|
Stato Solido Capitoli:
Appendici |
Note07DIATOMIC CHAIN (GEKO IMAGE and GLAZER animation) recall MONATOMIC concepts: normal modes, phonons, quantization, crystal momentum BZ, Heat Capacity exactly compare to Debye and Einstein Real materials different atoms, better model in 1 D [ o O o O o O ... a) draw unit cell b) draw as distance between equiv c) draw as size of unit cell d) draw again o O o O and draw alternative cell [ a) UNIT CELL = REPEATED MOTIF b) LATTICE CONSTANT = a = DISTANCE BETWEEN EQUIV ATOMS c) = SIZE of UNIT CELL d) UNIT CELL NOT UNIQUE] [ search normal modes, new things happen ALTERNATING or DIATOMIC CHAIN o O o O o O ... connect with springs draw distance a two probs: different masses, m1, m2 or different k, k1, k2, this easier other as homowork draw coordinates 0 = x, O = y x1,x2,x3, ... y1,y2,y3 ... dx, dy = DEVIATIONS FROM EQUILIBRIUM ] write Newton equations for dx and dy in colour [m ddot dxn = kappa2(dyn - dxn) (RIGHT) + kappa1(dyn-1 - dxn) (LEFT) m ddot dyn = kappa1 (dxn+1 - dyn) (RIGHT) + kappa2 (dxn- dyn) (LEFT) leave space for later [ WAVE ANSATZ (in OED! means mathematical guess) write in colour dxn = Ax e^{i wt - i k na} dyn = Ay e^{i wt - i k na} (can have independent origin since we care only about dy, dx) w>=0 k EITHER SIGN k <-> k + 2 pi/a are same e^{-i(k+2pi/a} na = e^{-i k na} USE PERIODIC B.Cs sistem of length L = Na, N UNIT CELLS] [k = (2pi/L) p with pi € Z NUMBER OF DISTINCT k (RANGE )/SPACING = 2pi/a/2pi/L = L/a = N = NUMEBR OF UNIT CELLS] go back to equation slide & plug in Ansatz carefully - w^2 m Ax e^{iwt-kna) = Ay (kappa2 e^{iwt-kna) + kappa1 e^{iwt-k(n-1)a)) - Ax (kappa1+kappa2) e^{iwt-kna) - w^2 m Ay e^{iwt-kna) = Ax (kappa1 e^{iwt-k(n+1)a) + kappa2 e^{iwt-kna)) - Ay (kappa1+kappa2)e^{iwt-kna)] [exponential on both sides drop out and a big matrix remains -m w^2 (Ax\\ Ay) = ( -(kappa1+kappa2) & kappa2 + kappa1 e^ika \\ kappa2+kappa1e^{-ika} & .(kappa1+kappa2) ) (Ax\\ Ay) eigenvalue equation, the eigenvalue is mw^2 get secular equation det (M-mw^2 1) (mw^2 - (kappa1+kappa2))^2 - |kappa2+kappa1 e^{ika}|^2 = 0 mw^2 = kappa1 + kappa2 pm |k2 + k1 e^{ika}|] (*) for each k two solution of w, pm [EACH k HAS TWO NORMAL MODELS count
(right becasue we have two masses per unit cell)] [Plot sketch in colour axis k w(k) with - pi/a pi/a BZ, draw the result, and then hint at periodic zone scheme why does it look as I have drawn it?] [convenient points k=0 -> |kappa2 + kappa1 e^{ika}| = kappa1 + kappa2 -> two solutions m w^2 = 0 or mw^2 = 2(kappa1 + kappa 2), i.e. w- = 0, w+ = sqrt(2(kappa1 + kappa2)/m)] low freq mode is expected: we know that sound is linear in k so w goes to 0 as k goes to 0 [expand w- near k=0 kappa2 + kappa1 e^{ika}| = sqrt(kappa2^2 + kappa1^2 + 2 kappa1 kappa2 cos ka)
replace cos ka = 1 - (ka)^2/2 = sqrt((kappa1+kappa2)^2 -kappa1kappa2 (ka)^2) = (kappa1+kappa2) sqrt(1 - kappa1kappa2/(kappa1+kappa2)^2 (ka)^2 ) ] [= (kappa1+kappa2 - kappa1kappa2/2(kappa1+kappa2) (ka)^2 plug back in the solution *, the first term cancels mw-^2 = kappa1kappa2/2(kappa1+kappa2) (ka)^2 ] i.e. w- = sqrt(kappa1kappa2 a^2/2m(kappa1+kappa2)) |k| the slope must be the sound velocity (ACOUSTIC MODE) we were able to obtain the same from thermodynamics vs = sqrt(B/mu) mu = 2m/a; B = kappa a spring constant in series is kappa = kappa1kappa2/(kappa1+kappa2) plug back to obtain vs = sqrt(kappa1kappa2 a^2/2m(kappa1+kappa2))] high frequency mode is called OPTICAL MODE write on sketch experiments on shining light on