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246 7 QUANTUM THEORY
−0.4 −0.2 0.2 0.4
−1.0
−0.5
0.5
1.0
x/L
ψ′n′
ψ′1
ψ′2
ψ′3
Figure 7.7
setting this equal to 1 gives N = (2/L)1/2 . Similarly for the sin functions,
again using Integral T.2
N2 ∫
L
0
sin2[2n′πx/L]dx = N2 [x/2 − (L/8n′π) sin(4n′πx/L)∣L/2L/2
= N2(L/2)
setting this equal to 1 gives N = (2/L)1/2 .
(f) �e probability density for all wavefunctions is symmetric about the cen-
ter of the box, which at this case is at x = 0. Hence, the average value of x
must be at the center, and so ⟨x⟩ = 0.
P7D.12 �e text de�nes the transmission probability as the ratio ∣A′∣2/∣A∣2, where the
coe�cients A and A′ are de�ned in [7D.15–268] and [7D.18–269] respectively.
Four equations are given in [7D.19a–269] and [7D.19b–269] for the �ve un-
known coe�cients of the full wavefunction.
(a) A+ B = C + D
(b) CeκW + De−κW = A′eikW
(c) ikA− ikB = κC − κD
(d) κCeκW − κDe−κW = ikA′eikW
�e coe�cient A′ is needed in terms of A alone, and hence B,C and D need to
be eliminated. B occurs only in (a) and (c). Solving these equations and setting
the results equal to each other yields
B = C + D − A = A− (κ/ik)C + (κ/ik)D
Solving this for C,
C = 2A+ D(κ/ik − 1)
κ/ik + 1
= 2Aik + D(κ − ik)
κ + ik