Numerical solution of the Schrödinger equation on...
Transcript of Numerical solution of the Schrödinger equation on...
Numerical solution of the Schrödinger
equation on unbounded domains
Maike Schulte
Institute for Numerical and Applied Mathematics
Westfälische Wilhelms-Universität Münster
Cetraro, September 2006
1
OUTLINE
Transparent boundary conditions (TBC)
OUTLINE 2
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
OUTLINE 2-A
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme
OUTLINE 2-B
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme
Approximation of DTBC
OUTLINE 2-C
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme
Approximation of DTBC
Simulation of quantum wave guides
OUTLINE 2-D
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme
Approximation of DTBC
Simulation of quantum wave guides
Extension of the DTBC to (nearly) arbitrary potentials
OUTLINE 2-E
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme
Approximation of DTBC
Simulation of quantum wave guides
Extension of the DTBC to (nearly) arbitrary potentials
more realistic simulations
OUTLINE 2-G
OUTLINE
Transparent boundary conditions (TBC)
Analytical transparent boundary conditions for the Schrödinger
equation
Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme
Approximation of DTBC
Simulation of quantum wave guides
Extension of the DTBC to (nearly) arbitrary potentials
more realistic simulations
solving the Schrödinger equation on circular domains
OUTLINE 2-H
TRANSPARENT BOUNDARY CONDITIONS (IN GENERAL )
Rn
ψ
Γ
Ω
Definition (TBC) :Consider a given whole-space initial value problem (IVP) on Rn
and Ω ⊂ Rn, Γ = ∂Ω. We are interested in the solution of the IVPon Ω. Therefore we need new artificial boundary conditions on Γ.We call these artificial BC transparent, if the solution of the IVBPon Ω corresponds to the whole-space solution of the IVP restrictedon Ω.
TRANSPARENT BOUNDARY CONDITIONS (IN GENERAL) 3
ANALYTICAL TBC FOR THE SCHRÖDINGER EQUATION
IVP: time-dependent Schrödinger equation (here: 1D)
i~∂
∂tψ(x, t) =
„
− ~2
2m∗
∂2
∂x2+ V (x, t)
«
ψ(x, t), x ∈ R, t > 0
ψ(x, 0) = ψI(x) ∈ L
2(R)
on a domain of interest Ω = x ∈ R|0 < x < X.
ANALYTICAL TBC FOR THE SCHRÖDINGER EQUATION 4
ANALYTICAL TBC FOR THE SCHRÖDINGER EQUATION
IVP: time-dependent Schrödinger equation (here: 1D)
i~∂
∂tψ(x, t) =
„
− ~2
2m∗
∂2
∂x2+ V (x, t)
«
ψ(x, t), x ∈ R, t > 0
ψ(x, 0) = ψI(x) ∈ L
2(R)
on a domain of interest Ω = x ∈ R|0 < x < X.
Assumptions:
• supp ψI ⊆ Ω
• potential V (., t) ∈ L∞(R), V (x, .) is piecewise continuous
• V constant on R\Ω (here: V (x, t) = 0 for x ≤ 0 and V (x, t) = VX forx ≥ X)
Goal:
Calculate the solution ψ(x, t) ∈ C on Ω with TBC at x = 0 and x = X.
ANALYTICAL TBC FOR THE SCHRÖDINGER EQUATION 4-A
DERIVATION OF TBC
0
ψ
x
V = const V = const
X
G G21
ψ I
Ω
x ∈ G : x ∈ G2 :
i~ψt = − ~2
2m∗ψxx + V (x, t)ψ i~ψt = − ~
2
2m∗ψxx + V (x, t)ψ
ψ(x, 0) = ψI(x) v(x, 0) = 0
ψx(0, t) = (T0ψ)(0, t) v(X, t) = Φ(t) t > 0, Φ(0) = 0
ψx(X, t) = (TXψ)(X, t) limx→∞
v(x, t) = 0
vx(0, t) = (TXΦ)(t)
DERIVATION OF TBC 5
Laplace-transformation on the exterior domains :
vxx(x, s) +2im∗
~
“
s+iVX
~
”
v(x, s) = 0 x > X
v(X, s) = Φ(s)
limx→∞
v(x, s) = 0
vx(X, s) = (TXΦ)(s)
solution: v(x, s) = e−i
+r
2im∗
~
`s+
iVX~
´(x−X)
Φ(s)
⇒ (TXΦ)(s) = −r
2m∗
~e−
iπ4
+
r
s+iVX
~Φ(s)
With the inverse Laplace-Transformation follows the analytical TBC
ψx(X, t) = −r
~
2πm∗e−i π
4 e−i
VX~
t d
dt
tZ
0
ψ(X, τ)eiVX τ
~
√τ − t
dτ.
