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molecular crystals or the electrostatic force in ionic crystals, however, it is more difficult to write down simple expressions for the covalent crystals. In covalent bonds, electrons are preferentially concentrated in regions connecting the nucleus, leaving some regions in the crystal with low charge concentration....
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that eiωT=1, which ensures that f(t) is periodic. A spatial function, f(x), with a periodicity a, into a Fourier series, ∞ ink x x + b 'e n −ink x x ) f (x) = ∑( n = −∞ a 'en f x ( ) = f x a , can be similarly expanded ( + ) where the wavevector, kx=2π/a, is the Fourier conjugate to spatial periodicity a. a …...
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For personal interests, you may want to compare the process with digital signal processing (DSP), where a is in analog to the sampling period, k corresponds to the detected frequency. The highest measurable frequency is inversely proportional to a. Now we will extend our discussion to three-dimensional cases. With ...
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G π n m n m • +i2 ( 1 1 + 2 2 +n m 3 3 ) = ∑ u eG G = ∑ u e G i • r G G r = u( ) Thus, we see that the new set of vectors introduced, (b1,b2,b3), which has a unit of m-1, are the corresponding Fourier conjugate to the real space lattice vector (a1,a2,a3). We can use (b1,b2,b3) to construct a new lattice cal...
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, the total amplitude at the detector is ) dV = Ae ( i k k r e i k r n( )e f ik • e f i − f )•r • −r r ) ik ( i k r r • − • s i d n( ) Ae dV (r d (r d n i(k r − ) − • r ) s f ∫ whole crystal ∫ whole crystal i r k r s d • − • ) ∫ whole crystal (k ∑ i = A e G f z i( + − G k k •r ) i f n e ...
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Consider the special set of crystal planes separated by a distance a as shown in the following figure and an incident wave (photon or electron) of wavelength λ at an angle θ. Constructive interference between waves reflected from crystal planes occurs when the phase difference of the waves scattered between two cons...
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0,1,2…) The wavefunction is Ψ , and the degeneracy is g(n)=1. (3) Rigid rotation The energy eigenvalues are n E l = 2= I 2 A A + 1) = hB A A + 1) ( ( (for |m|≤ A , A =0,1,2, …). For wavefunction Ψ nlm , the degeneracy is g(l)=2l+1. (4) Hydrogen atom 2 Mc1 2 2= n 13.6 eV n2 el = − En The wavefunction Ψnlms co...
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have energy dispersion as ω= vk = v 2 k x 2 + k y 2 k + z ; E n = hν(n + 1 2 ) , where v is sound velocity. Density of (quantum mechanical) states (DOS): (a) Electron ky π4 L π2 L 0 Volume of One Unit Cell k+dk dk π2 L π4 L π6 L kx kx 2.57 Fall 2004 – Lecture 9 kz k ky 61 In ab...
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2π = 1 2 2 )3/ 2 (E E ) c − 1/ 2 . E l e c t r o n D e n s i t y o f S t a t e s Ec EF E (b) Phonon According to the Debye model that assumes three modes (two transverse, one longitudinal) are identical, we have 2.57 Fall 2004 – Lecture 9 62 D( ) ω = 3 3 ∆N = 2 V ∆ω 2π k 2 dk 2 3ω = ...
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1 ⎛ 3π2 ⎜ in which the electron density n=N/V. E f n = ∫ E c (E − E c ) ) ( Ed E D 3 / 2 3 / 2 = f , For nonzero temperatures, Ef is replaced by the chemical potential. Nanostructures: z D For a thin film with thickness d, we have the energy as ( k E x k , y n , ) = 2 2 = k xy m 2 * + n 2 = 2π2 2...
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D E constraint (quantum dot, n, l, k as the quantum numbers), the density of states is just like jumps at different Enl k . − nl ( kz (n) D(E ) kx bulk (3D) quantum well (2D) n=3 n=2 n=1 E ky Chapter 4 Statistical thermal & energy storage In Lecture 2, we have mentioned that matter tends to occupy the lo...
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number density. The Boltzmann factor can only be applied to closed systems, while for an open system exchanging energy with the outside, the probability becomes P E Ni ) = Aexp[ −(E − µN ) / k T ] , where Ni is the particle number, chemical potential µ is the criteria for the equilibrium state of mass exchanging pr...
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) ν( k TB ] = 1 exp(− ∞ hν 2k T n=0 B )∑ Aexp( − n h ν k T B ) = 1 exp(− hν 1 B2k T 1− exp( − ) A A = exp( )(1 − exp( − hν 2k T B = 1. )) . ) hν k TB hν k T B And (P E) = exp( − )[1 − exp( − )] . nhν k T B hν k T B Then the average number of the phonons, or the occupancy of the quantum state is ...
