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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 14 "Fourier series" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "In the workshee
t " }{TEXT 19 11 "HeatEqn.mws" }{TEXT -1 553 " we have seen how a part
ial differential equation (the heat equation) in one space and one tim
e dimension admitted solutions in product form (separation of variable
s technique), and how the solution was approximated by expanding the s
patial part in a Fourier sine series. In this worksheet we generalize \+
the idea of representing functions defined on a finite interval (part \+
of the real axis) by means of an expansion in basis functions. The con
cepts can then be extended to basis functions defined as the eigenfunc
tions of some Sturm-Liouville operator." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 296 "We begin by defining our basic inte
rval on which the function is to be defined. This would normally be th
e interval on which solutions to a differential equation are seeked, i
.e., usually the unknown function (solution to the ODE) has specified \+
values at the edges of the interval. The choice of " }{TEXT 19 1 "a" }
{TEXT -1 5 " and " }{TEXT 19 1 "b" }{TEXT -1 155 " below is arbitrary,
 and depends on the problem posed. In quantum mechanics we could be as
king for particles to be located in a box bounded by these edges." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; a:=-2; b:=3;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG!\"#" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"bG\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "We
 define an inner product for real-valued functions (without weight fun
ction) assuming that the independent variable is called " }{TEXT 19 1 
"x" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "IP:=(
f,g)->int(expand(f*g),x=a..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#I
PGf*6$%\"fG%\"gG6\"6$%)operatorG%&arrowGF)-%$intG6$-%'expandG6#*&9$\"
\"\"9%F5/%\"xG;%\"aG%\"bGF)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
132 "This inner (dot) product will be used to calculate the norm of fu
nctions, and the projection of functions onto basis functions. The " }
{TEXT 19 6 "expand" }{TEXT -1 104 " function has been included to help
 Maple calculate integrals by breaking them into many simpler pieces.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 94 "The b
asis will be defined as the eigenfunctions of the following simple dif
ferential operator:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "DO:=
f->-diff(f,x$2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#DOGf*6#%\"fG6\"
6$%)operatorG%&arrowGF(,$-%%diffG6$9$-%\"$G6$%\"xG\"\"#!\"\"F(F(F(" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 209 "In order to determine the eigenf
unctions and eigenvalues of this differential operator on the interval
 [a,b] we have to specify some boundary conditions (BC). We anticipate
 (based on elementary Calculus) that " }{TEXT 19 8 "sin(k*x)" }{TEXT 
-1 5 " and " }{TEXT 19 8 "cos(k*x)" }{TEXT -1 49 " are candidates for \+
the eigenfunctions, and that " }{TEXT 19 3 "k^2" }{TEXT -1 83 " will a
ppear as an eigenvalue. The boundary conditions will impose which valu
es of " }{TEXT 19 1 "k" }{TEXT -1 15 " are permitted." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "In the example carried
 out in " }{TEXT 19 11 "HeatEqn.mws" }{TEXT -1 108 " we encountered tw
o types of BCs: we can fix the value of the basis function, or we can \+
fix its derivative. " }}{PARA 0 "" 0 "" {TEXT -1 22 "The eigenvalue pr
oblem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 34 "
DO(f) = lambda*f     ,  lambda=k^2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 37 "is homogeneous, which means that if  " }
{TEXT 19 4 "f(x)" }{TEXT -1 68 "  is a solution, then any multiple of \+
it will be a solution as well." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 59 "If we do not impose boundary conditions, \+
then the value of " }{TEXT 19 1 "k" }{TEXT -1 248 " will be unrestrict
ed, and the interval [a,b] becomes somehow irrelevant. This means that
 the functions are defined on the entire real axis (something which be
comes important in quantum mechanics when discussing plane-wave free-p
article solutions)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 86 "Let us start with the simple case of imposing that the \+
basis functions vanish at both " }{TEXT 19 1 "a" }{TEXT -1 5 " and " }
{TEXT 19 1 "b" }{TEXT -1 35 ". These BCs are called homogeneous." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "BC1:=f(a)=0;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$BC1G/-%\"fG6#!\"#\"\"!" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 22 "SL:=DO(f(x))=k^2*f(x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#SLG/,$-%%diffG6$-%\"fG6#%\"xG-%\"$G6$F-\"\"#!\"\"*&)
%\"kGF1\"\"\"F*F6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "How do we f
