Born-Oppenheimer Approximation
当研究电子运动时,忽略原子运动;当研究原子运动时,忽略电子运动。
原子运动影响晶体的热、磁、光、超导性质,晶格动力学正是研究原子运动的规律。
m u ¨ n = β ( u n + 1 − 2 u n + u n − 1 ) m \ddot u_n = \beta (u_{n+1} - 2u_n + u_{n-1}) mu¨n=β(un+1−2un+un−1)
Two Basic Hypothesises
简谐振动假设 Harmonic approximation (Harmonic limit): The restoring force on each atom is approximately proportional to its displacement.
F = − β x F= - \beta x F=−βx
最近邻假设 Nearest-neighbor limit: To calculate the forces simply, only nearest-neighbor interactions of atoms be considered.
如何计算β: Taylor’s series
Atom potential U( r ):
U ( r ) = U ( a ) + ( r − a ) 2 2 ( d 2 U d r 2 ) r = a + … U(r) = U(a) + \frac{(r-a)^2}{2} \left (\frac{d^2 U}{ d r^2} \right )_{r=a} + \dots U(r)=U(a)+2(r−a)2(dr2d2U)r=a+…
F = − d U ( r ) d r = − [ d 2 U ( r ) d r 2 ] ⋅ ( r − a ) = − β x F = - \frac{d U(r)}{dr} = - \left [ \frac{d^2 U(r)}{dr^2} \right ]\cdot (r-a) = - \beta x F=−drdU(r)=−[dr2d2U(r)]⋅(r−a)=−βx
∴ β = ( d 2 U d r 2 ) r = a \therefore \beta = \left ( \frac{d^2 U}{dr^2} \right) _{r=a} ∴β=(dr2d2U)r=a
β \beta β is the force constant between nearest-neighbor atoms. It will differ for longitudinal and transverse waves in 3D space. (β可以区分横波和纵波)
假设单原子链是无限延伸的,再空间中构成单原子环。这是一种近似,由于原子个数N很大,可以忽略首尾,从而得到周期性边界条件: u N + 1 = u 1 , u N + n = u n u_{N+1} = u_1, \ \ u_{N+n} = u_n uN+1=u1, uN+n=un
m u ¨ n = β ( u n + 1 − 2 u n + u n − 1 ) m \ddot u_n = \beta (u_{n+1} - 2u_n + u_{n-1}) mu¨n=β(un+1−2un+un−1)
u ¨ n = − ω 2 u n \ddot u_n = - \omega^2 u_n u¨n=−ω2un
u n = A e i ( k x n − ω t ) = A e i ( n a k − ω t ) u_n = A e^{i(kx_n-\omega t)} = Ae^{i(nak-\omega t)} un=Aei(kxn−ωt)=Aei(nak−ωt)
∴ m ω 2 = 2 β ( 1 − cos k a ) \therefore m \omega^2 = 2\beta (1-\cos{ka}) ∴mω2=2β(1−coska)
solve the equation above, then:
ω = 2 β m ∣ sin k a 2 ∣ \omega = 2\sqrt{\frac{\beta}{m}} \left | \sin{\frac{ka}{2}} \right | ω=2mβ sin2ka
k = 2 π N a m , m = 0 , ± 1 , ± 2 , … k=\frac{2\pi}{Na}m, \ \ \ m=0, \pm1, \pm2,\dots k=Na2πm, m=0,±1,±2,…
其中,N:单原子链的原子个数(长度),a:晶格常数
相速度(通常意义上的速度): v ϕ = λ T = ω λ 2 π = ω k = 2 β m ∣ sin k a 2 ∣ k v_{\phi} = \frac{\lambda}{T} = \frac{\omega \lambda}{2 \pi} = \frac{\omega }{k} = 2\sqrt{\frac{\beta}{m}}\frac{\left |\sin \frac{ka}{2} \right |}{k} vϕ=Tλ=2πωλ=kω=2mβk sin2ka
群速度(多列波叠加,波的包络前进的速度): v k = d ω d k = a β m cos k a 2 v_k = \frac{d \omega}{d k} = a \sqrt{\frac{\beta}{m}}\cos \frac{ka}{2} vk=dkdω=amβcos2ka
波速与波长有关的现象叫做色散。
k取值不连续,对应不同的振动模式。
ω 0 = β m , ω M = 2 ω 0 , ω m = 0 \omega_0 = \frac{\beta}{m}, \ \ \ \omega_M = 2\omega_0,\ \ \ \omega_m = 0 ω0=mβ, ωM=2ω0, ωm=0
晶体振动以格波的形式体现,表现为一种集体行为。一组格波代表一组振动模式,即一组 ( k , ω ) (k,\omega) (k,ω)的取值.
independent lattice wave
ω ( k ) = ω ( k + G ) , G = m ⋅ 2 π a i ^ \omega(k) = \omega(k+G), \ \ \ G=m\cdot \frac{2\pi}{a} \hat i ω(k)=ω(k+G), G=m⋅a2πi^
The movement of atoms with k or k+G is the same. So only consider k in 1st BZ.