[J. S. Papadakis (1982)]
DERIVATION OF TBC 6
Application of DTBC for the Schrödinger equation:
• Simulation of quantum transistors in quantum waveguides (withinhomogeneous DTBC for the 2D Schrödinger equation)
• Analyse steady states and transient behaviour
x0 X
0
Y
Vcontrol
ψ inc
y
ΩΓ Γ1 2
x0 X
0
Yinc
Γ Γ2
ψ
1
y
Ω
Vcontrol ψ pw
DERIVATION OF TBC 7
FORMER STRATEGIES :
• Discretization of the analytic TBCs with an numericalapproximation of the convolution integral [e.g. B. Mayfield(1989)]
⇒ only conditionally stable, not transparent!
FORMER STRATEGIES: 8
FORMER STRATEGIES :
• Discretization of the analytic TBCs with an numericalapproximation of the convolution integral [e.g. B. Mayfield(1989)]
⇒ only conditionally stable, not transparent!
• Create a buffer zone Θ of the length d with a complex potentialV (X) = W − iA around the computational domain Ω withDirichlet 0-BC at ∂Θ and absorbing boundary conditions on ∂Ω
[e.g. L. Burgnies (1997)]
ψ=0ψ=00 X
Ω ΘΘ
⇒ unconditionally stable, unphysical reflections at theboundary, huge numerical costs
FORMER STRATEGIES: 8-A
SUCCESSFUL STRATEGIES
• Family of absorbing BCs (also for the non-linear Schrödingerequation, wave equation)
[J. Szeftel (2005)]
• discretize the whole space problem with an unconditionally stablescheme (e.g. Crank-Nicolson finite difference scheme) and calculatenew discrete transparent boundary conditions for the full discretizedSchrödinger equation
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
|ψ|
DTBC: dx=0.00625, dt=2e−05, t=0.014
[A. Arnold, M. Ehrhardt (since 1995)]
SUCCESSFUL STRATEGIES 9
DERIVATION OF DISCRETE TRB
Discretize 2D Schrödinger equation:
• Crank-Nicolson scheme in time, tn = n∆t, n ∈ N, H := − 12∆ + V
Hamilton-Operator, Ω = [0,X] × [0, Y ]
„
1 +iH∆t
2
«
ψ(x, y, t+ ∆t) =
„
1 − iH∆t
2
«
ψ(x, y, t)
⇒„∂2
∂x2+∂2
∂y2
«`ψ
n+1(x, y) + ψn(x, y)
´
= gn+ 1
2 (x, y)`ψ
n+1(x, y) + ψn(x, y)
´+Wψ
n(x, y)
• compact 9-point scheme in space, xj = j∆x, yk = k∆y withj ∈ Z, 0 ≤ k ≤ K
• DTBC at x0 = 0 and xJ = X = J∆x with J ∈ Z
• 0-BC at y = 0 and y = Y = K∆y with K ∈ N
DERIVATION OF DISCRETE TRB 10
9-point discretization scheme:
(j,k)
(j+1,k+1)
(j+1,k)
“
D2x +D
2y +
∆x2 + ∆y2
12D
2x D
2y
”
ψn+ 1
2j,k
=
„
I +∆x2
12D
2x +
∆y2
12D
2y
« h
2Vn+ 1
2j,k ψ
n+ 12
j,k − 2iD+t ψ
nj,k
i
with
ψn+ 1
2j,k =
1
2
`ψ
n+1j,k + ψ
nj,k
´
D+t ψ
nj,k =
ψn+1j,k − ψn
j,k
∆t, n ≥ 0
D2x ψ
nj,k =
ψnj−1,k − 2ψn
j,k + ψnj+1,k
∆x2, j ∈ Z
D2y ψ
nj,k =
ψnj,k−1 − 2ψn
j,k + ψnj,k+1
∆y2, k ∈ N
DERIVATION OF DISCRETE TRB 11
Discrete Sine-Transformation in y-direction:
ψnj,m :=
1
K
K−1X
k=1
ψnj,k sin
„πkm
K
«
m = 0, . . . ,K
Motivation:Solve discrete stationary Schrödinger equation in 1D:
−1
2∆2
y χmj,k = E
mχ
mj,k, k = 0, . . . ,K
χmj,0 = χ
mj,K = 0.