ind the basis functions that satisfy this equation? We can't use the s
econd boundary condition, because that would force the boring trivial \+
solution " }{TEXT 19 6 "f(x)=0" }{TEXT -1 42 ". We have to use the cho
ice of eigenvalue " }{TEXT 19 6 "lambda" }{TEXT -1 31 " to find non-tr
ivial solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sol:=ds
olve(\{SL,BC1\},f(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"
fG6#%\"xG,&*&%$_C1G\"\"\"-%$sinG6#*&%\"kGF-F)F-F-F-*(F,F--%$tanG6#,$F2
\"\"#F--%$cosGF0F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "The choic
e of integration constant " }{TEXT 19 3 "_C1" }{TEXT -1 105 " is arbit
rary and defines the normalization of the solution. We impose now the \+
second boundary condition:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "fB:=subs(_C1=1,rhs(sol));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fB
G,&-%$sinG6#*&%\"kG\"\"\"%\"xGF+F+*&-%$tanG6#,$F*\"\"#F+-%$cosGF(F+F+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "BC2:=subs(x=b,fB)=0;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$BC2G/,&-%$sinG6#,$%\"kG\"\"$\"\"\"
*&-%$tanG6#,$F+\"\"#F--%$cosGF)F-F-\"\"!" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 198 "This is a nonlinear transcendental equation which admits
 infinitely many roots. To find the roots we have to employ a numerica
l solver. First, let us graph the expression whose roots we are seekin
g." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "plot(lhs(BC2),k=-5..5
,view=[-5..5,-4..4],discont=true);" }}{PARA 13 "" 1 "" {GLPLOT2D 696 
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3q$fBgw&ow\\F1$\"3OCP9mnNdGF-7$$\"\"&F*$\"3?]Ilj&ftd\"F--%'COLOURG6&%$
RGBG$\"*++++\"!\")$F*F*Fc]t-%+AXESLABELSG6$Q\"k6\"Q!Fh]t-%%VIEWG6$;F(F
h\\t;$!\"%F*$\"\"%F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Clearl
y, we need to consider " }{TEXT 19 6 "k >= 0" }{TEXT -1 48 " only. The
 functions are labelled by eigenvalue " }{TEXT 19 3 "k^2" }{TEXT -1 
57 ", which cannot distinguish positive from negative values." }}
{PARA 0 "" 0 "" {TEXT -1 168 "The poles that appear in-between the roo
ts can make life difficult for a systematic root search, but at least \+
it is clear where they occur, as they are associated with " }{TEXT 19 
8 "tan(2*k)" }{TEXT -1 23 ", i.e., they occur for " }{TEXT 19 10 "2*k \+
= m*Pi" }{TEXT -1 8 ", where " }{TEXT 19 9 "m=1,2,..." }}{PARA 0 "" 0 
"" {TEXT -1 9 "The root " }{TEXT 19 5 "k = 0" }{TEXT -1 109 " does not
 lead to an interesting solution (due to the homogeneous BCs imposed w
e don't have a constant term)." }}{PARA 0 "" 0 "" {TEXT -1 30 "The exp
ression is periodic in " }{TEXT 19 1 "k" }{TEXT -1 13 " with period " 
}{TEXT 19 4 "2*Pi" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "k1:=fsolve(BC2,k=0.0001..Pi/4-0.0001);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#k1G$\"+2`=$G'!#5" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 36 "k2:=fsolve(BC2,k=Pi/4+0.0001..Pi/2);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#k2G$\"+hqjc7!\"*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 38 "k3:=fsolve(BC2,k=Pi/2..3*Pi/4-0.0001);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#k3G$\"+#fb\\)=!\"*" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "k4:=fsolve(BC2,k=2..3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#k4G$\"+BTF8D!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "k5:=fsolve(BC2,k=3*Pi/4..5*Pi/4);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#k5G$\"+aEfTJ!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "k6:=fsolve(BC2,k=Pi+0.0001..5*Pi/4);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#k6G$\"+%=6*pP!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "evalf(k6-Pi),k1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$
\"*I&=$G'!\"*$\"+2`=$G'!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Map
le's solve command finds analytic expressions for the roots:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "sol:=solve(BC2,k);" }}{PARA 
12 "" 1 "" {XPPMATH 20 "6#>%$solG6,%#PiG\"\"!-%'arctanG6#,$*&,&\"#5\"
\"\"*&\"\"#F/-%%sqrtG6#\"\"&F/!\"\"#F/F1,&F7F/*&F7F/F2F/F/F6F7,$F(F6,&
-F)6#,$*&,&F.F/*&F1F/F2F/F/F7,&F7F/*&#F/F1F/*$F2F/F/F6F6F7F/F&F/,&F<F6
F&F6-F)6#,$*&F@F7,&FEF7#F/F1F6F6F7,$FGF6,&-F)6#,$*&F-F7,&#F6F1F/*&#F/F
1F/FEF/F6F6F7F/F&F/,&FOF6F&F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 17 "simplify(sol[3]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'arctanG6
#*&,&\"#5\"\"\"*&\"\"#F)-%%sqrtG6#\"\"&F)!\"\"#F)F+,&*$F,F)F)F)F)F0" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(sol);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6,$\"+aEfTJ!\"*$\"\"!F'$\"+2`=$G'!#5$!+2`=$G'F*$\"
+$fb\\)=F%$!+$fb\\)=F%$\"+hqjc7F%$!+hqjc7F%$\"+BTF8DF%$!+BTF8DF%" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "We will be content with using the
 first five basis states, but the periodicity allows one to construct \+
all other eigenvalues easily from the five numbers " }{TEXT 19 6 "k1..
