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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 56 "Quantum tunneling throu
gh an arbitrary potential barrier" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 481 "We are i
nterested in solving the Schroedinger wave equation with scattering bo
undary conditions for an arbitrary potential function in one dimension
. The boundary conditions (incident and reflected and transmitted wave
) require the solution to be complex-valued. Our attitude will be to a
pply knowledge about general solution to a second-order ordinary diffe
rential equations (two independent solutions are required, and then ma
tched to the physically imposed boundary conditions)." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 213 "Textbook treatments b
egin with simple potential shapes (square barrier) for which the solut
ion can be obtained exactly. For a symbolic solution along these lines
 we refer the reader to our book [Marko Horbatsch:  " }{TEXT 257 29 "Q
uantum Mechanics using Maple" }{TEXT -1 428 ", chapter 3.2 (Springer 1
995)]. Here we rely on numerical techniques included in Maple, namely \+
to solve initial-value problems for systems of first-order differentia
l equations. The tunneling problem requires us to solve a boundary-val
ue problem. This will be achieved by generating two independent unphys
ical solutions based on inital values, and then to match a linear comb
ination of these to the physical boundary conditions." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 468 "Maple6 has added capa
bility to dsolve[numeric] to handle complex-valued differential equati
ons. This permits to solve the Schroedinger equation for scattering bo
undary conditions directly. To remain compatible with Maple 5 we stay,
 however, with real-valued boundary conditions for the solution of the
 differential equation, and introduce the complex-valued nature of the
 physical solution in the step where linear combinations of two fundam
ental solutions are formed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 241 "We could begin with simple barrier penetration
. However, we assume that the reader is already familiar with the basi
cs of tunneling. Therefore, we proceed with the more interesting doubl
e-barrier. An example of a double-barrier potential is:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "V:=exp(-5*x^2)+exp(-5*(x-2)^2);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "plot(V,x=-2..5);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "We are interested in discovering \+
so-called transmission resonances. We can safely assume that the inter
action vanishes at " }{TEXT 259 1 "x" }{TEXT -1 8 "=-2 and " }{TEXT 
258 1 "x" }{TEXT -1 19 "=4 for our example." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "We work in units where hbar=" }
{TEXT 260 1 "m" }{TEXT -1 90 "=1. This simplifies relationships betwee
n momentum and wavenumber (they become identical)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "The wave number (momentum
) can be obtained from the non-relativistic energy as:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "kE:=En->sqrt(2*En);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 130 "We need to propagate two independent sol
utions, and then to apply matching conditions. The Schroedinger wave e
quation is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "SE:
=En->-1/2*diff(u(x),x$2)+(V-En)*u(x)=0;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 35 "The boundary condition states that:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "1) From the left we have \+
an incident free plane wave with wavenumber " }{TEXT 261 1 "k" }{TEXT 
-1 219 " and of known amplitude, and a reflected free plane wave with \+
unknown amplitude (whose magnitude squared will eventually become the \+
reflection probability; (the choice of side from which the wave comes \+
in is arbitrary)." }}{PARA 0 "" 0 "" {TEXT -1 92 "2) To the right of t
he potential region we have an outgoing free plane wave with wavenumbe
r " }{TEXT 262 1 "k" }{TEXT -1 153 " and unknown magnitude. The square
d magnitude of the transmitted wave becomes the transmission probabili
ty (if the incident wave is normalized to unity)." }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 53 "We know from (2) that the transmitted flux goes li
ke " }{TEXT 263 1 "T" }{TEXT -1 6 "exp(i " }{TEXT 264 2 "kx" }{TEXT 
-1 100 "). Thus, te ratio of the derivative of the physical solution t
o the solution itself has to equal  i " }{TEXT 265 1 "k" }{TEXT -1 1 "
." }}{PARA 0 "" 0 "" {TEXT -1 143 "Therefore, the strategy is to gener
ate two independent solutions, and to form the linear combination that
 satisfies the requested relationship." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 125 "The initial and final values for the
 independent varible range in the numerical integration of the differe
ntial equation are:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "x_i:
=-2; x_f:=4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "We pick an energy
 value:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "En:=1/2;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "An aside:" }}{PARA 0 "" 0 "" {TEXT 
-1 97 "Experts can fix an oversight in Maple6.0, if they wish to use c
omplex-valued boundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 93 "For
 complex-valued solutions in Maple6 override: (Robert Israel contribut
ed this fix to MUG)." }}{PARA 0 "" 0 "" {TEXT -1 90 "`dsolve/numeric/i
nit_y0`:=subs();numeric=complex(numeric),eval(`dsolve/numeric/init_y0`
));" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 65 "We pick two intial conditions that provide independent solution
s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "IC1:=u(x_i)=1,D(u)(x_
i)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "IC2:=u(x_i)=1,D(u)
(x_i)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "sol1:=dsolve(\{
SE(En),IC1\},u(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 20 "u1:=subs(sol1,u(x)):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 29 "u1p:=subs(sol1,diff(u(x),x)):" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 46 "To demonstrate that it calculates something...