The eigenfunctions χmj,k = sin
`πkm
K
´provide the energies
Em =
1
∆y2
“
1 − cos“πm
K
””
.
Hence follows for m = 0, . . . ,K
⇒ − 1
2∆y2
“
ψnj,k−1 − 2ψn
j,k + ψnj,k+1
”b
m=
1
∆y2
“
1 − cos“πm
K
” ”
ψnj,m.
DERIVATION OF DISCRETE TRB 12
Sine-Transformation of the discrete Schrödinger equation on theexterior domains j ≤ 0, j ≥ J yields for the modes m = 0, . . . , K:
Cmj+1ψ
n+1j+1,m + Cm
j−1ψn+1j−1,m +Rm
j ψn+1j,m
= (D − Cmj+1)ψ
nj+1,m + (D − Cm
j−1)ψnj−1,m + (Bm
j −Rmj )ψn
j,m.
DERIVATION OF DISCRETE TRB 13
Sine-Transformation of the discrete Schrödinger equation on theexterior domains j ≤ 0, j ≥ J yields for the modes m = 0, . . . , K:
Cmj+1ψ
n+1j+1,m + Cm
j−1ψn+1j−1,m +Rm
j ψn+1j,m
= (D − Cmj+1)ψ
nj+1,m + (D − Cm
j−1)ψnj−1,m + (Bm
j −Rmj )ψn
j,m.
Definition [Z - Transformation ]:The Z-Transformation of a sequence (ψn)n∈N is given by
Z ψn = Ψ(z) :=
∞∑
n=0
ψnz−n z ∈ C, |z| > 1.
DERIVATION OF DISCRETE TRB 13-A
One can show:
• Z“
ψn+1j,m
”
= −zψ0j,m + zΨm
j (z)
• ψ0J+1,k = ψ0
J−1,k = ψ0J,k = 0 for k = 0, . . . ,K
• Vj constant for j ≤ 1, j ≥ J − 1 ⇒ Cmj = Cm, Rm
j = Rm, Bmj = Bm
⇒ ΨJ+1(z) +
»Rz + R−B
C z + C −D
–
ΨJ(z) + ΨJ−1(z) = 0.
DERIVATION OF DISCRETE TRB 14
One can show:
• Z“
ψn+1j,m
”
= −zψ0j,m + zΨm
j (z)
• ψ0J+1,k = ψ0
J−1,k = ψ0J,k = 0 for k = 0, . . . ,K
• Vj constant for j ≤ 1, j ≥ J − 1 ⇒ Cmj = Cm, Rm
j = Rm, Bmj = Bm
⇒ ΨJ+1(z) +
»Rz + R−B
C z + C −D
–
ΨJ(z) + ΨJ−1(z) = 0.
This difference equation with constant coefficients is solved byΨj(z) = νj(z):
ν2(z) +
»Rz +R−B
C z + C −D
–
ν(z) + 1 = 0,
Physical background forces decay of the solution for j → ∞
⇒ |ν(z)| > 1 und ν(z)ΨJ(z) = ΨJ−1(z) → Z-transformed DTBC
DERIVATION OF DISCRETE TRB 14-A
Theorem [ DTBC for the 2D Schrödinger equation ]: Discretizethe 2D Schrödinger-Equation with the compact 9-point differencescheme in space and with the Crank-Nicolson scheme in time.Then the DTBC at xJ = J∆x and x0 = 0 for n ≥ 1 read
ψn1,m − s
(0)0,mψ
n0,m =
n−1∑
ν=1
s(n−ν)0,m ψν
0,m − am1 ψ
n−11,m ,
ψnJ−1,m − s
(0)J,mψ
nJ,m =
n−1∑
ν=1
s(n−ν)J,m ψν
J,m − amJ−1ψ
n−1J−1,m.
The convolution coefficients s(n)j,m can be calculated by
s(n)j,m = αm
j
(λmj )−n
2n− 1
[
Pn(µmj ) − Pn−2(µ
mj )
]
with the Legendre-Polynomials Pn ( P−1 ≡ P−2 ≡ 0).
DERIVATION OF DISCRETE TRB 15
Advantages of the new DTBC :
• no numerical reflections
• 3-point recursion for s(n)
• these DTBC have exactly the same structure like the DTBCcalculated with the 5-point scheme [A. Arnold, M. Ehrhardt]
• convergence: O(∆x4 + ∆y4 + ∆t2)
• same numerical effort like discretized analytical TRB
• s(n)j,m = O(n−3/2)
• CN-FD scheme with DTBC is unconditionally stable, withA := I + ∆x2
12 D2x + ∆y2
12 D2y follows:
D+t ||ψ||2A = 0 with ||ψ||2A := 〈ψ,Aψ〉. (1)
DERIVATION OF DISCRETE TRB 16
some drawbacks :
• DTBC are non-local in time→ high memory costs: the solution ψn
j,m has to be saved in
x0 and xJ for all time steps n = 1, 2, . . .
→ in each time step n = 1, 2, . . . you have to calculate K
convolutions of the length n
(FFT not useable, ⇒ O(Kn2))
DERIVATION OF DISCRETE TRB 17
some drawbacks :
• DTBC are non-local in time→ high memory costs: the solution ψn
j,m has to be saved in
x0 and xJ for all time steps n = 1, 2, . . .
→ in each time step n = 1, 2, . . . you have to calculate K
convolutions of the length n
(FFT not useable, ⇒ O(Kn2))
• These DTBC are given in the Sine-transformed form: BC ofone mode is a linear combination of all other boundary points
→ diagonal structure of the system matrix is destroyed
DERIVATION OF DISCRETE TRB 17-A
0 20 40 60 80 100
0
10
20
30
40
50
60
70
80
90
100
Sparsity pattern of the system matrix for J = K = 10
DERIVATION OF DISCRETE TRB 18
Example 1 :
free 2D Schrödinger equation on Ω = [0, 2] × [0, 2] with the initialdata:
ψI(x, y) = eikxx+ikyy−60
(
(x− 12 )
2+(y− 1
2 )2
)
, (x, y) ∈ Ω
0
50
100
020
4060
80100
120
0
0.2
0.4
0.6
0.8
1
x
it = 10
y
T = 10
0
50
100
020
4060
80100
120
0
0.2
0.4
0.6
0.8
1
x
it = 60
y
T = 60∆t
0
50
100
020
4060
80100
120
0
0.01
0.02
0.03
0.04
0.05
x
it = 160
y
T = 80∆t
DERIVATION OF DISCRETE TRB 19
*
APPROXIMATION OF DTBC
• Idea: Approximate the convolution coefficients s(n)j,m by a sum of
exponentials:
s(n) ≈ s(n) =L
∑
l=1
blq−nl , n ∈ N, |ql| > 1, L ≤ 40
APPROXIMATION OF DTBC 20
*
APPROXIMATION OF DTBC
• Idea: Approximate the convolution coefficients s(n)j,m by a sum of
exponentials:
s(n) ≈ s(n) =L
∑
l=1
blq−nl , n ∈ N, |ql| > 1, L ≤ 40
• bl, ql are calculated by the Padé - Approximation of
f(x) =2L−1∑
n=0
s(n)xn, |x| ≤ 1
[A.Arnold, M. Ehrhardt, I. Sofronov (2003)]
APPROXIMATION OF DTBC 20-A
*
Recursion formula for the convolution coefficients:n−1X
t=0
s(n−t)
ψt =
LX
l=1
c(n)l
with
c(n)l = q
−1l c
(n−1)l + blq
−1l ψ
n−1, n = 1, . . . , N
c(0)l = 0
APPROXIMATION OF DTBC 21
*
Recursion formula for the convolution coefficients:n−1X
t=0
s(n−t)
ψt =
LX
l=1
c(n)l
with
c(n)l = q
−1l c
(n−1)l + blq
−1l ψ
n−1, n = 1, . . . , N
c(0)l = 0
Advantages:
If you have calculated bl, ql once for a set ∆x,∆y,∆t, V , you’ll easilyderive b∗l , q
∗
l for any ∆x∗,∆y∗,∆t∗, V ∗ by
q∗
l =qla− b
a− qlb
b∗
l = blql
aa− bb
(a− qlb)(qla− b)
1 + q∗l1 + ql
Numerical effort: O(Kn2) → O(KLn)
Memory: O(Kn) → O(KL)
APPROXIMATION OF DTBC 21-A
*
Example 2 :
free 2D Schrödinger-Equation in Ω = [0, 1] × [0, 1]
Initial function:
ψI(x, y) = sin(πy)eikxx−60(x− 1
2 )2 , (x, y) ∈ Ω
00.2
0.40.6
0.81
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
y
x
IT=150
|psi
(x,y
)|, t
=T
L = 5 : T = 150∆t
00.2
0.40.6
0.81
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
y
x
IT=300
|psi
