Handout 21
Phonon Thermal Statistics and Heat Capacities
In this lecture you will learn:
• Phonon occupation statistics • Bose-Einstein distribution • Phonon density of states in 1D, 2D, and 3D • Phonon thermal energy and heat capacity of solids
Peter Debye Born: 1884 (Netherlands) Died: 1966 (Ithaca, NY)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A Single Lattice Wave Mode Consider the Hamiltonian of just a single lattice wave mode:
ˆ 1 H q aˆ q aˆ q 2
Its eigenstates, and the corresponding eigenenergies, are:
n where n 0,1,2,3......
ˆ 1 H n q aˆ q aˆ q n 2 1 q n n 2 En n
The state n corresponds to “n” phonons in the lattice wave mode
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
1 A Single Lattice Wave Mode in Thermal Equilibrium ˆ 1 H n q n n En n 2 Thermal Equilibrium
In thermal equilibrium, let P(n) be the probability that there are “n” phonons in this lattice wave mode
P(n) must be related to the energy corresponding to the “n” phonons: En qn1 2 (1) Pn e KT e KT
P(n) must be normalized properly: Pn 1 (2) n0
(1) and (2) give: qn q Pn e KT 1 e KT
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Bose-Einstein Distribution The probability distribution given by, qn q Pn e KT 1 e KT is called the Bose-Einstein distribution Average Phonon Number: One can calculate the average phonon number in equilibrium:
1 n n P n q KT n0 e 1 Average phonon number in any lattice wave mode depends on the phonon energy
Limiting Cases: 1 KT KT q n E q n KT q KT e 1 q
Classical 1 q KT equipartition KT q n e e q KT 1 theorem
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
2 Classical Equipartition Theorem
Every independent quadratic term in position or momentum in the expression for the energy of a system has an average value equal to KT/2 at temperature T
• Only holds when classical statistics apply - which is generally the case at high enough temperatures
Example: A Free Particle in 1D Example: A Free Particle in 3D 2 p2 1 p2 p p2 3 E x E KT E x y z E KT 2m 2 2m 2m 2m 2
Example: A Classical Simple Harmonic Oscillator in 1D p2 1 E x kx 2 E KT 2m 2 Example: A Single Lattice Wave Mode of a 1D Crystal
N NM E Pq,t P * q,t 2q U q,t U * q,t E KT 2M 2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Acoustic Phonons in 1D: Density of States x a1 a xˆ
Consider acoustic phonons in a N-primitive-cell 1D crystal of length L: L N a First BZ First we need to figure out how to convert a summation over all lattice wave modes of the form: qx in FBZ qx into an integral for the form: 2 a a N a a dqx a We now that there are N different allowed wavevector values in FBZ (in interval 2/a) So in interval dqx there must be (Na/2) dqx different wavevector values: a dq a dq Na x L x qx in FBZ a 2 a 2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
3 Acoustic Phonons in 1D: Density of States a dq a dq Na x L x 2 2 D D v qx in FBZ a a a Now we need to figure out how to convert an integral of the form: a dq L x a 2 into an integral over frequency of the form: qx ? Density of states First BZ L d g1D a a 0 We need to know the dispersion of the phonons. We approximate it by a linear function:
v qx Therefore: a dq a dq L D dq D 1 L x 2 L x x d L d a 2 0 2 0 d 0 v
The density of states function g1D() is the number of phonon modes per unit frequency interval per unit length: 1 g 1D v
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Acoustic Phonons in 1D: Debye Frequency We know that: v N D D a qx in FBZ Since: D L d g1D qx in FBZ 0 We must have: First qx D a BZ a L d g1D N (1) 0 Since: 1 v g D a 1D v It can be verified that (1) above holds
The frequency D is called the Debye frequency (after Peter Debye – Cornell University). It is chosen to ensure that the total number of phonon modes are conserved when going from q-space integrals to frequency domain integrals. In 1D this is automatic.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
4 Acoustic Phonons in 2D: Density of States Consider acoustic phonons in a N-primitive-cell 2D crystal of area A D LA We need to go from a q-space integral to a frequency integral: D D TA A d g2D q in FBZ 0 We need to know the dispersion for the 2 acoustic LA phonon bands. We assume that for both phonon bands the dispersion is linear: TA v q 1,2 for LA,TA For each phonon band we get: d 2q ? 2 q dq A 2 A 2 q in FBZ FBZ 2 0 2 A D dq D q d A d 2 g2D 2 0 d 0 2 v 2 2 v
The question is what is D ?
