( ) 2. ( ) is the Fourier transform of! ( x). ( ) ( ) ( ) = Ae i kx"#t ( ) = 1 2" ( )"( x,t) PC 3101 Quantum Mechanics Section 1

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1 1. 1D Schrödinger Eqution G chpters the Free Prticle V 0 "( x,t) i = 2 t 2m x,t = Ae i kxt "( x,t) x 2 where = k 2 2m. Normliztion must hppen: 2 x,t = 1 Here, however: " A 2 dx " " As this integrl represents the probbility tht the prticle cn be found, this must lwys be equl to 1 nything other thn this, nd it cnnot represent physicl model. 1.2 Wve Pckets Wve pckets hve to be used to void this. A single plne wve cnnot describe quntum prticle, but wve pcket cn. ( x) = g( k)e ikx dk g k " is the Fourier trnsform of ( x). xk " 1 xp " (uncertinty principle) x g( k)e ikx dk = = 1 2" ( p x ) = 1 ( x)e ip x x " dx 1 2" ipx " ( p)e dp 2"" The ltter prt is where the wve function depends on the momentum, rther thn the position or time. This is denoted by the tilde. = ( p) 2 dp P p This is the probbility tht the prticle hs momentum between p nd p + dp. 1.3 Time independent potentils "( x,t) i ( x,t) 2 " x,t = 2 t 2m + V x x 2 = T t Seprtion of vribles let x,t "( x,t) " ( x).

2 i dt T dt = d 2 " " 2m dx 2 Let E be the constnt here. i dt = ET (1.3-1) dt + V ( x ) 2 d 2 " 2m dx + V ( x )" = E" (1.3-2) 2 The solution to (1.3-1) is: = Ae iet T t For (1.3-2), d" V ( x )" ( x) = Ĥ" = E". 2m dx 2 This is the Time Independent Schrödinger Eqution. It is n eigenvlue problem involving the Hmilton opertor. The complete solution is: = A" ( x)e iet. must stisfy to correspond to physiclly meningful wve function. 1) ( x) must be single-vlued. 2) ( x) 2 must be finite. 3) ( x) must be continuous everywhere. x,t Condition x d is continuous everywhere except t n infinite discontinuity of the dx potentil. infinite wll 0 = ( 0) = ( 0) = d ( 0 ) dx dx 0 d Infinite squre well & V ( x) = 0 " x " '& x.

3 ( x) = ( x) = 0 d 2 + E 2m dx 2 1 = 0 ( x ) ek 1 x + B 1 e " k 1 x k 1 2 = m E For E > 0, k 1 is purely imginry. x = Acos kx + Bsin kx where k =. (") = 0, ( ) = 0. Acos k + Bsin k = 0 ( x = ) Acos k Bsin k = 0 ( x = ) If sin k = 0, then; k = n n = 0,±1,±2,... If cos k = 0, then: k = n n = ±1,±3,±5,... ( x) = & 2 x = 1 2 x dx = 1 " Acos n" 2 x n = 1,3,5,... Bsin n" 2 x n = 2,4,... A 2 n cos 2 2 x dx = 1 " B 2 n sin 2 2 x dx = 1 "

4 n x cos 2 2 dx = A = 1 E n = n2 8 2 m. A = B = 1? 1.5 Probbility current = ( x,t) 2 dx probbility density. P x,t the probbility of finding the prticle t time t between x nd x + dx. Probbility current: j( x,t) = * " 2mi "x " * & "x ' ( ) (1.5-1) Replcing (1.5-1) into the Schrödinger eqution, we obtin: P x,t j x,t + = 0 t x b P x,t " dx = b P( x,t) t t " This is the totl probbility in ",b. b " = "x j ( x,t )dx = j,t Exmple: t time t = 0, ip x x ( x) = ce ikx = ce j( b,t) j( x) = c 2 p x m For prticle trveling in the x-xis t velocity v, wht is the flux? j( x) = v where is the density of the prticle. ip x t Then plne wve ce is equivlent to bem of prticles with density c Finite Well We wnt to solve 2 d 2 " 2m dx + v ( x )" = E" 2 for = 0 " x " V x & V 0 x > Let region 1 = < x <, region 2 =<, region 3 =<.

5 e k 1 x + B 1 e " k 1 x = A 2 e k 2 x + B 2 e " k 2 x = A 3 e k 3 x + B 3 e " k 3 x Replcing into the TISE, we obtin: k 2 2 = k 2 3 = 2m ( V o E) k 1 2 = me If the energy is positive, then: k 1 = ±i = ±ik Conditions for x : = ( ) = (") " As the jump in the potentil is finite, rther thn infinite, we must require: d ( ) = d ( ) dx dx d (") = d ( " ) dx dx For the bound sttes E < V 0 k 3 = k 2 = e kx + B 1 e " kx = A 2 e x + B 2 e " x = A 3 e x + B 3 e " x ; 2m ( V 0 E) = " ( rel, positive number.) B 2 must be zero, otherwise tht exponentil will diverge. The sme for A 3. sin kx + B 1 cos kx = A 2 e "x = B 3 e "x We obtin non-trivil solutions only for certin vlues of k nd. Through little lgebr: tn k = tn k = 2m V 0 k 2 k k 2mV 0 k 2

6 For the unbound stte, E > V 0. k 2 2 = k 2 3 = 2m V 0 & " E k 2 = k 3 = ±i 2m ( V 0 E) = ±i" k 1 = ±i = ±ik We re using strem of prticles e i x of density 1 coming from ", going to +, nd seeing wht hppens when they rrive t the potentil well. Solutions: cos kx + b 1 sin kx = e i" x + Re i" x = Te i" x Incident bem of density 1 ( e i x ), with reflected bem of density R ( Rei" x ). Also, trnsmitted bem of density T ( Te i x ). The extr term in mkes the eqution inhomogenous. Now we hve four inhomogeneous equtions for R, A 1, B 1 nd T 1. We then find solutions for ny k, nd then for ny energy. We obtin: ie ( zi" k 2 " 2 )sin( 2k) R = 2k" cos( 2k) i( k 2 + " 2 )sin( 2k) e 2i" 2k" T = 2k" cos( 2k) i( k 2 " 2 )sin 2k R 2 nd T 2 re proportionl to the rtios of reflected nd trnsmitted flux of prticles to the incident flux. When E >> V 0, then ~ k, nd R 0. When E V o, then " 0 nd T 0. If 2k = n, then E = V 0 + n2 8m 2 R = 0,T = 1 This is clled trnsmission resonnce. 1.7 Potentil Brrier

7 E > V 0 Similr to finite potentil well. E < V 0 = e i" z + Re i" x e k 1 x + B 1 e " k 1 x = Te i" x where = k 1 2 = 2m V 0 E k 1 = 2m ( K E) = k ie ( 2i" k 2 + " 2 )sinh( 2k) R = 2k" cosh( 2k) i( k 2 " 2 )sinh 2k T = e 2i 2k" 2k" cosh( 2k) i( k 2 " 2 )sinh 2k 2 T 2 2 k = ( + k 2 )sinh 2 ( 2k) + ( 2 k) 2 (DNLT) So, s V 0 ", k " sinh2k ". Therefore, T 0. " sinh2k ". Therefore T 0. For 2k >> 1, T 2 6" 2 k 2 (" 2 + k ) 2 e4k. Exmple, V 0 = 2eV, 2 = 1Angstrom. (i) Electron 1 k.96 V 0 " E A. T (ii) Proton 1 k.96a, T 2 4 " V 0 " E much more likely to find n electron.

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