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Vk k 2 , (2 ) π in which V=L3 is the crystal volume. π 2 4 k V1 ∆ =N k∆ = The density of states is defined as the number of quantum states per unit interval of energy and per unit volume D E( ) = 1 ∆N V E ∆ = k 2 dk k 2 ∆k = 2π2 ∆E 2π2 dE . A factor that considers polarization of waves may be added (elect...
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. 1+ For phonons, the Pauli exclusion principle is no longer applicable. And the occupancy of the quantum state is changed into 1 k T B n = f ( ) ν = ν / h , −1 which is called Bose-Einstein distribution. e For molecules, similar results exist ) n = f E ( = 1 k T ( E −µ ) / B i e , −1 where µ is again t...
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E = 2= 2m 2 (k + k + k ); 2 y x 2 z k k k z y x , , = π / 2 n L . i 1 The Bose-Einstein distribution is presented in the following figure according to the temperature. For constrained electrons, we can find that the ground state (quantum numbers n l = = = ) has much larger occupancy than any other energy leve...
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.57 Fall 2004 – Lecture 10 69 Suppose we have Ω quantum states. We can treat each accessible quantum state as a system. The collection of these Ω systems is called an ensemble. Three ensembles are analyzed in the table. Quantum State 1 Quan...
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real quantum states. The principle of is no equal probability longer The valid. probability becomes − e E k T / B i P E ( ) i = Z = ∑ − e i Z / E k T B i The Helmholz free energy F U TS - ) = F T V N ( , , = - k T Z ln = B becomes potential in this case. the thermal Grand canoni...
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For a constrained particle, its energy is expressed as 2 E = =2 2m where k = ±2 (kx + k y + kz ) , πi L k / y , 2 2 x = ±2π / j L k z 2 , = ± πn L / (n, j, l=1,2…). The internal energy is u1 = E1 = ∑∑∑ ( E k k k , z x , y k x k y k z )× exp ⎛ E − µ ⎞ ⎜ − ⎟ k T ⎝ ⎠ B ∑∑∑ ( x z ) E k k k , ex...
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Broglie wavelength is λ= 2 = Bmk T 2 π . Let y = E = 1 1 k TB 1 ∂Z Z y ∂ . The energy expression can be rewritten as = − d dy 3 ln Z = k T , 2 B which is just the familiar energy expression. It is also a special case of the equipartition theorem, which states that at high temperature every degree of f...
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∑∑∑ =ω f ( )ω = ∫0 ∞ k x k y k z =ω f ( )ω D( )ω dω , where the density of states (two transverse electromagnetic waves) is 2.57 Fall 2004 – Lecture 10 72 D( ω = × 2 ) c 2 3 ;ω= k for light. 2 2 4πk dk ω π c = 3 ⎛ 2π⎞ ⎟ ⎜ ⎝ L ⎠ dω 2.57 Fall 2004 – Lectu...
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(ω, T) = ( ωD ω = ) = 2 3 π c ω3 ) − ⎤ ω/κ BT 1 ⎦ . ⎡exp(= ⎣ Recall the Planck’s law as c 1 c / λT λ5 (e 2 e = b,λ −1) , which has λ5 but in uω we have ω3 . Noticing uλ = ∆ω ∆λ uω and ω= ck = 2 c /π λ, we obtain uλ = dω dλ uω = 2πc λ 2 uω , which accounts for the power difference in above two ex...
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, which is different from original ωmax. For phonons, we have three polarizations (two transverse, one longitudinal waves). Similar to photons, we have 2 ω 2 2π vD 3 . D ω ( ) = dN Vdω = 3 • ω ωD ωmax Debye k π/a The internal energy is U = ∑∑∑ = ω f (T, ω) k x k y k z ωD = = ω f T, ω D ω dω ( ) ( ...
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) x , 2.57 Fall 2004 – Lecture 11 75 in which x = =ω k TB . At low temperatures, the upper limit of above integration θD / T → ∞ and thus CV ∝ T 3 , while in the high-temperature limit we have constant CV = 3 N V kB . Note: At high temperatures, in each di...
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y included in D(E), the energy dispersion E −E = c 2= * 2m 2.57 Fall 2004 – Lecture 11 76 E DOE Ec E E − c E * (k + k y + k ) 2 x 2 z 2 = E E c − =2 2m 2= k 2 2m * = k We can use the same idea as phonons to solve µ , N V 2 ∑∑∑ ( V k f E µ ) ,...
