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Hints and answers

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Hints and answers
27.10 HINTS AND ANSWERS
27.27
The Schrödinger equation for a quantum mechanical particle of mass m moving
in a one-dimensional harmonic oscillator potential V (x) = kx2 /2 is
2 d2 ψ
kx2 ψ
+
= Eψ.
2m dx2
2
For physically acceptable solutions, the wavefunction ψ(x)
must be finite at x = 0,
tend to zero as x → ±∞ and be normalised, so that |ψ|2 dx = 1. In practice,
these constraints mean that only certain (quantised) values of E, the energy of
the particle, are allowed. The allowed values fall into two groups: those for which
ψ(0) = 0 and those for which ψ(0) = 0.
Show that if the unit of length is taken as [2 /(mk)]1/4 and the unit of energy
is taken as (k/m)1/2 , then the Schrödinger equation takes the form
−
d2 ψ
+ (2E − y 2 )ψ = 0.
dy 2
Devise an outline computerised scheme, using Runge–Kutta integration, that will
enable you to:
(a) determine the three lowest allowed values of E;
(b) tabulate the normalised wavefunction corresponding to the lowest allowed
energy.
You should consider explicitly:
(i)
(ii)
(iii)
(iv)
the variables to use in the numerical integration;
how starting values near y = 0 are to be chosen;
how the condition on ψ as y → ±∞ is to be implemented;
how the required values of E are to be extracted from the results of the
integration;
(v) how the normalisation is to be carried out.
27.10 Hints and answers
27.1
27.3
27.5
27.7
27.9
27.11
27.13
5.370.
(a) ξ = 0 and f (ξ) = 0 in general; (b) ξ = b, but f (b) = 0 whilst f(b) = 0.
Interchange is formally needed for the first two steps, though in this case no error
will result if it is not carried out; x1 = −12, x2 = 2, x3 = −1, x4 = 5.
The quadratic equation
is z 2 + 4z +√1 = 0; α + β − 3 = x0 = α + β + 3.
√
−2 − 3, β must be zero for i > 0 and α must
With p = −2 + 3 and q = √
√ be
zero for i < 0; xi = 3(−2 + 3)i for i > 0, xi = 0 for i = 0, xi = −3(−2 − 3)i
for i < 0.
The error is 1% or less for |k| less than about 1.1.
Exact values (6 s.f.) for p = 1, 2, . . . , 6 are 1.570 796, 1.178 097, 0.981 748, 0.859 029,
0.773 126, 0.708 699. The Gauss–Chebyshev integration is in error by about 1%
when n = p.
Listed below are the relevant indefinite integrals F(y) of the distributions together
with the functions ξ = ξ(η):
√
(a) y 2 , ξ = η;
(b) y 3/2 , ξ = η 2/3 ;
(c) 12 {sin[πy/(2a)] + 1}, ξ = (2a/π) sin−1 (2η − 1);
(d) 12 exp y for y ≤ 0, 12 [2 − exp (−y)] for y > 0; ξ = ln 2η for 0 < η ≤ 12 ,
ξ = − ln[2(1 − η)] for 12 < η < 1.
1039
NUMERICAL METHODS
V = 80
−∞
40.5
41.8
46.7
48.4
46.7
16.8
20.4
16.8
41.8
40.5
∞
V =0
Figure 27.8 The solution to exercise 27.25.
27.15
27.17
27.19
27.21
27.23
27.25
27.27
1 − x2 /2 + x4 /8 − x6 /48; 1.0000, 0.9950, 0.9802, 0.9560, 0.9231, 0.8825; exact
solution y = exp(−x2 /2).
(b) a1 = 23/12, a2 = −4/3, a3 = 5/12.
(c) b1 = 5/12, b2 = 2/3, b3 = −1/12.
(d) ȳ(0.4) = 0.224 582, y(0.4) = 0.225 527 after correction.
4
(a) The error is 5h3 u(3)
n /12 + O(h ).
(b) α = −4, β = 5, µ = 4 and ν = 2
For λ positive the solutions are (boringly) monotonic functions of x. With y(0)
given, there are no real solutions at all for any negative λ!
(a) Setting A∆t = c∆x gives, for example, u(0, 2) = (1 − c)2 , u(1, 2) = 2c(1 − c),
u(2, 2) = c2 . For stability, 0 < c < 1.
(b) G(n, s) = [(1 − c) + cs]n for 0 ≤ p ≤ n.
(c) [n!(1 − c)n−p cp ]/[p!(n − p)!].
(d) When c = 1 and the difference equation becomes u(p, n + 1) = u(p − 1, n).
See figure 27.8.
If x = αy then
mk
2mE
d2 ψ
− α4 2 y 2 ψ + α2 2 ψ = 0.
dy 2
Solutions will be either symmetric or antisymmetric with ψ(0) = 0 but ψ (0) = 0
for the former and vice versa for the latter. Integration to a largish but finite
value of y followed by an interpolation procedure to estimate the values of E
that lead to ψ(∞) = 0 needs to be incorporated. Simple numerical integration
such as Simpson’s rule will suffice for the normalisation integral. The solutions
should be λ = 1, 3, 5, . . . .
1040
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