INTRODUCTION TO QUANTUM MECHANICS
CONTENTS THE WAVE FUNCTION THE TIME-INDEPENDENT SCHR¨ODINGER EQUATION FORMALISM QUANTUM MECHANICS IN THREE DIMENSIONS IDENTICAL PARTICLES SYMMETRIES AND CONSERVATION LAWS TIME-INDEPENDENT PERTURBATION THEORY THE VARIATIONAL PRINCIPLE THE WKB APPROXIMATION 10 SCATTERING 11 QUANTUM DYNAMICS 12 AFTERWORD CHAPTER (A) ?J?2 = 212 = 441 ?J2?= N X J2N(J) = (142) + (152) + 3(162) + 2(222) + 2(242) + 5(252) = 14(196 + 225 + 768 + 968 + 1152 + 3125) = = 459571 (B) J ?J = J ??J? 22 ?21 = 24 ?21 = 25 ?21 = ?2 = N X (?J)2N(J) = (?7)2 + (?6)2 + (?5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 14(49 + 36 + 75 + 2 + 18 + 80) = 14 = 18571 ? = ? 18571 = 4309 (C) ?J2???J?2 = 459571 ?441 = 18571 [AGREES WITH (B)] (A) ?X2?= Z H ? HX DX = ? H 5 ? H 45H2 ?? = 2H ? 5 = 02981H (B) P = 1 ? Z X+ X? ? HX DX = 1 ? Z ? ?? E??U2DU = A R? ? = A2 + 2? ?2 = ?X2???X?2 = A2 + 2? ?A2 = 2?; ? = ? 2? INTRODUCTION TO QUANTUM MECHANICS Z A A 3 + B ?A = |A|2 B 3 ? A = R B (B) X A A B ^ (C) AT X = A (D) P = Z A X2DX = |A|2 A 3 = A B P = X3DX + Z B A X(B ?X)2DX = B ( B A ) = 3 ?A2B2 + 3A3B = 4(B ?A)2 (B3 ?3A2B + 2A3) = 2A + B INTRODUCTION TO QUANTUM MECHANICS Z ? E?2?X ?2? ? X2E?2?XDX = 2? 2?2 (C) ?2 = ?X2???X?2 = 2?2 ; ? = H M <M + X 24H PROBABILITY OUTSIDE: FOR INTEGRATION BY PARTS, THE DI?ERENTIATION HAS TO BE WITH RESPECT TO THE INTEGRATION VARIABLE – IN THIS CASE THE DI?ERENTIATION IS WITH RESPECT TO T, BUT THE INTEGRATION VARIABLE IS X IT’S TRUE THAT ? ?T(X|?|2) = ?X ?T |?|2 + X ? ?T|?|2 = X ? ?T|?|2, BUT THIS DOES NOT ALLOW US TO PERFORM THE INTEGRATION: Z B A X ? ?T|?|2DX = Z B A ? ?T(X|?|2)DX ?= (X|?|2) B A INTRODUCTION TO QUANTUM MECHANICS E?2AMX2/?DX = 2|A|2 ? ? = 2A2MX2?, SO V (X) = 2MA2X2 INTRODUCTION TO QUANTUM MECHANICS 22(2AM/?) R ?? 2AM = ? 4AM ?P?= MD?X? DT = 0 ?P2?= Z ?? ? I ? ?X ? ? DX = 2AM? Z |?|2DX ?2AM ? Z X2|?|2DX = 2AM? 1 ?2AM ? ?X2? = 2AM? 1 ?2AM ? ? 4AM = 2AM? FROM MATH TABLES: ? = 3141592653589793238462643 · · · (A) P(0) = IN GENERAL, P(J) = N(J) N (B) MOST PROBABLE: 3 MEDIAN: 13 ARE ?4, 12 ARE ?5, SO MEDIAN IS 4 AVERAGE: ?J?= 25[0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3] = 25[0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 25 = 472 (C) ?J2?= 25[0 + 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3] = 25[0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 25 = 284 ?2 = ?J2???J?2 = 284 ?4722 = 284 ?222784 = 61216; ? = ? 61216 = 2474 INTRODUCTION TO QUANTUM MECHANICS (A) 2MV2 + V = E ? V(X) = R M (E ?V (X)) (B) T = Z B A 2KX2 DX = RM K Z B A P (2E/K) ?X2 DX TURNING POINTS: V = 0 ?E = V = 2KB2 ?B = P 2E/K; A = ?B T = RM K Z B ? B2 ?X2 DX = RM K SIN?1 X B B 0 = RM K SIN?1(1) = RM K ? = ? RM K ?(X) = ?P M K Q ? ? B2 ?X2 Z B A ?(X) DX = ? Z B ? ? = 1 ? ?(X) X B -B (C) ?X?= 0 ?X2?= ? ?X 2 SIN?1 X B B 2M + 2KX2 = E ?X = ± S 2M , INTRODUCTION TO QUANTUM MECHANICS SO ?(P) = ?P M K K R 2M = ? P 2ME ?P2 = R@X_D := ?T = ? ?X I? 2M ???? ?X ???? ?X ? = ?? ?XJ(X, T) ?DPAB DT = ? Z B A ? ?XJ(X, T)DX = ?[J(X, T)]|B A = J(A, T) ?J(B, T) QED PROBABILITY IS DIMENSIONLESS, SO J HAS THE DIMENSIONS 1/TIME, AND UNITS SECONDS?1 (B) HERE ?(X, T) = F(X)E?IAT, WHERE F(X) ?AE?AMX2/?, SO ? ??? ?X = FE?IAT DF DXEIAT = F DF DX, AND ???? ?X = F DF DX TOO, SO J(X, T) = 0 INTRODUCTION TO QUANTUM MECHANICS USE EQS [123] AND [124], AND INTEGRATION BY PARTS: D DT Z ? ?? ?? 1?2 DX = Z ? ?? ? ?T (?? 1?2) DX = Z ? ?? ??? ?T ?2 + ?? ?T DX = Z ? ?? ?I? 2M ?2?? ?X2 + I ?V ?? ?2 + ?? DX = ?I? 2M Z ? ?? ?2?? ?X2 ?2 ??? DX = ?I? 2M " ??? ? ?? ? Z ? ?? ??? ?X DX ??? ?X ? ?? + Z ? ?? ??? 