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z gap should be excluded and we should not count the corresponding wavevectors. (2) In a semiconductor, the phonons contribute much more to the specific heat than electrons. Electron contribution can only exceed that of the phonon at very low temperatures in the following figure. In the equation k = CvΛ 3 , there...
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/m2. 1 2 q has the 12 Q1->2 1 2 Q2->1 x Now the question is how to calculate the transmissivity τ and velocity v. For nanostructures, the interface characteristic length is comparable to wavelength so that the interface is important even in the classic viewpoint. 5.1 Plane waves & their interface reflection ...
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C k A B = k C ( + = 1 B k k− 1 = k Ak + 1 − ) C ; = A k 1 2 2k 1 k + 2 or 2 2 . i r x=0− = Ψ ' | t x=0+ , The flux term is (note A and k1 can be complex) * ⎞ ⎟ ⎠ ⎛ i= J = Re Ψ∇Ψ ⎜ ⎝ m ⎛ = i Re ⎜ Ae ⎝ m (ω − = −i t k x) 1 * * (− k A i 1 (ω − * ) t k x 1 i e ) |x=0 ⎞ ⎟ ⎠ = = = m = m...
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mE 2= 2 ( m E u − ) 2 = . i t r x=0 − = Ψ | The boundary conditions are applied + , (Ψ +' Ψ ') (Ψ + Ψ ) i x=0 which yields ; ( A B C k A B = k C + = 1 k− B k 1 = k Ak + 1 − ) C ; = A k 1 2 2k 1 k + or 2 2 2 . r − = Ψ ' | t + , x=0 x=0 The incoming flux term is (note A and k1 can be complex) ...
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The reflectivity is 2 R = − r / i J J = B A 2 = = k k 1 − 2 k1 + k2 E − E u − E + E u − 2 , and transmittivity is T = J t / J i = 2 C A * ) Re k2( k1 . For E>u, the equation gives reflectivity R ≠ 0 in both cases shown below, which is inconsistent with classical mechanics. x x When E<u, the equat...
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A measure of the capability of the material responding to incoming electric field is the electric polarization per unit volume, or the dipole moment per unit volume, P [C m-2], which is related to the electric field through the electric susceptibility, χ, P=εoχE, N-1m-2], and the electric where εo is the vacuum...
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2) ∇ × H = + J e ∂D ∂t D ∂ ∂t Without the term, the above equation is the Ampere law, which says an electric current induces a magnetic field. The term ∂D ∂t is the current due to the electron oscillation around the ion even though they are not free to move. It is also called displace current. This term is M...
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• k = µω2[εο(1+χ)+iσe/ω] = µεω2, where the first equation has the same form as the foregoing transmitted wave. In vacuum σe=χ =0, the second equation yields k 2 = ε µω2 , or 0 0 ω= k εµ 0 0 0 = c k , in which c0 is the light speed in vacuum. This is familiar energy dispersion relationship for photons. Compared ...
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2004 – Lecture 12 83 In the special case that a plane wave is traveling along the x-direction with the electric and magnetic fields pointing to the y- and z-direction, respectively. We also define the Poynting vector S (W/m2) as S=E×H, which represents the instan...
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n1 n2 Ei Hi θi θr k i x kr Er x Et θt kt z When an electric field is parallel to the plane of incidence, its conjugated magnetic field component, in this case pointing out of the paper, is perpendicular to the plane of incidence and is thus always parallel to the interface (refer to the figure). 2.57 ...
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(upward in the z direction) from the incoming and refraction waves. The subscript “//” means that the electric field is polarized parallel to the plane of incidence for the sketched TM transverse wave. Note: For EM waves, two transverse vibrating directions exist. In general, the resultant electric field vector wil...
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cosθ E + cosθ E n1sinθi=n2sinθt, = cosθ E i // i r // r t // t . The magnitude of the magnetic field, which is pointing out of the paper, is related to the electric field by n c µ o H y = E// . We can write the continuity of the tangential component of the magnetic field as n E − n E = n E 2 1 // i 1 // ...
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4 Lecture 13 Review of previous lectures 1. Energy transport between two points Q1->2 1 2 Q2->1 x 2. Plane waves & their interface reflection We are interested in the wave energy at points 1 and 2 on two sides of the interface. Transmission wave Reflection wave Incoming wave 2 1 x Energy barrier u 3. O...
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2 + n1 E// r E// i r// = = t // = 2n1 n2 + n1 Sr z Sr , = Si z Si , St z = Re ⎜ , Si z , = 2 r// ⎛ n2 cos θ ⎞ ⎝ n1 cos θi ⎠ t ⎟ t// 2 Discussions 1) Critical & total internal reflection A) n1<n2 The Snell law is applied, n1 sin θ = n2 sin θ . t i n1<n2 x n2 Note: The Snell law indicates the momentum cons...
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t leads to an exponential decay wave ⎡ ⎛ E// t exp ⎢ −iω⎜ t − ⎢ ⎝ ⎣ 2 n x sinθ + n z cos θt ⎞⎤ ⎟⎥ co ⎠⎥⎦ 2 t E= ⎛ ⎡ // t exp ⎢−iω⎜ t − ⎢⎣ ⎝ 2 ⎞ n x sinθt ⎟ co ⎠ − 2 n z a ⎤ ω ⎥ co ⎥⎦ , which is similar to the encountered evanescent wave. Two applications: a) Optical fiber An optical fiber has ...
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θB 3) Complex refractive index For a complex refractive index N n = + κ, the intensity of the wave decay as I i αx e∝ − , where α = 4πκ λ 0 . I e−∝ αx x For real n1 and complex N2, from n1 sinθ = N2 sin θ , we obtain complex θ and t i t sinθ = t n2 sinθ i n1 = + a bi , cosθ = 1− sin θt = + c di t 2 . ...
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displacement, i.e., K K d v dt In mechanics, we define the strain as ⎞ ⎟ , ⎟ ⎠ ⎛ u ∂ 1 + 2 ⎜⎜ xi ∂ ⎝ and the stress is ui ∂ x ∂ Sij = j j σ c S . = ⋅ Note: c is a fourth-order tensor and has 81 components. For phonons, we have two transverse waves and one longitudinal wave. The poynting vector for acou...
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very important for nanomaterials. Acoustic waves (or heat propagation) can be cut off by the interface, just as using a foil to cut off the radiation between two surfaces. 5.3 Wave propagation in thin films First let us consider the following structure. Many reflections and transmissions exist in this case. Summat...
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Lecture 13 93 (0) ⎞ ⎛ E x ⎜ ⎜ H (0) ⎝ y ⎛ Ei ⎞ ⎟ E r ⎠ ⎟⎟ = A⎜ ⎝ ⎠ x (0) ⎞ ⎛ E x ⎜ ⎜ H (0) ⎟⎟ ⎝ ⎠ y = M1 (d ) ⎞ ⎛ E x (d ) ⎟⎟ ⎜⎜ H y ⎠ ⎝ BEt For multilayers, M is determined by the chain rule as M = M M 2 " M , n where n is the number of layers. 1 For a single layer of film, ⎜ −1 ⎟ = A M BE...
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– Lecture 13 94 (2) Tunneling Back to the case in which θ i >θ . We have cr cosθ t = 1− sin θ t = i 2 2 n1 sin θ ⎞ ⎛ ⎟ ⎜ n2 ⎠ ⎝ i − 1 = ai , which causes decay in the second medium. If the second medium is very thin, a third medium attached to the second medium will have tunneling effects. The wave w...
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– Lecture 13 95 2.57 Nano-to-Macro Transport Processes Fall 2004 Lecture 14 Review of last lecture d R n1 n2 n3 φ In lecture 13, we talked about tunneling through a thin film, which can also happen to general EM waves and acoustic waves. If the second medium is very thin, a third medium attached to th...
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9 ~ 4 e m-1, 9 2.57 Fall 2004 – Lecture 14 96 and the characteristic length for d is 1/k~1 Å. Applications of tunneling 1) Scanning tunneling electron microscope (STM) TIP VOLTAGE TUNNELING GAP SAMPLE CURRENT FEEDBACK ...
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the other side of the surface, there will be tunneling between the surface and the probe and the topography is obtained as in a STM. Tunneling θ>θcr 2.57 Fall 2004 – Lecture 14 97 B) Atomic force microscope (AFM) A STM cannot be used to scan a dielectr...
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vacuum to avoid air conduction and convection effects. Large effective contact area 5.4 Bragg reflector In practice, the reflectivity and transmissivity of multilayers can be controlled quite accurately with various thin-film deposition techniques and the possibility in controlling spectral and directional proper...
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T V T C E L F E R 1 0.8 0.6 0.4 0.2 0 1 Stop Band 1.5 2 2.5 3 WAVENUMBER=1/WAVELENGTH (µm -1) It is interesting to compare the stop band with electron band gap in semiconductors. In the latter case, no electrons exist in the forbidden energy levels, while in the reflector no photons exist in the stop band....