All independent vibrations are described by k inside BZ.
(1)波矢端点落在布里渊区界面上,行波不再传播(驻波),相邻原子的振动方向相反,满足布拉格定律。
k = π a k=\frac{\pi}{a} k=aπ
(2)长波极限,相速度=群速度,与波长无关,此时的格波是一种弹性波,晶格可看作连续介质(声波)
λ ≫ a , v ϕ = v k = a β m \lambda \gg a, \ \ \ v_{\phi} = v_k = a \sqrt{\frac{\beta}{m}} λ≫a, vϕ=vk=amβ
{ M u ¨ n = β ( ν n + ν n − 1 − 2 u n ) m ν ¨ n = β ( u n + u n + 1 − 2 ν n ) \begin{cases} M \ddot u_n = \beta (\nu_n +\nu_{n-1} - 2u_n) \\ m \ddot \nu _n = \beta (u_n + u_{n+1} - 2\nu_n) \end{cases} {Mu¨n=β(νn+νn−1−2un)mν¨n=β(un+un+1−2νn)
u N + 1 = u 1 , ν N + n = ν n u_{N+1} = u_1, \ \ \ \nu_{N+n} = \nu_n uN+1=u1, νN+n=νn
u n = A e i ( n a k − ω t ) , ν n = B e i [ n + 1 2 a k − ω t ] u_n = A e^{i(nak - \omega t)}, \ \ \ \nu_n = Be^{i \left [ \frac{n+1}{2} ak - \omega t \right ]} un=Aei(nak−ωt), νn=Bei[2n+1ak−ωt]
代入运动方程可以得到线性齐次代数方程组:
{ ( 2 β − M ω 2 ) A − 2 β cos ( 1 2 a k ) B = 0 − 2 β cos ( 1 2 a k ) A + ( 2 β − m ω 2 ) B = 0 \begin{cases} (2\beta - M \omega^2) A - 2\beta \cos(\frac{1}{2}ak) B = 0 \\ -2\beta \cos(\frac{1}{2}ak) A + (2\beta - m \omega^2) B=0 \end{cases} {(2β−Mω2)A−2βcos(21ak)B=0−2βcos(21ak)A+(2β−mω2)B=0
线性齐次代数方程组有非零解,则系数行列式为0,从而得到:
∣ 2 β − M ω 2 − 2 β cos ( 1 2 a k ) − 2 β cos ( 1 2 a k ) 2 β − m ω 2 ∣ = 0 \begin{vmatrix} 2\beta - M \omega^2 & - 2\beta \cos(\frac{1}{2}ak) \\ -2\beta \cos(\frac{1}{2}ak) & 2\beta - m \omega^2 \end{vmatrix} = 0 2β−Mω2−2βcos(21ak)−2βcos(21ak)2β−mω2 =0
ω ± 2 = β ( M + m ) M m { 1 ± 1 − 4 M m ( M + m ) 2 sin 2 ( 1 2 a k ) } \omega_{\pm}^2 = \frac{\beta (M+m)}{ Mm } \left \{ 1 \pm \sqrt{ 1 - \frac{4Mm}{(M+m)^2} \sin^2{(\frac{1}{2}ak}) } \right \} ω±2=Mmβ(M+m){1±1−(M+m)24Mmsin2(21ak)}
ω ± 2 = β M m [ ( M + m ) ± M 2 + m 2 + 2 M m cos ( a k ) ] \omega_{\pm} ^2 =\frac{\beta}{Mm}[(M+m) \pm \sqrt{ M^2 + m^2 + 2Mm\cos{(ak)} }] ω±2=Mmβ[(M+m)±M2+m2+2Mmcos(ak)]
k = 2 π N a m , , m = 0 , ± 1 , ± 2 , … k=\frac{2\pi}{Na} m,\ \ \ ,m = 0,\pm 1, \pm 2, \dots k=Na2πm, ,m=0,±1,±2,…
一个k 对应两个ω: ω + , ω − \omega_+,\ \ \omega_- ω+, ω−
一维双原子链有两个格波:光学支(Optical Branch)、声学支(Acoustical Branch)
comparation with 1D monoatomic chain and 1D diatomic chain:
volume of 1st BZ: v b = ( 2 π ) 3 v a v_b = \frac{(2\pi)^3}{v_a} vb=va(2π)3
number of k in 1st BZ:
3D: ρ ( k ) = N ( 2 π ) 3 v a = V ( 2 π ) 3 \rho (k) = \frac{N}{ \frac{(2\pi)^3}{v_a} } = \frac{V}{(2\pi)^3} ρ(k)=va(2π)3N=(2π)3V
V: volume of the primitive unit cell
1D: ρ ( k ) = N a 2 π = L 2 π \rho (k) = \frac{Na}{2\pi} = \frac{L}{2\pi} ρ(k)=2πNa=2πL
2D: ρ ( k ) = S ( 2 π ) 2 \rho(k) = \frac{S}{(2\pi)^2} ρ(k)=(2π)2S
equivalence between a vibration mode and a harmonic oscillator
The energy of crystal corresponding to vibrarion mode k will be E k = T k + U k = 1 2 p k 2 + 1 2 ω 2 q k 2 E_k = T_k + U_k = \frac{1}{2}p_k^2+\frac{1}{2}\omega^2 q_k^2 Ek=Tk+Uk=21pk2+21ω2qk2
which is the energy of a harmonic oscillator with displacement q k q_k qk, momentum p k p_k pk and vibration frequency ω \omega ω.