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Acoustic Phonons in 2D: Debye Frequency
To find D we count and conserve the total number of phonon modes in each band: N q in FBZ LA D D A d N 2 0 2 v D TA 2 A D N 2 4 v LA
2 N D 4 v A TA Each phonon band has a different Debye frequency
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
5 Acoustic Phonons in 3D: Density of States Consider acoustic phonons in a N-primitive-cell 3D crystal of volume V
We need to go from a q-space integral to a frequency integral:
D V d g3D q in FBZ 0 We need to know the dispersion for the 3 acoustic phonon bands. We assume that for all 3 phonon bands the dispersion is linear:
v q 1,2,3 for LA,TA,TA
For each phonon band we get:
d 3q ? 4 q2 dq V 3 V 3 q in FBZ FBZ 2 0 2 V D dq D 2 q2 d V d 2 2 2 3 g 2 0 d 0 2 v 3D 2 3 2 v
The question is what is D ?
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Acoustic Phonons in 3D: Debye Frequency
To find D we count and conserve the total number of phonon modes in each band: N q in FBZ D 2 V d 3 3 N 0 2 v 3 D V 2 2 N 6 v 1 3 2 3 N D 6 v V
Each phonon band has a different Debye frequency
Silicon Phonon Bands Silicon: In Silicon the TA phonon velocity is 5.86 km/s. The corresponding Debye frequency is 13.4 THz. The LA phonon velocity is 8.44 km/s. The corresponding Debye frequency is 19.3 THz.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
6 Acoustic Phonons in 3D: Thermal Energy Consider acoustic phonons in a N-primitive-cell 3D crystal of volume V Also assume that all three acoustic phonon modes have the same velocity (for simplicity)
v q for LA,TA,TA D Then for each phonon band we have: 2 LA TA g3D 2 2 v 3 TA The energy u of the lattice per unit volume at temperature T can be written as:
3 degenerate phonon bands qy 1 q u 3 q nq x V q in FBZ D 3 d g3D KT 1 3 0 e 1 2 3 N D 6 v 3 D 3 V d 2 3 KT 2 v 0 e 1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Acoustic Phonons in 3D: Thermal Energy 3 D 3 u d 2 3 KT 2 v 0 e 1 D Define dimensionless variable “x” as: LA D x x TA KT D KT TA To get: 3 xD x 3 u KT 4 dx 2 3 x 2 v 0 e 1 qy qx Case I: KT D xD 1
3 x3 3 4 2 KT 4 u KT 4 dx KT 4 2 3 x 2 3 3 2 v 0 e 1 2 v 15 10 v
Specific Heat or du 2 2 K 4 T 3 Heat Capacity: C 3 3 Debye’s famous T law dT 5 v
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
7 Acoustic Phonons in 3D: Debye Temperature The low temperature limit: D KT D can also be written as: LA TA TA T D
Where D is the Debye temperature: qy D qx D K
The Debye frequency thus defines a natural temperature scale for the phonon energetics
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Silicon Heat Capacity Silicon Heat Capacity
C T 3
In silicon where the Debye frequency for TA phonons is 13.4 THz, the corresponding Debye temperature is 643 K. Silicon Phonon Bands The Debye frequency for LA phonons is 19.3 THz and the corresponding Debye temperature is 926 K D LA 19.3 THz DLA 926 K The T3 law for heat capacity holds well in Silicon for temperatures less than 50 K (much less than the Debye temperature of any phonon band) D TA 13.4 THz DTA 643 K
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
8 Acoustic Phonons in 3D: Classical Equipartition Theorem
Case II: KT T D D D
3 xD x 3 LA u KT 4 dx 2 3 x TA 2 v 0 e 1 TA
e x 1 x
x 3 qy 3 4 D 2 3 4 x u KT dx x KT D q 2 3 2 3 x 2 v 0 2 v 3 KT N 1 3 3 3 KT 2 3 N 2 3 D D 6 v 2 v V V
Physical explanation: There are N/V phonon modes per band per unit volume and each mode has energy equal to KT as per the classical equipartition theorem
du N Specific Heat: C 3 K Dulong and Petit Law (1819) dT V
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Silicon Heat Capacity Silicon Heat Capacity
The Heat capacity approaches 3(N/V)K as the temperature exceeds the Debye temperature of all acoustic phonon Silicon Phonon Bands bands
D LA 19.3 THz DLA 926 K
D TA 13.4 THz DTA 643 K
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
9 Optical Phonons in 1D: Einstein Model and Density of States Consider optical phonons in a N-primitive-cell 1D crystal of length L Let the optical phonon frequency be LO q We want to be able to write: ? L d g1D qx in FBZ 0 We suppose that all optical phonon modes in FBZ have the same frequency L O (i.e. the phonon band is completely flat - Einstein model):
g1D C LO What is C ? We know that: N First q in FBZ x BZ LO Therefore: N L d g1D N C 0 L Finally: N g qx 1D L LO 2 a N a a ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in 2D and 3D: Einstein Model Consider optical phonons in a N-primitive-cell 2D (or 3D) crystal of area A (or volume V)
For each optical phonon band we want to be able to write: ? ? A d g or V d g 2D 3D q in FBZ 0 q in FBZ 0 Silicon Phonon Bands We suppose that each optical phonon band is completely flat and every phonon mode in a band has the same frequency 1,2,3 for LO,TO,TO TO/TO N N g or g 2D A 3D V
On can check that the number of phonon modes per band is conserved: TA/TA A d g2D N or V d g3D N 0 0
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
10 Heat Capacity of Optical Phonons in 2D: Einstein Model Consider a material with two atoms per primitive cell in 2D There must be 2 optical phonons bands in 2D (LO and TO)
Suppose the optical phonon frequencies are L O and T O . Assuming Einstein model: N N g g 2DLO A LO 2DTO A LO Total energy per unit area in both the optical phonon modes is: 1 1 u LO q nq TO q nq A q in FBZ A q in FBZ d g d g 2DLO KT 2DTO KT 0 e 1 0 e 1 N N LO TO A eLO KT 1 A eTO KT 1
The heat capacity is: KT 2 KT 2 du N e LO N e TO C K LO K TO dT A KT 2 KT A KT 2 KT e LO 1 e TO 1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Classical Equipartition Theorem According to the canonical ensemble of statistical physics, a system at temperature T will have energy E with the probability given by: E 1 PE e KT Z The constant Z is determined by adding the probabilities for all possible states of the system and equating the result to unity
1D Example: Consider a free particle in 1D with the energy given by: p2 E x 2m
The probability that the particle at temperature T will have momentum px is then: p2 2m 1 x Pp e KT x Z We must have: dpx Ppx 1 Z 2 mKT
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
11 Appendix: Classical Equipartition Theorem
So we finally have for the probability distribution of the particle momentum: p2 2m 1 x Pp e KT x 2 mKT The average energy of the particle is then: 2 px 1 dpx Ppx KT 2m 2
General Proof: Consider a system whose total energy can be written in terms of various independent momenta and displacements as follows: 2 2 E a j p j bju j j j The probability that the system will have some specific values for all the displacements and momenta is: 2 2 a j p j b ju j j j 1 Pp , p ,...... u ,u ,..... e KT 1 2 1 2 Z
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Classical Equipartition Theorem 2 2 a j p j b ju j j j 1 Pp , p ,...... u ,u ,..... e KT 1 2 1 2 Z The constant Z is determined by requiring: dp j dur P p1, p2,...... u1,u2,..... 1 j r It then follows that the average value of any one particular quadratic term in the expression for the total energy of the system is: 2 2 1 anpn dp j dur anpn P p1, p2,...... u1,u2,..... KT j r 2 2 2 1 bnun dp j dur bnun P p1, p2,...... u1,u2,..... KT j r 2 The above results follow from the properties of standard Gaussian integrals
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
12