3 + 15A5|A|2, SO A = R DX = 5 + 35 ?42 + 3 · 5 · 8 · 7 INTRODUCTION TO QUANTUM MECHANICS DX = 4 · 4A2 · 3 = A2 (F) ?X = P ?X2???X?2 = R 7A2 = A ? 7 (G) ?P = P ?P2???P?2 = R A2 = R ? A (H) ?X?P = A ? 7 · R ? A = R 14?= R 3(48 × 10?26)(14 × 10?23)(3 × 10?10)2 = 24 K INTRODUCTION TO QUANTUM MECHANICS B ?T < P 2/5 FOR HELIUM (M = 4MP = 68 × 10?27 KG) AT 1 ATM = 10 × 105 N/M2: T < 3(17 × 10?27)(14 × 10?23)(10?2)2 = 62 × 10?14 K AT 3 K IT IS DE?NITELY IN THE CLASSICAL REGIME INTRODUCTION TO QUANTUM MECHANICS = C1(E?1) + C2(E?2) = E(C1?1 + C2?2) = E?3 THUS, (? + ??) AND I(? ???) – BOTH OF WHICH ARE REAL – SATISFY EQ 25 CONCLUSION: FROM ANY COMPLEX SOLUTION, WE CAN ALWAYS CONSTRUCT TWO REAL SOLUTIONS (OF COURSE, IF ? IS ALREADY REAL, THE SECOND ONE WILL BE ZERO) IN PARTICULAR, SINCE ? = + V (?X)?(?X) = E?(?X); SO IF V (?X) = V (X) THEN ?(?X) ALSO SATIS?ES EQ 25 IT FOLLOWS THAT ?+(X) ??(X) + ?(?X) (WHICH IS EVEN: ?+(?X) = ?+(X)) AND ??(X) ??(X) ??(?X) (WHICH IS ODD: ??(?X) = ???(X)) BOTH SATISFY EQ 25 BUT ?(X) = 2(?+(X) + ??(X)), SO ANY SOLUTION CAN BE EXPRESSED AS A LINEAR COMBINATION OF EVEN AND ODD SOLUTIONS QED ?X?= Z X|?|2DX = A Z A X SIN2 N? A X DX LET Y ?N? A X, SO DX = A N? DY; Y : 0 ?N? = A A N? 2 Z N? 4 ?Y SIN 2Y ?COS 2Y ?COS 2N? = A 2 (INDEPENDENT OF N) ?X2?= X2 SIN2 N? A X DX = A A N? 3 Z N? SIN 2Y ?Y COS 2Y ?N? COS(2N?) 3 ? ?P?= MD?X? DT = 0 (NOTE : EQ 133 IS MUCH FASTER THAN EQ 135) ?P2?= Z ?? N ? I D DX Z ?? N?N DX = 2MEN = N?? A ; ?X = A P = ?P2???P?2 = N?? A ; ?P = N?? A ??X?P = ? ?2 THE PRODUCT ?X?P IS SMALLEST FOR N = 1; IN THAT CASE, ?X?P = ? Z [|?1|2 + ?? 1?2 + ?? 2?1 + |?2|2]DX = 2|A|2 ?A = 1/ ? 2 (B) ?(X, T) = H ?1E?IE1T/?+ ?2E?IE2T/?I (BUT EN ?= N2?) = A SIN ? A X E?I?T + SIN 2? A X E?I4?T = ?AE?I?T SIN ? A X + SIN 2? A X E?3I?T |?(X, T)|2 = 2? A X = 2? A X + 2 SIN ? A X SIN 2? A X COS(3?T) (C) ?X?= Z X|?(X, T)|2DX = 2? A X + 2 SIN ? A X SIN 2? A X COS(3?T) DX INTRODUCTION TO QUANTUM MECHANICS # A 4 = Z A 2? A X DX Z A X SIN ? A X SIN 2? A X DX = X COS ? A X ?COS 3? A X DX = 9?2 COS 3? A X ?AX 3? SIN 3? A X A 9?2 ??X?= 9?2 COS(3?T) = A 9?2 COS(3?T) AMPLITUDE: = 03603(A/2); ANGULAR FREQUENCY: 3? = 3?2? 2MA2 (D) ?P?= MD?X? DT = M A (?3?) SIN(3?T) = 8? 3A SIN(3?T) (E) YOU COULD GET EITHER E1 = ?2?2/2MA2 OR E2 = 2?2?2/MA2, WITH EQUAL PROBABILITY P1 = P2 = 1/2 SO ?H?= FROM PROBLEM 25, WE SEE THAT ?(X, T) = ?AE?I?T SIN ? AX + SIN 2? A X E?3I?TEI? ; |?(X, T)|2 = A SIN2 ? AX + SIN2 2? A X + 2 SIN ? AX SIN 2? A X COS(3?T ??) ; AND HENCE ?X?= A 1 ? 9?2 COS(3?T ??) THIS AMOUNTS PHYSICALLY TO STARTING THE CLOCK AT A DI?ERENT TIME (IE, SHIFTING THE T = 0 POINT) IF ? = ? 2 , SO ?(X, 0) = A[?1(X) + I?2(X)], THEN COS(3?T ??) = SIN(3?T); ?X?STARTS AT A 2 IF ? = ?, SO ?(X, 0) = A[?1(X) ??2(X)], THEN COS(3?T ??) = ?COS(3?T); ?X?STARTS AT A 1 + ?A = ? A3 (B) CN = R (A ?X) SIN N? A X DX = A N? ? A N? SIN N? 2N? COS N? N? COS N? + A N? (N?)2 SIN N? = ( 0, N EVEN, (?1)(N?1)/2 (N?)2 , N ODD SO ?(X, T) = A X N=1,3,5, (?1)(N?1)/2 2MA2 (C) P1 = |C1|2 = 16 · = 09855 (D) ?H?= X |CN|2EN = 32 + 52 + MA2 INTRODUCTION TO QUANTUM MECHANICS DX = A2(A/2) = 1 ?A = R A FROM EQ 237, C1 = A R SIN ? A X DX = ? H COS ? ?COS I = M Z A (A) USING EQS 248 AND 260, A+?0 = ? 2?M? M? ?? 1/4 H ?? ?M? 2? 2X + M?X I E?M? 2?X2 = 2M?XE?M? 2?X2 (A+)2?0 = E?M? 2?X2 THEREFORE, FROM EQ 268, ?2 = ? 2(A+)2?0 = E?M? 2?X2 INTRODUCTION TO QUANTUM MECHANICS (B) S S S (C) SINCE ?0 AND ?2 ARE EVEN, WHEREAS ?1 IS ODD, R ?? 0?1DX AND R ?? 2?1DX VANISH AUTOMATICALLY THE ONLY ONE WE NEED TO CHECK IS R ?? 2?0 DX: Z ?? 