k5" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 30 "Let us graph the ba
sis states:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "plot([seq(su
bs(k=k||j,fB),j=1..5)],x=a..b,color=[red,blue,green,black,brown]);" }}
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pansion coefficients are obtained by projection:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 33 "c_n:=[seq(IP(f1,Bfn[j]),j=1..5)];" }}{PARA 
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}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "Five basis functions are suffici
ent to reach almost 1%-level accuracy!" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT 258 11 "Exercise 1:" }}{PARA 0 "" 0 "" 
{TEXT -1 120 "Change the power for the (x-a) and (x-b) factors and rep
eat the calculation with five basis states. What do you observe?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 11 "Exercise
 2:" }}{PARA 0 "" 0 "" {TEXT -1 170 "Increase the order of the expansi
on by finding the next five eigenvalues (you can use the periodicity),
 and eigenfunctions. Does the convergence behaviour become evident?" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 513 "This wa
s an example of applying linear algebra techniques to a function space
. We used an eigenvalue problem to define a set of eigenvectors (basis
 functions), and then we expanded some arbitrary vector (function) in \+
this basis. The economy is obvious: we can have approximate knowledge \+
of a function by knowing a set of numbers (the expansion coefficients)
. Often one can adapt a basis to a problem at hand, and, thus, limit t
he number of expansion coefficients required for a reasonably accurate
 representation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 72 "We demonstrate now what happens when we look outside the \+
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ft$\"3!\\'*QA2p$ofFi_n7$Fgft$\"3I(z&HbpZ(>'Fi_n7$F\\gt$\"39/NxA5DIkFi_
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45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
124 "We observe that the answer obtained from the Fourier series is pe
riodic, i.e., it repeats itself outside the interval [a,b]." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 11 "Exercise 3:" }
}{PARA 0 "" 0 "" {TEXT -1 177 "Change the interval [a,b] to cover a wi
der range (e.g., [-3,4]), repeat the calculation for the Fourier serie
s, and make your observation. Investigate the convergence behaviour." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 395 "For fu
nctions that do not vanish on the edge of the interval some interestin
g pheonomena can be observed. The textbooks on the subject develop the
 calculation of Fourier coefficients for simple periodic functions (sq
uare wave train, sawtooth function, etc.), for which a formula for the
 expansion coefficients can be derived. This permits to study converge
nce behaviour in a straightforward way." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 90 "Let us define the next wavenumbers f
or the Fourier basis. We found k5 to be related to k1." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot(lhs(BC2),k=0..9,view=[0..9,-4.
.4],discont=true);" }}{PARA 13 "" 1 "" {GLPLOT2D 696 299 299 
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hMT@:ViF47$Fir$!343YNq&)*4I&F47$F^s$!3S\"H(z0<d1WF47$Fcs$!3qv%z5clpI$F
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t$\"315LE48si#)Fet7$F^u$\"3U&pi]!fDD=F47$Fcu$\"3F=#3F32D$GF47$Fhu$\"3!
