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Phonon spectroscopy = Conditions for: in elastic scattering Constraints: Conservation laws of Momentum Energy In all interactions involving phonons, energy must be conserved and crystal momentum must be conserved to within a reciprocal lattice vector. Phonons have important effects on Solar cells Heat Capacity Revisited (now with phonons) Revisiting Heat Capacity In Chapters 1&2, g f(E) = 1 𝐸−𝜇 exp( )±1 𝑘 𝑇 Plus for fermions, minus for bosons Planck distribution Lattice Vibrational Contribution to the Heat Capacity The thermal energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution. Some other contributions: Conduction Electrons in metals & semiconductors. The magnetic ordering in magnetic materials. Classical Specific Heat (Spring Model) For phonon gas, classically expect a temperature independent result. Cv 3R 3 6.02 1023 (atoms / mole) 1.38 1023 ( J / K ) U= 3NkBT 3RT Dulong-Petit law states that specific heat of a given number of atoms of any solid is independent of temperature and is the same for all materials at high temperature! At low T quantum phenomena are important and must be taken into account. Specific heat tends to classical value at high temperatures. Thermal Energy & Heat Capacity Quantum Model According to Quantum Mechanics if a particle is constrained; • the energy of particle can only have special discrete energy values. • it cannot increase infinitely from one value to another. • it has to go up in steps. Treating Phonons like QM Oscillator Einstein Model = Heat capacity C can be found by differentiating the QM average phonon energy Pn n The energy of a harmonic oscillator and hence of a lattice mode of angular frequency at temperature T n What happens to as the temperature increases? The probability of the oscillator being in this level as given by the Boltzman factor exp( n / k BT ) Energy of a single oscillator 1 n n 2 Average energy _ Pn n n P n n 1 1 n exp n / kBT _ 2 2 n 0 1 exp n / kBT 2 n 0 1 z exp[(n ) ] 2 kBT n 0 Partition function What is dz/dT? z e / 2 k BT e3 / 2 kBT e5 z e / 2 k BT (1 e / k BT z e / 2 k BT (1 e / k BT 1 e2 ) / 2 k BT / k BT ..... ..... Will need soon According to the Binomial expansion for x«1 where xe h / k BT 1 x 1 x n 0 n 1 1 n exp n / k T B _ 2 2 n 0 1 exp n / k T B 2 n 0 Can be written as: 1 z k BT 2 (ln z ) z T T _ e / 2 k BT 2 k BT ln T 1 e / kBT _ k BT 2 / 2 k BT / k BT ln e ln 1 e T _ / k BT 2 k BT ln 1 e T 2 k T T B k B / k BT 2k k 2T 2 e 1 _ e 2 B B k BT 2 2 / k BT 4 k BT 1 e 2 1 e _ k BT 2 _ Finally, the result is 1 2 e x' (ln x) x x / k BT / k BT / k BT 1 _ 1 2 e / k BT 1 This is the Mean Phonon Energy. The first term in the above equation is the zero-point energy. As mentioned before even at 0ºK atoms vibrate in the crystal. The number of phonons is given by the Bose-Einstein distribution as Planck distribution = 1 f ( ) e kBT 1 _ (number of phonons) x (energy of phonon) = (second term in ) The second term in the mean energy is the phonon contribution to the thermal energy. k BT Mean energy of a harmonic oscillator as a function of T 1 2 T Low Temperature Limit kBT 1 2 e kBT 1 _ _ 1 2 Since exponential term gets bigger Zero Point Energy kBT Mean energy of a harmonic oscillator as a function of T 1 2 x2 e 1 x .......... 2! x T High Temperature Limit << kBT is assumed independent of frequency of oscillation by the Einstein method. This is the classical limit because the energy steps are now small compared with the energy of the harmonic oscillator. Notice that is the thermal energy of the classical 1D harmonic oscillator. e 1 k BT _ 1 2 1 1 k BT _ 1 k BT 2 kBT _ kBT Heat Capacity C (Einstein Model) • Heat capacity C can be found by differentiating the average phonon energy 1 2 e kBT 1 _ d Cv dT Let kB kBT e kBT e 2 1 k 2 kBT Cv kB 2 k T B 2 e e kBT eT Cv k B T e T 1 2 2 kBT 1 2 eT Cv k B T e T 1 2 where 2 k Specific heat in this approximation vanishes exponentially at low T and tends to classical value at high temperatures. Cv kB Area = 2 kB T These features are common to all quantum systems; the energy tends to the zero-point-energy at low T and to the classical value of the Boltzmann constant at high T. The difference between classical and Einstein models comes from zero point energy. Einstein Model Fails at Low Temp • Einstein model also gave correctly a specific heat tending to zero at absolute zero, but the temperature dependence near T= 0 did not agree with experiment. Why? • Taking into account the actual distribution of vibration frequencies in a solid this discrepancy can be accounted for Einstein was aware that getting the frequency of the actual oscillations would be different. He proposed this theory because it was a simple demonstration that QM could solve the specific heat problem. CV of Diamond Finding the 3d density of states D(E) for Photons g (k ) k 2 Good analogy for low temp phonons. Why? 2 g (kg(k) )dk gg(E) ( E )dE dE g (k ) g ( E ) dk g(k) g(E) Number of States N=gV (for Photons) How about for Phonons? N Factor of 3 from one longitudinal and 2 transverse When k is not small, DOS bigger Total energy determination tricky Gradient goes to zero at edges of Brillouin Zone Like electronic density of states, there will be a van Hove singularity in the phonon density of states. Another View of the Low Temperature Limit At low T’s only lattice modes having low frequencies can be excited from their ground states, since the Bose-Einstein function falls to zero for small values of Low frequency sound waves at low T 0 vs k vs k k a Vk 2 dk g 2 2 d k 1 dk 1 vs k vs d vs 2 V 2 vs 1 g 2 2 vs v s depends on the direction V2 1 V2 1 2 g g 3 2 3 2 3 2 vs 2 vL vT Debye approximation has two main steps 1. Approx. dispersion relation of any branch by a linear extrapolation 2. Ensure correct number of modes by imposing a cut-off frequency D, above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose D so that: Einstein Debye D approximation to the dispersion approximation to the dispersion vk g ( )d 3 N 0 V 1 2 ( )D3 3N 2 3 3 6 vL vT g ( ) 9N D3 2 V 1 2 D 2 ( 3 3 ) d 3N 2 2 vL vT 0 V 1 2 3N 9N ( ) 3 2 2 vL3 vT3 D3 D3 g () / 2 Accordingly, the Debye spectrum may be written as 9 N 2 3 g( ) D 0 for for Einstein Debye D D Actual Notes: The Debye spectrum is only an idealization of the actual situation obtaining in a solid; it may be compared with a typical spectrum. While for low-frequency modes (the so called acoustical modes) the Debye approximation is reasonably valid, there are serious discrepancies in the case of high-frequency modes ( the optical modes). For “averaged” quantities, such as the specific heat, the finer details of the spectrum are not very important. Though not as common, the longitudinal and the transverse modes of the solid should have their own cut-off frequencies, D,L and D,T say, rather than D. Accordingly, we should have: D ,L D ,T 2 d 2 d 0 V 2 2cL3 N and 0 2 2cT3 2 N having a common cut-off at Debye Lattice Energy and Heat Capacity 1 E ( / kBT )g ()d 2 e 1 0 The lattice vibration energy of becomes and, 9N E 3 D g ( ) 9N D3 2 D D 3 3 1 9 N 2 0 ( 2 e / kBT 1) d D3 0 2 d 0 e / kBT 1d D 9 9N E N D 3 8 D D 3d e / k BT 0 1 First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature. dE 9 N CD 3 dT D D 0 4 e / k BT d 2 2 kBT e / kBT 1 2 3 /T D T CD 9 NkB D 0 4 x e e x x 1 d kT dx k BT kT x define the Debye temperature x 2 dx D D kB How does CD limit at high and low temperatures? T >> D High temperature x2 x3 e 1 x 2! 3! 3 /T D T CD 9 NkB D x x (1 x) x (1 x) ~ 2 x 2 2 2 x x e 1 1 x 1 4 x 4 xe T >> 4 T D CD 9 Nk B D x 3 /T D 0 0 x 4e x e x 1 2 dx is small k BT x 2 dx 3Nk B T << D Low temperature For low temperature the upper limit of the integral is ~infinite; the integral is then a known integral of 4 4 /15 . 3 /T D T << T D CD 9 NkB D We obtain the Debye T 3 0 x 4e x e x 1 2 dx law in the form 12 Nk B 4 T CD 5 D 3 D D kB How good is the Debye approximation at low T? 1 / kT g d 2 e 1 0 1 d 2 2 k T Cv V 2 kB 3 3 B dT 15 vL vT 3 The lattice heat capacity of solids varies as T 3at low temperatures; Debye law. Excellent agreement for non-magnetic insulators! Motivation for Debye: The exact calculation of DOS is difficult for 3D. Debye obtained a good approximation to the heat capacity by neglecting the dispersion of the acoustic waves, i.e. assuming s k for arbitrary wavenumber. In a 1D crystal, Debye’s approximation gives the correct answer in both the high and low temperature limits. This approximation is still widely used today. Lattice heat capacity due to Debye scheme T CD 9 Nk B D 3 D / T 0 x 4e x e x 1 2 dx C 3NkBT Debye formula gives quite a good representation of the heat capacity of most solids, even though the actual phonondensity of states curve may differ appreciably from the Debye assumption. 