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "u1(x_f);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "u1p(x_f);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 24 "Now the second solution:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 61 "sol2:=dsolve(\{SE(En),IC2\},u(x),numeric,output=listp
rocedure);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "# not needed
 as we picked real-valued initial conditions #`dsolve/numeric/init_y0`
:=subs(numeric=complex(numeric),eval(`dsolve/numeric/init_y0`));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "sol2:=dsolve(\{SE(En),IC2\},
u(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "u2:=subs(sol2,u(x)):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 29 "u2p:=subs(sol2,diff(u(x),x)):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "u2(x_f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "
u2p(x_f);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "We generated the two
 independent solutions using real boundary conditions." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 270 "Now we assemble the
 complex-valued solution that satisfies the physical boundary conditio
n of only a transmitted wave existing to the right of x_f. We need a s
ingle mixing coefficient, and determine it by imposing that the ratio \+
of derivative over the solution equals i " }{TEXT 266 1 "k" }{TEXT -1 
1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "k:=kE(En);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "c2:='c2': c2:=solve((u1p(x_f
)+c2*u2p(x_f))/(u1(x_f)+c2*u2(x_f))=I*k,c2);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 126 "Now we can form the physical solution, which has arbit
rary normalization at this point. We graph the real and imaginary part
s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "P1:=plot('Re(u1(x)+c2
*u2(x))',x=-12..14,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 51 "P2:=plot('Im(u1(x)+c2*u2(x))',x=-12..14,color=red):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plots[display](P1,P2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 202 "This is a solution from which we \+
should be able to extract the physical information. To the right of th
e potential we have obviously a travelling plane wave. It is travellin
g to the right with momentum " }{TEXT 268 2 "k " }{TEXT -1 62 "(we rec
ognize the periodicity as about 2 Pi, corresponding to " }{TEXT 269 1 
"k" }{TEXT -1 135 "=1, and we see that the imaginary part precedes the
 real part, corresponding to motion to the right) and we can read off \+
the amplitude " }{TEXT 267 1 "T" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "On the left we must have
 the sum of an incident and reflected wave. This is characterized by u
nequal amplitudes for real and imaginary parts." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 138 "In the middle we have a \+
complicated connecting piece. This is where dsolve[numeric] and formin
g the linear combo has done the work for us!" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 150 "So far the incident wave has a
rbitrary normalization. Let us see whether we can figure the normaliza
tion out from the solution in region I (at x=x_i)." }}{PARA 0 "" 0 "" 
{TEXT -1 112 "We match now at the x_i a solution which has independent
 amplitudes for the incoming and reflected waves, named " }{TEXT 271 
1 "N" }{TEXT -1 5 " and " }{TEXT 270 1 "R" }{TEXT -1 14 " respectively
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "eq1:=N*exp(I*k*x_i)+R*
exp(-I*k*x_i)=u1(x_i)+c2*u2(x_i);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 65 "eq2:=I*(k*N*exp(I*k*x_i)-k*R*exp(-I*k*x_i))=u1p(x_i)+
c2*u2p(x_i);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=solve(
\{eq1,eq2\},\{N,R\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "as
sign(sol);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "We observe that " }
{TEXT 272 1 "N" }{TEXT -1 5 " and " }{TEXT 273 1 "R" }{TEXT -1 51 " ar
e amplitudes, and turn out to be complex-valued." }}{PARA 0 "" 0 "" 
{TEXT -1 59 "The properly normalized reflection coefficient now become
s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "r:=R/N;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 33 "and the transmission coefficient:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "t:=evalf((u1(x_f)+c2*u2(x_f)
)/exp(I*k*x_f)/N);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "The check \+
whether this all makes sense is to sum the magnitude squared for refle
