Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

27.10 HINTS AND ANSWERS


27.27 The Schr ̈odinger equation for a quantum mechanical particle of massmmoving
in a one-dimensional harmonic oscillator potentialV(x)=kx^2 /2is



^2


2 m

d^2 ψ
dx^2

+


kx^2 ψ
2

=Eψ.

For physically acceptable solutions, the wavefunctionψ(x) must be finite atx=0,
tend to zero asx→±∞andbenormalised,sothat


|ψ|^2 dx=1.Inpractice,
these constraints mean that only certain (quantised) values ofE, the energy of
the particle, are allowed. The allowed values fall into two groups: those for which
ψ(0) = 0 and those for whichψ(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 Schr ̈odinger equation takes the form

d^2 ψ
dy^2

+(2E′−y^2 )ψ=0.

Devise an outline computerised scheme, using Runge–Kutta integration, that will
enable you to:
(a) determine the three lowest allowed values ofE;
(b) tabulate the normalised wavefunction corresponding to the lowest allowed
energy.
You should consider explicitly:

(i) the variables to use in the numerical integration;
(ii) how starting values neary= 0 are to be chosen;
(iii) how the condition onψasy→±∞is to be implemented;
(iv) how the required values ofEare 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 5.370.
27.3 (a)ξ=0andf′(ξ)= 0 in general; (b)ξ=b, butf′(b) = 0 whilstf(b)=0.
27.5 Interchange is formally needed for the first two steps, though in this case no error
will result if it is not carried out;x 1 =− 12 ,x 2 =2,x 3 =− 1 ,x 4 =5.
27.7 The quadratic equation isz^2 +4z+1=0;α+β−3=x 0 =α′+β′+3.
Withp=−2+



3andq=− 2 −


3,βmust be zero fori>0andα′must be
zero fori<0;xi=3(−2+


3)ifori>0,xi=0fori=0,xi=−3(− 2 −


3)i
fori<0.
27.9 Theerroris1%orlessfor|k|less than about 1.1.
27.11 Exact values (6 s.f.) forp=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%
whenn=p.
27.13 Listed below are the relevant indefinite integralsF(y) of the distributions together
with the functionsξ=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 expyfory≤0,^12 [2−exp (−y)] fory>0;ξ=ln2ηfor 0<η≤^12 ,
ξ=−ln[2(1−η)] for^12 <η<1.
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