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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 64 "Harmonic and anharmonic m
otion in the microscopic, quantum world" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 300 "We will develop some intuition for \+
the description of motion in the microscopic, i.e., quantum domain. We
 start with the harmonic oscillator, for which we develop solutions to
 Schroedinger's equation in analytic form, and then we move on to solv
e the problem numerically for an anharmonic oscillator." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "Without further mot
ivation we state that the allowed energies for a quantum particle boun
d by a potential " }{TEXT 259 1 "V" }{TEXT -1 1 "(" }{TEXT 258 1 "x" }
{TEXT -1 57 ") is given by a differential-equation eigenvalue problem:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "V:=k/2*x^2;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "SE:=-h_^2/(2*m)*diff(psi(x),x$2)+V*
psi(x)=lambda*psi(x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "Let us m
ake life simple by adopting Bohr units (h_=m_e=e=1), and by choosing k
=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "h_:=1: m:=1: k:=1:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "SE;" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 146 "The physical constraints on this problem are that t
he wavefunction psi has to vanish at infinity, i.e., for large positiv
e and negative values of " }{TEXT 257 1 "x" }{TEXT -1 1 "." }}{PARA 0 
"" 0 "" {TEXT -1 112 "We will be interested only in such solutions, i.
e., we will discard solutions that do not vanish asymptotically." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "On the o
ther hand one can show that the solutions have to be either symmetric,
 or antisymmetric functions, i.e., " }}{PARA 0 "" 0 "" {TEXT -1 50 "ei
ther:  psi(x) = psi(-x), or:  psi(x) = - psi(-x)" }}{PARA 0 "" 0 "" 
{TEXT -1 76 "for the acceptable solutions. These conditions translate \+
into conditions at " }{TEXT 260 1 "x" }{TEXT -1 16 " = 0 as follows:" 
}}{PARA 0 "" 0 "" {TEXT -1 12 "either: psi(" }{TEXT 261 1 "x" }{TEXT 
-1 58 ") = const, D(psi)(0)=0 ; or:  psi(0)=0, D(psi)(0) = const." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 251 "The two \+
undetermined constants fix the amplitude of the solution, and we recog
nize that this amplitude is arbitrary from a mathematical point of vie
w. This is the consequence of solving a homogeneous problem. Let us st
art with the symmetric solutions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICs:=psi(0)=1,D(psi)(0)=0;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sol:=dsolve(\{SE,ICs\},ps
i(x));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "So, it's not as simple
 as that in Maple... Let us substitute a number for lambda (we are usi
ng special units, so the lambda - value had dimensions of energy." }}
{PARA 0 "" 0 "" {TEXT -1 209 "Let us try whether the energy can be zer
o, i.e., whether we can have a particle sitting at the bottom of the w
ell, which corresponds to having the mass simply at rest in the equili
brium position of the spring:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 44 "sol:=dsolve(\{subs(lambda=0,SE),ICs\},psi(x));" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 41 "plot(rhs(sol),x=-5..5,view=[-5..5,0..5]);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(subs(x=-1,rhs(sol
)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "We don't get an acceptabl
e answer for negative " }{TEXT 264 1 "x" }{TEXT -1 43 ", but presumabl
y we could use the positive-" }{TEXT 263 1 "x" }{TEXT -1 66 " results,
 and use the symmetry to extend the solution to negative " }{TEXT 262 
1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 40 "The reason for no
t plotting at negative " }{TEXT 267 1 "x" }{TEXT -1 33 " is that the s
olution has a sqrt(" }{TEXT 268 1 "x" }{TEXT -1 42 ") factor, which is
 imaginary for negative " }{TEXT 269 1 "x" }{TEXT -1 8 " values." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "The main \+
point is that the solution is not acceptable for large " }{TEXT 265 1 
"x" }{TEXT -1 97 ". It blows up. The question is: Can we find energy v
alues such that the solution is normalizable?" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Strategy: try lambda=1/4,
 and lambda=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsol
ve(\{subs(lambda=1/4,SE),ICs\},psi(x));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "plot(rhs(sol),x=-5..5,view=[-5..5,0..5]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsolve(\{subs(lambda=3/4,SE),I
Cs\},psi(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "plot((rhs(
sol)),x=-5..5,view=[-5..5,-5..5],numpoints=500);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 68 "Somewhere in-between there should be a solution th
at approaches the " }{TEXT 266 1 "x" }{TEXT -1 21 "-axis asymptoticall
y." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsolve(\{subs(la
mbda=1/2,SE),ICs\},psi(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
41 "plot(rhs(sol),x=-5..5,view=[-5..5,0..2]);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 219 "To complete the interpretation of the Schroedinger wa
ve equation result we compute the probability density for this state. \+
The square of the wave function is providing the probability for findi
ng a particle at location " }{TEXT 293 1 "x" }{TEXT -1 140 ". For this
 interpretation to hold we need to normalize the solution of the homog
eneous Schroedinger eigenvalue problem in the following way:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "int(rhs(sol)^2,x=-infinity..
infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "WF:=1/sqrt(%)
*rhs(sol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "int(WF^2,x=-i
nfinity..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 259 "Now we gr
aph the probability in such a way as to show that there is a finite pr
obability to find the quantum particle outside the classically allowed
 range. For this we plot the density superimposed on the potential whi
le using the energy value as a baseline:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "plot([V,1/2,WF^2+1/2],x=-2..2,color=[red,blue,green])
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "We see that the classical tu
rning point for a particle of energy 1/2 is at " }{TEXT 295 1 "x" }
{TEXT -1 10 " = 1, and " }{TEXT 294 1 "x" }{TEXT -1 12 " = -1. For |" 
}{TEXT 296 1 "x" }{TEXT -1 546 "| > 1 a classical particle is not to b
e found with this energy value, since the kinetic energy is positive (
by definition), and the potential energy exceeds the value of 1/2 in t
his range. Yet the quantum particle (which does not follow any traject
ory) has a finite probability to be found in the classically forbidden
 regime. This regime is called the quantum tunneling region. It is a f
eature of Schroedinger solutions of all finite potentials that quantum
 particles penetrate beyond the classical turning points for the given
 amount of energy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
256 "" 0 "" {TEXT -1 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 202 "
Search for the next two acceptable valuea of lambda. Hint: use the asy
mptotic behaviour of the solution to determine whether an energy (lamb
da) value can be found such that the solution merges with the " }
{TEXT 270 1 "x" }{TEXT -1 83 "-axis asymptotically. Tabulate the lambd
a-values together with the one found above." }}{PARA 0 "" 0 "" {TEXT 
-1 78 "Graph the probability distributions for the properly normalized
 wavefunctions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 61 "We proceed with the determination of antisymmetric s
olutions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICa:=psi(0)=0,
D(psi)(0)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsolve
(\{subs(lambda=1/2,SE),ICa\},psi(x));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "plot(rhs(sol),x=-5..5,view=[-5..5,0..2]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsolve(\{subs(lambda=5/2,SE),I
Ca\},psi(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot(rhs(s
ol),x=-5..5,view=[-5..5,-2..2]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "sol:=dsolve(\{subs(lambda=3/2,SE),ICa\},psi(x));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot(rhs(sol),x=-5..5,view=[
-5..5,-2..2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "We carry out th
e normalization of the state, and display the probability distribution
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "int(rhs(sol)^2,x=-infi
nity..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "WF:=1/s
qrt(%)*rhs(sol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "int(WF^
2,x=-infinity..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 259 "Now
 we graph the probability in such a way as to show that there is a fin
ite probability to find the quantum particle outside the classically a
llowed range. For this we plot the density superimposed on the potenti
al while using the energy value as a baseline:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 58 "plot([V,3/2,WF^2+3/2],x=-2.5..2.5,color=[red,b
lue,green]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 184 "We observe again
 the fact that the quantum particle penetrates into the classically fo
rbidden region. We also observe that the particle has zero probability
 to be found at the origin (" }{TEXT 297 1 "x" }{TEXT -1 220 "=0) when
 it is in an antisymmetric state. Given that there are no continuous t
rajectories for quantum particles this is not a problem, however: the \+
interpretation of the quantum probability distribution is the followin
g:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 467 "If
 repeated measurements are made on an ensemble of particles prepared i
n a given eigenstate (the ensemble can be simultaneous, or there could
 be one particle at a time with repeated measurements), then each time
 the probability is recorded (a histogram is made of the position for \+
the ensemble), the distribution will be an approximation to the above \+
curve. There is no meaning to to question as to how an individual quan
tum particle moved in such a way as to cross " }{TEXT 298 1 "x" }
{TEXT -1 3 "=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT 271 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 69 "Find tw
o more anti-symmetric solutions, and tabulate the eigenvalues." }}
{PARA 0 "" 0 "" {TEXT -1 98 "Combine these eigenvalues with the ones f
ound for the symmetric case, and observe how they behave." }}{PARA 0 "
" 0 "" {TEXT -1 79 "Graph the probability distributions using the prop
erly normalized wavefunction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 272 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 
40 "Change the value of the spring constant " }{TEXT 273 1 "k" }{TEXT 
-1 7 ". Pick " }{TEXT 275 1 "k" }{TEXT -1 7 "=2 and " }{TEXT 274 1 "k