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expressed as ∞ ∞ ∞ 1 2 v E τ f ( E , T 1 ∑ ∑ ∑ z1 1 12 3m ⎡ q → = ∑ ⎢ p=1 ⎣⎢V1 kx1 =−∞ k y1 =−∞ kz1 =0 ⎤ 1 )⎥ , ⎦⎥ where Te1 represents the temperature of the phonons coming towards the interface and f(Ε1,Te1) is the Bose-Einstein distribution for phonons at Te1, and τ12 is the phonon transmissivity from me...
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4 – Lecture 14 ksinθd φ kdθ kx 100 However, we also notice that velocity vz1 depends on k and transmissivity τ depend on (k . To deal with this, first we note k , k ) , x1 12 y1 x1 z1 dk dk dk = k sin θ θ ϕ = k dkd Ω = k dkd d x1 y1 z1...
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− q 1 2 ) → ( e2 ) 1 ( where we use q in the last step. This is obtained by considering the T e2 equilibrium status q12 =0 and Te2 =Te1 . In nonequilibrium situations, this relationship still holds true because we have the same expression for q → (T ) . This idea is comparable to the Kirchoff's Law in radiatio...
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T − T e1 R12 , where K is the contact thermal conductance (W/m2 K), R12 is the contact thermal resistance (m2 K/W). Note: (1) The value of τ12 ranges from 0 to 1. In the Landauer formalism, E1 ∂f ∂T )D E dE ( has the physical meaning of volumetric specific heat C. For normal materials, the magnitude of C is 10...
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. If the refractive index is n, we have D(E)~n3 and v~1/n (Snell’s law). Thus the heat flux is proportional to n2. 4 2 4 1 12 2.57 Fall 2004 – Lecture 14 102 When a surface wave (decays exponentially on both sides of the surface) exists, the maximum h...
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beyond one mean free path) calculated from the local internal energy u(T1) in the small region around point 1. Note: The relationship q = 12 ) is consistent with the Fourier’s law. K T ( − T 2 1 R12f(Te1) R21f(Te2) τ12f(Te1) f(Te1) u1(T1) u2f(T2) 2.57 Fall 2004 – Lecture 14 103 ...
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single nanotube (as one molecule) spectroscopy. Due to the unique properties, currently many applications are being attempted worldwide for CNs. 2.57 Fall 2004 – Lecture 15 104 3. Synthesis Three methods are utilized to grow CNs: (1) Arc Discharge In the following figure, two graphite rods (5-20 mm in diameter) a...
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catalysts and furnace temperature (H. Kataura et al., 2000). Image removed for copyright reasons. N. Wang et al. Nature 408, 50 (2000) The above figure shows the smallest SWNT with a diameter 0.42 nm. It is a (5,0) zigzag nanotube and has metallic electronic structure. Under 15 K it becomes a superconductor. Imag...
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iral Vectors : (n,m) R. Saito et al., Phys. Rev. B46, 1804 (1992) = dt = L a n + nm + m2 2 π L = Ch π Although there can be other ways to define the primitive lattice vectors in a hexagon sheet, the method presented in the above figure is widely used in all current publications. The chiral vector (equator of na...
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391 (1998) 59 Using a STM, the structure of a SWNT can be observed. In the right above figure, we can barely see the hexagons. SWNTs can be metallic or semiconductive. The transition happens when the conduction band and valence band touch each other at six K points in the k space. The energy contours are also drawn...
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Semiconductor R. Saito et al, Phys. Rev. B61, 2981(2000) The wave vector k for one-dimensional carbon nanotubes is shown in the two- dimensional Brillouin zone of graphite hexagon. In the direction of K1, discrete k values are obtained by periodic boundary conditions for the circumferential direction of the carbon ...
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the energy level can be shifted and tunneling will happen at discrete levels. In addition, quantized conductance is also observed in this experiment, which is a multiple of G0 = 2e 2 h = ( . 12 Ω ) −1 k 9 2.57 Fall 2004 – Lecture 15 110 Image removed for copyright reasons. W. Liang Harvard Univ. Ballistic tra...
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. dR = 2(m N = + nm) 2 + 2n dR if n− m = 3d ⎧ ⎨ ⎩ d d ⋅3 p otherwise “Physical Properties of Carbon Nanotubes” R. Saito, G. Dresselhaus, and M.S. Dresselhaus, Imperial College Press, (1998) 2N carbon atoms 6N phonon modes (three directions of vibrations) 2.57 Fall 2004 – Lecture 15 111 Image removed for ...