This tells us the equivalence between the energy of a vibration mode and the energy of a harmonic oscillator. Crystal lattice vibration energy of a monoatomic chain contain N atoms can be expressed as the summation of the energy of N harmonic oscillators.
A 3D monoatomic crystal: E = T + U = 1 2 ∑ k = 1 3 N ( p k 2 + ω k 2 q k 2 ) E = T+U = \frac{1}{2} \sum_{k=1}^{3N} (p_k^2 + \omega_k^2 q_k^2) E=T+U=21k=1∑3N(pk2+ωk2qk2)
Crystal lattice vibration energy of a 3D monoatomic crystal contain N atoms can be expressed as the summation of the energy of 3N harmonic oscillator, whose frequencies are just the frequencies of the 3N independent vibration modes.
( k , ω ) ⟷ ω ( k ) (k, \omega) \longleftrightarrow \omega (k) (k,ω)⟷ω(k)
独立格波 ⟷ \longleftrightarrow ⟷ 独立的振动模式 ⟷ \longleftrightarrow ⟷ 谐振子
Energy of a harmonic oscillator with angular frequency ω \omega ω: KaTeX parse error: Undefined control sequence: \E at position 1: \̲E̲= ( n + \frac{1…
Energy of a vibration mode or a lattice wave with angular frequency ω j ( k ) \omega_j (k) ωj(k): E j ( k ) = ( n j ( k ) + 1 2 ) ℏ ω j ( k ) , n j ( k ) = 0 , 1 , 2 , … E_j(\mathbf{k}) = (n_j(\mathbf{k}) + \frac{1}{2}) \hbar \omega_j (\mathbf{k}), \ \ \ n_j(\mathbf{k})= 0, 1, 2,\dots Ej(k)=(nj(k)+21)ℏωj(k), nj(k)=0,1,2,…
Total Energy of lattice vibration of a crystal with N unit cells, each cell having P atoms: E = ∑ j = 1 3 P ∑ k = 1 N E j ( k ) = ∑ j = 1 3 P ∑ k = 1 N ( n j ( k ) + 1 2 ) ℏ ω j ( k ) E = \sum_{j=1}^{3P} \sum_{k=1}^{N} E_j (\mathbf{k}) = \sum_{j=1}^{3P} \sum_{k=1}^{N} (n_j(\mathbf{k}) + \frac{1}{2}) \hbar \omega_j(\mathbf{k}) E=j=1∑3Pk=1∑NEj(k)=j=1∑3Pk=1∑N(nj(k)+21)ℏωj(k)
声子 phonon: ℏ ω j ( k ) \hbar \omega_j (\mathbf{k}) ℏωj(k)
光子 photon: h ν , ℏ ω h\nu, \ \ \hbar \omega hν, ℏω
E j ( k ) = ( n j ( k ) + 1 2 ) ℏ ω j ( k ) , n j ( k ) = 0 , 1 , 2 , … E_j(\mathbf{k}) = (n_j(\mathbf{k}) + \frac{1}{2}) \hbar \omega_j (\mathbf{k}), \ \ \ n_j(\mathbf{k})= 0, 1, 2,\dots Ej(k)=(nj(k)+21)ℏωj(k), nj(k)=0,1,2,…
k B k_B kB :Boltzmann constant
Average energy of harmonic oscillators with angular frequency ω j \omega_j ωj at temperature T: E ˉ i = 1 2 ℏ ω i + ℏ ω i 1 e ℏ ω i k B T − 1 = ( 1 2 + 1 e ℏ ω i k B T − 1 ) ℏ ω i \bar E_i = \frac{1}{2} \hbar \omega_i + \hbar \omega_i \frac{1}{e^{\frac{\hbar \omega_i}{k_B T} - 1}} = \left (\frac{1}{2} + \frac{1}{e^{\frac{\hbar \omega_i}{k_B T} - 1}} \right) \hbar \omega_i Eˉi=21ℏωi+ℏωiekBTℏωi−11=(21+ekBTℏωi−11)ℏωi