1 Lattice heat capacity of a solid as predicted by the Debye interpolation scheme 1 T / D Debye frequency and Debye temperature scale with the velocity of sound in the solid. So solids with low densities and large elastic moduli have high D . Debye energy D can be used to estimate the maximum phonon energy in a solid. Solid Ar D (K ) 93 Na Cs Fe Cu Pb C KCl 158 38 457 343 105 2230 235 Summary: Debye Specific Heat 1 c 3 Nk For T>>D V Same as a classical non-interacting gas For T<<D cT 1 4 T 3Nk V 5 4 3 Quantum effects important The assumptions of Debye theory are: • the crystal is harmonic • elastic waves in the crystal are non-dispersive • the crystal is isotropic (no directional dependence) • there is a high-frequency cut-off determined by the number of degrees of freedom Comparing Phonon and Electron contributions to heat capacity (for metals) Phonon contribution Electron contribution 1 d 2 2 k BT 2 Cv V kB 3 3 dT 15 v v L T 3 Cv= 𝜋2𝑘𝐵 𝑇 2 𝑇𝐹 T > Debye temp • If each unit cell contributes one or two electrons, then the ratio is about kBT/EF (small for most metals) • Thus, electrons do not contribute much to room temperature heat capacity (significant at low temp) Class Discussion Experimentally, the specific heat of EuO at very low temperatures is proportional to T3/2. • Is EuO a metal or an insulator? Why? • Any idea what type of elementary excitation this behavior results from? (you may not know) • What was the temperature dependence in metals? • Remember what materials Debye was good for? • Magnons with an energy dependence of k2 How good is the Debye approximation at low T? 1 d 2 2 k T Cv V 2 kB 3 3 B dT 15 vL vT The lattice heat capacity of solids thus varies as T 3 at low temperatures; this is 3 referred to as the Debye T law. Figure illustrates the excellent agreement of this prediction with experiment for a non-magnetic insulator. The heat capacity vanishes more slowly than the exponential behaviour of a single harmonic oscillator because the vibration spectrum extends down to zero frequency. 3 Reminder: High and Low Temperature Limits 3NkBT • Each of the 3N lattice modes of a crystal containing N atoms d C dT This result is true only if T >> C 3NkB kB At low T’s only lattice modes having low frequencies can be excited from their ground states; long Low frequency sound waves 0 k a vs k vs k Calculating Specific Heat Capacity 1 / kT g d 2 e 1 0 1 2 e 0 / kT Zero point energy = z 2 V 1 2 2 3 3 d 1 2 vL vT x V 1 2 3 z 2 3 3 / kT d 2 vL vT 0 e 1 3 V 1 2 k BT 4 z 2 3 3 3 2 vL vT 15 4 e 0 e 3 / kT 0 / kT 3 1 1 d d 0 kBT 3 kBT 3 x kT B dx x e 1 k BT k BT x d k BT dx 4 x3 0 e x 1dx 4 15 1 2 k BT d 2 2 Cv V kB 3 3 dT 15 vL vT 3 at low temperatures Example Calculate the temperature at which the photon energy density in a solid will exceed the phonon energy density in the same solid, if the solid has a Debye temperature of 300K, a phonon velocity of 400,000 cm/s and an index of refraction of n=2 and assuming the solid does not melt. Should we consider the high or low temperature regime? Heat capacity of the free electron gas • From the diagram of n(E,T) the change in the distribution of electrons can be resembled into triangles of height ½ g(EF) and a base of 2kBT so the area gives that ½ g(EF)kBT electrons increased their energy by kBT. The difference in thermal energy from the value at T=0°K E (T ) E (0) ~ 1 g ( EF )(k BT ) 2 2 • Differentiating with respect to T gives the heat capacity at constant volume, E Cv g ( EF )k B 2T T g (E) E E 2 V N g ( E ) dE N EF g ( EF ) 0 0 2 2 3 3 N 3N g ( EF ) 2 EF 2k B TF F F 3N Cv g ( EF )kB T k B 2T 2kBTF 2 V 2 2 3/ 2 1/ 2 (2 m ) E dE 3 3/ 2 1/ 2 (2 m ) E 3 V 3 2 3/ 2 (2 mE ) F 3 T 3 Cv Nk B 2 TF Heat capacity of Free electron gas