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significant to develop method for large-scale, cheap synthesis, and improve nanotube characterization and manipulation. Separation processes reported till now include: (1) Precipitation of SWNTs non-covalently functionalized with ODA (2) Ion-exchange liquid chromatography of ssDNA wrapped SWNTs (3) Alternating cur...
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− x k .) x 1 2→ t x Consider a plane traveling along x-direction Ae− i(ω t − kx) . The velocity v =ω / k is NOT the speed of signal or energy propagation. We see that the plane wave represented by above equation extends from minus infinite to plus infinite in both time and space. There is no start or finish and...
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For simplicity, let’s consider that the signal is an electromagnetic wave. We pick up only two of the Fourier components and consider their superposition, one at frequency ωο−∆ω/2 and another at frequency ωο+∆ω/2 propagating along positive x-direction [figure (a)]. The superposition of these two waves gives the elec...
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only exists in 2 cos (∆ω• t− ∆k x ) . (2) We need to calculate the time-averaged Poynting vector Re ( E H ) to get the real energy flux. k x o − o a JG JJG * ⎣ × • 1 2 This yields another wave propagating at the speed 2.57 Fall 2004 – Lecture 16 115 v = g x, ∆ω dω , = ∆ k dk which means that ...
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V=0 V max k k π/a For photons, the group velocity is constant (the following left figure). ω vg=c0 π/a n 1 k k (2) In most cases, the energy velocity is just the group velocity. However, for photons the group velocity can be larger than the speed of light in special cases. We can appreciate this from the...
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the group velocity is consistent with the de Broglie relation p = =k and our classical relation p=mvg, but not the phase velocity. 5.6.2 Loss of coherence (1) Inelastic scattering The scattering of electrons can be elastic in which the electrons merely change directions but have the same energy before and after th...
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that the effective bandwidth for thermal emission is ∆ν = k T / h . Using this effective bandwidth, one can B estimate that the coherence length ~ hc /(k T ) = B E 3 8* 6.6 E − 34 1.38E − 23*300 = 3E − 5 , which corresponds to A cT = 2167.8 µm.K, very close to the Wien’s displacement law. For electrons, lc = 100-1...
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layer is very thick [figure (c)]. If we alter the thickness of each layer randomly, the overlap still exists. The transmissivity is drawn in the following figure. the Maxwell equations, to calculate the Transmissivity Periodic structures Thicknesses are distributed more randomly Number of layers Note: Rando...
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energy of the quantum state, the system temperature and chemical potential. When the system is not at equilibrium, these distribution functions are no longer applicable. In the statistical description, we use nonequilibrium distribution functions, which depend not only on the energy and temperature of the system, bu...
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the degree of freedom of the whole system is 2n = 2m × N . These 2n variables form a 2n-dimensional space. The system at any instant can be described as one point in such a space. This space is called a phase space. The time evolution of the system, i.e., the time history of all the particles in the system, traces ...
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on this conservation requirement. Recall that in fluid mechanics or heat transfer, we often use the control volume method rather than tracing the trajectory of individual fluid particles. We could do the same for the points in phase space and exam a small control volume in phase space as shown in the figure. The ra...
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at t=0 Flow-line control volume r(i) (i) r (i)+∆ (i) r r The time evolution of f(N)(t,r(n),p(n)) in the phase space is governed by the Liouville equation, (N) n (N) (N) n ∂f ∂t + ∑ r × (i) ∂f ∂r i =1 (i) (i) ∂f + ∑ p × i =1 (i) ∂p = 0 which can be derived based on the fact that the traces of systems ...
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2 N . For simplicity in notation, we will drop the subscript 1 and understand (r,p) as the coordinates and momenta of one particle. Since f(N)(t,r(n),p(n)) represents the number density of systems having generalized coordinates (r(n),p(n)) in the ensemble, the one particle distribution function represents number de...
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the collision process is to solve the corresponding time-dependent Schrödinger equation for the combined system made of both particles. This is, however, usually very complicated and not practical. A simpler way to treat the collision is to use the perturbation method. This method considers the time-dependent inter...
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� f H 'Ψid r⎤ ⎦ 3 2 f * 2π ⎡ ⎣ = δ( E f − Ei ) 2 i Η ' f 2 M δ( E f − E ) if i = = 2π = 2π = if where d3r=dxdydz means integration over the whole volume of the system and M ≡ i H' f ≡ ∫ Ψf is called the scattering matrix. The Kronecker delta function, δ(Ef − Ei ) , implies that the energy must be con...
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k1d k' d k1' ( 1 d ⎜ ⎝ ∂t ⎠c ⎟ = − f ( , ∫ f r k1 ) , ) ( , r k t 3 3 3 ( + ∫ f ( , ', ) ( , 1', t W k',k1'→ k, k1 ) k1d k' f r k r k t d ) 3 3 3 d k ', 1 where the first term on the right hand side represents the rate of particles being scattered out of quantum states determined by k and k1, and the second ...
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∫ ) ,t 1 k V 3 (2π) 3 d = f . , ' 3 1 where the factor V 3 (2π) 9 appears in the conversion from summation over wavevector into integration over the phase space. , ⎟ = − f − fo (T , E, µ) τ(r, k) The integral-differential Boltzman equation is very difficult to solve in general. Most solutions rely on a ...
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Lecture 17 125 where the vector G accounts for the momentum imbalance between the initial and final states when k1 + k > π/a . We have learned that k1 + k > π/a (out of the first Brillouin zone) is meaningless and must be flipped over into the first Brillouin zone. This m...
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requirements. The participation of phonons in the process reduces the absorption or emission of photons, as shown in the above figure. Generally speaking, electrons will collide with electrons, phonons, and impurities in the crystals. For a normal process in which two phonons merge into one (shown in the figure bel...
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For the merging of two phonons into one, the energy conservation gives hν1 + hν2 = hν3 and a similar equation can be written for the process that one phonon splits into two. The momentum conservation during the three-phonon interaction processes takes a special form. For the phonon merging process, the momentum con...
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Energy Repulsion Harmonic Force Approximation Interatomic Distance Attraction Equilibrium Position 2.57 Fall 2004 – Lecture 17 128 When the particles have a nonzero average velocity u, the following displaced Maxwell velocity distribution...
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ons have larger wavelength and is more like a wave. The wave theory is partially included for some problems. For example, we have solved the wavefunctions of electrons in deriving the band structure in crystals. The wavelength is far larger than the interatomic distance. 2.57 Fall 2004 – Lecture 17 129 ...
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placed Maxwell velocity distribution (molecules) Note: The relaxation time is due to combined factors and can be evaluated numerically by adding all possible influence together. This idea can be used in calculating the band structure by the Boltzmann transport equation. k2,ν2 ν3=ν1+ν2 k3 k1,ν1 (a) ν1,k1 ν2,k2...
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⎛ ⎟ ⎜ ⎝ ∂t ⎠ s f − τ degrees, while Tp is only hundreds degrees. The two temperatures differ a lot. = − f − f0 τ We can understand the meaning of τ easily by neglecting the spatial non-uniformity of the distribution function. The Boltzmann transport equation becomes ∂f ∂t and thus f − fo = Ce So the relaxatio...
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04 – Lecture 18 131 ⎛ f = f −τ⎜ v • ∇ f + ⎝ r o o F m • ∇ ⎞ f ⎟ . ⎠ v o From the distribution function changing in long period compared with the relaxation time, we can calculate the flux of various quantities of interests (charge, momentum, and energy). Fourier Law ...
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ωf dω⎢ ⎢⎣ ⎩ = ky φ θ kz ) D ( ⎤ ⎫ ω sin θ θ⎬ d ⎥ d ϕ 4π ⎥⎦ ⎭ kx Note: Here θ varies from 0 to π because both positive and negative vx are considered. 2.57 Fall 2004 – Lecture 18 132 With f = fo −τ⎜ v • ∇r fo + ⎛ ⎝ ⎡ 2 π ⎧ q...
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⎫ ⎪ dθ⎬ ⎪⎭ ( τv2 C ω dθ ) } k = ∫ τv Cdω in the Fourier law. 2 1 3 Note: The equilibrium fo term drops out in the integral, which is consistent with our expectation. In the case that both τ and v (group velocity v the above expression recover to the kinetic relation = dω/ dk ) are independent of frequency, 1 ...
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is ⎛ fo(vx,vy,vz)=nP(vz,vy,vz) = n⎜ ⎜ ⎝ ⎡ 3 / 2 − m⎢(v x − u) + v 2 ⎞ ⎟ ⎟ B ⎠ m 2πκ T e ⎣ 2 y + v 2 ⎤ z ⎥ / 2κ B T ⎦ From f = f τ( v • ∇ f ) (F is zero because this is no external fields here and gravity is − r o o neglected), the distribution function is ∂ f ∂y z ) ⎜ ( x   τ v i + v j + v k ⋅ y f =...
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on the same mechanism. 2.57 Fall 2004 – Lecture 18 134 In the following lecture, we will discuss the Ohm’s Law and Wiedmann-Franz’s Law. The force acting on the electron by the external field is F = −eε , where e is the unit charge, and the charge of an ele...