T=0K: Zero-point energy: 零点能
E ˉ i = 1 2 ℏ ω i \bar E_i = \frac{1}{2}\hbar \omega_i Eˉi=21ℏωi
T ≫ 0 K T \gg 0K T≫0K: ℏ ω ≪ k B T \hbar \omega \ll k_B T ℏω≪kBT: 与温度有关的晶格振动能
E ˉ i = 1 2 ℏ ω + k B T ≈ k B T \bar E_i = \frac{1}{2} \hbar \omega + k_B T \approx k_B T Eˉi=21ℏω+kBT≈kBT
Total lattice vibration energy of a crystal: E = ∑ i E ˉ i = ∑ i ( 1 2 + 1 e ℏ ω i k B T − 1 ) = ∑ i 1 2 ℏ ω i + ∑ i ℏ ω i e ℏ ω i k B T − 1 = E 0 + E ( T ) E = \sum_i \bar E_i = \sum_i \left( \frac{1}{2} + \frac{1}{e^{\frac{\hbar \omega_i}{k_B T} - 1}} \right ) = \sum_i \frac{1}{2}\hbar \omega_i + \sum_i \frac{\hbar \omega_i}{e^{\frac{\hbar \omega_i}{k_B T} - 1}} =E_0 +E(T) E=i∑Eˉi=i∑(21+ekBTℏωi−11)=i∑21ℏωi+i∑ekBTℏωi−1ℏωi=E0+E(T)
The problem is that the summation is very difficult to obtain cause the number of independent lattice waves or equivalent harmonic oscillators is very large ( 1 0 23 10^{23} 1023). We need a new concept called “Density of state” to change the summation into integration.
将求和转变为积分
The density of states g(ω) is defined as the number of oscillators (or k) per unit
frequency interval.
态密度的定义:单位频率间隔谐振子(振动模式)的数量
g ( ω ) = d n d ω g(\omega) = \frac{dn}{d\omega} g(ω)=dωdn
Total number of oscillators: ∫ 0 ω m g ( ω ) d ω = d N p \int_0^{\omega_m} g(\omega )d\omega = dNp ∫0ωmg(ω)dω=dNp
where d is dimension, N is unit cell number and p is atom number in a cell.
E 0 = ∫ 0 ω m 1 2 ℏ ω g ( ω ) d ω E_0 = \int_0^{\omega_m} \frac{1}{2} \hbar \omega g(\omega )d\omega E0=∫0ωm21ℏωg(ω)dω
E ( T ) = ∫ 0 ω m ℏ ω e ℏ ω k B T − 1 g ( ω ) d ω E(T) = \int_0^{\omega_m} \frac{\hbar \omega}{e^{\frac{\hbar \omega}{k_B T} - 1}} g(\omega )d\omega E(T)=∫0ωmekBTℏω−1ℏωg(ω)dω
∴ E = E 0 + E ( T ) \therefore E = E_0 + E(T) ∴E=E0+E(T)
For one branch of the dispersion relation: d n = g j ( ω ) d ω = ρ j ( k ) d k dn=g_j(\omega) d\omega =\rho_j(k)dk dn=gj(ω)dω=ρj(k)dk
ρ j ( k ) = { L 2 π , 1D S 4 π 2 , 2D V 8 π 3 , 3D \rho_j(k) = \begin{cases} \frac{L}{2\pi}, & \text{1D} \\ \frac{S}{4\pi^2}, & \text{2D} \\ \frac{V}{8\pi^3}, & \text{3D} \end{cases} ρj(k)=⎩ ⎨ ⎧2πL,4π2S,8π3V,1D2D3D
g j ( ω ) = ρ j ( k ) d k d ω j g_j(\omega) = \rho_j(k) \frac{dk}{d\omega_j} gj(ω)=ρj(k)dωjdk
g ( ω ) = ∑ j = 1 d p g j ( ω ) g(\omega) = \sum_{j=1}^{dp} g_j(\omega) g(ω)=j=1∑dpgj(ω)
lattice vibration ~ harmonic oscillator ~ phonon