2?0 DX = ? M? 3/2 Z ? ?? ?2E??2D? = ?? ? M? ?? ? I D DX E??2D? = ?M?? ?? ?? ??? = M?? ? M? 3/2 Z ? ?? ?4E??2D? = 2? ??M? 3?? E??2D? = ?2M?? ?? = 3M?? (B) N = 0: ?X = P ?X2???X?2 = R ? 2M? ; ?P = P ?P2???P?2 = R M?? ; ?X?P = R ? 2M? R M?? = ? 2 (RIGHT AT THE UNCERTAINTY LIMIT)? N = 1: ?X = R 3? 2M? ; ?P = R 3M?? ; ?X?P = 3? 2 > ? 2 ? (C) ?T?= 2M?P2?= ? ? ? 4?? (N = 0) 4?? (N = 1) ? ? ?; ?V ?= 2M?2?X2?= ? ? ? 4?? (N = 1) ? ? ? ?T?+ ?V ?= ?H?= ? ? ? FROM EQ 270, X = R ? 2M? (A+ + A?), P = I R ?M? (A+ ?A?), SO ?X?= R ? 2M? Z ?? N(A+ + A?)?N DX BUT (EQ 267) A+?N = ? N + 1?N+1, A??N = ?N?N?1 SO ?X?= R ? 2M? ? N + = P (N ?1)N?N?2 SO ?X2?= ? 2M? 0 + N Z |?N|2DX + (N + 1) Z |?N|2 DX + = ? 2M? (2N + 1) = N + ? M? INTRODUCTION TO QUANTUM MECHANICS P2 = ??M? (A+ ?A?)2 = ??M? ? ? ?P2?= ??M? [0 ?N ?(N + 1) + 0] = ?M? (2N + 1) = N + M?? ?T?= ?P2/2M?= N + ?? ?X = P ?X2???X?2 = R N + R ? M? ; ?P = P ?P2???P?2 = R N + ? M??; ?X?P = N + DX = |A|2(9 + 0 + 0 + 16) = 25|A|2 ?A = 1/5 (B) ?(X, T) = H 3?0(X)E?IE0T/?+ 4?1(X)E?IE1T/?I = H 3?0(X)E?I?T/2 + 4?1(X)E?3I?T/2I (HERE ?0 AND ?1 ARE GIVEN BY EQS 260 AND 263; E0 AND E1 BY EQ 262) |?(X, T)|2 = 1 + 24?0?1 COS(?T) (WITH ?2 IN PLACE OF ?1 THE FREQUENCY WOULD BE (E2 ?E0)/?= [(5/2)?? ?(1/2)??]/?= 2?) (C) ?X?= 0 DX + 1 DX = 0 (SEE PROBLEM 211 OR 212), WHILE Z X?0?1 DX = RM? ?? R 2M? ? Z XE?M? 2?X2XE?M? 2?X2DX = R ? M? ? Z ? ?? X2E?M? ?X2DX = R = R ? 2M? SO ?X?= R M?? SIN(?T) INTRODUCTION TO QUANTUM MECHANICS ? COS(?T), V = 2M?2X2 ??V ?X = M?2X, SO ? ?V ?X = ?M?2?X?= ?M?2 R ?M? ? COS(?T), SO EHRENFEST’S THEOREM HOLDS (D) YOU COULD GET E0 = 2??, WITH PROBABILITY |C0|2 = 9/25, OR E1 = E??2/2, SO P = E??2DX = E??2D? CLASSICALLY ALLOWED REGION EXTENDS OUT TO: 0 = E0 = 2??, OR X0 = Q ? M?, SO ?0 = 1 P = ?? Z ? 5A3 = 3A1?3 + (A) D D? (E??2) = ?2?E??2; D D? E??2 = D D? (?2?E??2) = (?2 + 4?2)E??2; D D? = 8? + (?2 + 4?2)(?2?) E??2 = (12? ?8?3)E??2; D D? = 12 ?24?2 + (12? ?8?3)(?2?) E??2 = (12 ?48?2 + 16?4)E??2 H3(?) = ?E?2 D D? E??2 = ?12? + 8?3; H4(?) = E?2 D D? E??2 = 12 ?48?2 + 16?4 INTRODUCTION TO QUANTUM MECHANICS D? = 1440? ?1920?3 + 384?5 = 12(120? ?160?3 + 32?5) = (2)(6)H5 ? (D) D DZ (E?Z2+2Z?) = (?2Z + 2?)E?Z2+2Z?; SETTING Z = 0, H1(?) = 2? D DZ E?Z2+2Z?; SETTING Z = 0, H2(?) = ?2 + 4?2 D DZ AEIKX + BE?IKX = A(COS KX + I SIN KX) + B(COS KX ?I SIN KX) = (A + B) COS KX + I(A ?B) SIN KX = C COS KX + D SIN KX, WITH C = A + B; D = I(A ?B) C COS KX + D SIN KX = C EIKX + E?IKX + D EIKX ?E?IKX 2I = 2(C ?ID)EIKX + 2(C + ID)E?IKX = AEIKX + BE?IKX, WITH A = 2(C ?ID); B = 2(C + ID) INTRODUCTION TO QUANTUM MECHANICS AN 2I + BN ?AN 2I + BN E?IN?X/A LET C0 ?B0; CN = 2 (IA?N + B?N) , FOR N = ?1, ?2, ?3, THEN F(X) = ? X N=?? CNEIN?X/A QED (B) Z A ?A F(X)E?IM?X/ADX = ? X N=?? CN Z A ?A EI(N?M)?X/ADX BUT FOR N ?= M, Z A ?A EI(N?M)?X/ADX = EI(N?M)?X/A I(N ?M)?/A A ?A = EI(N?M)? ?E?I(N?M)? I(N ?M)?/A = (?1)N?M ?(?1)N?M I(N ?M)?/A = 0, WHEREAS FOR N = M, Z A ?A EI(N?M)?X/ADX = Z A ?A DX = 2A SO ALL TERMS EXCEPT N = M ARE ZERO, AND Z A ?A F(X)E?IM?X/A = 2ACM, SO CN = 2A Z A ?A F(X)E?IN?X/ADX QED (C) F(X) = ? X N=?? R? AF(K)EIKX = ? 2? X F(K)EIKX?K, INTRODUCTION TO QUANTUM MECHANICS WHERE ?K ?? A IS THE INCREMENT IN K FROM N TO (N + 1) F(K) = R ? A 2A Z A ?A F(X)E?IKXDX = ? 2? Z A ?A F(X)E?IKXDX (D) AS A ??, K BECOMES A CONTINUOUS VARIABLE, F(X) = ? 2? Z ? ?? F(K)EIKXDK; F(K) = E?2AXDX = 2|A|2 E?2AX ?2A ? A ?A = ?A (B) ?(K) = A ? 2? Z ? ?? E?A|X|E?IKX DX = A ? 2? Z ? ?? E?A|X|(COS KX ?I SIN KX)DX THE COSINE INTEGRAND IS EVEN, AND THE SINE IS ODD, SO THE LATTER VANISHES AND ?(K) = 2 A ? 2? Z ? E?AX COS KX DX = A ? 2? Z ? E?AX EIKX + E?IKX DX = A ? 2? Z ? E(IK?A)X + E?(IK+A)X DX = A ? 2? E(IK?A)X IK ?A + E?(IK+A)X ?(IK + A) ? IK ?A + = R A 2? 2A K2 + A2 (C) ?(X, T) = ? 2? 2? Z ? ?? ? Z ? ?? IS A SHARP NARROW SPIKE; POSITION IS ILL-DE?NED BUT MOMENTUM IS WELL-DE?NED INTRODUCTION TO QUANTUM MECHANICS (B) Z ? ?? E?(AX2+BX)DX = Z ? ?? E?Y2+(B2/4A) ?A DY = ?AEB2/4A Z ? ?? E?Y2DY = R? A EB2/4A ?(K) = ? 2? A Z ? ?? E?AX2E?IKXDX = ? 2? 2A ? 1/4 R? A E?K2/4A = (2?A)1/4 E?K2/4A ?(X, T) = Z ? ?? E?K2/4AEI(KX??K2T/2M) | {Z } E?[( 4A +I?T/2M)K2?IXK] DK = ?? Q 4A + I?T/2M E?X2/4( OR, WITH W ? R A 1 + ?2 , |?|2 = R X X T = T > (D) ?X?= Z ? ?? X|?|2DX = 0 (ODD INTEGRAND); ?P?= MD?X? DT = 0 ?X2?= R ? W Z ? ?? X2E?2W2X2DX = R ? W R ? 2W2 = Z ? ?? ???2? ?X2 DX INTRODUCTION TO QUANTUM MECHANICS = ?2BB(1 ?2BX2)E?BX2 ???2? D?X2 = ?2B|B|2(1 ?2BX2)E?(B+B?)X2; B + B?= A 1 + I? + A 1 ?I? = 2A 1 + ?2 = 2W2 |B|2 = R 2A ? ? 1 + ?2 = R ? W SO ???2? ?X2 = ?2B R ? W R ? 2W2 ?2B 2A = 1 ?(1 ?I?) = 1 + I? = A 2B, SO ?P2?= 2B?2 A 2B = ?2A ?X = 2W; ?P = ??A (E) ?X?P = 2W??A = ? P 1 + ?2 = ? (A) (?2)3 ?3(?2)2 + 2(?2) ?1 = ?8 ?12 ?4 ?1 = ?25 (B) COS(3?) + 2 = ?1 + 2 = 1 (C) 0 (X = 2 IS OUTSIDE THE DOMAIN OF INTEGRATION) INTRODUCTION TO QUANTUM MECHANICS (A) LET Y ?CX, SO DX = C DY IF C > 0, Y : ???? IF C < 0, Y : ???? Z ? ?? F(X)?(CX)DX = ? ? ? C R ? ??F(Y/C)?(Y)DY = CF(0) (C > 0); OR CF(0) (C < 0) IN EITHER CASE, Z ? ?? F(X)?(CX)DX = |C|F(0) = Z ? ?? F(X) |C|?(X)DX SO ?(CX) = |C|?(X) ? (B) Z ? ?? F(X) D? DXDX = F? ? ?? ? Z ? ?? DF DX?DX (INTEGRATION BY PARTS) = F(?) ? Z ? X2E?2M?X/?2DX = 2M? ?2 2M? ?M? ? ?M? ? 3 Z ? ?? E?M?|X|/?2 H ?2?(X) + M? ?2 E?M?|X|/?2I DX = M? ? E?2M?X/?2 DX = M? ? 2M? = M? ? ? 2M? M? ? = ? 2? 2 > ? 2 ? INTRODUCTION TO QUANTUM MECHANICS ??BOUND|?SCATTERING? = ?M? ? Z ?? EM?X/?2 AEIKX + BE?IKX DX + Z ? E?M?X/?2 FEIKX + GE?IKX DX = ?M? ? A Z ?? E( M? ?2 +IK)XDX + B Z ?? E( M? ?2 ?IK)XDX + F Z ? E(?M? ?2 +IK)XDX + G Z ? E(?M? ?2 ?IK)XDX = ?M? ? " A E( M? ?2 +IK)X M? ?2 + IK ?? + B E( M? ?2 ?IK)X M? ?2 ?IK ?? + F E(?M? ?2 +IK)X ?M? ?2 + IK ? + G E(?M? ?2 ?IK)X ?M? ?2 ?IK ? PUT F(X) = ?(X) INTO EQ 2103: F(K) = ? 2? Z ? ?? ?(X)E?IKXDX = ? 2? ?F(X) = ?(X) = ? 2? Z ? ?? ? 2? EIKXDK = (A) V(X) -A A X (B) FROM PROBLEM 21(C) THE SOLUTIONS ARE EVEN OR ODD LOOK ?RST FOR EVEN SOLUTIONS: ?(X) = ? ? ? AE??X (X > A), B(E?X + E??X) (?A < X < A), AE?X (X < ?A) CONTINUITY AT A : AE??A = B(E?A + E??A), OR A = B(E2?A + 1) INTRODUCTION TO QUANTUM MECHANICS 2AM?, SO E?Z = CZ ?1 PLOT BOTH SIDES AND LOOK FOR INTERSECTIONS: ?2M? ?2? + 1, 1 = 2M? ?2? ?1 ?2M? ?2? E?2?A; ?2? M? = 1 ?E?2?A, E?2?A = 1 ??2? M?, OR E?Z = 1 ?CZ 1/C 1/C Z INTRODUCTION TO QUANTUM MECHANICS MA ?C = 2 EVEN: E?Z = ? = ? ? ? AEIKX + BE?IKX (X < ?A) CEIKX + DE?IKX (?A < X < A) FEIKX (X > A) ? ? ? IMPOSE BOUNDARY CONDITIONS: (1) CONTINUITY AT ?A : AE?IKA + BEIKA = CE?IKA + DEIKA ??A + B = ?C + D, WHERE ? ?E?2IKA (2) CONTINUITY AT +A : CEIKA + DE?IKA = FEIKA ?F = C + ?D INTRODUCTION TO QUANTUM MECHANICS A = 4?A ?F A = = 16G4 + 8G2 + 2 + (8G2 ?2) COS ? ?8G SIN ? T = F A 8MA2 ?NO ODD BOUND STATE INTRODUCTION TO QUANTUM MECHANICS 1 = |?|2DX = Z ? A E?2?XDX = X 2 + 4L SIN 2LX A 2?E?2?X ? A = A + ? D = EQUATION 2158 ?Z0 = A ? ?2MV0 WE WANT ? = AREA OF POTENTIAL = 2AV0 HELD CONSTANT AS A ?0 THEREFORE V0 = ? 2A; Z0 = A ? P2M ? 2A = 0 BUT WE FOUND THAT Z0 ?= ? ?M?A HERE, SO ?A = (WHICH AGREES WITH EQ 2132) IN EQ 2172, V0 ?E ?T ?1 ?= 1+ V 2?2E , IN AGREEMENT WITH EQ 2144 INTRODUCTION TO QUANTUM MECHANICS MULTIPLY EQ 2168 BY SIN LA, EQ 2169 BY L COS LA, AND ADD: C SIN2 LA + D SIN LA COS LA = FEIKA SIN LA C COS2 LA ?D SIN LA COS LA = IK L FEIKA COS LA C = FEIKA SIN LA + IK L COS LA MULTIPLY EQ 2168 BY COS LA, EQ 2169 BY L SIN LA, AND SUBTRACT: C SIN LA COS LA + D COS2 LA = FEIKA COS LA C SIN LA COS LA ?D SIN2 LA = IK L FEIKA SIN LA D = FEIKA COS LA ?IK L SIN LA PUT THESE INTO EQ 2166: (1) AE?IKA + BEIKA = ?FEIKA SIN LA + IK L COS LA SIN LA + FEIKA COS LA ?IK L SIN LA COS LA = FEIKA COS2 LA ?IK L SIN LA COS LA ?SIN2 LA ?IK L SIN LA COS LA = FEIKA COS(2LA) ?IK L SIN(2LA) LIKEWISE, FROM EQ 2167: (2) AE?IKA ?BEIKA = ?IL K FEIKA SIN LA + IK L COS LA COS LA + COS LA ?IK L SIN LA SIN LA = ?IL K FEIKA SIN LA COS LA + IK L COS2 LA + SIN LA COS LA ?IK L SIN2 LA = ?IL K FEIKA SIN(2LA) + IK L COS(2LA) = FEIKA COS(2LA) ?IL K SIN(2LA) ADD (1) AND (2): 2AE?IKA = FEIKA 2 COS(2LA) ?I K L + L K SIN(2LA) , OR: F = E?2IKAA COS(2LA) ?I SIN(2LA) 2KL (K2 + L2) (CON?RMING EQ 2171) NOW SUBTRACT (2) FROM (1): 2BEIKA = FEIKA I L K ?K L SIN(2LA) ?B = ISIN(2LA) 2KL (L2 ?K2)F (CON?RMING EQ 2170) T ?1 = A F = COS(2LA) ?ISIN(2LA) 2KL (K2 + L2) | {Z } (2KL)2 [K4+2K2L2+L4?4K2L2]= 2 V 2 E(E + V0) = V 4E(E + V0) ?T ?1 = 1 + V 2A ? P 2M(E + V0) , CON?RMING EQ 2172 INTRODUCTION TO QUANTUM MECHANICS E < V0 ? = ? ? ? AEIKX + BE?IKX (X < ?A) CE?X + DE??X (?A < X < A) FEIKX (X > A) ? ? ? K = ? 2ME ? ; ? = P 2M(V0 ?E) ? (1) CONTINUITY OF ? AT ?A: AE?IKA + BEIKA = CE??A + DE?A (2) CONTINUITY OF ?? AT ?A: IK(AE?IKA ?BEIKA) = ?(CE??A ?DE?A) ?2AE?IKA = 1 ?I? K CE??A + 1 + I? K DE?A (3) CONTINUITY OF ? AT +A: CE?A + DE??A = FEIKA (4) CONTINUITY OF ?? AT +A: ?(CE?A ?DE??A) = IKFEIKA ?2CE?A = 1 + IK ? FEIKA; 2DE??A = 1 ?IK ? FEIKA 2AE?IKA = 1 ?I? K 1 + IK ? FEIKA E?2?A + 1 + I? K 1 ?IK ? FEIKA E2?A = FEIKA 1 + I K ? ?? K + E?2?A + 1 + I ? K ?K ? + E2?A = FEIKA E?2?A + E2?A + I(?2 ?K2) K? E2?A ?E?2?A BUT SINH X ?EX ?E?X , COSH X ?EX + E?X , SO = FEIKA 4 COSH(2?A) + I(?2 ?K2) K? 2 SINH(2?A) = 2FEIKA COSH(2?A) + I(?2 ?K2) 2K? SINH(2?A) T ?1 = A F | {Z } ? SINH2(2?A) = 1 + V = V 2M D2? DX2 + V0? = E? ?D2? DX2 = 0, SO ? = C + DX) INTRODUCTION TO QUANTUM MECHANICS (1) CONTINUOUS ? AT ?A : AE?IKA + BEIKA = C ?DA (2) CONTINUOUS ? AT +A : FEIKA = C + DA ?(25) 2DA = FEIKA ?AE?IKA ?BEIKA (3) CONTINUOUS ?? AT ?A : IK AE?IKA ?BEIKA = D (4) CONTINUOUS ?? AT +A : IKFEIKA = D ?(45) AE?2IKA ?B = F USE (4) TO ELIMINATE D IN (25): AE?2IKA + B = F ?2AIKF = (1 ?2IAK)F, AND ADD TO (45): 2AE?2IKA = 2F(1 ?IKA), SO T ?1 = A F = 1 + (KA)2 = 1 + 2ME ?2 A2 (YOU CAN ALSO GET THIS FROM EQ 2172 BY CHANGING THE SIGN OF V0 AND TAKING THE LIMIT E ?V0, USING SIN ? ?= ?) E > V0 THIS CASE IS IDENTICAL TO THE ONE IN THE BOOK, ONLY WITH V0 ??V0 SO T ?1 = 1 + V (A) ? = AEIKX + BE?IKX (X < 0) FE??X (X > 0) WHERE K = ? 2ME ? ; ? = P 2M(V0 ?E) ? (1) CONTINUITY OF ? : A + B = F (2) CONTINUITY OF ?? : IK(A ?B) = ??F ?A + B = ?IK ? (A ?B) ?A 1 + IK ? = ?B 1 ?IK ? R = B A 1 + (K/?)2 = 1 ALTHOUGH THE WAVE FUNCTION PENETRATES INTO THE BARRIER, IT IS EVENTUALLY ALL RE?ECTED (B) ? = AEIKX + BE?IKX (X < 0) FEILX (X > 0) WHERE K = ? 2ME ? ; L = P 2M(E ?V0) ? (1) CONTINUITY OF ? : A + B = F (2) CONTINUITY OF ?? : IK(A ?B) = ILF INTRODUCTION TO QUANTUM MECHANICS ?A + B = K L (A ?B); A 1 ?K L = ?B 1 + K L R = B A E F A K L + = A 2K/L K L + (K2 ?L2)2 = (A) ?(X) = AEIKX + BE?IKX (X < 0) FEILX (X > 0) WHERE K ? ? 2ME ? , L ? P 2M(E + V0) ? CONTINUITY OF ? ?A + B = F CONTINUITY OF ?? ?IK(A ?B) = ILF =? A + B = K L (A ?B); A 1 ?K L = ?B 1 + K L ; B A = ? 1 ?K/L 1 + K/L INTRODUCTION TO QUANTUM MECHANICS R = B A = L ?K L + K P 1 + V0/E + ?1 + 3 + 2 + 2M(2A)2 (IN AGREEMENT WITH EQ 230 FOR A WELL OF WIDTH 2A) THE SUBSTITUTION X ?(X + A)/2 TAKES EQ 231 TO R A SIN N? A (X + A) = R A SIN N?X 2A + N? = ? ? ? ? ? ? ? (?1)N/2Q A SIN N?X 2A (N EVEN), (?1)(N?1)/2Q A COS N?X 2A (N ODD) SO (APART FROM NORMALIZATION) WE RECOVER THE RESULTS ABOVE THE GRAPHS ARE THE SAME AS FIGURE 22, EXCEPT THAT SOME ARE UPSIDE DOWN (DI?ERENT NORMALIZATION) COS(/X/2A) SIN(2/X/2A) COS(3/X/2A) INTRODUCTION TO QUANTUM MECHANICS ?X A = 4 SIN 3?X A SO (EQ 231): ?(X, 0) = A RA 4?3(X) NORMALIZE USING EQ 220: |A|2 A 16 + 16A|A|2 = 1 ?A = ? 5A SO ?(X, 0) = ? 10 [3?1(X) ??3(X)] , AND HENCE (EQ 217) ?(X, T) = H 3?1(X)E?IE1T/???3(X)E?IE3T/?I |?(X, T)|2 = ? T ; SO ?X?= Z A X|?(X, T)|2DX = 10?X?1 + ? T Z A X?1(X)?3(X)DX, WHERE ?X?N = A/2 IS THE EXPECTATION VALUE OF X IN THE NTH STATIONARY STATE THE REMAINING INTEGRAL IS X SIN ?X A SIN 3?X A DX = X COS 2?X A ?COS 4?X A DX = A A 2? COS 2?X A + XA 2? SIN 2?X A ? A 4? COS 4?X A ? XA 4? SIN 4?X A A = 0 EVIDENTLY THEN, ?X?= = A 2 USING EQ 221, ?H?= |C1|2E1 + |C3|2E3 = 2MA2 + ?? = 2?N2, SO E?I(N2?2?/2MA2)(T+T ) = E?I(N2?2?/2MA2)TE?I2?N2, AND SINCE N2 IS AN INTEGER, E?I2?N2 = 1 THEREFORE ?(X, T + T) = ?(X, T) QED INTRODUCTION TO QUANTUM MECHANICS (B) THE CLASSICAL REVIVAL TIME IS THE TIME IT TAKES THE PARTICLE TO GO DOWN AND BACK: TC = 2A/V, WITH THE VELOCITY GIVEN BY E = (A) D? DX = A?A · 1, (0 < X < A/2) ?1, (A/2 < X < A) = A?A 1 ?2? X ?A (B) D2? DX2 = A?A ?2? X ?A A?A? X ?A A?A Z ??? X ?A MA?A ?? A (A) IN THE STANDARD NOTATION ? ? P M?/?X, ? ?(M?/??)1/4, ?(X, 0) = A(1 ?2?)2E??2/2 = A(1 ?4? + 4?2)E??2/2 IT CAN BE EXPRESSED AS A LINEAR COMBINATION OF THE ?RST THREE STATIONARY STATES (EQ 260 AND 263, AND PROBLEM 210): ?0(X) = ?E??2/2, ?1(X) = ? 2 ??E??2/2, ?2(X) = ? ? 2(2?2 ?1)E??2/2 SO ?(X, 0) = C0?0 + C1?1 + C2?2 = ?(C0 + ? 2?C1 + ? 2?2C2 ? ? 2C2)E??2/2 WITH (EQUATING LIKE POWERS) ? ? ? ? ? ? ? 2C2 = 4A ?C2 = ? 2A/?, ?(C0 ?C2/ ? 2) = A ?C0 = (A/?) + C2/ ? 2 = (1 + 2)A/? = 3A/? INTRODUCTION TO QUANTUM MECHANICS NORMALIZING: 1 = |C0|2 + |C1|2 + |C2|2 = (8 + 8 + 9)(A/?)2 = 25(A/?)2 ?A = ?/5 = C0 = 5 , C2 = 5 (B) YOU COULD GET 2??, PROBABILITY 25; 25 ?H?= 2?? + 2?? = ?? 50 (9 + 24 + 40) = 50?? (C) ?(X, T) = 5 ?1E?3I?T/2+ 5 ?1E?I?T + EVERYTHING IN SECTION 232 STILL APPLIES, EXCEPT THAT THERE IS AN ADDITIONAL BOUNDARY CONDITION: ?(0) = 0 THIS ELIMINATES ALL THE EVEN SOLUTIONS (N = 0, 2, 4, ), LEAVING ONLY THE ODD SOLUTIONS SO EN = N + (A) NORMALIZATION IS THE SAME AS BEFORE: A = 2A ? 1/4 (B) EQUATION 2104 SAYS ?(K) = ? 2? 2A ? 1/4 Z ? ?? E?AX2EILXE?IKXDX [SAME AS BEFORE, ONLY K ?K ?L] = (2?A)1/4 E?(K?L)2/4A ?(X, T) = Z ? ?? E?(K?L)2/4AEI(KX??K2T/2M) DK LET U ?K ?L, SO K = U + L AND DK = DU: ?(X, T) = Z ? ?? E?U2/4AEI[UX+LX?(?T/2M)(U2+2UL+L2)] DU = (2?A)1/4 EIL(X??LT 2M ) Z ? ?? E?U2( 4A +I ?T 2M)+IU(X??LT M ) DU INTRODUCTION TO QUANTUM MECHANICS USING THE HINT IN PROBLEM 221, THE INTEGRAL BECOMES 4A + I ?T 2M E(X??LT M ) 2/4( 4A +I ?T 2M) Z ? ?? E?Y2 DY = 2?A ? E?A(X??LT M ) 2/?2??, SO ?(X, T) = 2A ? 1/4 ? E?A(X??LT M ) 2/?2EIL(X??LT 2M) THE GAUSSIAN ENVELOPE (THE ?RST EXPONENTIAL) TRAVELS AT SPEED ?L/M; THE SINUSOIDAL WAVE (THE SECOND EXPONENTIAL) TRAVELS AT SPEED ?L/2M (C) |?(X, T)|2 = R 2A ? |?|2 E A(X??LT M ) 2H ?2 + I THE TERM IN SQUARE BRACKETS SIMPLI?ES: |?|4 [(??)2 + ?2] = 1 ?2I?T M + 1 + 2I?T M = |?|4 |?|2 = S 1 + 2IA?T M 1 ?2IA?T M = P 1 + ?2, WHERE (AS BEFORE) ? ?2?AT/M SO |?(X, T)|2 = R 2A ? ? 1 + ?2 E2A(X??LT M ) 2/(1+?2) = R ? WE?2W2(X??LT M ) WHERE (AS BEFORE) W ? P A/(1 + ?2) THE RESULT IS THE SAME AS IN PROBLEM 221, EXCEPT THAT X ? X ??L MT , SO |?|2 HAS THE SAME (?ATTENING GAUSSIAN) SHAPE – ONLY THIS TIME THE CENTER MOVES AT CONSTANT SPEED V = ?L/M (D) ?X?