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no external fields here and gravity is neglected), the distribution function is ⎛ ∂ f ⎝ ∂x ( x z ) ⎜   τ v i v j + v k ⋅ y ⎞ o  ⎟ ⎠ k = f − τv y o ∂ f ∂y ∂ f ∂z f = − o o  j + i + o  +  f ∂ f o ∂ y = f − τv o y ∂ f ∂ u o ∂u ∂ y . Note: We can also prove u = ∫ ∞ −∞ ∫ ∞ −∞ ∞ −∞ ∫ v fdv...
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z B ⎛ ⎜ ⎝ ∞ dv v e z ∫ y −∞ 2 2 − mv /(2κ T) y B dv ∞ '2 v x κ T −∞ B y ∫ ' 2 x − mv /(2κ T) e B ' dv . x Note: Two integrations are used to calculate the above equation: 2 2 − x e dx) = (∫−∞ +∞ 2 − x e dx)(∫−∞ +∞ 2 − y e dy) = ∫−∞ 2 ( 2 +∞ − x + y e ) dxdy = ∫0 +∞ 2 −r e 2π rdr yie...
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= o ⎛ − τ v ⎜ x ⎝ qε ∂f ∂E ⎞ df o o x . ⎟ dx m E v ∂ ∂ x ⎠ + Empty levels µ Filled levels 2.57 Fall 2004 – Lecture 19 E Ec Ev µ 137 For metals (left figure), the electrons fill part of the band, while the Fermi level normally lies within the band ga...
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θ ϕ above equation with 2π π +∞ ∫ ∫ 0 0 0 ∫ J = q q ∞ = − ∫ τv 3 0 ( 2 D E sin θ θ ϕ d d dE qv f D E yields ) ( ( D E ) x π 4 ⎛ ∂fo )⎜ ⎝ ∂x ∂fo ⎞ ⎟ d ∂E ⎠ + qεx E . Now we let Ec=0 be the reference energy level. And the distribution becomes f , ) o ( f E E T = , , 1 ⎛ Ek − (µ− Ec ) ⎞ exp ⎜ ⎟ ...
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E ∂ ∂ k , B = ∂f 1 0 y k T ∂ B . 2.57 Fall 2004 – Lecture 19 138 The flux is thus q ∞ ⎛ ∂µ E − µ ∂T k J q = − ∫ τv ⎜ − 3 0 ∂x T ⎝ ∂x − 2 q + εx ⎟ ⎞ ∂f 0 ⎠ ∂Ek D E d ( k ) Ek = ⎧ ∞ ⎪1 ⎨ ∫ ⎪3 ⎩ 0 2 qτv ∂f 0 ∂E k ( k ) D E dE ⎫ ⎪⎛ ∂µ k ⎬⎜ ⎪⎝ ∂x ⎭ − qε + ⎧ ∞ ⎞ ⎪1 ...
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+ v is the electrochemical potential. Note: ∂µ ∂x − qεx = ∂µ x ∂ + q dV dx = ∂ ⎛ µ ⎞ + v ⎟ . ⎜ x q ∂ ⎝ ⎠ The flux is often written as J = qnµε + Dq x e x ∂n ∂x where the first term corresponds to drift, the second term denotes diffusion; µe is called the mobility [m2V-1s-1] and D is the diffusivity [m2...
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we have 0 = L q 11 ⎜ ⎛ dΦ ⎞ ⎟ ⎝ dx ⎠ + L 21 ⎜ ⎛ dΤ ⎞ ⎟ ⎝ dx ⎠ , thus dΦ dx L dT = − 12 dx L 11 L 12 L11 = −S dT dx , where S= (VK-1) is called Seebeck coefficient. The Seebeck voltage is the steady­ state voltage accumulated under the open circuit condition. If the conductor is a uniform material suc...
https://ocw.mit.edu/courses/2-57-nano-to-macro-transport-processes-spring-2012/2e4ecaa5cf55f03bcefbc8ccce79aed6_MIT2_57S12_lec_notes_2004.pdf
1D nanowires -Quantum dot superlattice nanowires – model calculations -Newly emerging research directions -New methods for synthesis and assembly of nanowires ● Self assembled composite nanostructures ● New 3D crystalline materials with quantum dots ● New thermoelectric tools 2. Introduction to Thermoelectricity...
https://ocw.mit.edu/courses/2-57-nano-to-macro-transport-processes-spring-2012/2e4ecaa5cf55f03bcefbc8ccce79aed6_MIT2_57S12_lec_notes_2004.pdf