" }{TEXT -1 118 "=4 respectively, and determine the lowest 6 eigenvalu
es in each case. Draw your conclusions about the energy spectrum." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 278 11 "Exercise
 4:" }}{PARA 0 "" 0 "" {TEXT -1 38 "Keep the value of the spring const
ant " }{TEXT 277 1 "k" }{TEXT -1 27 "=1, but change the mass to " }
{TEXT 276 1 "m" }{TEXT -1 110 "=2 (in Bohr units). Determine the 6 low
est energy values, and draw your conclusions about the energy spectrum
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
85 "Let us go back to the general case, and show what the exact eigenf
unctions look like:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "V:=1
/2*m*omega^2*x^2;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "SE:=-h_^2/(2*m
)*diff(psi(x),x$2)+V*psi(x)=lambda*psi(x);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "alpha:=sqrt(m*omega/h_);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 16 "with(orthopoly);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "phi:=n->exp(-1/2*(alpha*x)^2)*H(n,alpha*x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs(psi(x)=phi(0),SE);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(%,lambda);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 66 "Clearly, the ground-state energy equals 1/2 ome
ga. Let's continue:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs
(psi(x)=phi(1),SE);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(%,lambda);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "For the first excited state we fi
nd that the energy equals 3/2 omega. Substitute different values of " 
}{TEXT 279 1 "n" }{TEXT -1 36 " and verify the eigenvalue spectrum." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "The onl
y remaining point concerns the normalization of the eigenfunctions. Th
e interpretation of the functions phi(" }{TEXT 281 1 "n" }{TEXT -1 2 "
, " }{TEXT 280 1 "x" }{TEXT -1 81 ") is that their square provides the
 probability to find the particle at location " }{TEXT 282 1 "x" }
{TEXT -1 27 ". Therefore the sum of phi(" }{TEXT 285 1 "x" }{TEXT -1 
2 ", " }{TEXT 284 1 "n" }{TEXT -1 23 ")^2 over all locations " }{TEXT 
283 1 "x" }{TEXT -1 134 ", i.e., the integral is chosen to equal one. \+
The functions as defined do not meet that requirement, and one has to \+
find an omega- and " }{TEXT 286 1 "n" }{TEXT -1 52 "-dependent factor \+
to achieve this goal. For example:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "assume(omega>0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "int(phi(5)^2,x=-infinity..infinity);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 44 "The normalization factor can be found to \+
be:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "c:=n->sqrt(sqrt(m*om
ega/(h_*Pi))/(2^n*n!));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "c
(5)^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "Therefore, when the ei
genfunction for the nth state is multiplied by c(n) it will lead to a \+
normalized probability distribution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT 305 22 "Anharmonic oscillators" }}
{PARA 0 "" 0 "" {TEXT -1 316 "For other potentials (e.g., an x^4 anhar
monic oscillator) the eigenvalues and eigenfunctions cannot be found i
n terms of simple functions. Nevertheless, we can apply numerical tech
niques to find approximate eigenvalues and eigenfunctions using dsolve
[numeric]. We need to apply an iteration in the energy eigenvalue:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 7 "V:=x^4;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "SE:=-1/2*diff(psi(x),x$2)+V*psi(x)=E_t*psi(x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "unprotect(Psi);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICs:=psi(0)=1,D(psi)(0)=0;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_t:=0.668;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,ICs\},psi(x),numer
ic,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
32 "plot('Psi(x)',x=0..4,-0.5..1.5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 171 "The energy trial value was determined by trial and error. An e
nergy value (and the corresponding eigenfunction) is considered accept
able when the function is close to the " }{TEXT 287 1 "x" }{TEXT -1 
95 "-axis for some range. This range can be extended by tuning the ene
rgy value to higher accuracy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 288 11 "Exercise 5:" }}{PARA 0 "" 0 "" {TEXT -1 
314 "Find approximate eigenenergies for two more eigenvalues (above th
e ground state). Observe the nodal structure of the wavefunctions and \+
compare with the exact harmonic oscillator result above. Calculate als
o several anti-symmetric eigenfunctions by changing the initial condit
ions for the integration appropriately." }}{PARA 0 "" 0 "" {TEXT -1 
106 "Compare the spacing between these eigenvalues with the spacing pr
operty found for the harmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT 292 11 "Exercise 6:" }}{PARA 0 "" 0 "" 
{TEXT -1 85 "Solve numerically for the lowest six eigenvalues and eige
nfunctions of the potential " }{TEXT 291 1 "V" }{TEXT -1 1 "(" }{TEXT 
290 1 "x" }{TEXT -1 3 ")=|" }{TEXT 289 1 "x" }{TEXT -1 99 "|. Discuss \+
the properties of the spacing of the eigenvalues as compared to the ha
rmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 428 "Note that a systematic method to improve the trial values for \+
the eigenenergies involves a bisection: one can systematically cut the
 energy range defined by two brackets (with a solution going to positi
ve infinity at large x for one bracketing value, and to negative infin
ity for the other) in half, and determine which half contains the asym
ptotically acceptable solution to reduce the bracket systematically by
 factors of two." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Here we do the exercise for the li
near potential, and write a bisection algorithm." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "V:=x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "SE
:=-1/2*diff(psi(x),x$2)+V*psi(x)=E_t*psi(x);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 15 "unprotect(Psi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "ICs:=psi(0)=1,D(psi)(0)=0;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 82 "By trial and error we find that the lowest eigenvalue i
s bracketed by 0.8 and 0.9:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "E_t:=0.825;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dso
lve(\{SE,ICs\},psi(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 32 "plot('Psi(x)',x=0..4,-0.5..1.5);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "We choose a sufficiently large val
ue of " }{TEXT 299 1 "x" }{TEXT -1 42 " for which we want the solution
 to vanish:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "X:=4;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "SEroot:=proc(lambda) local S
E,sol,Psi; global ICs,X,V;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "SE:=-
1/2*diff(psi(x),x$2)+V*psi(x)=lambda*psi(x);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 74 "sol:=dsolve(\{SE,ICs\},psi(x),numeric,output=listproc
edure): unprotect(Psi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=sub
s(sol,psi(x)):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Psi(X); end:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 35 "fsolve('SEroot(E_t)',E_t,0.8..0.9);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 43 "Unfortunately, it is not as easy as that.
.." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "We
 first write a basic bisection step: given a map f and a bracketing in
terval, we come up with a new bracket for the root which is half the s
ize:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 245 "Bisect:=proc(f,xL)
 local x1,x2,xh,f1,f2,fh; x1:=xL[1]: x2:=xL[2]: f1:=f(x1): f2:=f(x2): \+
if f1*f2>0 then RETURN(\"Same sign on both ends of the bracket, cannot
 proceed\") fi: xh:=0.5*(x1+x2): fh:=f(xh): if f1*fh<0 then [x1,xh] el
se [xh,x2] fi: end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Bise
ct(SEroot,[0.8,0.9]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Now we n
eed a driver, to drive the trial energy bracket to some desired tolera
nce." }}{PARA 0 "" 0 "" {TEXT -1 100 "In this procedure we limit ourse
lves to a maximum of 100 iterations. We define an internal variable " 
}{TEXT 19 2 "rL" }{TEXT -1 105 " to represent the bracket as a list. W
e cannot overwrite the list that was passed down originally, i.e.. " }
{TEXT 19 2 "xL" }{TEXT -1 71 " (Maple syntax). When successful we retu
rn the midpoint of the bracket." }}{PARA 0 "" 0 "" {TEXT -1 33 "We als
o catch the incidence when " }{TEXT 19 6 "Bisect" }{TEXT -1 49 " retur
ns with an error message and the result of " }{TEXT 19 2 "rL" }{TEXT 
-1 16 " is not of type " }{TEXT 19 4 "list" }{TEXT -1 1 "." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 268 "SEsolve:=proc(xL,tol) local i,rL; \+
rL:=xL: for i from 1 to 100 do: rL:=Bisect(SEroot,rL): if not type(rL,
list) then RETURN(rL) fi: if abs(rL[2]-rL[1]) < evalf(tol) then RETURN
(0.5*(rL[1]+rL[2])) else fi: od: RETURN(\"Did not converge in 100 iter
ations in SEsolve\"); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "SEsolve([0.8,0.9],0.001);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "
We should keep in mind that the number is only accurate to the level d
etermined by the variable " }{TEXT 19 3 "tol" }{TEXT -1 1 ":" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "SEsolve([0.8,0.9],0.00001);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "SEsolve([0.8,0.9],0.000
0001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "E_t:=%;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,ICs\},psi(x
),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "plot('Psi(x)',x=0..4,-0.5..1.5);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 59 "Note what can happen when the original bracket is \+
too wide:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,1
.0],0.001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1
,2.0],0.001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0
.1,3.0],0.001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve(
[0.1,4.0],0.001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolv
e([0.1,5.0],0.001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "SEso
lve([4.,5.0],0.001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEs
olve([0.1,6.0],0.001);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "The pr
ogram still requires some attention, but the process of finding the ei
genvalues is automated to some extent." }}{PARA 0 "" 0 "" {TEXT -1 
123 "Note that in the present form the program calculates the symmetri
c eigenfunctions (and the corresponding eigenvalues) only." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "How far does the
 particle in the ground state explore the classically forbidden region
 in this case?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E0:=SEsol
ve([0.1,2.0],0.00001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E_
t:=E0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,
ICs\},psi(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 48 "P1:=plot(E0+'Psi(x)^2',x=0..3,0..2,color=green):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P2:=plot([x,E0],x=0..3,col
or=[red,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plots[di
splay](P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "Note that the \+
green curve does not show a properly normalized probability distributi
on. It is simply based on a solution that satisfies psi(0)=1." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 300 15 "
Solved Problem:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 108 "Determine several low-lying eigenvalues and eigenfunctio
ns for a 1d potential with a Rydberg series, namely " }{TEXT 19 14 "-1
/sqrt(1+x^2)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "V:=-1/sqrt(
1+x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "X:=10;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E0:=SEsolve([-1.,-0.2],0.000
1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E_t:=E0;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "SE:=-1/2*diff(psi(x),x$2)+V*psi(x)=
E_t*psi(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(
\{SE,ICs\},psi(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 49 "P1:=plot(E0+'Psi(x)^2',x=0..5,-1..1,color=gree
n):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P2:=plot([V,E0],x=0.
.5,color=[red,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "pl
ots[display](P1,P2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eva
lf(Int('Psi(x)^2',x=0..5),5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "
We can estimate the normalization integral:" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 55 "dx:=0.05: N:=5/dx: evalf(2*dx*add(Psi(i*dx)^2,i=0.
.N));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "To obtain the actual pro
bability to find a particle at a given location " }{TEXT 301 1 "x" }
{TEXT -1 72 " one would need to divide the square of the wavefunction \+
by that number." }}{PARA 0 "" 0 "" {TEXT -1 113 "Let us calculate the \+
first excited state by defining a procedure with a choice of appropria
te boundary condition:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "S
Eroot:=proc(lambda) local SE,IC,sol,Psi; global X,V,f_sym;" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 49 "SE:=-1/2*diff(psi(x),x$2)+V*psi(x)=lambda*p
si(x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "if f_sym=S then IC:=psi(0
)=1,D(psi)(0)=0; elif f_sym=A then IC:=psi(0)=0,D(psi)(0)=1; fi;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "sol:=dsolve(\{SE,IC\},psi(x),numeri
c,output=listprocedure): unprotect(Psi):" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 12 "Psi(X); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "f_sy
m:=S;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E0:=SEsolve([-1.,-
0.2],0.0001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "f_sym:=A;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E1:=SEsolve([-1.,-0.2],0.
0001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICa:=psi(0)=0,D(p
si)(0)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E_t:=E1;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "SE:=-1/2*diff(psi(x),x$2)+V*
psi(x)=E_t*psi(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=
dsolve(\{SE,ICa\},psi(x),numeric,output=listprocedure):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "P1:=plot(E1+'Psi(x)^2',x=0..8,-1..1
,color=green):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P2:=plot(
[V,E1],x=0..8,color=[red,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "plots[display](P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 302 11 "Exercise 7:" }}{PARA 0 "" 0 "" {TEXT -1 132 "Calculate t
he next two energye levels (one symmetric, and one antisymmetric eigen
function). Allow for a sufficiently large value of " }{TEXT 19 2 "X " 
}{TEXT -1 104 "such that the solution is not constrained by forcing th
e zero value of the wavefunction at too small an " }{TEXT 303 2 "x " }
{TEXT -1 7 "value. " }{TEXT 19 1 "X" }{TEXT -1 79 " has to be chosen l
arge enough such that the final answer is independent of it." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Note that the \+
above potential is one-dimensional, but shares some feature of the hyd
rogen atom problem:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 111 "there are bound and unbound solutions; the bound solut
ions are characterized by negative energies; the unbound " }{TEXT 304 
1 "E" }{TEXT -1 31 ">0 solutions are not quantized." }}{PARA 0 "" 0 "
" {TEXT -1 60 "there is an infinite series of bound states (Rydberg se
ries)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "151" 
0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }