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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 46 "The matrix representatio
n of quantum mechanics" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 117 "We use the matrix representation of quantum mechani
cs to approximate the energy spectrum of an anharmonic oscillator." }}
{PARA 0 "" 0 "" {TEXT -1 119 "We make use of the particle-in-a-box bas
is. The box size has to be chosen sufficiently large so that the trunc
ation of " }{TEXT 264 1 "x" }{TEXT -1 39 "-space to a finite domain is
 justified." }}{PARA 0 "" 0 "" {TEXT -1 112 "An alternative would have
 been to choose a harmonic oscillator basis (as done in Quantum Mechan
ics using Maple)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 206 "We construct the eigenfunctions using the new LinearAlge
bra package of Maple 6. At the end we compare the approximate eigenfun
ction with a numerical solution to the differential-equation eigenvalu
e problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 279 "We solve the problem by calculating matrix elements in the coo
rdinate representation. Note, however, that a worksheet exists to carr
y out the calculations based on the commutation relations of quantum m
echanics. These worksheets are commut1.mws (Maple5), and define.mws (M
aple6)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
185 "We start with the basis. The box size will put limits on the accu
racy of the higher-lying eigenvalues/eigenfunctions. We also choose a \+
truncation size for the matrix eigenvalue problem." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 41 "restart; Digits:=15: with(LinearAlgebra):" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "X:=8; N:=20;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%\"XG\"\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%\"NG\"#?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "uB:=n->if is(
n,even) then sqrt(1/X)*sin(n*Pi*x/(2*X)) else sqrt(1/X)*cos(n*Pi*x/(2*
X)) end if:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "plot([uB(1),
uB(2),uB(3),uB(4)],x=-X..X,color=[red,blue,green,black]);" }}{PARA 13 
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hx$!0[^(H2UNNFG7$$\"0j;/,V)[gF/$!0y[+\"G$H`$FG7$$\"0(**\\<$)\\(3'F/$!0
?vIx)=FNFG7$$\"0I$eCO:EhF/$!0!e!G\"R>=NFG7$$\"0lm;$*3[;'F/$!0SF$>l&f]$
FG7$$\"0KLea>@C'F/$!0)G0DA!=Z$FG7$F]y$!0gkSx^[U$FG7$$\"0lmm`Vm\\'F/$!0
qUtvi*pKFG7$Fby$!0&=wj/$=0$FG7$$\"0+++neB$oF/$!0D*\\#Rul!GFG7$Fgy$!0vs
\"[4)y^#FG7$$\"0NL$ej&)frF/$!0OC.#)os;#FG7$F\\z$!09P.\">`y<FG7$F\\[n$!
05AXjgqP\"FG7$Faz$!0A\"e**3&R`*F27$Fd[n$!0QgQ3Y<\"[F2Fez-Fiz6&F[[lF*F*
F*-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6$;F(Ffz%(DEFAULTG" 1 2 0 1 10 
0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curv
e 3" "Curve 4" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "int(uB(2)
^2,x=-X..X);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 87 "The basis states are normalized; they rep
resent the modes that the box can accomodate (" }{TEXT 258 1 "n" }
{TEXT -1 86 "-1 counts the number of nodes inside the box, and the wav
es vanish at the box edges (+" }{TEXT 257 1 "X" }{TEXT -1 6 " and -" }
{TEXT 256 1 "X" }{TEXT -1 280 "). What are the eigenenergies? We have \+
kinetic energy only, and evaluate it only over the allowed range; the \+
result agrees with the textbook formula. The kinetic energy matrix wil
l be diagonal in our basis, as the basis functions are eigenfunctions \+
of the kinetic energy operator." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 49 "Tkin:=psi->int(psi*(-1/2*diff(psi,x$2)),x=-X..X);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%%TkinGR6#%$psiG6\"6$%)operatorG%&arrowGF(-%$i
ntG6$,$*&9$\"\"\"-%%diffG6$F1-%\"$G6$%\"xG\"\"#F2#!\"\"F:/F9;,$%\"XGF<
F@F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "[Tkin(uB(1)),Tk
in(uB(2)),Tkin(uB(3)),Tkin(uB(4))];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#7&,$*$)%#PiG\"\"#\"\"\"#F)\"$7&,$F%#F)\"$G\",$F%#\"\"*F+,$F%#F)\"#K" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "seq(n^2*Pi^2/8/X^2,n=1..4
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&,$*$)%#PiG\"\"#\"\"\"#F(\"$7&,$F
