Introduction to Quantum Mechanics, 3rd Edition Solution Manual

16 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Chapter 2 The Time-Independent Schrödinger Equation Problem 2.1 (a) Ψ(x, t) = ψ(x)e−i(E0 +iΓ)t/~ = ψ(x)eΓt/~ e−iE0 t/~ =⇒ |Ψ|2 = |ψ|2 e2Γt/~ . Z ∞ Z ∞ 2 2Γt/~ |Ψ(x, t)| dx = e |ψ|2 dx. −∞ −∞ The second term is independent of t, so if the product is to be 1 for all time, the first term (e2Γt/~ ) must also be constant, and hence Γ = 0. QED 2 2 ~ d ψ (b) If ψ satisfies Eq. 2.5, − 2m dx2 + V ψ = Eψ, then (taking the complex conjugate and noting that V and 2 2 ∗ ~ d ψ ∗ ∗ ∗ E are real): − 2m dx2 + V ψ = Eψ , so ψ also satisfies Eq. 2.5. Now, if ψ1 and ψ2 satisfy Eq. 2.5, so too does any linear combination of them (ψ3 ≡ c1 ψ1 + c2 ψ2 ): ~2 d2 ψ1 d2 ψ2 ~2 d2 ψ3 + V ψ3 = − c1 + c2 − + V (c1 ψ1 + c2 ψ2 ) 2m dx2 2m dx2 dx2 ~2 d2 ψ1 ~2 d2 ψ2 = c1 − + V ψ1 + c2 − + V ψ2 2m dx2 2m dx2 = c1 (Eψ1 ) + c2 (Eψ2 ) = E(c1 ψ1 + c2 ψ2 ) = Eψ3 . Thus, (ψ + ψ ∗ ) and i(ψ − ψ ∗ ) – both of which are real – satisfy Eq. 2.5. Conclusion: From any complex solution, we can always construct two real solutions (of course, if ψ is already real, the second one will be zero). In particular, since ψ = 21 [(ψ + ψ ∗ ) − i(i(ψ − ψ ∗ ))], ψ can be expressed as a linear combination of two real solutions. QED (c) If ψ(x) satisfies Eq. 2.5, then, changing variables x → −x and noting that d2 /d(−x)2 = d2 /dx2 , − ~2 d2 ψ(−x) + V (−x)ψ(−x) = Eψ(−x); 2m dx2 so if V (−x) = V (x) then ψ(−x) also satisfies Eq. 2.5. It follows that ψ+ (x) ≡ ψ(x) + ψ(−x) (which is even: ψ+ (−x) = ψ+ (x)) and ψ− (x) ≡ ψ(x) − ψ(−x) (which is odd: ψ− (−x) = −ψ− (x)) both satisfy Eq. 2.5. But ψ(x) = 21 (ψ+ (x) + ψ− (x)), so any solution can be expressed as a linear combination of even and odd solutions. QED CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 17 Problem 2.2 2 00 Given ddxψ2 = 2m ~2 [V (x) − E]ψ, if E < Vmin , then ψ and ψ always have the same sign: If ψ is positive(negative), 00 then ψ is also positive(negative). This means that ψ always curves away from the axis (see Figure). However, it has got to go to zero as x → −∞ (else it would not be normalizable). At some point it’s got to depart from zero (if it doesn’t, it’s going to be identically zero everywhere), in (say) the positive direction. At this point its slope is positive, and increasing, so ψ gets bigger and bigger as x increases. It can’t ever “turn over” and head back toward the axis, because that would require a negative second derivative—it always has to bend away from the axis. By the same token, if it starts out heading negative, it just runs more and more negative. In neither case is there any way for it to come back to zero, as it must (at x → ∞) in order to be normalizable. QED s x Problem 2.3 2 2 2 Equation 2.23 says ddxψ2 = − 2mE = A + Bx; ~2 ψ; Eq. 2.26 says ψ(0) = ψ(a) = 0. If E = 0, d ψ/dx = 0, so ψ(x) √ 2 2 ψ(0) = A = 0 ⇒ ψ = Bx; ψ(a) = Ba = 0 ⇒ B = 0, so ψ = 0. If E ~/2. X σp2 = hp2 i − hpi2 = nπ~ a ; nπ~ σp = . a ~ ∴ σx σp = 2 Problem 2.5 (a) |Ψ|2 = Ψ∗ Ψ = |A|2 (ψ1∗ + ψ2∗ )(ψ1 + ψ2 ) = |A|2 [ψ1∗ ψ1 + ψ1∗ ψ2 + ψ2∗ ψ1 + ψ2∗ ψ2 ]. Z Z √ 1 = |Ψ|2 dx = |A|2 [|ψ1 |2 + ψ1∗ ψ2 + ψ2∗ ψ1 + |ψ2 |2 ]dx = 2|A|2 ⇒ A = 1/ 2. (b) i 1 h Ψ(x, t) = √ ψ1 e−iE1 t/~ + ψ2 e−iE2 t/~ 2 1 =√ 2 (but En = n2 ω) ~ r 2 2π 1 −iωt 2π π π −iωt −i4ωt −3iωt x e + sin x e = √ e sin x + sin x e . sin a a a a a a π π 1 2π 2π sin2 x + sin x sin x e−3iωt + e3iωt + sin2 x a a a a a π π 1 2π 2π = sin2 x + sin2 x + 2 sin x sin x cos(3ωt) . a a a a a |Ψ(x, t)|2 = (c) Z x|Ψ(x, t)|2 dx Z π π 1 a 2π 2π = x sin2 x + sin2 x + 2 sin x sin x cos(3ωt) dx a 0 a a a a hxi = CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Z a π x sin x dx = a 0 ” 2 19 # a Z a x sin 2π cos 2π a2 x2 2π 2 a x a x = − − = x sin x dx. 4 4π/a 8(π/a)2 4 a 0 0 Z a Z π π 2π 1 a 3π x sin x cos x sin x dx = x − cos x dx a a 2 0 a a 0 a π ax π 3π 1 a2 a2 ax 3π cos = x + sin x − 2 cos x − sin x 2 π2 a π a 9π a 3π a 0 2 1 1 a a2 a2 8a2 1 − . = cos(π) − cos(0) − cos(3π) − cos(0) = − = − 2 π2 9π 2 π2 9 9π 2 1 a2 a2 16a2 a 32 cos(3ωt) = cos(3ωt) . + − 1 − a 4 4 9π 2 2 9π 2 ∴ hxi = 32 a = 0.3603(a/2); 9π 2 2 Amplitude: angular frequency: 3ω = 3π 2 ~ . 2ma2 (d) hpi = m a 32 8~ dhxi =m − 2 (−3ω) sin(3ωt) = sin(3ωt). dt 2 9π 3a (e) You could get either E1 = π 2 ~2 /2ma2 or E2 = 2π 2 ~2 /ma2 , with equal probability P1 = P2 = 1/2. So hHi = 1 5π 2 ~2 (E1 + E2 ) = ; 2 4ma2 it’s the average of E1 and E2 . Problem 2.6 From Problem 2.5, we see that Ψ(x, t) = √1 e−iωt a |Ψ(x, t)|2 = 1 a π ax sin 2 sin π ax + sin + sin2 2π a x 2π a x −3iωt iφ e e ; + 2 sin π ax sin 2π a x cos(3ωt − φ) ; 32 and hence hxi = a2 1 − 9π This amounts physically to starting the clock at a different time 2 cos(3ωt − φ) . (i.e., shifting the t = 0 point). If φ = π a , so Ψ(x, 0) = A[ψ1 (x) + iψ2 (x)], then cos(3ωt − φ) = sin(3ωt); hxi starts at . 2 2 a 32 If φ = π, so Ψ(x, 0) = A[ψ1 (x) − ψ2 (x)], then cos(3ωt − φ) = − cos(3ωt); hxi starts at 1+ 2 . 2 9π 20 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.7 ^(x,0) Aa/2 a a/2 (a) Z a/2 1 = A2 x2 dx + A2 Z a (a − x)2 dx = A2 a/2 0 x a/2 x3 3 0 a − (a − x)3 3 a/2 √ A a3 2 3 a3 A2 a3 = + = ⇒ A= √ . 3 8 8 12 a3 2 (b) √ Z a/2 Z a 22 3 nπ nπ √ x dx + (a − x) sin x dx x sin aa a 0 a a a/2 √ 2 a/2 2 6 nπ xa nπ a = 2 sin x − cos x a nπ a nπ a 0 2 a a a a nπ ax nπ nπ − +a − cos x x − x sin cos nπ a nπ a nπ a a/2 a/2 √ 2 2 2 2 2 6 a nπ nπ a a nπ − a cos = 2 sin − cos nπ + cos a nπ 2 2nπ 2 nπ nπ 2 2 a nπ a2 a2 nπ + cos sin + cos nπ − nπ 2 nπ 2nπ 2 √ √ ( 2 0, n even, 2 6 a nπ 4 6 nπ √ = 2 sin = sin = (n−1)/2 4 6 2 2 2 (−1) , (nπ) 2 (nπ) 2 a (nπ)2 n odd. r cn = √ r 4 6 2 X 1 nπ n2 π 2 ~2 So Ψ(x, t) = 2 (−1)(n−1)/2 2 sin x e−iEn t/~ , where En = . π a n=1,3,5,… n a 2ma2 (c) P1 = |c1 |2 = 16 · 6 = 0.9855. π4 (d) hHi = X |cn |2 En = 96 π 2 ~2 π 4 2ma2 1 1 1 1 + 2 + 2 + 2 + ··· 1 3 5 7 | {z } π 2 /8 = 48~2 π 2 6~2 = . 2 2 π ma 8 ma2 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 21 Problem 2.8 A 2 r Z a/2 2 dx = A (a/2) = 1 ⇒ A = 0 2 . a From Eq. 2.37, r Z a/2 π π i π i a/2 2 2 2h a 2h sin − cos − cos 0 = . c1 = A x dx = x = − cos a 0 a a π a π 2 π 0 P1 = |c1 |2 = (2/π)2 = 0.4053. Problem 2.9 ĤΨ(x, 0) = − Z ~2 ∂ 2 ~2 ∂ ~2 [Ax(a − x)] = −A (a − 2x) = A . 2m ∂x2 2m ∂x m 2 a x x3 a − Ψ(x, 0) ĤΨ(x, 0) dx = A x(a − x) dx = A m 0 m 2 3 0 3 2 3 2 3 2 a ~ a 30 ~ a 5~ = A2 − = 5 = m 2 3 a m 6 ma2 ∗ 2~ 2 Z a 2~ 2 (same as Example 2.3). Problem 2.10 (a) Using Eqs. 2.48 and 2.60, mω 1/4 − mω x2 1 d + mωx e 2~ −~ dx π~ 2~mω i mω 2 mω 1/4 h mω mω 1/4 mω 2 1 1 =√ −~ − 2x + mωx e− 2~ x = √ 2mωxe− 2~ x . 2~ 2~mω π~ 2~mω π~ mω 1/4 mω 2 d 1 2mω −~ + mωx xe− 2~ x (a+ )2 ψ0 = 2~mω π~ dx i mω 2 mω 1/4 2mω mω 2 1 mω 1/4 h mω 2 − 2~ x 2 −~ 1 − x 2x + mωx e x − 1 e− 2~ x . = = ~ π~ 2~ π~ ~ a+ ψ0 = √ Therefore, from Eq. 2.68, 1 mω 1/4 1 ψ2 = √ (a+ )2 ψ0 = √ 2 2 π~ mω 2 2mω 2 x − 1 e− 2~ x . ~ 22 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION (b) s0 s1 (c) Since ψ0 and ψ2Rare even, whereas ψ1 is odd, we need to check is ψ2∗ ψ0 dx: Z R s2 ψ0∗ ψ1 dx and R ψ2∗ ψ1 dx vanish automatically. The only one Z mω 2 mω ∞ 2mω 2 x − 1 e− ~ x dx ~ 2 π~ −∞ r Z ∞ Z mω 2 mω 2mω ∞ 2 − mω x2 e− ~ x dx − x e ~ dx =− 2π~ ~ −∞ −∞ r r r 2mω ~ mω π~ π~ − = 0. X =− 2π~ mω ~ 2mω mω 1 ψ2∗ ψ0 dx = √ r Problem 2.11 R (a) Note that ψ0 is even, and ψ1 is odd. In either case |ψ|2 is even, so hxi = x|ψ|2 dx = 0. Therefore hpi = mdhxi/dt = 0. (These results hold for any stationary state of the harmonic oscillator.) √ 2 2 From Eqs. 2.60 and 2.63, ψ0 = αe−ξ /2 , ψ1 = 2αξe−ξ /2 . So n = 0: hx2 i = α2 Z ∞ 2 x2 e−ξ dx = α2 −∞ 2 ~ mω 3/2 Z ∞ 2 1 ξ 2 e−ξ dξ = √ π −∞ Z ∞ ~ mω √ π ~ = . 2 2mω d2 −ξ2 /2 e e dξ hp i = ψ0 ψ0 dx = −~ α dξ 2 −∞ √ −ξ2 m~ω m~ω π √ m~ω 2 √ √ =− ξ − 1 e dξ = − − π = . 2 2 π −∞ π 2 Z ~ d i dx Z ∞ 2 2 r mω ~ −ξ 2 /2 n = 1: hx2 i = 2α2 Z ∞ 2 x2 ξ 2 e−ξ dx = 2α2 −∞ Z ∞ ~ mω 3/2 Z ∞ −∞ 2 ξ 4 e−ξ dξ = √ √ 2~ 3 π 3~ = . 2mω πmω 4 d2 −ξ 2 /2 hp i = −~ 2α ξe ξe dξ dξ 2 −∞ √ Z 2mω~ 3 √ π 3m~ω 2mω~ ∞ 4 2 −ξ 2 √ √ =− ξ − 3ξ e dξ = − π−3 = . 4 2 2 π −∞ π 2 2 2 r mω ~ −ξ 2 /2 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION (b) n = 0: r σx = p hx2 i − hxi2 = r σx σp = ~ 2mω r 3~ ; 2mω r p ~ ; σp = hp2 i − hpi2 = 2mω mω~ ~ = . 2 2 r m~ω ; 2 (Right at the uncertainty limit.)X n = 1: σx = r σp = 3m~ω ; 2 σx σp = 3 ~ ~ > .X 2 2 (c) 1   4 ~ω (n = 0)  1 2 ; hp i = hT i = 3  2m ~ω (n = 1) 4 hT i + hV i = hHi = 1   4 ~ω (n = 0)  1 . hV i = mω 2 hx2 i = 3  2 ~ω (n = 1) 4 1   2 ~ω (n = 0) = E0  3 2 ~ω (n = 1) = E1 , as expected.  Problem 2.12 From Eq. 2.70, r ~ ~mω (a+ + a− ), p = i (a+ − a− ), 2mω 2 r Z ~ hxi = ψn∗ (a+ + a− )ψn dx. 2mω r x= so But (Eq. 2.67) a+ ψn = So √ n + 1ψn+1 , a− ψn = √ nψn−1 . r Z Z √ √ ~ ∗ ∗ hxi = n + 1 ψn ψn+1 dx + n ψn ψn−1 dx = 0 (by orthogonality). 2mω dhxi ~ ~ hpi = m = 0. x2 = (a+ + a− )2 = a2+ + a+ a− + a− a+ + a2− . dt 2mω 2mω Z ~ ψn∗ a2+ + a+ a− + a− a+ + a2− ψn . But hx2 i = 2mω p √  2 √ √ a ψ = a n + 1ψ = n + 1 n + 2ψ = (n + 1)(n + 2)ψn+2 .  n+1 n+2 n + +  √ √  a a ψ = a √nψ = n nψ = nψ + − n + n−1 n n. p √ √ a a ψ = a n + 1ψ = n + 1) n + 1ψ = (n +  − + n − n+1 n  p 1)ψn . √ √ √  2 a− ψn = a− nψn−1 = n n − 1ψn−2 = (n − 1)nψn−2 . So Z Z ~ ~ 1 ~ 2 2 hx i = 0 + n |ψn | dx + (n + 1) |ψn | dx + 0 = (2n + 1) = n + . 2mω 2mω 2 mω 2 23 24 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION ~mω ~mω 2 (a+ − a− )2 = − a+ − a+ a− − a− a+ + a2− ⇒ 2 2 ~mω ~mω 1 hp2 i = − [0 − n − (n + 1) + 0] = (2n + 1) = n + m~ω. 2 2 2 p2 = − 1 1 hT i = hp /2mi = n+ ~ω . 2 2 r r p 1 ~ 2 2 σx = hx i − hxi = n + ; 2 mω 2 r p σp = hp2 i − hpi2 = n+ 1√ m~ω; 2 σx σp = n+ 1 ~ ~≥ .X 2 2 Problem 2.13 (a) Z 1= |Ψ(x, 0)|2 dx = |A|2 Z 9|ψ0 |2 + 12ψ0∗ ψ1 + 12ψ1∗ ψ0 + 16|ψ1 |2 dx = |A|2 (9 + 0 + 0 + 16) = 25|A|2 ⇒ A = 1/5. (b) Ψ(x, t) = i i 1h 1h 3ψ0 (x)e−iE0 t/~ + 4ψ1 (x)e−iE1 t/~ = 3ψ0 (x)e−iωt/2 + 4ψ1 (x)e−3iωt/2 . 5 5 (Here ψ0 and ψ1 are given by Eqs. 2.60 and 2.63; E0 and E1 by Eq. 2.62.) i 1 h 2 9ψ0 + 12ψ0 ψ1 eiωt/2 e−3iωt/2 + 12ψ0 ψ1 e−iωt/2 e3iωt/2 + 16ψ12 25 1 2 = 9ψ0 + 16ψ12 + 24ψ0 ψ1 cos(ωt) . 25 |Ψ(x, t)|2 = (With ψ2 in place of ψ1 the frequency would be (E2 − E0 )/~ = [(5/2)~ω − (1/2)~ω]/~ = 2ω.) (c) hxi = But R xψ02 dx = R Z Z Z 1 9 xψ02 dx + 16 xψ12 dx + 24 cos(ωt) xψ0 ψ1 dx . 25 xψ12 dx = 0 (see Problem 2.11 or 2.12), while r Z xψ0 ψ1 dx = r = mω π~ r 2mω ~ Z xe 2 mω √ 2 π2 π ~ 2 − mω 2~ x 1 2 r xe ~ mω 2 − mω 2~ x !3 r dx = r = 2 mω π ~ Z ∞ mω −∞ ~ . 2mω So 24 hxi = 25 r ~ cos(ωt); 2mω d 24 hpi = m hxi = − dt 25 r 2 x2 e− ~ x dx mω~ sin(ωt). 2 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 25 Ehrenfest’s theorem says dhpi/dt = −h∂V /∂xi. Here r dhpi 24 mω~ 1 ∂V =− ω cos(ωt), V = mω 2 x2 ⇒ = mω 2 x, dt 25 2 2 ∂x so r r ~ ∂V 24 ~mω 2 2 24 − = −mω hxi = −mω cos(ωt) = − ω cos(ωt), ∂x 25 2mω 25 2 so Ehrenfest’s theorem holds. (d) You could get E0 = 12 ~ω, with probability |c0 |2 = 9/25, or E1 = 32 ~ω, with probability |c1 |2 = 16/25. Problem 2.14 r r r Z Z ∞ 2 mω ∞ −ξ2 mω ~ e dx = 2 e−ξ dξ. ψ0 = e , so P = 2 π~ π~ x0 π~ mω ξ0 q ~ Classically allowed region extends out to: 12 mω 2 x20 = E0 = 12 ~ω, or x0 = mω , so ξ0 = 1. mω 1/4 2 P =√ π Z ∞ −ξ 2 /2 √ 2 e−ξ dξ = 2(1 − F ( 2)) (in notation of CRC Table) = 0.157. 1 Problem 2.15 −2(5−1) −2(5−3) 4 n = 5: j = 1 ⇒ a3 = (1+1)(1+2) a1 = − 43 a1 ; j = 3 ⇒ a5 = (3+1)(3+2) a3 = − 51 a3 = 15 a1 ; j = 5 ⇒ a7 = 0. So a1 4 4 3 5 3 5 H5 (ξ) = a1 ξ − 3 a1 ξ + 15 a1 ξ = 15 (15ξ − 20ξ + 4ξ ). By convention the coefficient of ξ 5 is 25 , so a1 = 15 · 8, and H5 (ξ) = 120ξ − 160ξ 3 + 32ξ 5 (which agrees with Table 2.1). −2(6−0) −2(6−2) n = 6: j = 0 ⇒ a2 = (0+1)(0+2) a0 = −6a0 ; j = 2 ⇒ a4 = (2+1)(2+2) a2 = − 23 a2 = 4a0 ; j = 4 ⇒ a6 = −2(6−4) 2 8 8 6 2 4 6 (4+1)(4+2) a4 = − 15 a4 = − 15 a0 ; j = 6 ⇒ a8 = 0. So H6 (ξ) = a0 − 6a0 ξ + 4a0 ξ − 15 ξ a0 . The coefficient of ξ 8 a0 ⇒ a0 = −15 · 8 = −120. H6 (ξ) = −120 + 720ξ 2 − 480ξ 4 + 64ξ 6 . is 26 , so 26 = − 15 Problem 2.16 (a) 2 2 2 2 d d d −ξ2 −ξ 2 (e ) = −2ξe ; e−ξ = (−2ξe−ξ ) = (−2 + 4ξ 2 )e−ξ ; dξ dξ dξ 3 2 2 d d −ξ 2 2 −ξ 2 2 e = (−2 + 4ξ )e = 8ξ + (−2 + 4ξ )(−2ξ) e−ξ = (12ξ − 8ξ 3 )e−ξ ; dξ dξ 4 2 2 d d −ξ 2 3 −ξ 2 2 3 e = (12ξ − 8ξ )e = 12 − 24ξ + (12ξ − 8ξ )(−2ξ) e−ξ = (12 − 48ξ 2 + 16ξ 4 )e−ξ . dξ dξ 3 4 2 d d ξ2 −ξ 2 3 ξ2 H3 (ξ) = −e e = −12ξ + 8ξ ; H4 (ξ) = e e−ξ = 12 − 48ξ 2 + 16ξ 4 . dξ dξ 26 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION (b) H5 = 2ξH4 − 8H3 = 2ξ(12 − 48ξ 2 + 16ξ 4 ) − 8(−12ξ + 8ξ 3 ) = 120ξ − 160ξ 3 + 32ξ 5 . H6 = 2ξH5 − 10H4 = 2ξ(120ξ − 160ξ 3 + 32ξ 5 ) − 10(12 − 48ξ 2 + 16ξ 4 ) = −120 + 720ξ 2 − 480ξ 4 + 64ξ 6 . (c) dH5 = 120 − 480ξ 2 + 160ξ 4 = 10(12 − 48ξ 2 + 16ξ 4 ) = (2)(5)H4 . X dξ dH6 = 1440ξ − 1920ξ 3 + 384ξ 5 = 12(120ξ − 160ξ 3 + 32ξ 5 ) = (2)(6)H5 . X dξ (d) 2 d −z2 +2zξ (e ) = (−2z + 2ξ)e−z +2zξ ; setting z = 0, H1 (ξ) = 2ξ. dz d dz 2 d dz 3 (e (e −z 2 +2zξ −z 2 +2zξ d −z 2 +2zξ )= (−2z + 2ξ)e dz 2 −z 2 +2zξ = − 2 + (−2z + 2ξ) e ; setting z = 0, H2 (ξ) = −2 + 4ξ 2 . d 2 −z 2 +2zξ )= − 2 + (−2z + 2ξ) e dz 2 2 = 2(−2z + 2ξ)(−2) + − 2 + (−2z + 2ξ) (−2z + 2ξ) e−z +2zξ ; setting z = 0, H3 (ξ) = −8ξ + (−2 + 4ξ 2 )(2ξ) = −12ξ + 8ξ 3 . Problem 2.17 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). ikx ikx e + e−ikx e − e−ikx 1 1 C cos kx + D sin kx = C +D = (C − iD)eikx + (C + iD)e−ikx 2 2i 2 2 = Aeikx + Be−ikx , with A = 1 1 (C − iD); B = (C + iD). 2 2 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.18 ~k2 Equation 2.95 says Ψ = Aei(kx− 2m t) , so i h ~k2 ~k2 ~k2 ~k2 i~ ∂Ψ∗ i~ ∗ ∂Ψ J = Ψ −Ψ = |A|2 ei(kx− 2m t) (−ik)e−i(kx− 2m t) − e−i(kx− 2m t) (ik)ei(kx− 2m t) 2m ∂x ∂x 2m = i~ ~k 2 |A|2 (−2ik) = |A| . 2m m It flows in the positive (x) direction (as you would expect). Problem 2.19 (a) ∞ X bn inπx/a einπx/a − e−inπx/a + e + e−inπx/a 2i 2 n=1 n=1 ∞ ∞ X X an bn an bn inπx/a = b0 + + e + − + e−inπx/a . 2i 2 2i 2 n=1 n=1 f (x) = b0 + ∞ X an Let c0 ≡ b0 ; cn = 12 (−ian + bn ) , for n = 1, 2, 3, . . . ; cn ≡ 12 (ia−n + b−n ) , for n = −1, −2, −3, . . . . Then f (x) = ∞ X cn einπx/a . QED n=−∞ (b) Z a f (x)e −imπx/a dx = −a Z a ∞ X n=−∞ Z a ei(n−m)πx/a dx. But for n 6= m, cn −a a ei(n−m)πx/a dx = −a ei(n−m)πx/a ei(n−m)π − e−i(n−m)π (−1)n−m − (−1)n−m = = = 0, i(n − m)π/a −a i(n − m)π/a i(n − m)π/a whereas for n = m, Z a Z a ei(n−m)πx/a dx = dx = 2a. −a −a So all terms except n = m are zero, and Z a Z a 1 f (x)e−inπx/a dx. f (x)e−imπx/a = 2acm , so cn = 2a −a −a (c) f (x) = ∞ X n=−∞ r π1 1 X F (k)eikx = √ F (k)eikx ∆k, 2a 2π QED 27 28 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION where ∆k ≡ π is the increment in k from n to (n + 1). a r F (k) = 2 1 a π 2a Z a f (x)e −ikx −a 1 dx = √ 2π Z a f (x)e−ikx dx. −a (d) As a → ∞, k becomes a continuous variable, 1 f (x) = √ 2π Z ∞ 1 F (k)eikx dk; F (k) = √ 2π −∞ Z ∞ f (x)e−ikx dx. −∞ Problem 2.20 (a) Z ∞ 2 2 Z ∞ |Ψ(x, 0)| dx = 2|A| 1= −∞ ∞ e−2ax dx = 2|A|2 0 √ |A|2 e−2ax = ⇒ A = a. −2a 0 a (b) A φ(k) = √ 2π Z ∞ A e−a|x| e−ikx dx = √ 2π −∞ 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 Z ∞ Z ∞ A A −ax φ(k) = 2 √ e cos kx dx = √ e−ax eikx + e−ikx dx 2π 0 2π 0 (ik−a)x ∞ Z ∞ e A A e−(ik+a)x = √ e(ik−a)x + e−(ik+a)x dx = √ + −(ik + a) 0 2π 0 2π ik − a r A −1 1 A −ik − a + ik − a a 2a = √ + =√ = . −k 2 − a2 2π k 2 + a2 2π ik − a ik + a 2π (c) r Z Z 2 ~k2 1 a3 ∞ 1 a3/2 ∞ 1 i(kx− ~k 2m t) dk = Ψ(x, t) = √ 2 e ei(kx− 2m t) dk. 2 2 2 2 2π −∞ k + a π −∞ k + a 2π p (d) For large a, Ψ(x, 0) is a sharp narrow spike whereas φ(k) ∼ = 2/πa is broad and flat; position p is welldefined but momentum is ill-defined. For small a, Ψ(x, 0) is a broad and flat whereas φ(k) ∼ = ( 2a3 /π)/k 2 is a sharp narrow spike; position is ill-defined but momentum is well-defined. CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 29 Problem 2.21 (a) 1 = |A| 2 Z ∞ e −2ax2 2 r dx = |A| −∞ π ; 2a A= 2a π 1/4 . (b) Z ∞ e −(ax2 +bx) Z ∞ dx = e −∞ −y 2 +(b2 /4a) −∞ 1 φ(k) = √ A 2π Z ∞ −ax2 −ikx e e −∞ 1 1 Ψ(x, t) = √ 2π (2πa)1/4 1 2 1 √ dy = √ eb /4a a a 1 dx = √ 2π Z ∞ 2a π 2 1/4 r Z ∞ e −y 2 r dy = −∞ π b2 /4a e . a 2 π −k2 /4a 1 e = e−k /4a . 1/4 a (2πa) 2 t/2m) e−k /4a ei(kx−~k | {z } dk −∞ 1 2 e−[( 4a +i~t/2m)k −ixk] =√ 1/4 −ax2 /(1+2i~at/m) √ 2 1 π e 1 2a q p e−x /4( 4a +i~t/2m) = . 1/4 π 1 2π(2πa) 1 + 2i~at/m + i~t/2m 4a (c) r 2 Let θ ≡ 2~at/m. Then |Ψ| = 2 2 2a e−ax /(1+iθ) e−ax /(1−iθ) p . The exponent is π (1 + iθ)(1 − iθ) r 2 2 ax2 ax2 −2ax2 2a e−2ax /(1+θ ) 2 (1 − iθ + 1 + iθ) 2 √ − − = −ax = ; |Ψ| = . (1 + iθ) (1 − iθ) (1 + iθ)(1 − iθ) 1 + θ2 π 1 + θ2 r r a 2 −2w2 x2 2 Or, with w ≡ , |Ψ| = we . As t increases, the graph of |Ψ|2 flattens out and broadens. 2 1+θ π |^| 2 |^|2 x t=0 t>0 x (d) Z ∞ hxi = x|Ψ|2 dx = 0 (odd integrand); hpi = m −∞ r 2 hx i = 2 w π Z ∞ 2 −2w2 x2 x e −∞ r dx = 2 1 w π 4w2 r dhxi = 0. dt π 1 = . 2w2 4w2 2 hp i = −~ 2 Z ∞ −∞ Ψ∗ ∂2Ψ dx. ∂x2 30 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 2 Write Ψ = Be−bx , where B ≡ 2a π 1/4 √ 1 a and b ≡ . 1 + iθ 1 + iθ 2 ∂2Ψ ∂ −bx2 = B −2bxe = −2bB(1 − 2bx2 )e−bx . 2 ∂x ∂x ∗ 2 ∂2Ψ a a 2a = −2b|B|2 (1 − 2bx2 )e−(b+b )x ; b + b∗ = = 2w2 . + = 2 d∂x 1 + iθ 1 − iθ 1 + θ2 r r r 2 2 2 2a 2 2 1 2 ∗∂ Ψ √ = |B| = = −2b w. So Ψ w(1 − 2bx2 )e−2w x . 2 2 π 1+θ π ∂x π Ψ∗ r 2 hp i = 2b~ 2 r = 2b~ 2 2 w π Z ∞ 2 w π r 2 −∞ b But 1 − =1− 2w2 hp2 i = 2b~2 2 (1 − 2bx2 )e−2w x dx π 1 − 2b 2 2 2w 4w a 1 + iθ a = ~2 a. 2b r π 2w2 1 + θ2 2a σx = 1 ; 2w 2 = 2b~ =1− b 1− 2w2 √ σp = ~ a. ~p ~p ~ 1 √ 1 + θ2 = 1 + (2~at/m)2 ≥ . X ~ a= 2w 2 2 2 Closest at t = 0, at which time it is right at the uncertainty limit. Problem 2.22 (a) (−2)3 − 3(−2)2 + 2(−2) − 1 = −8 − 12 − 4 − 1 = −25. (b) cos(3π) + 2 = −1 + 2 = 1. (c) 0 (x = 2 is outside the domain of integration). . 