ction and transmission:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "
abs(r)^2+abs(t)^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Within num
erical accuracy we preserve the norm; therefore we interpret as reflec
tion and transmission probabilities:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "abs(r)^2,abs(t)^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 122 "Now we turn the calculation into a procedure: for given energy
 it calculates the reflection and transmission coefficients." }}{PARA 
0 "" 0 "" {TEXT -1 66 "Our previously developed commands are used step
-by-step as before:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "Tun
nel:=proc(En) local IC1,IC2,sol1,sol2,u1,u1p,u2,u2p,k,c2,N,R,r,t,sol,e
q1,eq2; global V,kE,x_i,x_f;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "IC1
:=u(x_i)=1,D(u)(x_i)=1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "IC2:=u(x
_i)=1,D(u)(x_i)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "sol1:=dsolve(
\{SE(En),IC1\},u(x),numeric,output=listprocedure);" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "u1:=subs(sol1,u(x));" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 29 "u1p:=subs(sol1,diff(u(x),x));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 61 "sol2:=dsolve(\{SE(En),IC2\},u(x),numeric,output=listprocedure);
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u2:=subs(sol2,u(x));" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 29 "u2p:=subs(sol2,diff(u(x),x));" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 10 "k:=kE(En);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 62 "c2:=solve((u1p(x_f)+c2*u2p(x_f))/(u1(x_f)+c2*u2(x_f))=I*k,c2);" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "N:='N': R:='R':" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 55 "eq1:=N*exp(I*k*x_i)+R*exp(-I*k*x_i)=u1(x_i)+c2*u
2(x_i);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "eq2:=I*(k*N*exp(I*k*x_i)
-k*R*exp(-I*k*x_i))=u1p(x_i)+c2*u2p(x_i);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "sol:=solve(\{eq1,eq2\},\{N,R\});" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "assign(sol);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "r
:=R/N;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "t:=evalf((u1(x_f)+c2*u2(x
_f))/exp(I*k*x_f)/N);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "#print(\" \+
R= \",abs(r)^2,\" T= \",abs(t)^2,\" R+T= \",abs(r)^2+abs(t)^2,\" norm \+
equal to 1.0 ?\");" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "[abs(r)^2,abs
(t)^2]; end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "We check it for t
he previously tested energy value:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "Tunnel(1/2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 
"Now we are ready to scan the energy range from zero to above the peak
 height of the barrier. The results are accumulated in lists for subse
quent plotting." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "Tr:=[]:
 Rf:=[]: for iE from 1 to 60 do: En:=iE/20; res:=Tunnel(En); Tr:=[op(T
r),[En,res[2]]]: Rf:=[op(Rf),[En,res[1]]]: od:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 48 "P1:=plot(Rf,color=red): P2:=plot(Tr,color=blue
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plots[display](P1,P2,
labels=[\"E\",\"\{t,r\}\"]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "O
bserve the interesting features:" }}{PARA 0 "" 0 "" {TEXT -1 128 "1) F
or sufficiently high energies only transmission occurs. Note, however,
 that classically this situation would be reached for " }{TEXT 274 1 "
E" }{TEXT -1 3 ">1!" }}{PARA 0 "" 0 "" {TEXT -1 7 "2) For " }{TEXT 
275 1 "E" }{TEXT -1 640 "=0.66 perfect transmission and no reflection \+
is observed. This is the phenomenon of a transition resonance. The res
onance structure has a width associated with it. Most physical phenome
na that have a lifetime associated with them (unstable elementary part
icles, alpha decay, nuclear gamma lines, atomic optical transitions, e
xcitations in semiconductors, etc.) can be understood as processes whe
re the wave equation describing the matter (or radiation) field involv
es a tunneling process. Broad resonances are associated with short lif
etimes, sharp resonances with long lifetimes, according to Heisenberg'
s energy-time uncertainty relation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 157 "The transmission resonance is quite bro
ad. We can make the walls of the two barriers thicker, and observe wha