$#F(\"$G\",$F$#\"\"*F*,$F$#F(\"#K" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
44 "Now let us define our potential of interest:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 7 "V:=x^4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"VG*$)%\"xG\"\"%\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "We de
fine matrix elements where, again, space is restricted to a finite int
erval. This is imposed by our choice of basis! We allow two different \+
functions in bra and ket." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
48 "Vpot:=(phi,psi)->int(expand(phi*psi*V),x=-X..X);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%%VpotGR6$%$phiG%$psiG6\"6$%)operatorG%&arrowGF)-%$
intG6$-%'expandG6#*(9$\"\"\"9%F5%\"VGF5/%\"xG;,$%\"XG!\"\"F<F)F)F)" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "We are ready to assemble the hami
ltonian matrix:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "HM:=Matr
ix(N):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We make use of the sym
metry of the hamiltonian matrix: it allows to save almost a factor of \+
2 in computation time." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "
for i from 1 to N do: for j from 1 to i do: HM[i,j]:=Vpot(uB(i),uB(j))
; if i=j then HM[i,j]:=HM[i,j]+ Tkin(uB(i)); fi: od: od:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "for i from 1 to N do: for j from i+
1 to N do: HM[i,j]:=HM[j,i]: od: od:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "print(SubMatrix(HM,1..4,1..4));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%'RTABLEG6$\")w\\(\\\"-%'MATRIXG6#7&7&,&*&,(*$)%#PiG\"
\"%\"\"\"F3\"$?\"F3*&\"#?F3)F1\"\"#F3!\"\"F3*$F0F3F9#\"%'4%\"\"&*&#F3
\"$7&F3F7F3F3\"\"!,$*&,&!#:F3*&F8F3F7F3F3F3*$F0F3F9!%WhFA7&FA,&*&,(\"#
:F3*&\"#5F3F7F3F9*&F8F3F0F3F3F3*$F0F3F9#\"%[?F=*&#F3\"$G\"F3F7F3F3FA,$
*&,&!#?F3*&\"\"$F3F7F3F3F3*$F0F3F9#!'s58\"#F7&FBFA,&*&,(*$F7F3!#g\"#SF
3*&FjnF3F0F3F3F3*$F0F3F9#F<\"$N\"*&#\"\"*F@F3F7F3F3FA7&FAFWFA,&*&,(F_o
!#S*&\"#KF3F0F3F3FMF3F3*$F0F3F9#FVF=*&#F3F_pF3F7F3F3" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 251 "All integrals were calculated in closed form. \+
The matrix has a sparse structure: the chosen potential respects parit
y symmetry, and the basis consists of even/odd functions. Therefore, o
ne could solve the problem of the even and odd eigenfunctions of " }
{TEXT 259 1 "V" }{TEXT -1 100 " separately. We do keep it general, so \+
that one may insert a non-symmetric potential function above." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "We canno
t expect the matrix diagonalization to be carried out symbolically. Th
erefore, we switch to floating-point evaluation:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 19 "HMf:=map(evalf,HM):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "evals:=Eigenvalues(HMf, output='list'):" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 26 "ev_s:=sort(map(Re,evals));" }}{PARA 12 "" 1 "
" {XPPMATH 20 "6#>%%ev_sG76$\"3b3+g2WX3n!#=$\"3<8%oza,)4C!#<$\"3D7E8t/
aA`F+$\"3%[*)HAF%e%f*F+$\"3_V'QMk_41#!#;$\"3cu-*o7Jj@$F2$\"3/y9H7_!))
\\'F2$\"3%oGL^&[hp))F2$\"35;+h]x#)Q;!#:$\"3vy%pcoV].#F;$\"3lo:#RG!R0NF
;$\"3C(***zT#H`2%F;$\"3o*o#4d([Cm'F;$\"3)yHsu4/SQ(F;$\"3+\"***GG\"*>g6
!#9$\"3#p_5M'RDS7FH$\"3')G*y=Od*))=FH$\"3-8D?0`Cj>FH$\"3'e<kyZpO\"HFH$
\"3_3$>F:i4'HFH" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 11 "Exercise 1:" 
}}{PARA 0 "" 0 "" {TEXT -1 173 "Carry out matrix diagonalizations with
 submatrices of the N-by-N Hamiltonian matrix, and observe the behavio
ur of the low-lying eigenvalues as a function of truncation size." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "
Now the eigenfunctions. We need a sorting procedure to arrange the res
ult from the eigenvector calculation in proper order." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "VE:=Eigenvectors(HMf,output='list')
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "Vp:=[seq([Re(VE[i][1])
,VE[i][2],map(Re,VE[i][3])],i=1..nops(VE))]:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 344 "Min:=proc(x,y); if type(x,numeric) and type(y,n
umeric) then if x<=y then RETURN(true): else RETURN(false): fi; elif t
ype(x,list) and type(y,list) and type(x[1],numeric) and type(y[1],nume
ric) then if x[1]<=y[1] then RETURN(true): else RETURN(false): fi; eli
f convert(x,string)<=convert(y,string) then RETURN(true): else RETURN(
false): fi: end:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "VEs:=
sort(Vp,Min):\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 252 "Suppose that \+
we would like to see the eigenfunctions corresponding to the four lowe
st-lying eigenvalues. The eigenvector for a given eigenvalue contains \+
the expansion coefficients for the expansion of the eigenstate in term
s of the chosen basis states." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "for i from 1 to 4 do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "psi0
:=add(VEs[i][3][j]*uB(j),j=1..N):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
40 "No:=1/sqrt(int(expand(psi0^2),x=-X..X));" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "phi_a[i]:=add(No*VEs[i][3][j]*uB(j),j=1..N): od:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot([seq(phi_a[i],i=1..4)],
x=-5..5,color=[red,blue,green,black]);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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ur functions should have 0,1,2,3 nodes. The additional nodes at large \+
" }{TEXT 261 1 "x" }{TEXT -1 44 " are clearly an artefact of the calcu
lation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 73 "Let us check the accuracy of these approximate eigenfunct
ions one by one:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "ev_s[1];
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"3b3+g2WX3n!#=" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 57 "We can construct a shooting-method solution: we
 will use " }{TEXT 19 6 "dsolve" }{TEXT -1 60 " in numeric mode to int
egrate out the Schroedinger equation." }}{PARA 0 "" 0 "" {TEXT -1 59 "
For the even-parity state we choose boundary conditions as " }{TEXT 
19 17 "u(0)=1, D(u)(0)=0" }{TEXT -1 32 "; and for the odd-parity state
s " }{TEXT 19 17 "u(0)=0, D(u)(0)=1" }}{PARA 0 "" 0 "" {TEXT -1 128 "W
e use the energy as a trial value to ensure that the wavefunction vani
shes at the 'boundary' (where the potential grows large)." }}{PARA 0 "
" 0 "" {TEXT -1 153 "This 'boundary' value is state-dependent. One ite
rates the procedure by hand. One can also write a bisection algorithm \+
to automate the eigenvalue search." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "IC:=u(0)=1,D(u)(0)=0;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#ICG6$/-%\"uG6#\"\"!\"\"\"/--%\"DG6#F(F)F*" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 11 "E_t:=0.668;" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 39 "SE:=-1/2*diff(u(x),x$2)+(V-E_t)*u(x)=0;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 55 "sol:=dsolve(\{SE,IC\},u(x),numeric,output=listprocedu
re):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u_s:=subs(sol,u(x)):" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_tG$\"$o'!\"$" }}{PARA 11 "" 1 "" 
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{PARA 0 "" 0 "" {TEXT -1 99 "Now we change the normalization on the re
sult of the matrix diagonalization to make the comparison:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "P2:=plot(phi_a[1]/subs(x=0,phi_a[1]
),x=0..4,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "pl
ots[display](P1,P2);" }}{PARA 13 "" 1 "" {GLPLOT2D 695 225 225 
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Fjq$\"0t'Gh,JXIF27$$\"0nm\"fBIY7Fjq$\"0UqL\"Qx`DF27$$\"0LLLO[kL\"Fjq$
\"0t=62ul-#F27$$\"0LLL&Q\"GT\"Fjq$\"0j?'>,\">j\"F27$$\"0++D2X;]\"Fjq$
\"0#*RszbVB\"F27$$\"0LLLvv-e\"Fjq$\"0r@xhh9P*F/7$$\"0++D2Ylm\"Fjq$\"0V
c<catn'F/7$$\"0++v\"ep[<Fjq$\"0[sBQ$oDYF/7$$\"0LL$e/TM=Fjq$\"0[^][$RnH
F/7$$\"0LLeDBJ\">Fjq$\"0VUyz]h#=F/7$$\"0nm;kD!)*>Fjq$\"0YJ!zOIP%*F\\x7
$$\"0nm\"f`@'3#Fjq$\"06\\2kNVT$F\\x7$$\"0++vw%)H;#Fjq$\"0/f)>Iv*H#Ffy7
$$\"0nm;$y*eC#Fjq$!0P[yk#*ee\"F\\x7$$\"0+++9b:L#Fjq$!0qU4^h/?#F\\x7$$
\"0++]5a`T#Fjq$!0VN;EQ=+#F\\x7$$\"0++D\"RV'\\#Fjq$!09X>+w@R\"F\\x7$$\"
0++]@fke#Fjq$!07Rp<C0N&Ffy7$$\"0LLL&4NnEFjq$\"0$z?0$z)p?Ffy7$$\"0+++:?
Pv#Fjq$\"0T&f$pV%>%)Ffy7$$\"0nm\"zM)>$GFjq$\"0VGnL=<A\"F\\x7$$\"0+++(f
a<HFjq$\"0\\65#ze49F\\x7$$\"0LLeg`!)*HFjq$\"0op#*[+\")Q\"F\\x7$$\"0++D
G2A3$Fjq$\"0(=$=w#*>?\"F\\x7$$\"0LLL)G[kJFjq$\"0k7q!\\g0\"*Ffy7$$\"0++
D\"yh]KFjq$\"0avgyuv[&Ffy7$$\"0nmm)fdLLFjq$\"0M54yL\"y>Ffy7$$\"0nm;q7%
=MFjq$!05^1.,/?\"Ffy7$$\"0LLe#pa-NFjq$!0N%o)\\$fTOFfy7$$\"0+++ad)zNFjq
$!0yV7<%z=^Ffy7$$\"0LL$GUYoOFjq$!0yfe?oU(eFfy7$$\"0nmm5:xu$Fjq$!0&H'*3
'=%pdFfy7$$\"0++D28A$QFjq$!0AwpBMx*\\Ffy7$$\"0++vS)38RFjq$!0y?!\\k?5QF
fy7$$\"\"%F)$!0%=G`4^wAFfy-F][l6&F_[lF(F($\"*++++\"!\")-%+AXESLABELSG6
%Q\"x6\"Q!6\"%(DEFAULTG-%%VIEWG6$;F(F^[mF^\\m" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 
0 "" 0 "" {TEXT 262 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 192 "C
heck the accuracy of the eigenenergies of the first and second excited
 states, and compare the graphs of the eigenfunctions as obtained by m
atrix diagonalization versus numerical integration." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 11 "Exercise 3:" }}{PARA 
0 "" 0 "" {TEXT -1 102 "Check the results obtained for different choic
es of the box size that defines the free-particle basis." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 478 "It is pos
sible to assess the accuracy of the matrix diagonalization without ref
erring to the numerical calculation. One can calculate the expectation
 value of the Hamiltonian in the coordinate representation for the giv
en eigenstate. This agrees with the eigenvalue obtained from the matri
x diagonalization. Then, one can ask about the distribution of energie
s in the state. The deviation from the average is calculated from the \+
matrix elements of the square of the Hamiltonian." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 209 "The functions to calcula
te these are defined below. However, they work only when the truncatio
n size of the matrix is relatively small. Otherwise too much memory is
 consumed in the evaluation of the integrals." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 65 "Havg:=psi->int(simplify(psi*(-1/2*diff(psi,x$2)+
V*psi)),x=-X..X);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%HavgGR6#%$psiG
6\"6$%)operatorG%&arrowGF(-%$intG6$-%)simplifyG6#*&9$\"\"\",&-%%diffG6
$F3-%\"$G6$%\"xG\"\"##!\"\"F=*&%\"VGF4F3F4F4F4/F<;,$%\"XGF?FEF(F(F(" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "E1_avg:=Havg(phi_a[1]);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'E1_avgG$\"-f2WX3n!#7" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "H2me:=psi->int(simplify((-1/2*diff(
psi,x$2)+V*psi)^2),x=-X..X);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%H2m
eGR6#%$psiG6\"6$%)operatorG%&arrowGF(-%$intG6$-%)simplifyG6#*$),&-%%di
ffG6$9$-%\"$G6$%\"xG\"\"##!\"\"F=*&%\"VG\"\"\"F8FBFBF=FB/F<;,$%\"XGF?F
FF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Edev:=sqrt(abs(E
1_avg^2-H2me(phi_a[1])));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%EdevG$
\"0]=s0FAm#!#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 283 "The deviation \+
is much larger than what we would have expected from our comparison wi
th the numerical value. Nevertheless, the number is proportional to th
e accuracy for the given state: when the matrix size is increased the \+
deviation from the average of the energy decreases for all " }{TEXT 
280 1 "x" }{TEXT -1 49 ". The large deviation is likely to come from t
he " }{TEXT 281 1 "x" }{TEXT -1 124 "-range where the approximate wave
 function is oscillating. This can be verified by restricting the calc
ulation to a smaller " }{TEXT 282 1 "x" }{TEXT -1 61 "-range where the
 true ground-state eigenfunction is non-zero." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "Another way to explore t
he accuracy of the matrix diagonalization result is to graph the local
 energy in position space. For an eigenstate the function " }{TEXT 
284 1 "f" }{TEXT -1 1 "(" }{TEXT 283 1 "x" }{TEXT -1 57 ") = (H*psi)/p
si should be equal to the eigenvalue at all " }{TEXT 285 1 "x" }{TEXT 
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "f:=psi->(-1/2*diff(psi,x$2))/psi+V;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"fGR6#%$psiG6\"6$%)operatorG%&arrowGF(,&*&-%%diff
G6$9$-%\"$G6$%\"xG\"\"#\"\"\"F1!\"\"#F8F6%\"VGF7F(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot(f(phi_a[1]),x=0..3,view=[0..3,
0..2]);" }}{PARA 13 "" 1 "" {GLPLOT2D 693 223 223 {PLOTDATA 2 "6%-%'CU
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$$\"0](=<WtMFFfp$\"0aFF+*ehwF_u7$$\"0v=UHTIu#Ffp$\"0Pk@,sGi(F_u7$$\"0+
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5GFfp$\"0!z@Rn5&o(F_u7$$\"0+vV!)fT(GFfp$\"0FWJi@8,)F_u7$$\"0+DcI;[$HFf
p$\"0e(>bF%3V)F_u7$$\"\"$F)$\"0Zlh#R%\\&*)F_u-%'COLOURG6&%$RGBG$\"#5!
\"\"F(F(-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6$;F(Fg_m;F($\"\"#F)" 1 2 
0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 451 "This graph should have a sobering
 effect for any enthusiasm that may have developed after obtaining a r
easonably accurate eigenvalue, and an approximate eigenfunction that a
ppeared to follow the numerically calculated one: The local energy cal
culation is a highly sensitive quantity. In particular, we notice that
 the basis-state representation in the 'particle-in-a-box' basis has d
ifficulties with obtaining the correct fall-off of the wavefunction." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 444 "We sho
uld not be deterred too much by this mixture of success and disappoint
ment: we should be aware of the fact that eigenenergies are much easie
r to obtain than accurate eigenfunctions, and proceed with cautiously.
 Matrix diagonalization is often the only tool available to solve the \+
Schroedinger equation. Usually we also have the possibility to improve
 matters by adjusting the basis to the problem at hand, and thereby re
ducing the errors." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 286 11 "Exercise 4:" }}{PARA 0 "" 0 "" {TEXT -1 218 "Graph the l
ocal energy for matrix diagonalization solutions differing by matrix s
ize, and observe how the better converged calculations approximate the
 eigenvalue locally. Repeat the calculation for a smaller value of " }
{TEXT 287 1 "X" }{TEXT -1 7 " (with " }{TEXT 288 1 "X" }{TEXT -1 112 "
 larger than the point at which the numerically obtained wavefunction \+
goes to zero), and make your observations." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "We carry out the calcul
ation for the first excited state:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "ev_s[2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"3<8%oza
,)4C!#<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "IC:=u(0)=0,D(u)(
0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6$/-%\"uG6#\"\"!F*/--%
\"DG6#F(F)\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_t:=2.
392;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "SE:=-1/2*diff(u(x),x$2)+(V-
E_t)*u(x)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "sol:=dsolve(\{SE,IC
\},u(x),numeric,output=listprocedure):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "u_s:=subs(sol,u(x)):" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_t
G$\"%#R#!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&-%%diffG6$-%
\"uG6#%\"xG-%\"$G6$F-\"\"##!\"\"F1*&,&*$)F-\"\"%\"\"\"F9$\"%#R#!\"$F3F
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "We observe that the numerically o
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45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 124 "We have shifted the graph of the eigenfunction to indi
cate the location of the classical turning point (the crossing of the \+
" }{TEXT 265 1 "x" }{TEXT -1 168 "-axis with the potential curve). The
 plot shows size of the tunneling region, as well as the fact that the
 region where the numerical eigenfunction touches down to the " }
{TEXT 266 1 "x" }{TEXT -1 293 "-axis (it diverges for large x, due to \+
the finite accuracy of the trial eigenenergy, and due to numerical pro
blems) is well inside the classically forbidden region for the given s
tate. For higher-lying states one needs to go to larger x-values for t
he 'touchdown' point. Eventually, for large " }{TEXT 267 1 "n" }{TEXT 
-1 86 " the basis set for the matrix diagonalization will be inappropr
iate due to the finite " }{TEXT 268 1 "X" }{TEXT -1 7 "-value." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 714 "
It is important to explore the accuracy of states higher than the grou
nd states of the even- and odd-parity sectors. These states are harder
 to calculate, since one needs to find the locations of nodes that are
 not pre-determined. The eigenstates calculated from the matrix diagon
alization method are all mutually orthogonal. Given their approximate \+
nature, however, one cannot claim that the second excited state is ort
hogonal to the exact ground state. Experience shows that the accuracy \+
of the higher-lying eigenvalues (and their eigenfunctions) is substant
ially less than what was found for the two lowest states (for the same
 hamiltonian matrix). One can try to beat the problem by increasing th
e matrix size." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 92 "We demonstrate the problem on the example of the first ex
citation in the even-parity sector:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "ev_s[3];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"3D7E8t/
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "We observe that the numerically o
btained eigenvalues are slightly below the ones obtained from the matr
ix diagonalization." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "P2:=
plot(phi_a[3]/subs(x=0,phi_a[3]),x=0..4,color=blue):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 35 "P3:=plot(V-E_t,x=0..3,color=green):" }}}
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9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3
" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 549 "We observe that the third e
igenvalue is not approximated well by the 20-by-20 matrix diagonalizat
ion. This could be improved upon by simply choosing a smaller box. The
 numerical results for the wavefunction combined with a demonstration \+
of the vicinity of the classical turning point (beyond which the tail \+
of the wavefunction indicates that tunneling takes place) indicate tha
t the box size can be chosen to be substantially smaller. This would l
ead to more accurate results, as the basis functions would be more fle
xible in the region of interest." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 295 96 "Solution of \+
the numerical problem by bisection for the general case of non-symmetr
ic potentials." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 30 "We start the integration from " }{TEXT 272 1 "x" }{TEXT 
-1 3 "= -" }{TEXT 271 1 "s" }{TEXT -1 5 " and " }{TEXT 270 2 "x " }
{TEXT -1 2 "= " }{TEXT 269 1 "s" }{TEXT -1 108 ", and try to match the
m at the origin, by requesting that u'(0)/u(0) match from the left and
 from the right." }}{PARA 0 "" 0 "" {TEXT -1 418 "The boundary conditi
on at these two points distinguishes even- and odd-parity states. For \+
even-parity states the derivatives of the wavefunction are opposite to
 each other, for odd-parity states they are identical. The choices det
ermine the normalization of the propagated solution. Given that we mat
ch  u'(0)/u(0) this should not matter at all. Let us illustrate what h
appens as we integrate from the outside inwards." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 6 "s:=3.;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
sG$\"\"$\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eta:=3*10^
(-4):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "IC1:=u(s)=0,D(u)(s
)=eta: IC2:=u(-s)=0,D(u)(-s)=-eta:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_t:=4.
697;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "SE:=-1/2*diff(u(x),x$2)+(V-
E_t)*u(x)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "sol1:=dsolve(\{SE,I
C1\},u(x),numeric,output=listprocedure): u_1:=subs(sol1,u(x)):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "sol2:=dsolve(\{SE,IC2\},u(x),numeri
c,output=listprocedure): u_2:=subs(sol2,u(x)):" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$E_tG$\"%(p%!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
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COLOURG6&%$RGBG$\"#5!\"\"Fa`mFa`m-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6
$;F(F_`m;$Fh`mF*$\"\"\"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 96 "A matc
hing method relies on the fact that we wish the following properties f
or an eigenfunction:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 45 "normalizability, i.e., vanishing function at " }{TEXT 
274 1 "s" }{TEXT -1 6 " and -" }{TEXT 273 1 "s" }{TEXT -1 1 ";" }}
{PARA 0 "" 0 "" {TEXT -1 44 "continuous and differentiable solution fo
r -" }{TEXT 277 1 "s" }{TEXT -1 3 " < " }{TEXT 276 1 "x" }{TEXT -1 3 "
 < " }{TEXT 275 1 "s" }{TEXT -1 1 ";" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 199 "As the Schroedinger equation is of sec
ond order, we can produce two linearly independent solutions that both
 depend on the same trial value for the eigenenergy. One depends on in
itial conditions at -" }{TEXT 279 1 "s" }{TEXT -1 40 ", the other on t
wo intial conditions at " }{TEXT 278 1 "s" }{TEXT -1 267 ". We calcula
te the difference between u'(0)/u(0) for the integrations from the lef
t and from the right. We claim that this parameter crosses zero at an \+
energy eigenvalue, and therefore we can invoke a root search. The latt
er is carried out by the bisection algorithm." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 78 "Match:=proc(E_t) local SE,eta,IC1,IC2,sol1,sol2,
u_1,up_1,u_2,up_2; global V,s;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "S
E:=-1/2*diff(u(x),x$2)+(V-E_t)*u(x)=0; eta:=3*10^(-4):" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 52 "IC1:=u(s)=0,D(u)(s)=eta: IC2:=u(-s)=0,D(u)(-s)
=-eta:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "sol1:=dsolve(\{SE,IC1\},u
(x),numeric,output=listprocedure): u_1:=subs(sol1,u(x)):" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 79 "sol2:=dsolve(\{SE,IC2\},u(x),numeric,output=l
istprocedure): u_2:=subs(sol2,u(x)):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 61 "up_1:=subs(sol1,diff(u(x),x)): up_2:=subs(sol2,diff(u(x),x)):" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "up_1(0)/u_1(0)-up_2(0)/u_2(0); end
:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "The bisection algorithm is \+
written with two procedures: a basic bisection step, and a driver whic
h reports on progress:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "
Bis1:=proc(x1,x2,x3,f1,f2,f3);\nif evalf(f1*f3) < 0 then\nRETURN([x1,x
3,f1,f3]);\nelse\nRETURN([x3,x2,f3,f2]); fi;\nend:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 482 "Bisect:=proc(f,a,b) local res,x1,x2,x3,f1,
f2,f3,i;x1:=a: x2:=b: f1:=evalf(f(x1)): f2:=evalf(f(x2)):\nif evalf(f1
*f2)>0 then RETURN(\"No bracketed root\",f1,f2);\nelse\nx3:=0.5*(x1+x2
); f3:=evalf(f(x3)); fi;\nfor i from 1 to 50 do:\nres:=Bis1(x1,x2,x3,f
1,f2,f3);\nx1:=res[1]; f1:=res[3]; x2:=res[2]; f2:=res[4]; x3:=0.5*(x1
+x2);\nif abs(x1-x2) < 10^(-2) then print(\"Reached level 2\",x3); fi;
 if abs(x1-x2) < 10^(-7) then RETURN(x3) fi;\nf3:=evalf(f(x3)); od:\nR
ETURN(\"Loop exhausted\", x3); end:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "s:=3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sG\"\"$" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Bisect(Match,4.65,4.75);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\")](op%!\"(
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"*]7`p%!\")
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"+]P4'p%!\"*
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\",]P%['p%!#5
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"-](ozmp%!#6
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\".]PMxnp%!#7
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"/](=<Eop%!#
8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"0]7yv,op%!
#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"0D\"y]&*y
'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"0(oHa
cz'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"01R
Dg#z'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"0
(zTGTz'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$
\"0Ud8*[z'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26
\"$\"0:FGF&z'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~
26\"$\"0-iNY&z'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~leve
l~26\"$\"0f%>o`z'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~le
vel~26\"$\"0JyeT&z'p%!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"0JyeT&z
'p%!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Let us pick a non-symme
tric potential and calculate the eigenvalues:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "V:=x^4-2*x^
3-x^2/2+2/2*x+1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,,*$)%\"xG\"
\"%\"\"\"F**&\"\"#F*)F(\"\"$F*!\"\"*&#F*F,F**$)F(F,F*F*F/F(F*F*F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "P0:=plot(V,x=-5..5,view=[-2.
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{MPLTEXT 1 0 5 "s:=3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sG\"\"$" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Bisect(Match,0,1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\")vVB!*!\")" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Reached~level~26\"$\"*D1R+*!\"*" }}
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~level~26\"$\"0z(\\mX=$**)!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q0Rea
ched~level~26\"$\"0!\\SuL=$**)!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$Q
0Reached~level~26\"$\"0N^/(R=$**)!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$Q0Reached~level~26\"$\"08GCn$=$**)!#:" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"08GCn$=$**)!#:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "E
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "To graph the solution we use the \+
obtained best eigenvalue, and integrate the Schroedinger equation once
 more:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eta:=1*10^(-6):" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "IC1:=u(s)=0,D(u)(s)=eta: \+
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8 "E_t:=E1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "SE:=-1/2*diff(u(x),x
$2)+(V-E_t)*u(x)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "sol1:=dsolve
(\{SE,IC1\},u(x),numeric,output=listprocedure): u_1:=subs(sol1,u(x)):
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "sol2:=dsolve(\{SE,IC2\},u(x),nu
meric,output=listprocedure): u_2:=subs(sol2,u(x)):" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$E_tG$\"08GCn$=$**)!#:" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#SEG/,&-%%diffG6$-%\"uG6#%\"xG-%\"$G6$F-\"\"##!\"\"F1
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E1,x=-3..3,color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 
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Fgq$\"0r:_8_j@&F17$Fjq$\"0))y$\\\")*R^&F17$Faam$\"0\")[\\s&f6cF17$F]r$
\"0,BR;<!zcF17$$\"0]7G:h+G\"F1$\"02p\")f>9q&F17$Fiam$\"0@\\3E\"p7dF17$
$\"0D1R0!>G8F1$\"0_!*H[ZRr&F17$$\"0]P4-LUM\"F1$\"0z!zpE=7dF17$$\"0voz)
fFg8F1$\"0FmwvEtq&F17$F`r$\"0\"4n!)[J*p&F17$Fabm$\"0-39P?*RcF17$Fcr$\"
03TDRJF`&F17$Fibm$\"0@kp*eko`F17$Ffr$\"0Z%[`^F_^F17$Facm$\"0+%e!yaI*[F
17$Fir$\"06*RaEZ\"f%F17$Ficm$\"01m&G=]PUF17$F\\s$\"0N;kE\"f`QF17$Fadm$
\"0!)eM>gaY$F17$F_s$\"0*))G_#f&pIF17$Fidm$\"0G\"zJ'fsm#F17$Fbs$\"0#[KP
#exF#F17$Faem$\"0**zPToI\">F17$Fes$\"0TOj&G6x:F17$Fiem$\"0cDPL$G)H\"F1
7$Fhs$\"0\\b^B#p]5F17$F[t$\"09c-=\"HUgF-7$F^t$\"0CL)f)*\\rLF-7$Fat$\"0
X&Q*[m>h\"F-7$Fdt$\"0^\"*3)*zWU'F`im7$Fgt$\".y%3i2!p&Fbhm-Fjt6&F\\uFbf
lFbflFcfl-%+AXESLABELSG6%Q\"x6\"Q!6\"%(DEFAULTG-%%VIEWG6$;F(Fgt;$!\"\"
F*$\"\"'F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "C
urve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 233 "We have chosen the initial condition for solution 2 to b
e such that a positive ground-state wavefunction is calculated. Soluti
on 1 (which is integrated left-to-right), is small, and cannot be seen
 on this scale (it is covered by the " }{TEXT 289 1 "x" }{TEXT -1 60 "
-axis). Both solutions are normalized through the choice of " }{TEXT 
19 3 "eta" }{TEXT -1 135 ". It is interesting to observe how the wavef
unction reaches through the classically forbidden regime to extend ove
r the second minimum." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT 290 11 "Exercise 5:" }}{PARA 0 "" 0 "" {TEXT -1 115 "Calculat
e some of the excited eigenstates for this potential by supplying othe
r bracketing values to the procedure " }{TEXT 19 6 "Bisect" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 291 11 "
Exercise 6:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 167 "Modify the
 potential function such that there be one broad, and one sharp minimu
m with approximately equal depth by choosing different coeffients for \+
the monomials in " }{TEXT 292 1 "x" }{TEXT -1 162 ". Calculate the gro
und state and observe its shape. When does it become double-humped, an
d when does the second hump (over the narrower potential well) disappe
ar?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
293 23 "Matrix diagonalization:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 172 "We can get an idea about the location of
 the low-lying eigenvalues by carrying out the matrix diagonalization.
 We simply copy the lines from the beginning of the worksheet." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "HM:=Matrix(N):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 114 "We make use of the symmetry of the hamil
tonian matrix: it allows to save almost a factor of 2 in computation t
ime." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "for i from 1 to N \+
do: for j from 1 to i do: HM[i,j]:=Vpot(uB(i),uB(j)); if i=j then HM[i
,j]:=HM[i,j]+ Tkin(uB(i)); fi: od: od:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "for i from 1 to N do: for j from i+1 to N do: HM[i,j]
:=HM[j,i]: od: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "print
(SubMatrix(HM,1..4,1..4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABL
EG6$\")/vU9-%'MATRIXG6#7&7&,&*&,(*$)%#PiG\"\"%\"\"\"\"&V@\"*&\"'+[CF3)
F1\"\"#F3!\"\"\"(gXZ\"F3F3*$F0F3F9#F3\"#:*&#F3\"$7&F3F7F3F3,$*&,&*$F7F
3\"%\\6\"&S-\"F9F3*$F0F3F9#!$c#\"#F,$*&,&!$G\"F3*&\"#<F3F7F3F3F3*$F0F3
F9!$?(,$*&,&FD\"&D(G\"&;[$F9F3*$F0F3F9#F@\"&vo\"7&FA,&*&,(\"&g@*F3*&\"
&+7'F3F7F3F9*&F4F3F0F3F3F3*$F0F3F9F<*&#F3\"$G\"F3F7F3F3,$*&,&FD\"%v&*
\"&s)zF9F3*$F0F3F9#\"$o(\"$D',$*&,&!%C5F3*&\"$`\"F3F7F3F3F3*$F0F3F9#!%
gDFJ7&FKFao,&*&,(FD!'+[C\"'SQ;F3*&\"'(G4\"F3F0F3F3F3*$F0F3F9#F3\"$N\"*
&#\"\"*F@F3F7F3F3,$*&,&FD\"&n(=\"'+O:F9F3*$F0F3F9#!%O:\"%,C7&FSFjoFaq,
&*&,(FD!&+`\"*&F4F3F0F3F3\"%gdF3F3*$F0F3F9F<*&#F3\"#KF3F7F3F3" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "The Hamiltonian matrix is no longe
r sparse, as the potential is not symmetric." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 19 "HMf:=map(evalf,HM):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "evals:=Eigenvalues(HMf, output='list'):" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 26 "ev_s:=sort(map(Re,evals));" }}{PARA 12 "" 1 "
" {XPPMATH 20 "6#>%%ev_sG76$\"3!RlTH\"f;P#*!#=$\"3)34hE9=r)>!#<$\"3@ty
lV)3z#QF+$\"3yrN@!4\\+'oF+$\"3sy1f+8>78!#;$\"3qM(>%>U!Hk#F2$\"3X@>\"f/
vDa%F2$\"3)3Yw>`A)f#)F2$\"38\\eQ05qK7!#:$\"3mFZ#)pj#*G?F;$\"3w)*y8;Q*z
v#F;$\"3]:^a\"39AA%F;$\"3KUO\\-gP&R&F;$\"3$Rf2--3e#yF;$\"3S0@XGYD&e*F;
$\"3\")3'pIPBBL\"!#9$\"3e(zZHanVe\"FJ$\"3M(3neX[N7#FJ$\"3'[!R0Ev%eZ#FJ
$\"3['ocLkvnp$FJ" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "VE:=Eig
envectors(HMf,output='list'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 66 "Vp:=[seq([Re(VE[i][1]),VE[i][2],map(Re,VE[i][3])],i=1..nops(VE))
]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "VEs:=sort(Vp,Min):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 1 to 4 do:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "psi0:=add(VEs[i][3][j]*uB(j),j=1..N
):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "No:=1/sqrt(int(expand(psi0^2)
,x=-X..X));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "phi_a[i]:=add(No*VEs
[i][3][j]*uB(j),j=1..N): od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 66 "plot([seq(phi_a[i],i=1..4)],x=-5..5,color=[red,blue,green,black]
);" }}{PARA 13 "" 1 "" {GLPLOT2D 695 232 232 {PLOTDATA 2 "6(-%'CURVESG
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45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 138 "We observe that the matrix diagonalizati
on gets the right idea about the ground-state eigenfunction (the eigen
value is off by nearly 3 %)." }}{PARA 0 "" 0 "" {TEXT -1 83 "The first
 excited state has a bigger hump over the shallower part of the potent
ial." }}{PARA 0 "" 0 "" {TEXT -1 180 "The higher states are not unlike
 the quartic harmonic oscillator eigenstates, but shifted to the right
. For higher eigenenergies the particles explore mostly the steep rise
 in the " }{TEXT 19 3 "x^4" }{TEXT -1 23 "-part of the potential." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 294 11 "Exercise
 7:" }}{PARA 0 "" 0 "" {TEXT -1 181 "Choose a potential with a deep an
d narrow potential well, and make a detailed comparison between the gr
ound state as obtained by matrix-diagonalization, and by the numerical
 method." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0
 0" 46 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }
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