(x,y
)|, t
=T
T = 300∆t
00.2
0.40.6
0.81
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
y
x
IT=500
|psi
(x,y
)|, t
=T
T = 500∆t
00.2
0.40.6
0.81
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
y
x
IT=150
|psi
(x,y
)|, t
=T
L = 20 : T = 150∆t
00.2
0.40.6
0.81
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
y
x
IT=300
|psi
(x,y
)|, t
=T
T = 300∆t
00.2
0.40.6
0.81
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
y
x
IT=500
|psi
(x,y
)|, t
=T
T = 500∆t
APPROXIMATION OF DTBC 22
SIMULATION OF QUANTUM WAVEGUIDES
x0 X
0
Y
Vcontrol
ψ inc
y
ΩΓ Γ1 2
40.5nm32nm
20nm
20nm
• Incoming wave at x = 0:
ψinc(0, y, t) = sin(πy)e
−iEt~ , E = 29.9meV
• inhomogeneous DTBC at x = 0, DTBC at x = X
SIMULATION OF QUANTUM WAVEGUIDES 23
Little trick to suppress oscillations in time:
ψinc oscillates like e−iEt
~ in time.
E big:
fast oscillation of the solution in time
small time step size is necessary
high numerical effort for the analysis of steady state and long-timebehaviour
Define
ϕ(x, y, t) := e−iωt
ψ(x, y, t) mit ω = −E~.
ϕ solve the modified Schrödinger equation
i~ϕt = − ~2
2m∗(ϕxx + ϕyy) + (V − ω~)
| z
=:V
ϕ
SIMULATION OF QUANTUM WAVEGUIDES 24
0 1 2 3 4 5 6 7 8 9 100
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
t [10−12 s]
V=0V=−E
f1(t) = ||ψ1(x, y, t) − ψref (x, y, t)||2f2(t) = ||ψ2(x, y, t) − ψref (x, y, t)||2
ψ1 is calculated with V = 0, ψ2 with V = −E and ψref is a numerical
reference solution, which has been calculated with high accuracy.
SIMULATION OF QUANTUM WAVEGUIDES 25
0
20
40
60
0 10 20 30 40 50 60
0
0.5
1
1.5
2
2.5
3
y
Initial function: X=1, Y=0.86667, dx=0.016667, dy=0.016667, dt=0.0002, V=0
x
|psi
(x,y
)|, t
=0
T = 0∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
2.5
3
y [nm]
Quantenwellen.m, solution after T = −1.7998 ps, it = 1000
|ψ(x
,y,T
)|
T = 1000∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
2.5
3
y [nm]
Quantenwellen.m, solution after T = −0.1998 ps, it = 9000
|ψ(x
,y,T
)|
T = 9000∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
2.5
3
y [nm]
Quantenwellen.m, solution after T = 0.0202 ps, it = 10100
|ψ(x
,y,T
)|
T = 10100∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
2.5
3
y [nm]
Quantenwellen.m, solution after T = 0.0802 ps, it = 10400
|ψ(x
,y,T
)|
T = 10400∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
2.5
3
y [nm]
Quantenwellen.m, solution after T = 1.9002 ps, it = 19500
|ψ(x
,y,T
)|
T = 19500∆t
SIMULATION OF QUANTUM WAVEGUIDES 26
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS
Drawbacks of the simulations:
• Hard walls (zero Dirichlet BCs) and edges are not practicable inindustry!
→ Geometry of computational domain shall be realized by
potentials also in the simulations.
→ How to choose the incoming wave and the initial function then?
• Potentials are NOT constant in the exterior domains!→ Decoupling of the modes for V (x, y) 6= const. after Sine-
Transformation is not possible
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS 27
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS
Drawbacks of the simulations:
• Hard walls (zero Dirichlet BCs) and edges are not practicable inindustry!
→ Geometry of computational domain shall be realized by
potentials also in the simulations.
→ How to choose the incoming wave and the initial function then?
