So, now we have K values for Kx , KY, KZ • Kx , KY, KZ = ; +/- 2π/L, +/- 4π/L, … +/- Nπ/L • So we have one allowed K value per volume (2π/L)3 in K-space, which we can state as: V L = 2π 8π • It is more convenient now to look at what is happening in K-space (reciprocal space) K-space picture • In K-space, we are going to look at all the possible modes from K = to some K=Kmax Since we can look in the x, y and z-directions, this is a sphere: KZ Kmax KY KX How many K values are possible within this sphere? Total no of modes Unit of K-space N/(vol of sphere) = (L/2π)3 N/(4πK3/3) = (L/2π)3 So : N = (L/2π)3 (4πK3/3) (each dot is separated by 2π/L, and represents a wave with wavevectors (Kx, Ky, Kz)) Density of States: 3D • Now we can put this all together We can calculate the density of states by taking the derivative of this: dN d L = D(ω ) = πK dω dω 2π 3 L d L dK d = πK = πK 2π dω 2π dω dK L dK = π 3K dω 2π VK dK Again, we can get dω/dK from dispersion curve = 2π dω The Debye Model • • Now we can talk about the Debye Model Remember, what we want to solve is: U = ∑∑ κ p hωκ , p exp(hω / kT ) − U = ∑ ∫ dω ( D (ω )) p • p hω exp(hω / kT ) − And to this, we need to use the density of states, which means we need to know something about the dispersion curve VK dK D(ω )dω = 2π dω (In 3D) The Debye Model • • • Peter Debye made the approximation that the low lying lattice excitations had a linear relationship of ω= vκ (v is a constant – speed of sound) This is true for the lowest energies (long wavelength vibrations) Using this, we can calculate the density of states: VK dK V (ω / v) d D(ω ) = (ω / v) = 2 dω 2π dω 2π V (ω / v) Vω D(ω ) = = v 2π v 2π ω It appears linear in this region The Debye Model • • • • The second assumption he made was that this relationship was true up to a certain cut-off frequency ωD Why did he this? There are N acoustic modes for N atoms in the crystal So, he just said that once you run out of possible modes, then you are out of possible K values (you have used all possible wavelengths) We can write our cut-off frequency in terms of a cut-off in K-space (a max size of our sphere) using ωD = v KD, and using what we had before: Size of Debye sphere L 3 L 4 3 3 = π π ( ω / ) N = K v D D 2π 2π N ω D = 6π v V The Debye Model • Now what we have to is add up all the possible modes from frequency to a maximum frequency ωD: 1/ ωD • N N = 6π v → ω D = 6π v V V Our total thermal energy is now: U = ∑ ∫ dωD(ω ) < n(ω ) > hω P U =∑ P ωD Vω hω ∫0 dω 2π 2v ehω / kT − (in 3D) The Debye Model • • • We will now assume, for simplicity, that all polarizations have the same energy dependence We know that this isn’t true from looking at dispersion curve data, but the energies tend to be close anyways This leaves us with this integral to solve: ωD Vω hω U = ∫ dω hω / kT −1 2π v e (3 polarizations) (phonon dispersion curve for La2CuO4)