= Z ? ?? X|?(X, T)|2DX LET Y ?X ?VT, SO X = Y + VT = Z ? ?? (Y + VT) R ? WE?2W2Y2DY = 4W2 + 0 + (VT)2 = 4W2 + ?LT M X ??LT M + IL ? = AX2 + BX + C ?, INTRODUCTION TO QUANTUM MECHANICS 2 ?LT M (" A + A 2?AT M " ?1 ?4IA?T M + X = ?X2???X?2 = ? ?LT M 4W2 ??X = EQUATION 225 ??(X) = A SIN KX + B COS KX, 0 ?X ?A, WITH K = ? 2ME/?2 EVEN SOLUTIONS: ?(X) = ?(?X) = A SIN(?KX) + B COS(?KX) = ?A SIN KX + B COS KX (?A ?X ?0) BOUNDARY CONDITIONS ? ? ? ? CONTINUOUS AT 0 : B = B (NO NEW CONDITION) ?? DISCONTINUOUS (EQ 2128 WITH SIGN OF ? SWITCHED): AK + AK = 2M? ?2 B ?B = ?2K M?A ? ?0 AT X = A : A SIN(KA) + ?2K M?A COS(KA) = 0 ?TAN(KA) = ??2K M? ?(X) = A SIN KX + ?2K M? COS KX (0 ?X ?A); ?(?X) = ?(X) / 2/ 3/ TAN(KA) -H K M_ KA INTRODUCTION TO QUANTUM MECHANICS FROM THE GRAPH, THE ALLOWED ENERGIES ARE SLIGHTLY ABOVE KA = N? DX , OR = ?2M ?2 [(EM ?V )?M?N ??M(EN ?V )?N] = ?2M ?2 (EM ?EN)?M?N ? INTRODUCTION TO QUANTUM MECHANICS 2M D2? DX2 = E? (WHERE X IS MEASURED AROUND THE CIRCUMFERENCE), OR D2? DX2 = ?K2?, WITH K ? ? 2ME ? , SO ?(X) = AEIKX + BE?IKX BUT ?(X + L) = ?(X), SINCE X + L IS THE SAME POINT AS X, SO AEIKXEIKL + BE?IKXE?IKL = AEIKX + BE?IKX, AND THIS IS TRUE FOR ALL X IN PARTICULAR, FOR X = 0 : (1) AEIKL + BE?IKL = A + B AND FOR X = ? 2K : AEI?/2EIKL + BE?I?/2E?IKL = AEI?/2 + BE?I?/2, OR IAEIKL ?IBE?IKL = IA ?IB, SO (2) AEIKL ?BE?IKL = A ?B ADD (1) AND (2): 2AEIKL = 2A EITHER A = 0, OR ELSE EIKL = 1, IN WHICH CASE KL = 2N? (N = 0, ±1, ±2, ) BUT IF A = 0, THEN BE?IKL = B, LEADING TO THE SAME CONCLUSION SO FOR EVERY POSITIVE N THERE ARE TWO SOLUTIONS: ?+ N (X) = AEI(2N?X/L) AND ?? N (X) = BE?I(2N?X/L) (N = 0 IS OK TOO, BUT IN THAT CASE THERE IS JUST ONE SOLUTION) NORMALIZING: R L 0 |?±|2DX = 1 ?A = B = 1/ ? L ANY OTHER SOLUTION (WITH THE SAME ENERGY) IS A LINEAR COMBINATION OF THESE ?± N (X) = (N = 0, 1, 2, 3, ) THE THEOREM FAILS BECAUSE HERE ? DOES NOT GO TO ZERO AT ?; X IS RESTRICTED TO A ?NITE RANGE, AND WE ARE UNABLE TO DETERMINE THE CONSTANT K (IN PROBLEM 244) INTRODUCTION TO QUANTUM MECHANICS (A) (I) B = 0 ?ORDINARY ?NITE SQUARE WELL EXPONENTIAL DECAY OUTSIDE; SINUSOIDAL INSIDE (COS FOR ?1, SIN FOR ?2) NO NODES FOR ?1, ONE NODE FOR ?2 X X S S S (III) FOR B ?A, SAME AS (II), BUT WAVE FUNCTION VERY SMALL IN BARRIER REGION ESSENTIALLY TWO ISOLATED ?NITE SQUARE WELLS; ?1 AND ?2 ARE DEGENERATE (IN ENERGY); THEY ARE EVEN AND ODD LINEAR COMBINATIONS OF THE GROUND STATES OF THE TWO SEPARATE WELLS S S 2M(2A)2 = H 2MA2 = H (AGAIN, SLIGHTLY BELOW THIS) INTRODUCTION TO QUANTUM MECHANICS E +V E +V ? A = P 2M(32?2/MA2) ? A = 8 REFERRING TO THE ?GURE (PROBLEM 229), AND NOTING THAT (5/2)? = 785 < Z0 < 3? = 942, WE SEE THAT THERE ARE THREE BOUND STATES (B) LET I1 ? Z A 2L SIN LX COS LX A 2L SIN LA COS LA + SIN2 LA 2? = 2?|D|2 H ?A ?? L SIN LA COS LA + SIN2 LA I BUT (REFERRING AGAIN TO PROBLEM 229) ?/L = ?COT LA, SO = ?A + COT LA SIN LA COS LA + SIN2 LA = |D|2 (1 + ?A) 2? INTRODUCTION TO QUANTUM MECHANICS 0 ?Z2 (PAGE 80), SIN2 LA = SIN2 Z = 1 + COT2 Z = 0?2E?2I?T + ? INTRODUCTION TO QUANTUM MECHANICS 2M ?2? ?X2 + ? + 2M?2X2? = 0E?2I?T + ? = 0 COS2(?T)] = RM? ??E?M? ?(X?X0 COS ?T)2 THE WAVE PACKET IS A GAUSSIAN OF ?XED SHAPE, WHOSE CENTER OSCILLATES BACK AND FORTH SINUSOIDALLY, WITH AMPLITUDE X0 AND ANGULAR FREQUENCY ? (C) NOTE THAT THIS WAVE FUNCTION IS CORRECTLY NORMALIZED (COMPARE EQ 260) LET Y ?X ?X0 COS ?T : ?X?= Z X|?|2DX = Z (Y + X0 COS ?T)|?|2DY = 0 + X0A COS ?T Z |?|2DY = X0 COS ?T ?P?= MD?X? DT = ?MX0? SIN ?T D?P? DT = ?MX0?2 COS ?T V = ? ?T|X ?VT| ?I(E + 2MV2) ? ?; ? ?T|X ?VT| = ?V, IF X ?VT > V, IF X ?VT < WE CAN WRITE THIS IN TERMS OF THE ?-FUNCTION (EQ 2146): 2?(Z) ?1 = 1, IF Z > ?1, IF Z < , SO ? ?T|X ?VT| = ?V[2?(X ?VT) ?1] I??? ?T = IM?V ? [2?(X ?VT) ?1] + E + ? [?] INTRODUCTION TO QUANTUM MECHANICS ? ?X|X ?VT| + IMV ? ? ? ?X|X ?VT| = {1, IF X > VT; ?1, IF X < VT} = 2?(X ?VT) ?1 = ?M? ?2 [2?(X ?VT) ?1] + IMV ? ? ?2? ?X2 = ?M? ?2 [2?(X ?VT) ?1] + IMV ? 2M ?M? ?2 [2?(X ?VT) ?1] + IMV ? 2?2 + 2M? = 1 ? ?H?= Z ? ?? ??H?DX BUT H? = I??? ?T , WHICH WE CALCULATED ABOVE [?] = Z IM?V ? [2?(Y) ?1] + E + |?|2DY = E + 2MV2 (NOTE THAT [2?(Y)?1] IS AN ODD FUNCTION OF Y) INTERPRETATION: THE WAVE PACKET IS DRAGGED ALONG (AT SPEED V) WITH THE DELTA-FUNCTION THE TOTAL ENERGY IS THE ENERGY IT WOULD HAVE IN A STATIONARY DELTA-FUNCTION (E), PLUS KINETIC ENERGY DUE TO THE MOTION ( ?0 = 2A ? 1/4 ? E?A(X+ ?IMG ? X + EXP ?IMGT ? X + , INTRODUCTION TO QUANTUM MECHANICS D? DT + X + GT ?A X + D? DT #) E?A(X+ 2 D? DT # ?0 BUT D? DT = + 2IA2? ?4M X + IMGT ? EXP ?IMGT ? X + ?X = ?2A X + ?0, SO ?? ?X = ?2A X + /?2 ?IMGT ? ?, AND HENCE ?2? ?X2 = ?2A ?2 ? + ?2A X + /?2 ?IMGT ? ?? ?X = ( ?2A ?2 + ?2A X + 2M ?2A X + /?2 ?IMGT ? ?4M X + + MG X + # THE TERMS QUADRATIC IN X + CANCEL, AS DO THE LINEAR TERMS, AND SO DO THOSE OF ZEROTH ORDER THIS CON?RMS THAT ? SATIS?ES THE SCHR¨ODINGER EQUATION TO CALCULATE THE EXPECTATION VALUE OF X, ?RST NOTE THAT INTRODUCTION TO QUANTUM MECHANICS |?|2 = ?0|2 = R 2A ? |?|2 E ?(X+ 2 GT2) BUT |?|4 , SO (LETTING Y ?X+ 2GT2) ?X?= R 2A ? Z ? ?? XE?2A(X+ 2 GT2) 2/|?|4 DX = R 2A ? E?2AY2/|?|4 DY = R 2A ? 2M A SECH(AX) TANH2(AX) + SECH2(AX) BUT (TANH2 ? + SECH2 ?) = SINH2 ? COSH2 ? + COSH2 ? = SINH2 ? + Z ? ?? SECH2(AX)DX = |A|2 A TANH(AX) ? ?? = A|A|2 =?A = RA 2 INTRODUCTION TO QUANTUM MECHANICS 2M ?K = E?K QED AS X ?+?, TANH AX ?+1, SO ?K(X) ?A IK ?A IK + A EIKX, WHICH REPRESENTS A TRANSMITTED WAVE R = 0 T = IK ?A IK + A (A) (1) FROM EQ 2136: F + G = A + B (2) FROM EQ 2138: F ?G = (1 + 2I?)A ?(1 ?2I?)B, WHERE ? = M?/?2K SUBTRACT: 2G = ?2I?A + 2(1 ?I?)B ?B = 1 ?I? (I?A + G) MULTIPLY (1) BY (1 ?2I?) AND ADD: 2(1 ?I?)F ?2I?G = 2A ?F = 1 ?I? (A + I?G) S = 1 ?I? I? AND 2171 GIVE S21 = E?2IKA COS 2LA ?I (K2+L2) 2KL SIN 2LA ; S11 = I (L2?K2) 2KL SIN 2LA E?2IKA COS 2LA ?I (K2+L2) 2KL SIN 2LA SO INTRODUCTION TO QUANTUM MECHANICS S = E?2IKA COS 2LA ?I (K2+L2) 2KL SIN 2LA I (L2?K2) 2KL SIN 2LA (A) B = S11A + S12G ?G = , M22 = M = CONVERSELY: G = M21A + M22B ?B = ; S12 = S = , TR = |S12|2 = |M22|2 (B) A B C D F G X M M AEIKA + BE?IKA INTRODUCTION TO QUANTUM MECHANICS (1) FE2IKA + G = AE2IKA + B (2) FE2IKA ?G = AE2IKA ?B + I 2M? ?2K AE2IKA + B ADD (1) AND (2): 2FE2IKA = 2AE2IKA + I2M? ?2K AE2IKA + B ?F = 1 + IM? ?2K A + IM? ?2K E?2IKAB = M11A + M12B SO M11 = (1 + I?); M12 = I?E?2IKA; ? ?M? ?2K SUBTRACT (2) FROM (1): 2G = 2B ?2I?E2IKAA ?2I?B ?G = (1 ?I?)B ?I?E2IKAA = M21A + M22B SO M21 = ?I?E2IKA; M22 = (1 ?I?) M = (1 + I?) I?E?2IKA ?I?E2IKA (1 ?I?) (D) M2 = (1 + I?) I?E?2IKA ?I?E2IKA (1 ?I?) ; TO GET M1, JUST SWITCH THE SIGN OF A: M1 = (1 + I?) I?E2IKA ?I?E?2IKA (1 ?I?) M = M2M1 = [1 + 2I? + ?2(E?4IKA ?1)] 2I?[COS 2KA ?? SIN 2KA] ?2I?[COS 2KA ?? SIN 2KA] [1 ?2I? + ?2(E4IKA ?1)] T = TL = TR = |M22|2 ? T ?1 = [1 + 2I? + ?2(E?4IKA ?1)][1 ?2I? + ?2(E4IKA ?1)] = 1 ?2I? + ?2E4IKA ??2 + 2I? + 4?2 + 2I?3E4IKA ?2I?3 + ?2E?4IKA ??2 ?2I?3E?4IKA + 2I?3 + ?4(1 ?E?4IKA ?E4IKA + 1) = 1 + 2?2 + ?2(E?4IKA + E4IKA) ?2I?3(E?4IKA ?E4IKA) + 2?4 ??4(E?4IKA + E4IKA) = 1 + 2?2 + 2?2 COS 4KA + 2I?32I SIN 4KA + 2?4 ?2?4 COS 4KA = 1 + 2?2(1 + COS 4KA) ?4?3 SIN 4KA + 2?4(1 ?COS 4KA) = 1 + 4?2 COS2 2KA ?8?3 SIN 2KA COS 2KA ?4?4 SIN2 2KA T =