t the effect is on the transmission probability." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 31 "V:=exp(-8*x^6)+exp(-8*(x-2)^6);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "plot(V,x=-2..5);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 124 "Tr:=[]: Rf:=[]: for iE from 1 to 120 do:
 En:=iE/40; res:=Tunnel(En); Tr:=[op(Tr),[En,res[2]]]: Rf:=[op(Rf),[En
,res[1]]]: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "P1:=plot(
Rf,color=red): P2:=plot(Tr,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "plots[display](P1,P2,labels=[\"E\",\"\{t,r\}\"]);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "The transmission resonance is sh
arper now, and occurs just below the classical border value for the en
ergy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 
"Exercise: Observe what happens when the gap is widened between the po
tential humps. Be careful with the values of x_i, x_f, when changes ar
e made to the potential." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 139 "We can include an option in Tunnel to graph the s
olution. Perhaps this will help to understand why reflection and trans
mission comes about." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "Tu
nnel:=proc(En,iplot) local IC1,IC2,sol1,sol2,u1,u1p,u2,u2p,k,c2,N,R,r,
t,sol,eq1,eq2; global V,kE,x_i,x_f,PL;" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 26 "IC1:=u(x_i)=1,D(u)(x_i)=1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "IC2:=u(x_i)=1,D(u)(x_i)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "s
ol1:=dsolve(\{SE(En),IC1\},u(x),numeric,output=listprocedure);" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u1:=subs(sol1,u(x));" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "u1p:=subs(sol1,diff(u(x),x));" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 61 "sol2:=dsolve(\{SE(En),IC2\},u(x),numeric,output=
listprocedure);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u2:=subs(sol2,u(
x));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u2p:=subs(sol2,diff(u(x),x)
);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "k:=kE(En);" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 62 "c2:=solve((u1p(x_f)+c2*u2p(x_f))/(u1(x_f)+c2*u2(x_
f))=I*k,c2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "N:='N': R:='R':" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "eq1:=N*exp(I*k*x_i)+R*exp(-I*k*x_i)
=u1(x_i)+c2*u2(x_i);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "eq2:=I*(k*N
*exp(I*k*x_i)-k*R*exp(-I*k*x_i))=u1p(x_i)+c2*u2p(x_i);" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 28 "sol:=solve(\{eq1,eq2\},\{N,R\});" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 12 "assign(sol);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 7 "r:=R/N;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "t:=evalf((u1(x_f)+c
2*u2(x_f))/exp(I*k*x_f)/N);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "if \+
iplot=2 then print(\" R= \",abs(r)^2,\" T= \",abs(t)^2,\" R+T= \",abs(
r)^2+abs(t)^2,\" norm equal to 1.0 ?\") fi;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "if iplot=1 then" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
154 "PL:=plot([V,'Re((u1(x)+c2*u2(x))/N)','Im((u1(x)+c2*u2(x))/N)'],x=
x_i-10..x_f+10,color=[red,blue,green],title=cat(\"Energy= \",convert(e
valf(En,3),string))):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "fi;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "[abs(r)^2,abs(t)^2]; end:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Tunnel(0.8,1);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plots[display](PL);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 97 "Can we run an animation as a function of \+
energy? It takes a while to compute, but it can be done." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "PPL:='PPL': for iE from 1 to 120 do
: En:=iE/40; res:=Tunnel(En,1); PPL[iE]:=PL: od:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 53 "plots[display](seq(PPL[i],i=1..120),insequenc
e=true);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 294 "The possibilities to
 explore scattering from different potentials with this worksheet appe
ar to be unlimited! Transmission resonances appear to play a significa
nt role in newly developed semiconductor devices, and may prove to pro
vide opportunities for future generations of computing hardware." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "31 0 0" 10 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }