Chapter 2 Intro to Math Techniques for Quantum Mechanics

Transcription

1 Fll 4 Chem 356: Itroductory Qutum Mechcs Chpter Itro to Mth Techques for Qutum Mechcs... Itro to dfferetl equtos... Boudry Codtos... 5 Prtl dfferetl equtos d seprto of vrbles... 5 Itroducto to Sttstcs... 3 Chpter Itro to Mth Techques for Qutum Mechcs Itro to dfferetl equtos Fucto y = y( x) s to stsfy dfferetl equto d y dy 5 6y dx dx + = x () For ths type of Dfferetl equto (more lter), try soluto dy The = λe dx d y λx = λ e dx d y = λ e dx λx λx y = e λx Substtute to dfferetl equto () Or x x x λ e λ 5λe λ + 6e λ = x λ 5λ + 6= ( λ 3)( λ ) = Solutos: λ = 3, λ = 3x e : e x 5e x + 6e x = x e : 4e x e x + 6e x = But: y ler combto of solutos s lso soluto

2 Fll 4 Chem 356: Itroductory Qutum Mechcs yx ( ) = ce + ce 3x x 9c e 3x + 4c e x 5c e 3x c e x 6c e 3x + 6c e x ( d y dx ) ( 5 dy dx ) (6y) Let us try other oe d y y dx + = x λx λx λ e + e = λ + = λ =± x Geerl Soluto: ce + ce x Altertve wy to wrte: x e = cos x+ s x x e = cos x+ s( x) = cos x s x yx ( ) = ( c + c)cos x+ c ( c)sx = d cos x+ d s x defe d = c + c, d = ( c c) choose d, d rel Verfy: d d cos x = ( s x) = cos x dx dx d d s x = (cos x) = s x dx dx d y y dx + =, s expected Type of solutos x e λ, e λ x, cos λ x, s λ x, λ rel, λ > Chpter Itro to Mth Techques for Qutum Mechcs 3

3 Fll 4 Chem 356: Itroductory Qutum Mechcs Whe does ths work? 3 dy d y d y 3 3 = cb y+ c + c + c +... dx dx dx () Costt coeffcets frot of y d ts dervtves d y xy ot: dx + = () Ler fucto y d y dy ot: + y = dx dx (3) Homogeeous equto y ot: c x + c y x + c y = f (x) 3 For homogeeous dfferetl equto (lst cse 3): Fd prtculr soluto y = P( x) dd to ths the geerl soluto of homogeeous equto. For more complcted dfferetl equtos (e. Not homogeeous DE wth costt coeffcets) solutos re ofte hrd to fd! My trcks of the trde! Use symbolc mth progrm (t kows my of the trcks)! Numercl pproches (ofte work very esly pcture of soluto) Chpter Itro to Mth Techques for Qutum Mechcs 4

4 Fll 4 Chem 356: Itroductory Qutum Mechcs Boudry Codtos Let us cosder our orgl dfferetl equto. d y dy 5 y dx dx + = 3x yx ( ) = ce + ce Now mpose further codtos. Eg: y () = c + c = dy dx x= yx ( ) e e = x 3c + c = c = c 3c c = c = 3x x = Stsfes both DE d boudry codtos c = Soluto s completely specfed f oe supples s my codtos s oe hs free coeffcets c, c,. the soluto So recpe s very smple lwys ler set of equtos Try yx ( ) = e λx d work t out! Prtl dfferetl equtos d seprto of vrbles Cosder problem of vbrtg strg (eg. gutr, vol) Chpter Itro to Mth Techques for Qutum Mechcs 5

5 Fll 4 Chem 356: Itroductory Qutum Mechcs strg) We wt to descrbe the mpltude uxt (, ) Dfferetl Equto wth prtl dervtves: u u ( xt, ) = ( xt, ) x v t v : Velocty of wve propgto strg, relted to sprg costt, (s soud of the Boudry Codto: u(, t) = u(, t) = t uxt: (, ) fucto of vrbles use prtl dervtves I mth we typclly do ot wrte s kept costt ( cotrst to thermodymcs) u x t How to solve PDE (prtl dfferetl equto)? Try soluto u(x,t) = X (x)t (t) Smple product of fucto of x d fucto of t Boudry Codto: X() = X( ) = Substtute trl fucto to PDE Dvde both sdes by X( x) T( t ): d X( x) d T Tt () = Xx ( ) dx v dt d X( x) d T = X( x) dx v T( t) dt oly depeds o x oly depeds o t Lke f( x) = g( t). It s cler tht ths should be true for ll x, t f( x) = g( t ) f( x ) s costt = gt ( ) f( x ) s costt, should be sme gt () = f( x) gt () s costt = f( x ) Chpter Itro to Mth Techques for Qutum Mechcs 6

6 Fll 4 Chem 356: Itroductory Qutum Mechcs f( x) = g( t) xt, C oly be true f both fuctos re costt! I.e. The sme costt Let us cll ths costt, the seprto costt d X k X( x) k (for lter smplcty) = X() = X( ) = dx dt = ktt () o boudry codtos v dt Now we c use techques dscussed before ( k s costt) Try X( x) = e λx λx λx λ e = k e λ = k, λ = k X( x) = ce + c e kx Note: k could be mgry m k = ( m) =+ m > kx However we kow tht the strg wll oscllte, d hece c tcpte Usg wht we dd before ce = ( c + c )cos kx+ ( c c )skx kx kx + ce = d cos kx+ d s kx Ths s geerl soluto. Now cosder boudry codtos. x = : dcos k + ds k d+ d = d = x= : d s( k ) = d = (flt strg possblty) or s( k ) = kx e, wth k rel Whe s s(x) =? x =±, ± π, ± π, ± 3 π... x k = π π k = π x Geerl soluto: X( x) = d s π k = Chpter Itro to Mth Techques for Qutum Mechcs 7

7 Fll 4 Chem 356: Itroductory Qutum Mechcs Ths s soluto for x t prtculr vlue for k = π Now cosder correspodg soluto for Tt () d T dt Smlr equto s before: v dt dt π = Tt () = ω T (t) ω = πv T (t) = c sω (t) + c cos(ω t) If we combe ths wth X we get u(x,t) = A s π x πvt s + B π x πvt s cos Ths s soluto for y vlue of A, B d y vlue of =, ±, ±, ± 3 Most geerl soluto (we c restrct to o- egtve ): u(x,t) = s π x A s(ω t) + B cos(ω t) =,,,3... You c verfy tht ths deed stsfes PDE (qute some work). How c we terpret ths? s π x : Chpter Itro to Mth Techques for Qutum Mechcs 8

8 Fll 4 Chem 356: Itroductory Qutum Mechcs πx πx s = s : Sme soluto, restrct =,,,3 ( <, othg extr) So strg vbrtes s ler combto of modes, ech of the modes osclltes tme t dfferet frequecy. s π x ω = πv = ω All multples of fudmetl frequecy πv = ω Ths defes the ptch of the soud ω The other modes re clled overtoes Meg of coeffcets A, B? t = B cos πv t t= = B u(x,t = ) = B s π x The tl shpe of fucto Chpter Itro to Mth Techques for Qutum Mechcs 9

9 Fll 4 Chem 356: Itroductory Qutum Mechcs du dt A cos πvt The tl velocty of the strg. t= = A ; du(x,t) dt t= = A s( π x ) Dfferet strumets, gutrs, vols, cello dfferet A, B How you ttck the strg determes the tl shpe/velocty compre Chese zther: ht the sre dfferet spots or twg the strg Noe of ths ffects the ptch ω the geerl hrmoc Perod s(ωt) = sω(t + T ) s(ωt + π ) s(ωt) π = ωt T = π ω ; ω = π T gulr frequecy Itroducto to Sttstcs We wll see tht qutum mechcs s essetlly sttstcl theory. We c predct the results d ther dstrbuto from lrge umber of repeted expermets oly. We cot predct (eve prcple) the outcome of dvdul expermet. Let us therefore tlk bout sttstcs usg smple exmple: the dce If you throw the dce, ech throw wll yeld the result,, 3, 4, 5 or Chpter Itro to Mth Techques for Qutum Mechcs 3

10 Fll 4 Chem 356: Itroductory Qutum Mechcs If you throw the dce my tmes, sy 6 tmes, we mght get Totl 6 For fr dce, ech umber hs equl chce, d so we sy the probblty to throw for exmple 3 s. Ths s reflected by the ctul umbers we got the exmple. 6 = = P I the lmt tht we throw very lrge (fte) tmes, we get closer d closer to 6 P = oly hs meg for my repeted expermets N totl P = We mght cll the ctul outcome of expermet, here =,, 3, 4, 5, 6 The verge s gve by P N tot For dce: ( ) = = The verge vlue does ot eed to be possble outcome of dvdul throw. We re lso terested the vrce of the results. We cll the verge A or A. The the vrce s gve by: (both re used) σ = P( A) A Let us tke dce wth 5 sdes to mke the umbers eser =,, 3, 4, 5, P = 5 Chpter Itro to Mth Techques for Qutum Mechcs 3

11 Fll 4 Chem 356: Itroductory Qutum Mechcs A =3 [( 3) ( 3) (3 3) (4 3) (5 3) σ A = ] 5 = [ ] = 5 Stdrd Devto σ A = Chpter Itro to Mth Techques for Qutum Mechcs 3

12 Fll 4 Chem 356: Itroductory Qutum Mechcs I c wrte the vrce dfferetly s P A P A A ( ) = ( + ) = P A P + A P = + A AA AA = A A σ A Let us check for the 5 fce dce: ( A = ) 5 = ( ) 5 = (55) = 5 A = 33 = 9 A A A = = σ (s before) Of course: We proved ths s true mthemtclly! Ths cocludes (for ow) dscusso of dscrete sttstcs. Now cosder the cse of cotuous dstrbuto. For exmple desty dstrbuto. If we ormlze ρ(x)dx = dm = the mss betwee x d x + dx ( dmeso) ρ(x)dx = M totl mss b Also ρ(x)dx = mss betwee pots d b P(x)dx = M ρ(x)dx probblty to fd frcto of the totl mss betwee x d x + dx We c defe verge posto Chpter Itro to Mth Techques for Qutum Mechcs 33

13 Fll 4 Chem 356: Itroductory Qutum Mechcs x = xp(x) dx x = x P(x)dx σ x = x x Exmple: tke box betwee d Uform Px ( ) = < x < = elsewhere ) P(x)dx = dx = x ormlzed ) xp(x)dx = x dx 3) x P(x)dx = x dx = x = = x3 3 4) σ x = x x = 3 = = 3 = Alwys More complcted dstrbutos re possble of course Fmous s the Guss dstrbuto x P(x) = ce prmeter (wll be σ ) x c : ormlzto costt P(x) = c = x xe π The x = = Odd fucto f( x) = f( x) π Chpter Itro to Mth Techques for Qutum Mechcs 34

14 Fll 4 Chem 356: Itroductory Qutum Mechcs x = π x x e dx π = = π σ = s dvertsed x (ths ws the reso to defe the Guss lke ths) See book for dscusso tegrls Chpter Itro to Mth Techques for Qutum Mechcs 35