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hF47$F\\w$\"39-K[s\"Rh%oF47$Faw$\"39/#)QRzp`uF47$Ffw$\"3cFzIFkQdzF47$F
[x$\"3-LTO`pHb$)F47$F`x$\"3o!*ysq%G0p)F47$Fex$\"3]`:&)R%\\S*))F47$Fjx$
\"3&)\\D$ywx#3!*F47$Fdy$\"3^CWwn\\+?!*F47$F^z$\"3\"3*3.<j@M*)F47$Fhz$
\"3'yhB!>ech()F47$F][l$\"3')f2r)))e!*[)F47$Fb[l$\"3IY(=w@qd8)F47$Fg[l$
\"3ko7CP&f#zwF47$F\\\\l$\"3kK$GVj1=;(F47$Fa\\l$\"3/7A&=7g1c'F47$Ff\\l$
\"3)eq<)e%)>+fF47$F[]l$\"3AjhToU(QC&F47$F`]l$\"3sA&4_6UBW%F47$Fe]l$\"3
3$[Mt*p3*o$F47$Fj]l$\"35%\\h&36GdGF47$F_^l$\"3Ozw&3\"4:U?F47$Fd^l$\"3[
fAt/HRb6F4-Fj^l6&F\\_lF*F*F*-%+AXESLABELSG6$Q\"x6\"Q!F_[o-%%VIEWG6$;F(
Fd^l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 
0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 110 "What happens at the left-hand side is known as the Gibbs
 phenomenon. Given that all basis functions vanish at " }{TEXT 19 3 "x
=a" }{TEXT -1 8 " and at " }{TEXT 19 3 "x=b" }{TEXT -1 436 ", the appr
oximate Fourier series representation cannot be accurate there. The Fo
urier expansion tries to approximate a step function behaviour: at the
 endpoints the function is assumed to rise immediately to the non-zero
 value with infinite slope. Fourier expansions with a large number of \+
terms are able to produce a large slope, but the price is paid by the \+
wiggling around the true function even some distance away from the bou
ndary." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 
11 "Exercise 4:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 52 "Use the analytic expression found for the values of " }
{TEXT 19 4 "k||n" }{TEXT -1 44 " to push the expansion to high order (
e.g., " }{TEXT 19 8 "n_max=50" }{TEXT -1 35 ") and observe the Gibbs p
henomenon." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 260 35 "Solution of a differential equation" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 287 "So far we have not calcu
lated anything particularly useful given that the functions to be repr
esented could be easily expressed as elementary expressions. We will s
how now that the Fourier series expansion allows to convert differenti
al equation problems into linear algebra assignments." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "Let us define an inho
mogeneous linear second-order differential equation with boundary cond
itions specified at the ends of the interval [a,b]." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 51 "DE:=diff(u(x),x$2)+1/4*x*diff(u(x),x)+x^2
*u(x)=x^4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#DEG/,(-%%diffG6$-%\"u
G6#%\"xG-%\"$G6$F-\"\"#\"\"\"*(#F2\"\"%F2F-F2-F(6$F*F-F2F2*&)F-F1F2F*F
2F2*$)F-F5F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "BCd:=u(a)=0
,u(b)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$BCdG6$/-%\"uG6#!\"#\"\"
!/-F(6#\"\"$F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "solDE:=ds
olve(\{DE,BCd\},u(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&solDEG6\"
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "Maple can't solve this one. \+
We can generate a numerical solution to this boundary value problem us
ing the " }{TEXT 19 15 "dsolve[numeric]" }{TEXT -1 12 " integrator." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "solDE:=dsolve(\{DE,BCd\},u
(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "U:=subs(solDE,u(x)):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "plot(U(x),x=a..b);" }}{PARA 13 "" 1 "" {GLPLOT2D 652 