1 + iθ a (1 − iθ) = = , so 2 2 2b (e) σx σp = CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 31 Problem 2.23 If c > 0, y : −∞ → ∞. If c 0); or f (x)δ(cx)dx = R  1 R −∞ 1 ∞ 1 −∞ c ∞ f (y/c)δ(y)dy = − c −∞ f (y/c)δ(y)dy = − c f (0) (c . X 2 2 2mα ~ 32 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.25 hψbound |ψscattering i Z 0 √ Z ∞ 2 2 mα emαx/~ Aeikx + Be−ikx dx + = e−mαx/~ F eikx + Ge−ikx dx ~ −∞ 0 Z 0 √ Z 0 Z ∞ Z ∞ mα mα mα − mα − mα ( ( ( ( 2 +ik )x 2 −ik )x 2 +ik )x 2 −ik )x ~ ~ ~ ~ = A e dx + B e dx + F e dx + G e dx ~ −∞ −∞ 0 0 # ” mα mα mα mα √ 0 0 ∞ ∞ e( ~2 −ik)x e(− ~2 +ik)x e(− ~2 −ik)x mα e( ~2 +ik)x + B mα +F +G = A mα mα mα ~ − ~2 + ik 0 − ~2 − ik 0 ~2 + ik −∞ ~2 − ik −∞ √ √ mα A B F G mα A + G B+F = + − − = + mα mα mα mα ~ − mα − mα ~ ~2 + ik ~2 − ik ~2 + ik ~2 − ik ~2 + ik ~2 − ik ” # ” # √ √ mα mα mα mα mα ~2 (A + G + B + F ) + ik(B + F − A − G) ~2 − ik (A + G) + ~2 + ik (B + F ) = = mα 2 mα 2 ~ ~ + k2 + k2 2 2 ~ ~ But Equation 2.136 says (A + G + B + F ) = 2(A + B), and Equation 2.137 says ik(B + F − A − G) = −(2mα/~2 )(A + B), so ” # √ 2mα mα mα ~2 2(A + B) − ~2 (A + B) hψbound |ψscattering =i = 0. X mα 2 ~ + k2 2 ~ Problem 2.26 Z ∞ 1 1 Put f (x) = δ(x) into Eq. 2.103: F (k) = √ δ(x)e−ikx dx = √ . 2π −∞ 2π Z ∞ Z ∞ 1 1 1 √ eikx dk = ∴ f (x) = δ(x) = √ eikx dk. QED 2π 2π −∞ 2π −∞ Problem 2.27 V(x) (a) -a a x (b) From Problem 2.1(c) the solutions are even or odd. Look first for even solutions:  −κx (x > a),  Ae ψ(x) = B(eκx + e−κx ) (−a < x < a),  κx Ae (x a),  Ae ψ(x) = B(eκx − e−κx ) (−a < x < a),  −Aeκx (x < −a). Continuity at a : Ae−κa = B(eκa − e−κa ), or A = B(e2κa − 1). 2mα Discontinuity in ψ : −κAe − B(κe + κe ) = − 2 Ae−κa ⇒ B(e2κa + 1) = A ~ 2mα 2mα 2mα 2κa 2κa 2κa e + 1 = (e − 1) −1 =e − 1 − 2 + 1, 2 2 ~ κ ~ κ ~ κ 0 1= −κa −κa κa 2mα −1 , ~2 κ 2mα −2κa ~2 κ ~2 κ 2mα −2κa −2κa − 1 − e ; = 1 − e , e = 1 − , or e−z = 1 − cz. ~2 κ ~2 κ mα mα 1 1/c 1/c z 34 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION This time there may or may not be a solution. Both graphs have their y-intercepts at 1, but if c is too large (α too small), there may be no intersection (solid line), whereas if c is smaller (dashed line) there will be. (Note that z = 0 ⇒ κ = 0 is not a solution, since ψ is then non-normalizable.) The slope of e−z (at z = 0) is −1; the slope of (1 − cz) is −c. So there is an odd solution ⇔ c ~2 /2ma. Conclusion: One bound state if α ≤ ~2 /2ma; two if α > ~2 /2ma. s s -a -a a a x Even α= 1 ~2 ⇒c= . ma 2 x Odd Even: e−z = 12 z − 1 ⇒ z = 2.21772, Odd: e−z = 1 − 12 z ⇒ z = 1.59362. E = −0.615(~2 /ma2 ); E = −0.317(~2 /ma2 ). α= ~2 ⇒ c = 2. Only even: e−z = 2z − 1 ⇒ z = 0.738835; 4ma E = −0.0682(~2 /ma2 ). (c) (i) There is one bound state (even); c is huge, so z is small, so e−z ≈ 1 = cz − 1, which means z = 2/c, or 2κa = 2(2amα/~2 ) ⇒ κ = (2mα/~2 ). E=− ~2 κ2 2mα2 = − 2 . 2m ~ This makes sense: the two delta-functions coincide, so there is really just one delta-function, with “strength” 2α. Putting this into Equation 2.130 we recover the answer in the box. (ii) There are two bound states, one even and one odd; c is small, so z is huge, and e−z ≈ 0. For the even case, 0 = cz − 1 ⇒ z = 1/c ⇒ κ = (mα/~2 ). For the odd case, 0 = 1 − cz, which leads to the mα2 . Any linear combination of the two 2~2 will be an eigenstate (with the same energy); the sum (properly normalized) would represent a particle in the delta-function out at large positive x, and the difference would be a particle in the delta-function at large negative x (the other—distant—delta-function becomes irrelevant), so it makes sense that we get two states, each with the energy of a particle in a single delta-function well. same result: the two states are degenerate, each with energy − Problem 2.28  ikx   Ae + Be−ikx (x < −a)  ψ = Ceikx + De−ikx (−a < x a) (1) Continuity at −a : Ae−ika + Beika = Ce−ika + Deika ⇒ βA + B = βC + D, where β ≡ e−2ika . (2) Continuity at +a : Ceika + De−ika = F eika ⇒ F = C + βD. CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 35 −ika (3) Discontinuity in ψ 0 at −a : ik(Ce−ika − Deika ) − ik(Ae−ika − Beika ) = − 2mα + Beika ) ~2 (Ae ⇒ βC − D = β(γ + 1)A + B(γ − 1), where γ ≡ i2mα/~2 k. ika (4) Discontinuity in ψ 0 at +a : ikF eika − ik(Ceika − De−ika ) = − 2mα ) ~2 (F e ⇒ C − βD = (1 − γ)F. To solve for C and D, add (2) and (4) : 2C = F + (1 − γ)F ⇒ 2C = (2 − γ)F. subtract (2) and (4) : 2βD = F − (1 − γ)F ⇒ 2D = (γ/β)F. add (1) and (3) : 2βC = βA + B + β(γ + 1)A + B(γ − 1) ⇒ 2C = (γ + 2)A + (γ/β)B. subtract (1) and (3) : 2D = βA + B − β(γ + 1)A − B(γ − 1) ⇒ 2D = −γβA + (2 − γ)B. Equate the two expressions for 2C : (2 − γ)F = (γ + 2)A + (γ/β)B. Equate the two expressions for 2D : (γ/β)F = −γβA + (2 − γ)B. Solve these for F and B, in terms of A. Multiply the first by β(2 − γ), the second by γ, and subtract: 2 β(2 − γ)2 F = β(4 − γ 2 )A + γ(2 − γ)B ; (γ /β)F = −βγ 2 A + γ(2 − γ)B . 4 F = . ⇒ β(2 − γ)2 − γ 2 /β F = β 4 − γ 2 + γ 2 A = 4βA ⇒ A (2 − γ)2 − γ 2 /β 2 Let g ≡ i/γ = ~2 k i F 4g 2 ; φ ≡ 4ka, so γ = , β 2 = e−iφ . Then: = . 2mα g A (2g − i)2 + eiφ Denominator: 4g 2 − 4ig − 1 + cos φ + i sin φ = (4g 2 − 1 + cos φ) + i(sin φ − 4g). |Denominator|2 = (4g 2 − 1 + cos φ)2 + (sin φ − 4g)2 = 16g 4 + 1 + cos2 φ − 8g 2 − 2 cos φ + 8g 2 cos φ + sin2 φ − 8g sin φ + 16g 2 = 16g 4 + 8g 2 + 2 + (8g 2 − 2) cos φ − 8g sin φ. T = F A 2 = 8g 4 (8g 4 + 4g 2 + 1) + (4g 2 − 1) cos φ − 4g sin φ , where g ≡ ~2 k and φ ≡ 4ka. 2mα Problem 2.29  −κx  (x > a)  Fe  In place of Eq. 2.154, we have: ψ(x) = D sin(lx) (0 < x < a) .   −ψ(−x) (x < 0) Continuity of ψ : F e−κa = D sin(la); continuity of ψ 0 : −F κe−κa = Dl cos(la). q p Divide: − κ = l cot(la), or − κa = la cot(la) ⇒ z02 − z 2 = −z cot z, or − cot z = (z0 /z)2 − 1. Wide, deep well: Intersections are at π, 2π, 3π, etc. Same as Eq. 2.160, but now for n even. This fills in the rest of the states for the infinite square well. Shallow, narrow well: If z0 < π/2, there is no odd bound state. The corresponding condition on V0 is V0 < π 2 ~2 ⇒ no odd bound state. 8ma2 36 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION / z0 2/ z Problem 2.30 Z ∞ 2 2 Z a 2 2 Z ∞ −2κx cos lx dx + |F | e |ψ| dx = 2 |D| dx 0 a a ∞ 1 x 1 a sin 2la e−2κa = 2 |D|2 + |F |2 − e−2κx + sin 2lx = 2 |D|2 + + |F |2 . 2 4l 2κ 2 4l 2κ 0 a 1=2 0 sin(2la) cos2 (la) + . But F = Deκa cos la (Eq. 2.152), so 1 = |D|2 a + 2l κ Furthermore κ = l tan(la) (Eq. 2.157), so cos la 2 sin la cos la cos3 la 1 = |D|2 a + + = |D|2 a + (sin2 la + cos2 la) 2l l sin la l sin la 1 1 eκa cos la 1 = |D|2 a + D= p F =p = |D|2 a + . , . l tan la κ a + 1/κ a + 1/κ Problem 2.31 √ Equation 2.158 ⇒ z0 = ~a 2mV0 . We want α = area of potential = 2aV0 held constant as a → 0. Therefore p √ α α V0 = 2a ; z0 = ~a 2m 2a = ~1 mαa → 0. So z0 is small, and the intersection in Fig. 2.17 occurs at very small z. Solve Eq. 2.159 for very small z, by expanding tan z: q p 2 (z /z) − 1 = (1/z) z02 − z 2 . tan z ∼ z = = 0 Now (from Eqs. 2.149, 2.151 and 2.158) z02 −z 2 = κ2 a2 , so z 2 = κa. But z02 −z 2 = z 4 1 ⇒ z ∼ = z0 , so κa ∼ = z02 . 1 mα 1√ ∼ But we found that z0 = ~ mαa here, so κa = ~2 mαa, or κ = ~2 . (At this point the a’s have canceled, and we can go to the limit a → 0.) √ −2mE mα m2 α2 mα2 = 2 ⇒ −2mE = . E = − (which agrees with Eq. 2.132). ~ ~ ~2 2~2 √ V02 α 2mV0 . But V0 = 2a In Eq. 2.172, V0 E ⇒ T −1 ∼ sin2 2a , so the argument of the sine is small, = 1+ 4EV ~ 0 2 V0 2a mα2 −1 ∼ 2 m and we can replace sin by : T = 1 + 4E ~ 2mV0 = 1 + (2aV0 ) 2~2 E . But 2aV0 = α, so T −1 = 1 + 2~ 2E , in agreement with Eq. 2.144. CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 37 Problem 2.32 Multiply Eq. 2.168 by sin la, Eq. 2.169 by 1l cos la, and add: ik C sin2 la + D sin la cos la = F eika sin la ika C = F e sin la + cos la . C cos2 la − D sin la cos la = ikl F eika cos la l Multiply Eq. 2.168 by cos la, Eq. 2.169 by 1l sin la, and subtract: ik C sin la cos la + D cos2 la = F eika cos la ika sin la . D = F e cos la − C sin la cos la − D sin2 la = ikl F eika sin la l Put these into Eq. 2.166: (1) Ae −ika + Be ika ik ik ika = −F e sin la + cos la sin la + F e sin la cos la cos la − l l ik ik 2 ika 2 = Fe cos la − sin la cos la − sin la − sin la cos la l l ik = F eika cos(2la) − sin(2la) . l ika Likewise, from Eq. 2.167: ik ik il ika sin la + cos la cos la + cos la − sin la sin la (2) Ae − Be = − F e k l l il ika ik ik 2 2 = − Fe sin la cos la + cos la + sin la cos la − sin la k l l il ika ik il ika = − Fe sin(2la) + cos(2la) = F e cos(2la) − sin(2la) . k l k k l Add (1) and (2): 2Ae−ika = F eika 2 cos(2la) − i + sin(2la) , or: l k −ika ika e−2ika A F = (confirming Eq. 2.171). Now subtract (2) from (1): 2 2 cos(2la) − i sin(2la) 2kl (k + l ) l k sin(2la) 2 2Beika = F eika i − sin(2la) ⇒ B = i (l − k 2 )F (confirming Eq. 2.170). k l 2kl T −1 = A F 2 = cos(2la) − i sin(2la) 2 (k + l2 ) 2kl But cos2 (2la) = 1 − sin2 (2la), so 2 −1 T = 1 + sin (2la) √ 2 = cos2 (2la) + (k 2 + l2 )2 −1 (2lk)2 | {z } sin2 (2la) 2 (k + l2 )2 . (2lk)2 =1+ (k 2 − l2 )2 sin2 (2la). (2kl)2 (k2 −l2 )2 1 1 4 2 2 4 [k4 +2k2 l2 +l4 −4k2 l2 ]= (2kl) 2 [k −2k l +l ]= (2kl)2 . (2kl)2 p 2m(E + V0 ) 2a p 2mV0 ; so (2la) = 2m(E + V0 ); k 2 − l2 = − 2 , and ~ ~ ~ 2m 2 2 2 V (k 2 − l2 )2 V 0 0 ~2 . = = 2 (2kl)2 4E(E + V0 ) 4 2m E(E + V ) 0 ~2 V02 2a p ∴ T −1 = 1 + sin2 2m(E + V0 ) , confirming Eq. 2.172. 4E(E + V0 ) ~ But k = 2mE , l= ~ 38 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.33  ikx   Ae + Be−ikx (x < −a)  ψ = Ceκx + De−κx (−a < x a) E < V0 . √ k= 2mE ; κ= ~ p 2m(V0 − E) . ~ (1) Continuity of ψ at −a: Ae−ika + Beika = Ce−κa + Deκa . (2) Continuity of ψ 0 at −a: ik(Ae−ika − Beika ) = κ(Ce−κa − Deκa ). κ κa κ −κa Ce + 1+i De . ⇒ 2Ae−ika = 1 − i k k (3) Continuity of ψ at +a: Ceκa + De−κa = F eika . (4) Continuity of ψ 0 at +a: κ(Ceκa − De−κa ) = ikF eika . ik ik κa ika −κa ⇒ 2Ce = 1 + F e ; 2De = 1− F eika . κ κ 2Ae −ika −2κa ik iκ ik e2κa iκ ika e 1+ Fe + 1+ 1− F eika = 1− k κ 2 k κ 2 k κ κ k F eika − − = 1+i + 1 e−2κa + 1 + i + 1 e2κa 2 κ k k κ ika 2 2 Fe (κ − k ) 2κa = 2 e−2κa + e2κa + i e − e−2κa . 2 kκ x −x ex + e−x e −e , cosh x ≡ , so But sinh x ≡ 2 2 F eika (κ2 − k 2 ) = 2 sinh(2κa) 4 cosh(2κa) + i 2 kκ (κ2 − k 2 ) ika = 2F e cosh(2κa) + i sinh(2κa) . 2kκ 2 (κ2 − k 2 )2 sinh2 (2κa). But cosh2 = 1 + sinh2 , so (2κk)2 V02 (κ2 − k 2 )2 2a p 2 2 1 + sinh 2m(V − E) , T −1 = 1 + 1 + sinh (2κa) = 0 (2κk)2 4E(V0 − E) ~ | {z } T −1 = A F = cosh2 (2κa) + F 2 2 where F = 4 4 2 2 2 2 2 4κ k + k + κ − 2κ k (κ + k ) = = (2κk)2 (2κk)2 2m(V0 −E) 2mE ~2 + ~2 2 2m(V0 −E) 4 2mE ~2 ~2 = V02 . 4E(V0 − E) (You can also get this from Eq. 2.172 by switching the sign of V0 and using sin(iθ) = i sinh θ.)  ikx    Ae + Be−ikx (x < −a) (−a < x a) (In central region − ~2 d 2 ψ d2 ψ + V0 ψ = Eψ ⇒ = 0, so ψ = C + Dx.) 2 2m dx dx2 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 39 (1) Continuous ψ at −a : Ae−ika + Beika = C − Da. (2) Continuous ψ at +a : F eika = C + Da. ⇒ (2.5) 2Da = F eika − Ae−ika − Beika . (3) Continuous ψ 0 at −a : ik Ae−ika − Beika = D. (4) Continuous ψ 0 at +a : ikF eika = D. ⇒ (4.5) Ae−2ika − B = F. Use (4) to eliminate D in (2.5): Ae−2ika + B = F − 2aikF = (1 − 2iak)F , and add to (4.5): 2Ae−2ika = 2F (1 − ika), so T −1 = 2 A F = 1 + (ka)2 = 1 + 2mE 2 a . ~2 (You can also get this from Eq. 2.172 by changing the sign of V0 and taking the limit E → V0 , using sin ∼ = .) This case is identical to the one in the book, only with V0 → −V0 . So E > V0 . T −1 V02 =1+ sin2 4E(E − V0 ) 2a p 2m(E − V0 ) . ~ Problem 2.34 (a) ψ= Aeikx + Be−ikx (x 0) √ where k = p 2mE ; κ= ~ 2m(V0 − E) . ~ (1) Continuity of ψ : A + B = F. (2) Continuity of ψ 0 : ik(A − B) = −κF. ik ik ik ⇒ A + B = − (A − B) ⇒ A 1 + = −B 1 − . κ κ κ R= B A 2 = |(1 + ik/κ)|2 1 + (k/κ)2 = = 1. 2 |(1 − ik/κ)| 1 + (k/κ)2 Although the wave function penetrates into the barrier, it is eventually all reflected. (b) ψ= Aeikx + Be−ikx (x 0) √ where k = (1) Continuity of ψ : A + B = F. (2) Continuity of ψ 0 : ik(A − B) = ilF. 2mE ; l= ~ p 2m(E − V0 ) . ~ 40 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION k k k ⇒ A + B = (A − B); A 1 − = −B 1 + . l l l 2 (1 − k/l)2 (k − l)2 (k − l)4 = = . (1 + k/l)2 (k + l)2 (k 2 − l2 )2 √ p 2m 2m √ 2m 2 2 Now k − l = 2 (E − E + V0 ) = V ; k − l = [ E − E − V0 ], so 0 ~ ~2 ~ √ √ ( E − E − V 0 )4 R= . V02 R= B A = (c) vi dt vi vt dt vt From the diagram, T = Pt /Pi = |F |2 vt /|A|2 vi , where Pi is the probability of finding the incident particle in the box corresponding to the time interval dt, and Pt is the probability of finding the transmitted particle in the associated box to the right of the barrier. r √ 2 vt E − V0 E − V0 F √ But = (from Eq. 2.98). So T = . Alternatively, from Problem 2.18: vi E A E r 2 2 ~k 2 ~l 2 Jt F E − V0 F l Ji = |A| ; Jt = |F | ; T = = . = m m Ji A k A E For E V0 , F = A + B = A + A T = F A 2 T +R= l = k 2k k+l 2 √ √ √ √ l 4kl 4kl(k − l)2 4 E E − V0 ( E − E − V0 )2 = = = . k (k + l)2 (k 2 − l2 )2 V02 4kl (k − l)2 4kl + k 2 − 2kl + l2 k 2 + 2kl + l2 (k + l)2 + = = = = 1. X (k + l)2 (k + l)2 (k + l)2 (k + l)2 (k + l)2 Problem 2.35 (a) ψ(x) = Aeikx + Be−ikx (x 0) √ where k ≡ 2mE , l≡ ~ Continuity of ψ ⇒ A + B = F =⇒ Continuity of ψ 0 ⇒ ik(A − B) = ilF k k k = −B 1 + ; A + B = (A − B); A 1 − l l l B =− A p 2m(E + V0 ) . ~ 1 − k/l 1 + k/l . CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 41 √ !2 √ 2 2 B l−k E + V0 − E √ = R= = √ A l+k E + V0 + E !2 √ p 2 2 1 + V0 /E − 1 1+3−1 2−1 1 = p = . = = √ 2 + 1 9 1+3+1 1 + V0 /E + 1 (b) The cliff is two-dimensional, and even if we pretend the car drops straight down, the potential as a function of distance along the (crooked, but now one-dimensional) path is −mgx (with x the vertical coordinate), as shown. V(x) x -V0 (c) Here V0 /E = 12/4 = 3, the same as in part (a), so R = 1/9, and hence T = 8/9 = 0.8889. Problem 2.36 Start with Eq. 2.25: ψ(x) = A sin kx + B cos kx. This time the boundary conditions are ψ(a) = ψ(−a) = 0: A sin ka + B cos ka = 0; −A sin ka + B cos ka = 0. ( Subtract : A sin ka = 0 ⇒ ka = jπ or A = 0, Add : B cos ka = 0 ⇒ ka = (j − 21 )π or B = 0, (where j = 1, 2, 3, . . .). If B = 0 (so A 6= 0), k = jπ/a. In Rthis case let n ≡ 2j (so n is an even√integer); then k = nπ/2a, a ψ = A sin(nπx/2a). Normalizing: 1 = |A|2 −a sin2 (nπx/2a) dx = |A|2 a ⇒ A = 1/ a. If A = 0 (so B 6= 0), k = (j − 21 )π/a. In R athis case let n ≡ 2j − 1 (n is an odd√integer); again k = nπ/2a, ψ = B cos(nπx/2a). Normalizing: 1 = |B|2 −a cos2 (nπx/2a)dx = |B|2 a ⇒ B = 1/ a. 2 2 2 2 2 k n π ~ In either case Eq. 2.24 yields E = ~2m = 2m(2a) 2 (in agreement with Eq. 2.30 for a well of width 2a). The substitution x → (x + a)/2 takes Eq. 2.31 to q  2 nπx n/2  (−1) sin (n even), r r  a 2a  nπx nπ 2 nπ (x + a) 2 sin = sin + = q  a a 2 a 2a 2  (−1)(n−1)/2 2 cos nπx (n odd). a 2a So (apart from normalization) we recover the results above. The graphs are the same as Figure 2.2, except that some are upside down (different normalization). cos(/x/2a) sin(2/x/2a) cos(3/x/2a) 42 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.37 Use the trig identity sin 3θ = 3 sin θ − 4 sin3 θ to write 3 sin πx a 3 = sin 4 πx a Normalize using Eq. 2.20: |A|2 1 − sin 4 r a 3 3πx 1 . So (Eq. 2.31): Ψ(x, 0) = A ψ1 (x) − ψ3 (x) . a 2 4 4 a 9 1 5 4 + = a|A|2 = 1 ⇒ A = √ . 2 16 16 16 5a So Ψ(x, 0) = √110 [3ψ1 (x) − ψ3 (x)] , and hence (Eq. 2.17) i 1 h Ψ(x, t) = √ 3ψ1 (x)e−iE1 t/~ − ψ3 (x)e−iE3 t/~ . 10 1 E3 − E1 9ψ12 + ψ32 − 6ψ1 ψ3 cos t ; so 10 ~ Z a Z a E3 − E1 1 3 9 2 hxi1 + hxi3 − cos t hxi = xψ1 (x)ψ3 (x)dx, x|Ψ(x, t)| dx = 10 10 5 ~ 0 0 |Ψ(x, t)|2 = where hxin = a/2 is the expectation value of x in the nth stationary state. The remaining integral is Z Z πx 3πx 2πx 4πx 2 a 1 a x sin sin dx = x cos − cos dx a 0 a a a 0 a a 2 2 a a 2πx xa 2πx a 4πx xa 4πx 1 = 0. cos + sin − cos − sin = a 2π a 2π a 4π a 4π a 0 Evidently then, 9 a 1 a a hxi = + = . 10 2 10 2 2 Using Eq. 2.21, 2 2 hHi = |c1 | E1 + |c3 | E3 = 9 10 π 2 ~2 + 2ma2 1 10 9π 2 ~2 9π 2 ~2 = . 2ma2 10ma2 Problem 2.38 (a) According to Eq. 2.39, the most general solution to the time-dependent Schrödinger equation for the infinite square well is ∞ X 2 2 2 Ψ(x, t) = cn ψn (x)e−i(n π ~/2ma )t . n=1 2 2 2 2 2 2 2 n2 π 2 ~ n2 π 2 ~ 4ma2 T = = 2πn2 , so e−i(n π ~/2ma )(t+T ) = e−i(n π ~/2ma )t e−i2πn , and since n2 is 2 2 π~ 2ma 2ma 2 an integer, e−i2πn = 1. Therefore Ψ(x, t + T ) = Ψ(x, t). QED Now CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 43 (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 r r 1 2E 2m 2 ⇒ Tc = a . E = mv ⇒ v = 2 m E (c) The two revival times are equal if r 2m 4ma2 =a , π~ E E= or π 2 ~2 E1 = . 2 8ma 4 Problem 2.39 (a) √ √ dΨ 2 3 2 3 a 1, (0 < x < a/2) = √ · = √ 1 − 2θ x − . −1, (a/2 < x < a) dx 2 a a a a (b) √ √ d2 Ψ 4 3 2 3 a a = √ − 2δ x − = − √ δ x− . dx2 2 2 a a a a (c) ~2 hHi = − 2m √ Z √ 4 3 2 3~2 ∗ a 2 · 3 · ~2 a 6~2 ∗ √ Ψ − √ dx = = = . X Ψ δ x− 2 2 m·a·a ma2 a a ma a | {z } √ 3/a Problem 2.40 (a) In the standard notation ξ ≡ p mω/~ x, α ≡ (mω/π~)1/4 , 2 2 Ψ(x, 0) = A(1 − 2ξ)2 e−ξ /2 = A(1 − 4ξ + 4ξ 2 )e−ξ /2 . It can be expressed as a linear combination of the first three stationary states (Eq. 2.60 and 2.63, and Problem 2.10): 2 ψ0 (x) = αe−ξ /2 , ψ1 (x) = So Ψ(x, 0) = c0 ψ0 + c1 ψ1 + c2 ψ2 = α(c0 +  √  α√2c2 = 4A α 2c1 = −4A  √  α(c0 − c2 / 2) = A √ √ 2 2 αξe−ξ /2 , 2ξc1 + √ 2 α ψ2 (x) = √ (2ξ 2 − 1)e−ξ /2 . 2 2 2ξ 2 c2 − √12 c2 )e−ξ /2 with (equating like powers) √ ⇒ c2 = 2 2A/α, √ ⇒ c1 = −2 2A/α, √ ⇒ c0 = (A/α) + c2 / 2 = (1 + 2)A/α = 3A/α. 44 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Normalizing: 1 = |c0 |2 + |c1 |2 + |c2 |2 = (8 + 8 + 9)(A/α)2 = 25(A/α)2 ⇒ A = α/5 = c0 = (b) You could get 3 , 5 c1 = − √ 2 2 , 5 c2 = 1 mω 1/4 . 5 π~ √ 2 2 . 5 1 9 3 8 5 8 ~ω, probability ; ~ω, probability ; ~ω, probability . 2 25 2 25 2 25 9 hHi = 25 1 8 3 8 5 ~ω 73 ~ω + ~ω + ~ω = (9 + 24 + 40) = ~ω. 2 25 2 25 2 50 50 (c) " # √ √ √ √ 3 2 2 2 2 −iωt/2 2 2 −3iωt/2 2 2 −5iωt/2 −iωt/2 3 −iωt −2iωt Ψ(x, t) = ψ0 e − ψ1 e + ψ2 e =e ψ0 − ψ1 e + ψ2 e . 5 5 5 5 5 5 To change the sign of the middle term we need e−iωT = −1 (then e−2iωT = 1); evidently ωT = π, or T = π/ω. Problem 2.41 Everything in Section 2.3.2 still applies, except that there is an additional boundary condition: ψ(0) = 0. This eliminates all the even solutions (n = 0, 2, 4, . . .), leaving only the odd solutions. So En = n+ 1 2 ~ω, n = 1, 3, 5, . . . . Problem 2.42 (a) Normalization is the same as before: A = 2a 1/4 . π (b) Equation 2.104 says 1 φ(k) = √ 2π 2a π 1/4 Z ∞ 2 e−ax eilx e−ikx dx [same as before, only k → k − l] = −∞ 1 1 Ψ(x, t) = √ 2π (2πa)1/4 Z ∞ 2 2 e−(k−l) /4a ei(kx−~k t/2m) dk −∞ Let u ≡ k − l, so k = u + l and dk = du: Z ∞ 2 2 2 1 1 Ψ(x, t) = √ e−u /4a ei[ux+lx−(~t/2m)(u +2ul+l )] du 2π (2πa)1/4 −∞ Z ∞ 2 1 ~t ~lt ~lt 1 1 il(x− 2m ) =√ e e−u ( 4a +i 2m )+iu(x− m ) du. 2π (2πa)1/4 −∞ 2 1 e−(k−l) /4a . (2πa)1/4 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 45 Using the hint in Problem 2.21, the integral becomes √ Z ∞ 2 2√ 2 1 ~lt 2 ~t 2 a −a(x− ~lt 1 m ) /γ q e(x− m ) /4( 4a +i 2m ) e−y dy = π, e γ ~t 1 −∞ + i 4a 2m so Ψ(x, t) = 2a π 1/4 2 2 ~lt 1 −a(x− ~lt m ) /γ eil(x− 2m ) . e γ The gaussian envelope (the first exponential) travels at speed ~l/m; the sinusoidal wave (the second exponential) travels at speed ~l/2m . (c) r 2 |Ψ(x, t)| = h i 2 1 2a 1 a(x− ~lt + (γ ∗1 )2 m ) γ2 e . 2 π |γ| The term in square brackets simplifies: 1 1 1 1 2i~t 2i~t 2 ∗ 2 2 + = [(γ ) + γ ] = 1 − + 1 + = . γ2 (γ ∗ )2 |γ|4 |γ|4 m m |γ|4 s p 2ia~t 2ia~t 2 1+ |γ| = 1− = 1 + θ2 , m m where (as before) θ ≡ 2~at/m. So r r 2 2 2a 2 −2w2 (x− ~lt 1 /(1+θ 2 ) 2a(x− ~lt 2 ) m m ) . √ |Ψ(x, t)| = = e we π 1 + θ2 π p ~l where (as before) w ≡ a/(1 + θ2 ). The result is the same as in Problem 2.21, except that x → x − m t , so |Ψ|2 has the same (flattening gaussian) shape – only this time the center moves at constant speed v = ~l/m. (d) Z ∞ hxi = x|Ψ(x, t)|2 dx. Let y ≡ x − vt, so x = y + vt. −∞ Z ∞ = r (y + vt) −∞ 2 −2w2 y2 ~l we dy = vt = t. π m (The first integral is trivially zero; the second is 1 by normalization.) hpi = m 2 dhxi = ~l. dt Z ∞ hx i = r 2 (y + vt) −∞ 2 −2w2 y2 1 1 we dy = + 0 + (vt)2 = + 2 π 4w 4w2 ~lt m (The first integral is same as in Problem 2.21). Z ∞ 2 ∂Ψ 2a ~lt 2 2 ∗∂ Ψ = − 2 x− + il Ψ, hp i = −~ Ψ dx; ∂x2 ∂x γ m −∞ 2 ∂2Ψ 2a 2a ~lt = − Ψ + − x − + il Ψ = Ax2 + Bx + C Ψ, 2 2 2 ∂x γ γ m 2 . 46 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION where A≡ 2a C≡− 2 + γ 2 2a γ2 2a γ2 2 B≡− , 2 ~lt m 2 + 2a γ2 2 4ial γ2 2~lt 4ial 4ial − 2 =− 4 , m γ γ ~lt m 1 − l2 = − 4 (2aγ 2 + l2 ). γ Z ∞ Ψ∗ [Ax2 + Bx + C]Ψ dx = −~2 [Ahx2 i + Bhxi + C] " # 2 ! ~2 1 ~lt ~lt = − 4 4a2 + − 4ial − (2aγ 2 + l2 ) γ 4w2 m m " (" # 2 2 #) 2~at 4ia2 ~t 4ia~t ~at ~2 2 a+a − 2a − + l −1 − =− 4 +4 γ m m m m hp i = −~ 2 −∞ ~2 = − 4 (−aγ 4 − l2 γ 4 ) = ~2 (a + l2 ). γ σx2 = hx2 i − hxi2 = 1 + 4w2 ~lt m 2 − ~lt m 2 = 1 1 ⇒ σx = ; 4w2 2w √ σp2 = hp2 i − hpi2 = ~2 a + ~2 l2 − ~2 l2 = ~2 a, so σp = ~ a. (e) σx and σp are same as before, so the uncertainty principle still holds. Problem 2.43 √ Equation 2.25 ⇒ ψ(x) = A sin kx + B cos kx, 0 ≤ x ≤ a, with k = 2mE/~2 . Even solutions: ψ(x) = ψ(−x) = A sin(−kx) + B cos(−kx) = −A sin kx + B cos kx (−a ≤ x ≤ 0).   ψ continuous at 0 : B = B (no new condition). Boundary ~2 k ψ 0 discontinuous (Eq. 2.128 with sign of α switched): Ak + Ak = 2mα 2 B ⇒ B = mα A. ~ conditions  2 2 ψ → 0 at x = a : A sin(ka) + ~mαk A cos(ka) = 0 ⇒ tan(ka) = − ~mαk . ~2 k ψ(x) = A sin kx + cos kx (0 ≤ x ≤ a); ψ(−x) = ψ(x). mα / tan(ka) 2/ 3/ ka -h2k m_ CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 47 From the graph, the allowed energies are slightly above ka = nπ (n = 1, 3, 5, . . .) so 2 En & n 2 π 2 ~2 (n = 1, 3, 5, . . .). 2m(2a)2 These energies are somewhat higher than the corresponding energies for the infinite square well (Eq. 2.30, with a → 2a). As α → 0, the straight line (−~2 k/mα) gets steeper and steeper, and the intersections get closer to nπ/2; the energies then reduce to those of the ordinary infinite well. As α → ∞, the straight line approaches 2 2 2 π ~ horizontal, and the intersections are at nπ (n = 1, 2, 3, . . .), so En → n2ma – these are the allowed energies for 2 the infinite square well of width a. At this point the barrier is impenetrable, and we have two isolated infinite square wells. Odd solutions: ψ(x) = −ψ(−x) = −A sin(−kx) − B cos(−kx) = A sin(kx) − B cos(kx) (−a ≤ x ≤ 0).   ψ continuous at 0 : B = −B ⇒ B = 0. ψ 0 discontinuous: Ak − Ak = 2mα Boundary conditions ~2 (0) (no new condition).  ψ(a) = 0 ⇒ A sin(ka) = 0 ⇒ ka = nπ 2 (n = 2, 4, 6, . . .). ψ(x) = A sin(kx), (−a < x 0, in contradiction to Equation [F]. Conclusion: ψn (x) must have at least one node between x1 and x2 . Problem 2.46 − ~2 d2 ψ d2 ψ = Eψ (where x is measured around the circumference), or = −k 2 ψ, with k ≡ 2 2m dx dx2 √ 2mE , so ~ ψ(x) = Aeikx + Be−ikx . But ψ(x + L) = ψ(x), since x + L is the same point as x, so Aeikx eikL + Be−ikx e−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 = π : 2k Aeiπ/2 eikL + Be−iπ/2 e−ikL = Aeiπ/2 + Be−iπ/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 = 2nπ (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: ψn+ (x) = Aei(2nπx/L) and RL ψn− (x) = Be−i(2nπx/L) (n = 0 is ok too, but in that case there is just one solution). Normalizing: 0 |ψ± |2 dx = √ 1 ⇒ A = B = 1/ L. Any other solution (with the same energy) is a linear combination of these. 1 2n2 π 2 ~2 ψn± (x) = √ e±i(2nπx/L) ; En = (n = 0, 1, 2, 3, . . .). mL2 L The theorem fails because here ψ does not go to zero at ∞; x is restricted to a finite range, and we are unable to determine the constant K (in Problem 2.44). CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 49 Problem 2.47 (a) (i) b = 0 ⇒ ordinary finite square well. Exponential decay outside; sinusoidal inside (cos for ψ1 , sin for ψ2 ). No nodes for ψ1 , one node for ψ2 . s1 s2 -a -a a a x x (ii) Ground state is even. Exponential decay outside, sinusoidal inside the wells, hyperbolic cosine in barrier. First excited state is odd – hyperbolic sine in barrier. No nodes for ψ1 , one node for ψ2 . s1 s2 -(b/2+a) -b/2 -(b/2+a) -b/2 b/2 b/2+a x b/2 b/2+a x (iii) For b a, same as (ii), but wave function very small in barrier region. Essentially two isolated finite square wells; ψ1 and ψ2 are degenerate (in energy); they are even and odd linear combinations of the ground states of the two separate wells. s1 s 2 b/2 -b/2 -(b/2+a) b/2+a x -(b/2+a) -b/2 (b) From Eq. 2.160 we know that for b = 0 the energies fall slightly below 2 2 π ~ h E1 + V0 ≈ 2m(2a) 2 = 4 4π 2 ~2 E2 + V0 ≈ 2m(2a) 2 = h ) where h ≡ π 2 ~2 . 2ma2 For b a, the width of each (isolated) well is a, so E1 + V0 ≈ E2 + V0 ≈ π 2 ~2 = h (again, slightly below this). 2ma2 b/2 b/2+a x 50 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION E+V0 h E 2 +V0 E 1+V0 h/4 b 2 [Within each well, ddxψ2 = − 2m ~2 (V0 + E)ψ, so the more curved the wave function, the higher the energy.] (c) In the (even) ground state the energy is lowest in configuration (i), with b → 0, so the electron tends to draw the nuclei together, promoting bonding of the atoms. In the (odd) first excited state, by contrast, the electron drives the nuclei apart. Problem 2.48 (a) Let V0 ≡ 32~2 /ma2 . This is just like the odd bound states for the finite square well, since they are the ones that go to zero at the origin. Referring to the solution to Problem 2.29, the wave function is ( p D sin lx, l ≡ 2m(E + V0 )/~ (0 < x a), and the boundary conditions at x = a yield − cot z = with p (z0 /z)2 − 1 √ p 2m(32~2 /ma2 ) 2mV0 a= a = 8. ~ ~ Referring to the figure (Problem 2.29), and noting that (5/2)π = 7.85 < z0 < 3π = 9.42, we see that there are three bound states. z0 = (b) Let a x 1 1 2 a = |D| I1 ≡ |ψ| dx = |D| sin lx dx = |D| − sin lx cos lx − sin la cos la ; 2 2l 2 2l 0 0 0 −2κx ∞ Z ∞ Z ∞ e e−2κa I2 ≡ |ψ|2 dx = |F |2 e−2κx dx = |F |2 − = |F |2 . 2κ 2κ a a a Z a 2 2 Z a 2 2 2 But continuity at x = a ⇒ F e−κa = D sin la, so I2 = |D|2 sin2κla . Normalizing: h i a 1 sin2 la 1 κ 1 = I1 + I2 = |D| − sin la cos la + = |D|2 κa − sin la cos la + sin2 la 2 2l 2κ 2κ l But (referring again to Problem 2.29) κ/l = − cot la, so 1 (1 + κa) = |D|2 κa + cot la sin la cos la + sin2 la = |D|2 . 2κ 2κ 2 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 51 So |D|2 = 2κ/(1 + κa), and the probability of finding the particle outside the well is P = I2 = 2κ sin2 la sin2 la = . 1 + κa 2κ 1 + κa We can express this in terms of z ≡ la and z0 : κa = sin2 la = sin2 z = p z02 − z 2 (page 80), 1 1 = = 1 + (z0 /z)2 − 1 1 + cot2 z z z0 2 ⇒P = z02 (1 + z2 p . z02 − z 2 ) So far, this ispcorrect for any bound state. In the present case z0 = 8 and z is the third solution to − cot z = (8/z)2 − 1, which occurs somewhere in the interval 7.85 < z 0 v, if x − vt 0 −1, if z vt; −1, if x < vt} = 2θ(x − vt) − 1. ∂x imv mα = − 2 [2θ(x − vt) − 1] + Ψ. ~ ~ ∂2Ψ = ∂x2 − mα imv [2θ(x − vt) − 1] + ~2 ~ 2 Ψ− 2mα ∂ θ(x − vt) Ψ. ~2 ∂x ∂ But (from Problem 2.23(b)) ∂x θ(x − vt) = δ(x − vt), so − ~2 ∂ 2 Ψ − αδ(x − vt)Ψ 2m ∂x2 ! 2 imv mα ~2 + αδ(x − vt) − αδ(x − vt) Ψ = − − 2 [2θ(x − vt) − 1] + 2m ~ ~ ~2 m 2 α 2 m2 v 2 mv mα 2 =− [2θ(x − vt) − 1] − − 2i [2θ(x − vt) − 1] Ψ {z } 2m ~4 | ~2 ~ ~2 1 2 mα 1 mvα ∂Ψ = − 2 + mv 2 + i [2θ(x − vt) − 1] Ψ = i~ (compare [F]). X 2~ 2 ~ ∂t (b) mα −2mα|y|/~2 e (y ≡ x − vt). ~2 Z 2mα ~2 mα ∞ −2mαy/~2 Check normalization: 2 2 e dy = 2 = 1. ~ ~ 2mα 0 |Ψ|2 = Z ∞ X ∂Ψ , which we calculated above [F]. ∂t −∞ Z imαv 1 1 2 = [2θ(y) − 1] + E + mv |Ψ|2 dy = E + mv 2 . ~ 2 2 hHi = Ψ∗ HΨdx. But HΨ = i~ (Note that [2θ(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 ( 21 mv 2 ). Problem 2.51 2a π 1/4 p 1 −a(x+ 1 gt2 )2 /γ 2 2 e ; γ = 1 + 2ia~t/m. γ ∂Ψ ∂Ψ0 img 1 2 mgt 1 2 = + Ψ0 − x + gt exp −i x + gt , ∂t ∂t ~ 2 ~ 6 Ψ0 = 54 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION " 1/4 ( 2 #) 2 2 2 1 1 2 1 2a 1 2 2 dγ 1 dγ + − 2 x + gt gt − a x + gt − 3 e−a(x+ 2 gt ) /γ − 2 γ dt γ γ 2 2 γ dt # " 2 1 1 2agt 2a 1 dγ dγ x + gt2 + 3 x + gt2 Ψ0 . − 2 = − γ dt γ 2 γ 2 dt ∂Ψ0 = ∂t But 2a π dγ 1 2ia~ ia~ = = , so dt 2γ m γm " 2 # 1 2 ∂Ψ0 −ia~ 2agt 2ia2 ~ 1 2 x + gt + 4 = − 2 x + gt Ψ0 , and hence ∂t γ2m γ 2 γ m 2 " 2 # 1 2 2agt 1 2 img 1 2 ∂Ψ ia~ 2ia2 ~ x + gt + 4 = − 2 − 2 x + gt − x + gt Ψ. ∂t γ m γ 2 γ m 2 ~ 2 [F] Meanwhile ∂Ψ0 imgt 1 2 imgt ∂Ψ = − Ψ0 exp − x + gt ∂x ∂x ~ ~ 6 ∂Ψ0 1 2 = −2a x + gt /γ 2 Ψ0 , so ∂x 2 ∂Ψ 1 2 imgt 2 = −2a x + gt /γ − Ψ, and hence ∂x 2 ~ ( 2 ) ∂2Ψ 2a 1 2 imgt ∂Ψ 1 2 imgt 2a 2 2 = − 2 Ψ + −2a x + gt /γ − = − 2 + −2a x + gt /γ − Ψ. ∂x2 γ 2 ~ ∂x γ 2 ~ ~2 ∂ 2 Ψ +VΨ= − 2m ∂x2 ( ) 2 ~2 1 2 imgt ~2 a 2 − −2a x + gt /γ − + mgx Ψ. mγ 2 2m 2 ~ [F F] So the time-dependent Schrödinger equation is satisfied if i~ times the square bracket in Equation [F] is equal to the curly bracket in Equation [F F]: a~2 2ia~gt − 2 γ m γ2 1 x + gt2 2 2 1 2 1 2 x + gt + mg x + 2 gt 2 2 2 ~2 1 2 imgt ? ~ a 2 = − −2a x + gt /γ − + mgx. mγ 2 2m 2 ~ 2a2 ~2 − 4 γ m I have cancelled the first terms on either side, and also the mgx terms. This leaves 2ia~gt − γ2 1 x + gt2 2 2 1 2 mg 2 t2 x + gt + 2 2 " # 2 2 2 ~ 4a 4iamgt m2 g 2 t2 1 2 1 2 ? =− x + gt + x + gt − . 2m γ 4 2 ~γ 2 2 ~2 2a2 ~2 − 4 γ m The terms quadratic in x + 12 gt2 cancel, as do the linear terms, and so do those of zeroth order. This confirms that Ψ satisfies the Schrödinger equation. To calculate the expectation value of x, first note that CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION r 2 2 |Ψ| = Ψ0 | = r hxi = r = 2a 1 −(x+ 12 gt2 )2 γ12 + (γ ∗1 )2 e . But π |γ|2 2a 1 π |γ|2 Z ∞ 2a 1 π |γ|2 r 2 xe −2a(x+ 21 gt2 ) /|γ|4 dx = −∞ 1 − gt2 2 Z ∞ 2 55 1 1 (γ ∗ ) + γ 2 2 + = = , so (letting y ≡ x+ 12 gt2 ) 2 ∗ 2 4 γ (γ ) |γ| |γ|4 2a 1 π |γ|2 Z ∞ −∞ 1 2 −2ay2 /|γ|4 y − gt e dy 2 4 e−2ay /|γ| dy. −∞ 1 π/2a |γ|2 , so hxi = − gt2 . This is precisely the classical motion under free fall—as we should 2 have anticipated from Ehrenfest’s theorem. The integral is p Problem 2.52 V(x) (a) x (b) dψ0 = −Aa sech(ax) tanh(ax); dx d2 ψ0 = −Aa2 − sech(ax) tanh2 (ax) + sech(ax) sech2 (ax) . 2 dx ~2 d 2 ψ 0 ~2 a2 − sech2 (ax)ψ0 2 2m dx m ~2 a2 ~2 = Aa2 − sech(ax) tanh2 (ax) + sech3 (ax) − A sech3 (ax) 2m m ~2 a2 A = − sech(ax) tanh2 (ax) + sech3 (ax) − 2 sech3 (ax) 2m ~2 a2 =− A sech(ax) tanh2 (ax) + sech2 (ax) . 2m sinh2 θ 1 sinh2 θ + 1 But (tanh2 θ + sech2 θ) = + = = 1, so 2 2 cosh θ cosh θ cosh2 θ Hψ0 = − =− 1 = |A| 2 ~2 a2 ψ0 . QED 2m Z ∞ Evidently E = − ~2 a2 . 2m ∞ 2 = |A|2 =⇒ A = sech (ax)dx = |A| tanh(ax) a a −∞ −∞ 2 2 1 r a . 2 56 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION s(x) x (c) dψk A = (ik − a tanh ax)ik − a2 sech2 ax eikx . dx ik + a A d2 ψk = ik (ik − a tanh ax)ik − a2 sech2 ax − a2 ik sech2 ax + 2a3 sech2 ax tanh ax eikx . dx2 ik + a 2 ~2 a2 A −~ ik 2 ~2 d2 ψk + V ψ = −k − iak tanh ax − a2 sech2 ax + ik sech2 ax − k 2 2m dx ik + a 2m 2m ~2 a3 ~2 a2 − sech2 ax tanh ax − sech2 ax(ik − a tanh ax) eikx m m = Aeikx ~2 ik 3 − ak 2 tanh ax + ia2 k sech2 ax + ia2 k sech2 ax ik + a 2m −2a3 sech2 ax tanh ax − 2ia2 k sech2 ax + 2a3 sech2 ax tanh ax ~2 k 2 Aeikx ~2 2 k (ik − a tanh ax) = ψk = Eψk . QED ik + a 2m 2m ik − a ikx As x → +∞, tanh ax → +1, so ψk (x) → A e , which represents a transmitted wave. ik + a = R = 0. ik − a T = ik + a 2 = −ik − a −ik + a ik − a ik + a = 1. Problem 2.53 (a) (1) From Eq. 2.136: F + G = A + B. (2) From Eq. 2.138: F − G = (1 + 2iβ)A − (1 − 2iβ)B, where β = mα/~2 k. 1 Subtract: 2G = −2iβA + 2(1 − iβ)B ⇒ B = (iβA + G). Multiply (1) by (1 − 2iβ) and add: 1 − iβ 1 1 iβ 1 2(1 − iβ)F − 2iβG = 2A ⇒ F = (A + iβG). S = . 1 − iβ 1 − iβ 1 iβ (b) For an even potential, V (−x) = V (x), scattering from the right is the same as scattering from the left, with x ↔ −x, A ↔ G, B ↔ F (see Fig. 2.21): F = S11 G + S12 A, B = S21 G + S22 A. So S11 = S22 , S21 = S12 . (Note that the delta-well S matrix in (a) has this property.) In the case of the finite square well, Eqs. 2.170 and 2.171 give S21 = 2 e−2ika 2 2 +l ) sin 2la cos 2la − i (k 2kl ; S11 = 2 −k ) i (l 2kl sin 2la e−2ika 2 2 +l ) sin 2la cos 2la − i (k 2kl . So CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION S= e−2ika 2 2 +l ) sin 2la cos 2la − i (k 2kl 57 ! 2 −k2 ) i (l 2kl sin 2la 1 . 2 −k2 ) 1 i (l 2kl sin 2la Problem 2.54 (a) B = S11 A + S12 G ⇒ G = 1 S11 1 (B − S11 A) = M21 A + M22 B ⇒ M21 = − , M22 = . S12 S12 S12 S22 (S11 S22 − S12 S21 ) S22 (B − S11 A) = − A+ B = M11 A + M12 B. S12 S12 S12 1 det S S22 − det(S) S22 ⇒ M11 = − , M12 = . M= . Conversely: −S11 1 S12 S12 S12 F = S21 A + S22 G = S21 A + G = M21 A + M22 B ⇒ B = 1 M21 1 (G − M21 A) = S11 A + S12 G ⇒ S11 = − ; S12 = . M22 M22 M22 (M11 M22 − M12 M21 ) M12 M12 (G − M21 A) = A+ G = S21 A + S22 G. M22 M22 M22 1 det M M12 −M21 1 ⇒ S21 = ; S22 = . S= . M22 M22 M22 det(M) M12 F = M11 A + M12 B = M11 A + [It happens that the time-reversal invariance of the Schrödinger equation, plus conservation of probability, ∗ ∗ requires M22 = M11 , M21 = M12 , and det(M) = 1, but I won’t use this here. See Merzbacher’s Quantum Mechanics. Similarly, for even potentials S11 = S22 , S12 = S21 (Problem 2.53).] Rl = |S11 |2 = M21 M22 2 , Tl = |S21 |2 = det(M) M22 2 , Rr = |S22 |2 = M12 M22 2 , Tr = |S12 |2 = 1 . |M22 |2 (b) A C F B D G M2 M1 x F C C A F A A = M2 , = M1 , so = M2 M1 =M , with M = M2 M1 . QED G D D B G B B (c) ψ(x) = Aeikx + Be−ikx (x a) . Continuity of ψ : Aeika + Be−ika = F eika + Ge−ika 2mα 0 Discontinuity of ψ : ik F eika − Ge−ika − ik Aeika − Be−ika = − 2mα Aeika + Be−ika . ~2 ψ(a) = − ~2 58 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION (1) F e2ika + G = Ae2ika + B. 2ika (2) F e2ika − G = Ae2ika − B + i 2mα +B . ~2 k Ae Add (1) and (2): 2F e2ika = 2Ae2ika + i mα 2mα mα 2ika A + i 2 e−2ika B = M11 A + M12 B. Ae + B ⇒ F = 1 + i ~2 k ~2 k ~ k So M11 = (1 + iβ); M12 = iβe−2ika ; β ≡ mα . ~2 k Subtract (2) from (1): 2G = 2B − 2iβe2ika A − 2iβB ⇒ G = (1 − iβ)B − iβe2ika A = M21 A + M22 B. 2ika So M21 = −iβe (1 + iβ) iβe−2ika . −iβe2ika (1 − iβ) ; M22 = (1 − iβ). M= (d) M2 = (1 + iβ) iβe−2ika (1 + iβ) iβe2ika ; to get M , just switch the sign of a: M = . 1 1 −iβe2ika (1 − iβ) −iβe−2ika (1 − iβ) M = M2 M1 = [1 + 2iβ + β 2 (e−4ika − 1)] 2iβ[cos 2ka − β sin 2ka] . −2iβ[cos 2ka − β sin 2ka] [1 − 2iβ + β 2 (e4ika − 1)] T = Tl = Tr = 1 ⇒ |M22 |2 T −1 = [1 + 2iβ + β 2 (e−4ika − 1)][1 − 2iβ + β 2 (e4ika − 1)] = 1 − 2iβ + β 2 e4ika − β 2 + 2iβ + 4β 2 + 2iβ 3 e4ika − 2iβ 3 + β 2 e−4ika − β 2 − 2iβ 3 e−4ika + 2iβ 3 + β 4 (1 − e−4ika − e4ika + 1) = 1 + 2β 2 + β 2 (e−4ika + e4ika ) − 2iβ 3 (e−4ika − e4ika ) + 2β 4 − β 4 (e−4ika + e4ika ) = 1 + 2β 2 + 2β 2 cos 4ka + 2iβ 3 2i sin 4ka + 2β 4 − 2β 4 cos 4ka = 1 + 2β 2 (1 + cos 4ka) − 4β 3 sin 4ka + 2β 4 (1 − cos 4ka) = 1 + 4β 2 cos2 2ka − 8β 3 sin 2ka cos 2ka − 4β 4 sin2 2ka T = 1 1 + 4β 2 (cos 2ka − β sin 2ka)2 Problem 2.55 I’ll just show the first two graphs, and the last two. Evidently K lies between 0.9999 and 1.0001 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 59 Problem 2.56 The correct values (in Eq. 2.73) are K = 2n + 1 (corresponding to En = (n + 12 )~ω). I’ll start by “guessing” 2.9, 4.9, and 6.9, and tweaking the number until I’ve got 5 reliable significant digits. The results (see below) are 3.0000, 5.0000, 7.0000. (The actual energies are these numbers multiplied by 12 ~ω.) 60 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 61 Problem 2.57 2 ~ ψ 00 = Eψ, or, with the correct energies (Eq. 2.30) and a = 1, ψ 00 + (nπ)2 ψ = The Schrödinger equation says − 2m 0. I’ll start with a “guess” using 9 in place of π 2 (that is, I’ll use 9 for the ground state, 36 for the first excited state, 81 for the next, and finally 144). Then I’ll tweak the parameter until the graph crosses the axis right at x = 1. The results (see below) are, to five significant digits: 9.8696, 39.478, 88.826, 157.91. (The actual energies are these numbers multiplied by ~2 /2ma2 .) 62 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.58 (a) The total energy is simply N times the ground state energy of the infinite square well: Ea = N π 2 ~2 . 2ma2 (b) Filling the lowest N energy levels of the infinite square well (with width N a) gives Eb = X N n=1 N n2 π 2 ~2 π 2 ~2 X 2 = n . 2m(N A)2 2mN 2 a2 n=1 The sum is N (N + 1)(2N + 1)/6, so Eb = N 1 1 + + 3 2 6N π 2 ~2 . 2ma2 π 2 ~2 π 2 ~2 = . 2ma2 3ma2 (c) Ea − Eb ∆E = ≈ N N N − (N/3) N CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 63 (d) The binding energy of hydrogen (13.6 eV) is ~2 /2ma2B , where aB = 0.529 Å is the Bohr radius, so 2 ∆E 2 2 aB 2 2 0.529 π = π Ebinding = (13.6) eV = 1.6 eV. N 3 a 3 4 Problem 2.59 (a) − ~2 d2 ψ + mgx ψ = E ψ 2m dx2 dψ dψ dz dψ = =a ; dx dz dx dz − (x ≥ 0; ψ = 0 for x ≤ 0). 2 d2 ψ dz d2 ψ 2d ψ = a . = a dx2 dz 2 dx dz 2 √ z z ~2 2 d2 ψ ~2 2 √ 00 z√ 2m 2m + mg a ψ = Eψ ⇒ − a a y + mg a y = E a y ⇒ −y 00 + 2 2 mg y = 2 2 E y. 2m dz 2 a 2m a ~ a a ~ a Let 2m 1 mg = 1, or a ≡ ~2 a2 a 2m2 g ~2 1/3 and ≡ 2m 2m E= 2 ~2 a2 ~ ~2 2m2 g 2/3 E= Then −y 00 + zy = y. X DSolve@- y ”@xD + x y@xD ä s y@xD, y@xD, xD Out[1]= 88y@xD Æ AiryAi@- s + xD C@1D + AiryBi@- s + xD C@2D<< In[2]:= Plot@AiryAi@xD, 8x, – 15, 5<D In[1]:= 0.4 0.2 Out[2]= -15 -10 -5 5 -0.2 -0.4 In[3]:= Plot@AiryBi@xD, 8x, – 15, 5<D 5 4 3 Out[3]= 2 1 -15 -10 -5 FindRoot@AiryAi@xD == 0, 8x, – 2<D Out[4]= 8x Æ – 2.33811< In[5]:= FindRoot@AiryAi@xD == 0, 8x, – 12.8<D In[4]:= 5 Out[5]= 8x Æ – 12.8288< In[6]:= NIntegrate@HAiryAi@xDL ^ 2, 8x, – 2.338107410459767`, •<D Out[6]= 0.491697 In[7]:= NIntegrate@HAiryAi@xDL ^ 2, 8x, – 12.828776752865757`, •<D Out[7]= 1.14018 2 m~2 g 2 1/3 E. 64 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION FindRoot@AiryAi@xD == 0, 8x, – 2<D Out[12]= 8x Æ – 2.33811< In[13]:= FindRoot@AiryAi@xD == 0, 8x, – 12.8<D In[12]:= Out[13]= 8x Æ – 12.8288< In[14]:= NIntegrate@HAiryAi@xDL ^ 2, 8x, – 2.338107410459767`, •<D Out[14]= 0.491697 In[15]:= NIntegrate@HAiryAi@xDL ^ 2, 8x, – 12.828776752865757`, •<D Out[15]= 1.14018 In[16]:= Pone@x_D := H0.4916966179009774`L ^ H- 1 ê 2L * AiryAi@x – 2.338107410459767`D In[17]:= Plot@Pone@xD, 8x, 0, 16<D 0.7 0.6 0.5 Out[17]= 0.4 0.3 0.2 0.1 5 10 15 10 15 In[18]:= Pten@x_D := H1.1401837256117164`L ^ H- 1 ê 2L * AiryAi@x – 12.828776752865757`D In[19]:= Plot@Pten@xD, 8x, 0, 16<D 0.4 0.2 Out[19]= 5 -0.2 -0.4 0.1 5 10 15 Pten@x_D := H1.1401837256117164`L ^ H- 1 ê 2L * AiryAi@x – 12.828776752865757`D Plot@Pten@xD, 8x, 0, 16<D 0.4 2 Problem 2.59 copy.nb CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 0.2 In[20]:= 65 NIntegrate@Pone@xD * Pten@xD, 8x, 0, •<D 5 : NIntegrate::ncvb NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near 8x< = 81.72352<. NIntegrate 15 obtained -9.19403 ¥ 10-17 and 1.5861121045064646`*^-16 for the integral and error estimates. á -0.2 Out[20]= 10 – 9.19403 ¥ 10-17 -0.4 (b) NIntegrate@Pone@xD * Pten@xD, 8x, 0, • 0.5~ X. NIntegrate::ncvb : x 1 p x p NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near 8x< = 81.72352 0.5~ X. 10-17 and 1.5861121045064646`*^-16 forxthep integral and error estimates. á – 9.19403 ¥ 10-17 (See print-out.) NIntegrate@x HPone@xDL ^ 2, 8x, 0, •<D 1.55874 NIntegrate@x ^ 2 HPone@xDL ^ 2, 8x, 0, •<D 2.9156 2.9155980068599967` – H1.558738273638599`L ^ 2 Problem 2.59 copy.nb 0.697089 NIntegrate@x HPten@xDL ^ 2, 8x, 0, •<D 8.55252 NIntegrate@x ^ 2 HPten@xDL ^ 2, 8x, 0, •<D 87.7747 87.77467357595424` – H8.552517834822023`L ^ 2 3.8248 NIntegrate@- ‰ Pone@xD HPone '@xDL, 8x, 0, •<D NIntegrate::ncvb : NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near 8x< = 81.2858<. NIntegrate obtained 0. + 6.07153 ¥ 10-18 ‰ and 7.766131095614155`*^-17 for the integral and error estimates. á 0. + 6.07153 ¥ 10-18 ‰ NIntegrate@- Pone@xD HPone ''@xDL, 8x, 0, •<D 0.779369 NIntegrate@- Pten@xD HPten ''@xDL, 8x, 0, •<D 4.27626 0.7793691368188985` 0.882819 4.276258918045443` 2.06791 0.6970889478066314` * 0.8828188584409027` 0.615403 3.8248022512288737` * 2.067911728784728` 7.90935 3 66 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 1 (Equation 1.43). Here T v(x) r 2 1 2 E = mv + mgx ⇒ v = (E − mgx) and 2 m (c) ρC (x) = 1 2 E gT = h = ⇒T = 2 mg s 2E , mg 2 so mg 1 mg (2m2 g/~2 a2 ) 1 a 1 ρC (x) = p = p = q = p → p . 2 g/~2 a2 ) 2 2 ~2 a2 ~2 a2 2 E(E − mgx) ( − (2m ( − a) 2 ( − 1) 2 − mgx 4 2m Problem 2.59 copy.nb 2m For ψ10 , = 12.82877 (Out[13] on page 64). The graphs are Plot@H4 * 12.828776752865757` H12.828776752865757` – xLL ^ H- 1 ê 2L, 8x, 0, 12.5<, PlotRange Æ 80, .26<D 0.25 0.20 0.15 0.10 0.05 2 4 6 8 10 12 2 4 6 8 10 12 4 6 8 10 12 Plot@HPten@xDL ^ 2, 8x, 0, 12.5<, PlotRange Æ 80, .26<D 0.25 0.20 0.15 0.10 0.05 Show@%74, %75D 0.25 0.20 0.15 0.10 0.05 2 Comment: Well, they agree, in a kind of averaged sense. CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 67 Problem 2.60 (a) q n kg m2 kg m4 kg m2 p n+p+q 2 n+2q −(n+2q) E0 = (~) (m) (α) = (kg) = (kg) (m ) (s) = . s s2 s2 So n + p + q = 1, n + 2q = 1, n + 2q = 2. The last two are incompatible. Evidently there is, on purely dimensional grounds, no possible formula for E0 . n p q (b) Let ψλ (x) ≡ ψ(y), where y ≡ λx. Then 2 2 d2 ψλ (x) d2 ψ(y) 2 dy β β 2 d ψ(y) 2 2 = =λ = λ − 2 ψ(y) + κ ψ(y) = −λ2 2 2 ψλ (x) + λ2 κ2 ψλ (x), y dx2 d dx dy 2 y λ x β d2 ψλ (x) + 2 ψλ (x) = (λκ)2 ψλ (x) = (κ0 )2 ψλ (x). So ψλ (x) is a solution to Equation 2.190, with κ0 ≡ λκ. dx2 x −2mE −2mE 0 0 2 2 2 2 = (κ ) = λ κ = λ ⇒ E 0 = λ2 E. The corresponding energy is given by ~2 ~2 (c) or f@x_D := A b – H1 ê 4L , k xF x BesselKB‰ FullSimplifyBf ''@xD + b x^2 f@xD – k ^ 2 f@xD ä 0F True h@x_D := x BesselK@4 ‰, xD Plot@h@xD, 8x, 0, 5<D 0.004 0.003 0.002 0.001 1 2 3 4 5 0.0006 0.0008 0.0010 -0.001 -0.002 Plot@h@xD, 8x, 0, .001 0 && – 2 < ImA – 1 + 4 b E < 2F 68 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION s So A=κ 2 sinh(πg) . πg (d) From the first plot on page 67 we see that the highest zero crossing occurs at κx ≈ 1.7; to find the exact value, use FindRoot (see print-out below). We want this to occur at x = = 1, so κ = 1.69541. h@x_D := x BesselK@4 ‰, xD FindRoot@h@xD ä 0, 8x, 1.7<D 8x Æ 1.69541 + 0. ‰= 1, [email protected]` xD, 0D Plot@j@xD, 8x, 0, 6<D 0.004 0.003 0.002 0.001 1 2 3 4 5 6 The parameter β ≡ 2mα/~2 is dimensionless, so we may as well eliminate ~ (in favor of β, m, and α) in 4 q 2 (m)r = (kg)p+q (m)4q+r (s)−2q = kgsm , so the dimensional analysis. This leaves E0 = mp αq r = (kg)p kgsm 2 2 p + q = 1, 4q + r = 2, q = 1, ⇒ p = 0, r = −2. On dimensional grounds, therefore, the expression for E0 has to be of the form E0 = −α/2 times some function of β. As → 0 (for fixed values of m and α), E0 → −∞, indicating once again that there is no ground state for this system. Problem 2.61 (a) From Equation 2.197: N = 1 : j = 1 : −λψ2 + (2λ)ψ1 − λψ0 = Eψ1 ( j = 1 : −λψ2 + (2λ)ψ1 − λψ0 = Eψ1 N =2: j = 2 : −λψ3 + (2λ)ψ2 − λψ1 = Eψ2   j = 1 : −λψ2 + (2λ)ψ1 − λψ0 = Eψ1 N = 3 : j = 2 : −λψ3 + (2λ)ψ2 − λψ1 = Eψ2   j = 3 : −λψ4 + (2λ)ψ3 − λψ2 = Eψ2 (b) Denote the eigenvalues by Ẽn : ⇒ H = (2λ). ⇒ H= 2λ −λ . −λ 2λ  ⇒  2λ −λ 0 H = −λ 2λ −λ . 0 −λ 2λ CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 69 8~2 n 2 π 2 ~2 π 2 ~2 2~2 = . The exact energies are E = , so E = ; the agreement n 1 2m(a/2)2 2ma2 2ma2 2ma2 2 is not too bad: 8 ≈ π = 9.87. N = 1 : Ẽ1 = 2λ = N =2: 2λ − Ẽ −λ det −λ 2λ − Ẽ = 0 ⇒ (2λ − Ẽ)2 − λ2 = 0 ⇒ 2λ − Ẽ = ±λ. 9~2 ~2 = . This is better: 9 is closer to π 2 than 8 was. 2m(a/3)2 2ma2 3~2 27~2 Ẽ2 = 2λ + λ = 3λ = = . The exact answer has 4π 2 = 39.5 instead of 27. 2m(a/3)2 2ma2 Ẽ1 = 2λ − λ = λ = N =3:   2λ − Ẽ −λ 0 √ det  −λ 2λ − Ẽ −λ  = 0 ⇒ (2λ − Ẽ)3 − 2λ2 (2λ − Ẽ) = 0 ⇒ 2λ − Ẽ = 0 or 2λ − Ẽ = ± 2 λ. 0 −λ 2λ − Ẽ √ √ √ √ √ 16(2 − 2)~2 (2 − 2)~2 = . This is better yet: 16(2 − 2) = 9.37. Ẽ1 = 2λ − 2 λ = λ(2 − 2) = 2m(a/4)2 2ma2 2~2 32~2 = . Improving: the exact answer is 4π 2 = 39.5 instead of 32. Ẽ2 = 2λ = 2m(a/4)2 2ma2 √ √ √ √ √ 16(2 + 2)~2 (2 + 2)~2 Ẽ3 = 2λ + 2 λ = λ(2 + 2) = = ; 16(2 + 2) = 54.6 ≈ 9π 2 = 88.8. 2m(a/4)2 2ma2 (c) h = Table@If@i == j, 2 l, 0D, 8i, 10<, 8j, 10<D k = Table@If@i ä j + 1, – l, 0D, 8i, 10<, 8j, 10<D m = Table@If@i ä j – 1, – l, 0D, 8i, 10<, 8j, 10<D p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 10<, 8j, 10<D P = MatrixForm@%D 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l -l 0 0 0 0 0 0 0 0 -l 2 l l=1 1 EIG = Eigenvalues@N@pDD 83.91899, 3.68251, 3.30972, 2.83083, 2.28463, 1.71537, 1.16917, 0.690279, 0.317493, 0.0810141< 70 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 2 Ten.nb ~2 = To get the eigenvalues, multiply by λ = 2m(a/11)2 In[16]:= Out[16]= 121 * EIG ê Hp ^ 2L 121 π2 E1 : 848.0462, 45.147, 40.5767, 34.7056, 28.0092, 21.0302, 14.3339, 8.46272, 3.89242, 0.993221< EVE Thus theIn[17]:= lowest five= Eigenvectors@N@pDD eigenvalues, in units of E1 , are 0.9932, 3.8924, 8.4627, 14.3339, 21.0302, as compared to the exactOut[17]= values 1, 4, 9, 16, 25. -Doing the same for N =-100: 880.120131, 0.23053, 0.322253, 0.387868, 0.422061, – 0.422061, – 0.120131<, 8- 0.23053, 0.387868, – 0.422061, h = 0.387868, Table@If@i- 0.322253, == j, 2 l, 0.23053, 0D, 8i, 100<, 8j, 100<D 0.322253, – 0.120131, – 0.120131, 0.322253, – 0.422061, 0.387868, – 0.23053<, 8- 0.322253, 0.422061, – 0.23053, – 0.120131, 0.387868, – 0.387868, 0.120131, In[23]:= – 0.422061, – 0.322253, – 0.120131, k =0.23053, Table@If@i ä j + 1, 0.322253<, – l, 0D, 8i,80.387868, 100<, 8j, 100<D 0.422061, – 0.23053, – 0.23053, 0.422061, – 0.120131, – 0.322253, 0.387868<, In[24]:= m = Table@If@i ä j – 1, – l, 0D, 8i, 100<, 8j, 100<D 80.422061, – 0.120131, – 0.387868, 0.23053, 0.322253, – 0.322253, – 0.23053, 0.387868, 0.120131, – 0.422061<, 80.422061, 0.120131, – 0.387868, In[25]:= p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 100<, 8j, 100<D – 0.23053, 0.322253, 0.322253, – 0.23053, – 0.387868, 0.120131, 0.422061<, 80.387868, – 0.322253, 0.120131, 0.422061, 0.23053, – 0.23053, – 0.422061, In[26]:= l = 1 – 0.120131, 0.322253, 0.387868<, 80.322253, 0.422061, 0.23053, – 0.120131, Out[26]= 1 – 0.387868, – 0.387868, – 0.120131, 0.23053, 0.422061, 0.322253<, 8- 0.23053, – 0.387868, – 0.422061, – 0.322253, – 0.120131, 0.120131, = Eigenvalues@N@pDD In[27]:= EIG0.322253, 0.422061, 0.387868, 0.23053<, 80.120131, 0.23053, 0.322253, 0.387868, 0.422061, 0.422061, 0.387868, 0.322253, 0.23053, 0.120131<< In[28]:= 10 201 * EIG ê Hp ^ 2L In[22]:= ListLinePlot@EVE@@1DD, PlotRange Æ 80.5, 0.5<D 84133.31, 4130.31, 4125.32, 4118.33, 4109.36, 4098.41, 4085.5, 4070.64, 4053.84, 4035.11, 4014.49, 3991.97, 3967.6, 3941.39, 3913.36, 3883.55, 3851.98, 3818.69, 3783.7, 3747.04, 3708.77, 3668.9, 3627.49, 3584.57, 3540.18, 3494.36, 0.4 3447.16, 3398.63, 3348.81, 3297.75, 3245.5, 3192.11, 3137.63, 3082.12, 3025.62, 2968.2, 2909.9, 2850.79, 2790.92, 2730.35, 2669.14, 2607.35, 2545.03, 0.2 2482.25, 2419.07, 2355.55, 2291.76, 2227.74, 2163.57, 2099.3, 2035.01, 1970.74, 1906.57, 1842.55, 1778.75, 1715.24, 1652.06, 1589.28, 1526.96, 1465.17, 1403.96, 1343.39, 1283.52, 1224.41, 1166.11, 1108.69, 1052.19, Out[18]= 996.679, 942.201, 888.81, 836.559, 785.499, 735.679, 687.147, 639.95, 594.133, 2 4 6 8 10 549.742, 506.819, 465.405, 425.541, 387.265, 350.614, 315.624, 282.328, 250.76, 220.948, 192.922, 166.71, 142.336, 119.824, 99.1963, 80.4724, -0.2 63.6704, 48.8067, 35.8956, 24.9496, 15.9794, 8.99347, 3.99871, 0.999919< In[18]:= Out[28]= -0.4 This time the lowest five eigenvalues are 0.9999, 3.9987, 8.9934, 15.9794, 24.9496. (d) Always, ψ0 = ψN +1 = 0; might as wellPlotRange set λ = 1 for this 0.5<D part. Æ 80, In[19]:= ListLinePlot@EVE@@10DD, 0.5 N = 1 : 2ψ1 = Ẽψ1 , Ẽ1 = 2. Up to normalization: 0.4 0.3 Out[19]= 0.2 0.1 2 4 6 8 10 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION N =2: 2 −1 −1 2 ψ1 ψ1 = Ẽ ⇒ 2ψ1 − ψ2 = Ẽψ1 and − ψ1 + 2ψ2 = Ẽψ2 . ψ2 ψ2 n = 1 : Ẽ1 = 1 ⇒ 2ψ1 − ψ2 = ψ1 ⇒ ψ2 = ψ1 n = 2 : Ẽ2 = 3 ⇒ 2ψ1 − ψ2 = 3ψ1 ⇒ ψ2 = −ψ1      2 −1 0 ψ1 ψ1 N = 3 : −1 2 −1 ψ2  = Ẽ ψ2  ⇒ 2ψ1 − ψ2 = Ẽψ1 , −ψ1 + 2ψ2 − ψ3 = Ẽψ2 , −ψ2 + 2ψ3 = Ẽψ3 . 0 −1 2 ψ3 ψ3 √ √ √ n = 1 : Ẽ1 = 2 − 2 ⇒ 2ψ1 − ψ2 = (2 − 2)ψ1 ⇒ ψ2 = 2ψ1 ; √ √ −ψ1 + 2ψ2 − ψ3 = (2 − 2)ψ2 ⇒ ψ1 + ψ3 = 2ψ2 = 2ψ1 ⇒ ψ3 = ψ1 n = 2 : Ẽ2 = 2 ⇒ 2ψ1 − ψ2 = 2ψ1 ⇒ ψ2 = 0; −ψ1 + 2ψ2 − ψ3 = 2ψ2 ⇒ ψ3 = −ψ1 √ √ √ n = 3 : Ẽ3 = 2 + 2 ⇒ 2ψ1 − ψ2 = (2 + 2)ψ1 ⇒ ψ2 = − 2ψ1 ; √ √ −ψ1 + 2ψ2 − ψ3 = (2 + 2)ψ2 ⇒ ψ1 + ψ3 = − 2ψ2 = 2ψ1 ⇒ ψ3 = ψ1 71 72 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION For N = 10 we get EVE = Eigenvectors@N@pDD ListLinePlot@EVE@@10DD, PlotRange Ø 80, 0.5<D 0.5 0.4 0.3 0.2 0.1 2 4 6 8 10 ListLinePlot@EVE@@9DD, PlotRange Ø 8- 0.5, 0.5<D 0.4 0.2 2 4 6 8 10 -0.2 -0.4 ListLinePlot@EVE@@8DD, PlotRange Ø 8- 0.5, 0.5 80, 0.2 8- 0.2, 0.2 8- 0.2, 0.2<D 0.2 0.1 20 -0.1 -0.2 40 60 80 100 73 74 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.62 ~2 (N + 1)2 = (N + 1)2 V0 (here N = 100); 2ma2 πx πj∆x πj j Vj = bV0 sin = bV0 sin = bV0 sin a a N +1 Vj πj b vj = sin = . λ (N + 1)2 N +1 λ= 2 (here b = 500); 2p62.nb Factoring out λ, the diagonal elements of H are In[8]:= ListLinePlot@EVE@@98DD, PlotRange Æ 8.1, 0.2<D b 2 + vj = 2 + sin (N + 1)2 0.20 πj N +1 500 =2+ sin 10201 πj 101 . 0.15 In[1]:= Out[8]= In[2]:= In[3]:= h = Table@If@i == j, 2 + H500 ê 10 201L Sin@p j ê 101D, 0D, 8i, 100<, 8j, 100<D 0.10 k = Table@If@i ä j + 1, – 1, 0D, 8i, 100<, 8j, 100<D 0.05 20 40 – 1, 0D, 60 8i, 100<, 80 8j, 100<D 100 m = Table@If@i ä j – 1, p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 100<, 8j, 100<D -0.05 In[4]:= -0.10 In[5]:= In[9]:= EVE = Eigenvectors@N@pDD EIG = Eigenvalues@N@pDD 10 201 * EIG ListLinePlot@EVE@@100DD, PlotRange Æ 80, 0.2<D Out[10]= 841 255., 41 158.1, 41 063.4, 40 968.8, 40 869.4, 40 758.8, 40 632., 40 486.7, 40 322.1, 0.20 40 138.4, 39 935.6, 39 714.1, 39 474., 39 215.7, 38 939.5, 38 645.5, 38 334.2, 38 005.7, 37 660.6, 37 299., 36 921.3, 36 528., 36 119.3, 35 695.8, 35 257.7, 34 805.6, 34 339.9, 33 860.9, 33 369.3, 32 865.4, 32 349.7, 31 822.8, 31 285.1, 0.15 30 737.3, 30 179.7, 29 613., 29 037.7, 28 454.3, 27 863.4, 27 265.6, 26 661.5, 26 051.7, 25 436.7, 24 817.1, 24 193.6, 23 566.7, 22 937., 22 305.2, 21 671.9, 21 037.6, 20 403.1, 19 768.8, 19 135.5, 18 503.7, 17 874.1, 17 247.2, 16 623.6, 16 004., 15 389., 14 779.2, 14 175.1, 13 577.3, 12 986.5, 12 403.1, 11 827.8, Out[6]= 0.10 11 261.1, 10 703.5, 10 155.6, 9617.98, 9091.08, 8575.44, 8071.55, 7579.91, 7100.98, 6635.24, 6183.13, 5745.1, 5321.57, 4912.95, 4519.64, 4142.02, 3780.47, 3435.34, 3106.97, 2795.68, 2501.8, 2225.62, 1967.43, 1727.51, 1506.15, 1303.63, 0.05 1120.24, 956.36, 812.457, 689.191, 590.237, 499.854, 476.163, 304.8, 304.66< In[10]:= In[6]:= So the lowest three energies are 304.66 V0 , 304.8 V0 , and 476.163 V0 . Notice that the ground state is almost 20 40 two separated 60 100 barrier in between them, and the particle can degenerate—essentially we have wells80with a huge be either in the left one or in the right one (or the even and odd linear combinations thereof). In[7]:= ListLinePlot@EVE@@99DD, PlotRange Æ 8- .2, 0.2<D 0.2 0.1 Out[7]= 20 40 60 80 100 In[1]:= In[2]:= h = Table@If@i == j, 2 + H500 ê 10 201L Sin@p j ê 101D, 0D, 8i, 100<, 8j, 100<D k = Table@If@i ä j + 1, – 1, 0D, 8i, 100<, 8j, 100<D In[3]:= m = Table@If@i ä j – 1, – 1, 0D, 8i, 100<, 8j, 100<D In[4]:= p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 100<, 8j, 100<D In[5]:= EVE = Eigenvectors@N@pDD CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION In[6]:= ListLinePlot@EVE@@100DD, PlotRange Æ 80, 0.2<D 0.20 0.15 Out[6]= 0.10 0.05 In[7]:= 20 40 60 80 100 20 40 60 80 100 ListLinePlot@EVE@@99DD, PlotRange Æ 8- .2, 0.2<D 0.2 0.1 Out[7]= -0.1 2 2p62.nb -0.2 In[8]:= ListLinePlot@EVE@@98DD, PlotRange Æ 8- .1, 0.2<D 0.20 0.15 0.10 Out[8]= 0.05 20 40 60 80 100 -0.05 -0.10 Notice that the central barrier pushes the wave function out to the wings. = Eigenvalues@N@pDD In[9]:= EIG In[10]:= Out[10]= 10 201 * EIG 841 255., 41 158.1, 41 063.4, 40 968.8, 40 869.4, 40 758.8, 40 632., 40 486.7, 40 322.1, 40 138.4, 39 935.6, 39 714.1, 39 474., 39 215.7, 38 939.5, 38 645.5, 38 334.2, 38 005.7, 37 660.6, 37 299., 36 921.3, 36 528., 36 119.3, 35 695.8, 35 257.7, 34 805.6, 34 339.9, 33 860.9, 33 369.3, 32 865.4, 32 349.7, 31 822.8, 31 285.1, 30 737.3, 30 179.7, 29 613., 29 037.7, 28 454.3, 27 863.4, 27 265.6, 26 661.5, 26 051.7, 25 436.7, 24 817.1, 24 193.6, 23 566.7, 22 937., 22 305.2, 21 671.9, 21 037.6, 20 403.1, 19 768.8, 19 135.5, 18 503.7, 17 874.1, 17 247.2, 16 623.6, 16 004., 15 389., 14 779.2, 14 175.1, 13 577.3, 12 986.5, 12 403.1, 11 827.8, 11 261.1, 10 703.5, 10 155.6, 9617.98, 9091.08, 8575.44, 8071.55, 7579.91, 7100.98, 6635.24, 6183.13, 5745.1, 5321.57, 4912.95, 4519.64, 4142.02, 3780.47, 75 76 CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION Problem 2.63 X ∂ 1 ∂Z 1 X ln(Z) = − =− (−En )e−βEn = En P (n). X ∂β Z ∂β Z n n 1 (b) Geometric series: 1 + x + x2 + x3 + · · · = . Here x = e−β~ω : 1−x (a) − Z= ∞ X 1 e−β(n+ 2 )~ω = e−β~ω/2 ∞ X e(−β~ω)n = e−β~ω/2 n=0 n=0 1 . X 1 − e−β~ω (c) β~ω ln Z = ln e−β~ω/2) − ln 1 − e−β~ω = − − ln 1 − e−β~ω , 2 ~ω 1 − e−β~ω + 2e−β~ω ~ω 1 + e−β~ω ∂ ~ω ~ωe−β~ω = − ⇒ Ē = . X ln Z = − − −β~ω −β~ω ∂β 2 1−e 2 1−e 2 1 − e−β~ω (d) ∂ Ē ∂ Ē dβ 1 ∂ Ē = =− . ∂T ∂β dT kB T 2 ∂β 1 − e−β~ω −~ωe−β~ω − 1 + e−β~ω ~ωe−β~ω ~ω ∂ Ē ~ω −2~ωe−β~ω (~ω)2 eβ~ω = = = − 2 2. ∂β 2 2 (1 − e−β~ω )2 (1 − e−β~ω ) (eβ~ω − 1) 1 eβ~ω C=3 (~ω)2 2. 2 kB T (eβ~ω − 1) Or, using β = 1/kB T and ~ω = kB θE , C=3 (e) In[12]:= 1 kB T 2 2 (kB θE ) eθE /T 2 = 3kB eθE /T − 1 θE T 2 eθE /T eθE /T − 1 2 . X Plot@3 * Hx ^ H- 2LL * Exp@1 ê xD ê HExp@1 ê xD – 1L ^ 2, 8x, 0, .7<, PlotRange Æ 80, 2.6<D 2.5 2.0 1.5 Out[12]= 1.0 0.5 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Incidentally, comparing the graphs suggests that x = 0.7 corresponds to T = 1000 K, so θE = 1000/0.7 K = 1400 K. Then ~ω = (1400kB )K = (1400)(8.6 × 10−5 ) eV = 0.12 eV. CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION 77 Problem 2.64 (a) Plugging into the differential equation we have ∞ X an n (n − 1) xn−2 − n=2 ∞ X an n (n − 1) xn − 2 n=2 ∞ X an n xn + ` (` + 1) n=1 ∞ X an xn = 0 . n=0 Reindexing the sums so that all the powers of x match we have ∞ X ap+2 (p + 2) (p + 1) xp − p=0 ∞ X ap p (p − 1) xp − 2 p=2 ∞ X ap p xp + ` (` + 1) p=1 ∞ X a p xp = 0 . p=0 We can then combine the sums (extending the second and third sums to begin at p = 0) to get ∞ X {(p + 2) (p + 1) ap+2 − [p (p − 1) + 2 p − ` (` + 1)] ap } xp = 0 . p=0 From this we read off the recursion relation p (p + 1) − ` (` + 1) ap . (p + 2) (p + 1) ap+2 = (b) For large values of p, ap+2 ≈ p ap . p+2 with the (approximate) solution ap ≈ C . p and this gives the behavior f (x) ≈ C X1 p xp ≈ log 1 1−x which diverges at x = 1. (There will be finite corrections coming from the low values of p, but these cannot fix the divergence.) (c) For ` = 0 we need the even function (a1 = 0) and the recursion relation gives a2 = 0 so that P0 (x) = a0 . For ` = 1 we need the odd function (a0 = 0) and the recursion relation gives a3 = 0 so that P1 (x) = a1 x . For ` = 2 we need the even function again and the recursion relation gives a2 = −3 a0 and a4 = 0 so that P2 (x) = a0 1 − 3 x2 . For ` = 3 we need the odd function again and the recursion relation gives a3 = −5/3 a1 and a5 = 0 so that 5 3 P3 (x) = a1 x − x . 3