material and induce vibrations, light has w = c|k| with c HUGE >>vs draw a steep linear dispersion immagne to absorb a photon and create a phonon, match up frequency and momentum the only point is up there. Ma il processo non avviene veramente e la ragione è che dovrebbe anche conservare il momento angolare, ma il fotone ha spin 1 e il fonone ha spin 0. Quindi non esiste un processo semplice in cui un solo fotone è assorbito e un solo fonone emesso, perchè violerebbe la conservazione del momento angolare. Esistono però processi più complicati, ad esempio assorbire due fotoni ed emettere un fonone (ok perchè il momento angolare dei due fotoni può cancellarsi e così si conserva). [ ACOUSTIC MODE is any mode WHERE w~k at SMALL k OPTICAL MODE is any mode WHERE w~constant at SMALL k let's look at EIGENVECTORS close to k=0 is - (kappa1+kappa2)(1 & -1 \\ -1 & 1) for w=0 EIGEN (Ax \\ Ay) = (1 \\ 1) for w = sqrt(kappa1kappa2 a^2/2m(kappa1+kappa2)) |k| EIGEN (Ax \\ Ay) = (1 \\ -1) so acoustic mode atoms mooving in same direction at k~0, for optical moving in opposite directions this explains why the energy of optical is high and, motion is compressing a lot the springs, whereas in acoustic it is virtually not compressing at all, therefore energy goes to zero. [GENERAL M ATOMS in UNIT CELL M MODES for each k, 1 IS ACOUSTIC, only one wy to move all atoms in the same direction, IN CONCERT M-1 are OPTICAL in d DIMENSIONS dM MODES at EACH k d are ACOUSTIC d(m-1) OPTICAL count degrees of freedoms d per mass per cell. e.g. in 3 dimension there are three directions to move, one longitudinal and two transverse, each of them gets an AOCUSTIC MODE, the rest is optical [other intersting point CONSIDER BZ BOUNDARY k=pi/a where |kappa1 + e^{ika} kappa1 | = |kappa2 -kappa1| w^2 = 1/m (kappa1+kappa2 pm |kappa2 -kappa1|) either w = sqrt(2kappa1/m) or w = sqrt(2kappa2/m) draw two poins assume k1>k2, two modes never cross (dxn\\dyn) = (Ax\\Ay) e^{iwt} e^{-ikna} the last at BZ is a factor e^-ipi/ana = (-1)^n so alternate masses of the same kindmove in opposite directions SHOW BY GLAZER First ACOUSTIC k~0 move together then OPTICAL k~0 move opposite then k = BZ two cells move opposite to each other and only one kappa counts because the other sping is unchainged now switch to acoustic, smaller frequency and it is the other spring that is getting compressed, download the program and play redraw the picture from -2pi/a 2pi/a in extended zone, then erase original optical only one mode at each wave vector k the original was REDUCED ZONE SCHEME 1st BZ 2ND BZ put optical WHEN IS IT CONVENIENT suppose kappa1very close to kappa2 redraw with very small gap if kappa1 = hkappa2 MONATOMIC CHAIN, but now a' =a/2 BZ k € [-pi/a',pi/a'] = [-2pi/a,2pi/a] making kappa1 slightly different from kappa2 defines a new ZB opens a gap at the newly defined ZONE BOUNDARY ADD VAN DER WAALS BONS MOLECULAR FLUCTUATING DIPOLE BONDS OCCUR in INACTIVE CHEMICAL SPECIES (noble gasses, molecules e.g. N2 O2, etc.) draw two noble gas atoms. Assume one has a dipole at an instant, vectors: E = -p1/4pi eps0 r^3 -> p2 = chi E U = -|p1| |p2| /4pi eps0 r^3 = - |p1|^2 chi /(4pi eps0 r^3)^2 propto -1/r^6 so Force propto 1/r^7 but draw sphere with e- and nucleus + <vec p> = -e<vec R> = 0 however <p^2> != 0 if keep angles into account it is still attractive (see exercise) much weaker force than covalent ionic or H-BOND but still strong enough to make He liquify and N solidify GEKOS sticks to the wall, even glass walls. |