299 299 {PLOTDATA 2 "6%-%'CURVESG6$7^o7$$!\"#\"\"!$F*F*7$$!+HU,\"*=!\"
*$\"+9yz([\"F/7$$!+$H'='z\"F/$\"+1cvCHF/7$$!+gBa*o\"F/$\"+FttCYF/7$$!+
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+_$)oVCFP7$$\"+$p)=M8F/$\"+ntJOBFP7$$\"+*=]@W\"F/$\"+W//$=#FP7$$\"+]$z
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\"+?u2![*F/7$$\"+OzV4?F/$\"+A`FxyF/7$$\"+nAFj?F/$\"+Y`9tiF/7$$\"+E67:@
F/$\"+ZKZYZF/7$$\"+&)*pp;#F/$\"+ZE.eKF/7$$\"+JH**>AF/$\"+W$otz\"F/7$$
\"+ye,tAF/$\"+8Te>UFbt7$$\"+fO=yBF/$!+YTSh>F/7$$\"+E>#[Z#F/$!+$)Rf?OF/
7$$\"+26?IDF/$!+Y$>:H%F/7$$\"+(G!e&e#F/$!+K-@LZF/7$$\"+(=Y.h#F/$!+iq<_
[F/7$$\"+'37^j#F/$!+'R%)3#\\F/7$$\"+N]\\ZEF/$!+=F3O\\F/7$$\"+&)z()fEF/
$!+,=PQ\\F/7$$\"+M4EsEF/$!+fQmF\\F/7$$\"+&)Qk%o#F/$!+?Z)Q!\\F/7$$\"+)
\\\\5r#F/$!+;TE4[F/7$$\"+8^XPFF/$!+R\")eaYF/7$$\"+F2'Qw#F/$!+tH!)RWF/7
$$\"+UjE!z#F/$!+in<lTF/7$$\"+FM\"3%GF/$!+7h-vMF/7$$\"+60O\"*GF/$!+C(*z
vDF/7$$\"\"$F*F+-%'COLOURG6&%$RGBG$\"#5!\"\"F+F+-%+AXESLABELSG6$Q\"x6
\"Q!Fg`l-%%VIEWG6$;F(Fj_l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
217 "Now we produce a basis-expansion solution. We generate a matrix f
rom the left-hand side of the equation by acting with the differential
 operator on basis function Bfn[j], while multiplying with Bfn[i] and \+
integrating." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "lhs(DE);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,(-%%diffG6$-%\"uG6#%\"xG-%\"$G6$F*\"
\"#\"\"\"*(#F/\"\"%F/F*F/-F%6$F'F*F/F/*&)F*F.F/F'F/F/" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "Nb:=9;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#NbG\"\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "with(L
inearAlgebra):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "A:=Matrix
(Nb,Nb):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "for i from 1 to
 Nb do: for j from 1 to Nb do: A[i,j]:=IP(Bfn[i],subs(u(x)=Bfn[j],lhs(
DE))): od: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalm(A);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7+7+$\"+JOMqa!#5$\"*8%
*Hn*!\"*$!+)4'QP6F-$!+&fG<()*!#6$\"+*)\\r+GF*$!*o`$*p$F*$!++O$[M\"F*$
\"*mTn3#F*$\"+)pdYZ)F27+$\"*zg'R$)F-$\"+O$zc7$F*$!+kLo#4\"F-$!+hU7f9F-
$\"+pM?&R\"F*$!+Q_r\\UF*$!*jzI\"fF*$\"+hf'RB#F*$\"*5x!oNF*7+$!+\")4'Qi
(F*$!+eO$o_)F*$!+;\"[a[\"F-$\"+7E'R;\"F-$!+>w5c;F-$\"+@(ptm\"F*$\"+T5o
%G&F*$!*A=\"zvF*$!+F!H/$HF*7+$!+c`RQXF2$!+)*fdCzF*$\"*ta5@)F-$!+g1P(=%
F-$!+o_#HA\"F-$\"+1U&e\"=F-$\"+Q#R0*=F*$!+v1\"y9'F*$!+lgx\"**)F27+$\"*
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g>F-$\"+(oy94#F*$\"+(zrc#pF*7+$!*aDyq#F2$!+AC:(*\\F2$\"+aPOSLF2$\"+L?a
ehF*$!+%=46K(F*$!+'z2R?\"!\")$\"+:MHI8F-$!++/\"y4#F-$!+`$f2G#F*7+$\"+&
o(*\\8\"F2$\"*$[U\"4$F2$\"*$y5oMF2$\"+KcpN>F2$!+Q;n@]F*$\"+HdRToF*$!+a
Q>;<F`r$!+I?/#Q\"F-$!+005JAF-7+$!*z1%HXF2$!+3b+FVF2$\"+$3a\"[6F2$\"+s,
c)=&F2$\")`l>SF*$!+KpCNQF*$!+=Nv`jF*$!+BVw2BF`r$\"+h`BL9F-7+$!+T:MvFF2
$!+`$[s5\"F2$\"+c+r&>)F2$\"+**y9&3#F2$!+kD/56F*$\"+KoS#>\"F2$!*.0Ni#F-
$\"+K_wheF*$!+!4#[yHF`r" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "
Ai:=MatrixInverse(A);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#AiG-%'RTAB
LEG6$\")+[%R\"-%'MATRIXG6#7+7+$!30*z$y]%='H<!#<$\"3a-1Up,\"Qx\"F0$!3lE
3*pU%HMa!#=$!3e)>rNce&R!)F5$\"3YX%R`b;/Q#F5$!3GS>R$H#)3<#F5$!3i$RElvv'
)['!#>$\"37d2'y#**z4lF>$\"3>N]9\\:x;?F>7+$\"3&Gk.0QqS_\"F0$!38,vIxv!*H
6F0$!3P#H<SY9!4;F5$\"3p_$)*[!R#oA$F5$\"3k&R4J>=w2\"F>$\"3Aaq*R?1(3\")F
>$!3f*ok5K\\'=N!#?$!3+!3j;bdJY#F>$\"3s7>%HU?(*H#FR7+$!33$4xp(R$pj$F5$!
3]v*HaY9VB\"F5$!30!*H:I$RR%QF5$!3kpBP47XftF>$\"38lnH@@P-#)F>$!3:39Es<g
+@F>$!3ivq9P$Q'o?F>$\"3OXdj==ahdFR$\"3ilxV.5K;nFR7+$!3&)H#Qr3_Xu$F5$\"
3_[xI!*GPa<F5$!3k!*)yVU#43_F>$!3#HgFU\"p^cKF5$\"3-T!48aZv#fF>$!3mWana9
reiF>$!3w!pmY')p)>:F>$\"3wz$)HR*Qcu\"F>$\"3;l,T9@HwWFR7+$\"3XgC:bT9NqF
>$\"3%Qt$=7df)=\"FR$\"3o'HcQkVuj$F>$\"3s%*=r2%3\"=QF>$!3=@DJ)f')HW\"F5
$\"3')>+V:NNgAF>$\"3?e$>zwK=&>F>$!3@!)e8pR[LdFR$!3A[?BKV(y`&FR7+$!3-,9
KsyjLIF>$\"3[C>l)\\0bF\"F>$!3CQ'\\JT<vO&FR$!3Mn9rro(30#F>$\"3%y]EsU$)3
D\"F>$!3aB\"3b^'zr))F>$!3!yZ[%pvz%)))FR$\"3U=j8)Q.VU*FR$\"3Dhs!y_JH8#F
R7+$!3@F\"\\N3k*HVFR$\"3_-u_$3^/m\"FR$!3gPkJ)4Hat\"FR$!3^$)\\=aVFMFFR$
\"3*o%>E$[6B\"\\FR$!3&)*=T%H#z@^%FR$!3m(H$*z*=]YfF>$\"3vZ$=H&R,HUFR$\"
3LGfHq_j@[FR7+$!3;@1i8\\7PHFR$\"3wF'3+%f?d<FR$\"3`v`.0mzJ?!#@$!3GAN`H[
C(*yFgt$!3%)yG%y&y&zF#Fgt$\"3)))>c'Hw,V8FR$\"3k-;$H%>u#z\"FR$!3=&46,Q*
feVF>$!3O%)=U`-3`AFR7+$!3zcS+v(o\"=^Fgt$!3yT&yu8w*G9FR$!3M*\\D(G.?kkFg
t$\"3y0Z1'*pvrc!#A$\"3]P,i%zIUO&Fgt$\"3uH'Qi'>\"Rp\"F]v$\"3+BALkuppZFg
t$!3Gf_kU([K$*)Fgt$!3+#R$Ra*\\PO$F>" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 100 "Now we form a right-hand-side vector using a projection of the
 rhs of the ODE onto the basis states:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "bvec:=Vector(Nb):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 55 "for i from 1 to Nb do: bvec[i]:=IP(rhs(DE),Bfn[i]): o
d:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(bvec);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7+$\"*v`:*)*!\")$\"+Q!=\"f
8F)$!+lH.4>F)$!+)HkmR\"F)$\"+W,\\W;F)$!+^5k>5F)$!+*H>2G\"F)$\"+&4.(zy!
\"*$\"+#[P3.\"F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "cvec:= \+
Ai . bvec;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%cvecG-%'RTABLEG6$\")_
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0 "" {MPLTEXT 1 0 35 "solAP:=add(cvec[i]*Bfn[i],i=1..Nb):" }}}{EXCHG 
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{EXCHG {PARA 0 "" 0 "" {TEXT 262 11 "Exercise 5:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 289 "Pick a linear differenti
al equation of your choice with appropriate boundary conditions and co
mpare numerical solutions as provided by Maple with the Fourier-series
 solution. Observe the efficiency of the Fourier method: the solution \+
is specified using a small number of real coefficients." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 56 "Solution of a diffe
rential equation - eigenvalue problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 344 "Here we apply the Fourier-series exp
ansion technique to a Schroedinger equation. This will result in a mat
rix eigenvalue problem. We will continue to use our asymmetric interva
l [a,b], but apply it to a situation where a symmetric interval would \+
be more appropriate. We cosnsider the bound states in a potential that
 has an inverted bell shape:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "Vpot:=-2/(1+(2*x)^2)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Vp
otG,$*&\"\"\"F'*$),&F'F'*&\"\"%F')%\"xG\"\"#F'F'F/F'!\"\"!\"#" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(Vpot,x=a..b);" }}{PARA 
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&%$RGBG$\"#5!\"\"$F*F*F`bl-%+AXESLABELSG6$Q\"x6\"Q!Febl-%%VIEWG6$;F(Fe
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"Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "We will ask the foll
owing question: Given that the kinetic energy can be represented as" }
}{PARA 0 "" 0 "" {TEXT 19 29 "Tkin = - 1/2 diff(psi(x),x$2)" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "what are the al
lowed energy eigenvalues for " }{TEXT 19 9 "Tkin+Vpot" }{TEXT -1 67 "?
 [for physicists: we have chosen units in which Planck's constant " }
{TEXT 19 6 "h=2*Pi" }{TEXT -1 179 ", and the mass of the particle equa
ls unity, and a third constant was chosen so that the strength of the \+
potential energy is dimensonless]. The kinetic energy operator represe
nts " }{TEXT 19 9 "p^2/(2*m)" }{TEXT -1 43 " using the association (pr
escription) that " }{TEXT 19 21 "p = - I*h/(2*Pi)*d/dx" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "SE:=-1/2*diff(psi(x),x$2)+Vpot*psi(
x) = E*psi(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&-%%diffG6$-
%$psiG6#%\"xG-%\"$G6$F-\"\"##!\"\"F1*(F1\"\"\",&F5F5*&\"\"%F5)F-F1F5F5
!\"#F*F5F3*&%\"EGF5F*F5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "We are
 looking for bound solutions, i.e., " }{TEXT 19 5 "E < 0" }{TEXT -1 
174 ". The ODE problem is homogeneous (eigenvalue problems always are)
. This problem cannot be solved easily by dsolve[numeric], because the
 energy eigenvalue is unknown a priori." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 46 "We generate a matrix representation \+
as before:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "Nb:=5;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NbG\"\"&" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "A:=Matrix(Nb,Nb):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 121 "for i from 1 to Nb do: for j from 1 to Nb do: A[i,j]
:=IP(Bfn[i],subs(psi(x)=Bfn[j],lhs(SE))): print(i,j,A[i,j]);  od: od:
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"\"F#^$$!+<,ooL!#5$F'!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"\"\"\"#^$$\"+,.\"H&G!#5$\"#Z!#7" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"\"\"\"$^$$!+***3W$G!#5$!$@\"!#7" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"\"\"\"%^$$\"+er(Qg$!#5$\"#8!#7" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"\"\"\"&^$$!+nUPq>!#6$F(F(" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"#\"\"\"^$$\"+,.\"H&G!#5$!\")!#7" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"#F#^$$\"+;:\\(Q&!#5$\"\"\"!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"#\"\"$^$$\"+A'o'4v!#6$!\"%!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"#\"\"%^$$!+\"frtj#!#5$!#:!#6" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"#\"\"&^$$!+O%)zB5!#5$\"#L!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"$\"\"\"^$$!+'**3W$G!#5$!$5#!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"$\"\"#^$$\"+A(o'4v!#6$!)3GDU!#>" }
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{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"$\"\"%^$$\"+!z6\"H=!#5$\"#A!#6" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"$\"\"&^$$!+)>,%*[\"!#5$!\"#F(" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"%\"\"\"^$$\"+hr(Qg$!#5$!\"#!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"%\"\"#^$$!+q:PPE!#5$F#!#6" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"%\"\"$^$$\"+r<6H=!#5$!\"&!#6" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"%F#^$$\"+GM&Hx#!\"*$\"\"'!#5" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"%\"\"&^$$!+eh`sq!#6$\"$6$!#7" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"&\"\"\"^$$!+mVPq>!#6$\"\"!F*" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"&\"\"#^$$!+%\\)zB5!#5$!\"#!#6" }}
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{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"&F#^$$\"+%3q!yY!\"*$!*?)>TG!#>" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalm(A);" }}{PARA 12 "" 1 "
" {XPPMATH 20 "6#-%'matrixG6#7'7'^$$!+<,ooL!#5$F+!#7^$$\"+,.\"H&GF+$\"
#ZF-^$$!+***3W$GF+$!$@\"F-^$$\"+er(Qg$F+$\"#8F-^$$!+nUPq>!#6$F@F@7'^$F
/$!\")F-^$$\"+;:\\(Q&F+$\"\"\"F-^$$\"+A'o'4vF@$!\"%F-^$$!+\"frtj#F+$!#
:F@^$$!+O%)zB5F+$\"#LF-7'^$$!+'**3W$GF+$!$5#F-^$$\"+A(o'4vF@$!)3GDU!#>
^$$\"+Ze+1:!\"*$\"\"!Feo^$$\"+!z6\"H=F+$\"#AF@^$$!+)>,%*[\"F+$!\"#F+7'
^$$\"+hr(Qg$F+$F_pF-^$$!+q:PPEF+$\"\"%F@^$$\"+r<6H=F+$!\"&F@^$$\"+GM&H
x#Fco$\"\"'F+^$$!+eh`sqF@$\"$6$F-7'^$$!+mVPq>F@Fdo^$$!+%\\)zB5F+$F_pF@
^$$!*7,%*[\"Fco$F^qF+^$$!*EOD2(F+$\"#fF@^$$\"+%3q!yYFco$!*?)>TGF_o" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "What do we notice?" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "1) the imaginary par
t in the matrix elements looks like spurious noise" }}{PARA 0 "" 0 "" 
{TEXT -1 125 "2) the matrix is almost symmetric [the small deviations \+
are a result of using non-symmetric boundary conditions in the basis]
" }}{PARA 0 "" 0 "" {TEXT -1 75 "Before we proceed with finding the ei
genvalues we correct the two problems:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "A:=map(Re,A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
AG-%'RTABLEG6$\")k&Rt\"-%'MATRIXG6#7'7'$!+<,ooL!#5$\"+,.\"H&GF0$!+***3
W$GF0$\"+er(Qg$F0$!+nUPq>!#67'F1$\"+;:\\(Q&F0$\"+A'o'4vF9$!+\"frtj#F0$
!+O%)zB5F07'$!+'**3W$GF0$\"+A(o'4vF9$\"+Ze+1:!\"*$\"+!z6\"H=F0$!+)>,%*
[\"F07'$\"+hr(Qg$F0$!+q:PPEF0$\"+r<6H=F0$\"+GM&Hx#FJ$!+eh`sqF97'$!+mVP
q>F9$!+%\\)zB5F0$!*7,%*[\"FJ$!*EOD2(F0$\"+%3q!yYFJ" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 26 "A:=0.5*(A + Transpose(A));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"AG-%'RTABLEG6$\")%)z-=-%'MATRIXG6#7'7'$!3E+++<
,ooL!#=$\"35+++,.\"H&GF0$!3v****\\(**3W$GF0$\"3l****\\fr(Qg$F0$!3#****
*\\;VPq>!#>7'F1$\"3I+++;:\\(Q&F0$\"3*******>no'4vF9$!3v****\\!ertj#F0$
!35+++l%)zB5F07'F3F=$\"3&******p%e+1:!#<$\"37++]!y6\"H=F0$!3/+++f6S*[
\"F07'F5F?FG$\"37+++GM&Hx#FF$!3O******3i`sqF97'F7FAFIFN$\"3y*****R3q!y
YFF" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Eigenvalues(A);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6$\")+vC=-%'MATRIXG6#7'7#^$$
!3/V$oD?(3^_!#=$\"\"!F17#^$$\"3CB@]\\.#*\\G!#<F07#^$$\"339cg/5%)3hF/F0
7#^$$\"39LPrmym!p%F6F07#^$$\"3Z>/[-paK:F6F0" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 143 "We have found one bound state. To determine how reliab
le the energy estimate is we have to increase the matrix size and repe
at the calculation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 19 12 "E0 = -.52511" }{TEXT -1 14 " from 4-by-4 (" }{TEXT 19 
4 "Nb=4" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT 19 12 "E0 = -.54981" 
}{TEXT -1 14 " from 9-by-9 (" }{TEXT 19 4 "Nb=9" }{TEXT -1 1 ")" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 11 "Exercise
 6:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 143 "C
arry out a convergence analysis. Concentrate on the lowest-lying eigen
value (the ground state). The higher-lying states are artificial state
s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 11 "Ex
ercise 7:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
173 "Deepen the potential well by increasing the constant that pre-mul
tiplies the potential. Re-compute the ground state. Are there addition
al bound states in the potential well?" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 29 "To compute the eigenfunction " }
{TEXT 19 8 "psi_0(x)" }{TEXT -1 33 " associated with the eigenenergy \+
" }{TEXT 19 2 "E0" }{TEXT -1 102 " we have to look at the eigenvector \+
and assemble the superposition. The syntax is somewhat cumbersome." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "EV:=Eigenvectors(A);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#EVG6$-%'RTABLEG6$\")gwC=-%'MATRIXG6
#7'7#^$$!3/V$oD?(3^_!#=$\"\"!F47#^$$\"3CB@]\\.#*\\G!#<F37#^$$\"339cg/5
%)3hF2F37#^$$\"39LPrmym!p%F9F37#^$$\"3Z>/[-paK:F9F3-F'6$\"([Yf%-F+6#7'
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "n0:=1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#n0G\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
10 "EV[1][n0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#^$$!3/V$oD?(3^_!#=$
\"\"!F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "cf:=seq(EV[2][j,
n0],j=1..Nb);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#cfG6'^$$\"3>+++%>o
sL*!#=$\"\"!F+^$$!3:+++!>8U%HF)F*^$$\"3/+++;)GG`\"F)F*^$$!3++++$*HpS8F
)F*^$$\"3C+++w(f23$!#@F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 
"psi0:=add(Re(cf[j]*Bfn[j]),j=1..Nb):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "plot(psi0^2,x=a..b);" }}{PARA 13 "" 1 "" {GLPLOT2D 
400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7_o7$$!\"#\"\"!$\"3)='***>4Hg$Q
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to the square of the ground-state wavefunction which (when normalized \+
properly) carries the probability information of where to find particl
es in position space (if they are in the ground state). Observe carefu
lly how the wings of the distribution are in the classically forbidden
 regime:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "plot([Vpot,psi0
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" {TEXT -1 165 "The baseline for the wavefunction graph intersects the
 potential energy at the classical turning point (all kinetic energy i
s converted into potential energy there)." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "The asymmetry is caused by the bas
is. For larger values of " }{TEXT 19 2 "Nb" }{TEXT -1 92 " the ground \+
state is calculated to be symmetric (as it should be for a symmetric p
otential)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 42 "The state is actually properly normalized:" }}}{EXCHG {PARA 0 "
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ciently deep (and wide) potentials additional bound states appear. The
 next bound state (first excited state) in a symmetric potential is an
ti-symmetric (has odd parity). The probability density (square of the \+
wavefunction) has a node at " }{TEXT 19 3 "x=0" }{TEXT -1 64 " in this
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