• Potentials are NOT constant in the exterior domains!→ Decoupling of the modes for V (x, y) 6= const. after Sine-
Transformation is not possible
Discrete Schrödinger equation:5-point scheme in space, Crank-Nicolson in time:
i~D+t ψ
nj,k = − ~
2
2m∗
`D
2x +D
2y
´ψ
n+ 12
j,k + Vn+ 1
2j,k ψ
n+ 12
j,k
Calculate new Eigenfunctions, which take the potential into account!
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS 27-A
Solve the eigenvalue equation on the exterior domains (j ≤ 0, j ≥ J):
− 1
2∆y2
`χ
nj,k−1,m − 2χn
j,k,m + χnj,k+1,m
´+ V
n+ 12
k χnj,k,m = E
nj,mχ
nj,k,m
with
∆y
KX
k=0
|χnj,k,m|2 = 1 and χ
nj,0,m = χ
nj,K,m = 0
for 0 ≤ k, m ≤ K, n > 0.
Transformation w.r.t. the eigenfunctions
ψnj,m = ∆y
K−1X
k=1
χnj,k,mψ
nj,k, 0 ≤ m ≤ K.
yields
i~D+t ψ
nj,m = − ~
2
2m∗D
2x ψ
n+ 12
j,m + Enj,mψ
n+ 12
j,m
same structure as the sine-transformed Schrödinger equation![N. Ben Abdallah, M.S. (2005)]
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS 28
Example 3: 2D channel
0 10 20 30 40 500
10
20
30
40
50
60
70
80
90
100
110Potential V(0,y)
yk
V(0
,yk)
potential V (y) = 400y(1 − y)
0 10 20 30 40 50 60 70 80 90 1000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Eigenfunction χ for j=0, |χ j,k,m
|2 of the m = 1st mode with the Eigenvalue λ = 65.3211
yk
|χ j,
k,m
|2
eigenfunction χ of m = 1
initial function:
ψIj,k = e
ikxj∆xχ
0j,k,m
incoming wave:
ψincj,k,n = e
ikxj∆xχ
0j,k,me
−iExn∆t with Ex =1 − cos(kx∆x)
∆x2
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS 29
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
y
t = 0
x
|ψ(x
,y,t)
|2
T = 0∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
y
t = 50 ⋅ ∆ t = 0.01
x
|ψ(x
,y,t)
|2
T = 50∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
y
t = 150 ⋅ ∆ t = 0.03
x
|ψ(x
,y,t)
|2
T = 150∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
y
t = 250 ⋅ ∆ t = 0.05
x
|ψ(x
,y,t)
|2
T = 250∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
y
t = 500 ⋅ ∆ t = 0.1
x
|ψ(x
,y,t)
|2
T = 500∆t
0
0.5
1
00.2
0.40.6
0.81
0
0.5
1
1.5
2
y
t = 1700 ⋅ ∆ t = 0.34
x
|ψ(x
,y,t)
|2
T = 1700∆t
EXTENSION OF THE DTBC TO MORE ARBITRARY POTENTIALS 30
SCHRÖDINGER EQUATION ON CIRCULAR DOMAINS
Solve the Schrödinger equation (in polar coordinates)
iψt = −1
2
„1
r(rψr)r +
1
r2ψθθ
«
+V (r, θ, t)ψ, r > 0, 0 ≤ θ ≤ 2π, t > 0
on a circular domain Ω = [0, R] × [0, 2π] with TBC at x = R.
Problems:
• solve a second order difference equation with varying coefficients:
ajΨJ+1(z) + bj(z)ΨJ(z) + cjΨJ−1(z) = 0
• calculation of the convolution coefficients for the DTBC
→ "recursion from infinity"
• singularity at r = 0 → not equidistant offset-grid rj = rj+ 12
• approximation of the convolution coefficients and the -sum
[A.Arnold, M. Ehrhardt, M. S., I. Sofronov (2006)]
SCHRÖDINGER EQUATION ON CIRCULAR DOMAINS 31
Example 4:free Schrödinger equation on unit disc Ω1 = [0, 1] × [0, 2π]
ψI(r, θ) =
1√αxαy
e2ikxr cos θ+2ikyr sin θ−
(r cos θ)2
2αx−
(r sin θ)2
2αy
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1
0
1
0
5
10
15
20
initial function on Ω1 = [0,1]x[0,2π], ∆r = 1/128, ∆θ = 2π /128
|ψ(r
,θ,0
)|
initial function
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1
0
1
0
1
2
3
4
5
6
7
solution on Ω1 = [0,1]x[0,2π] at time t= 0.125, ∆t = ∆r = 1/128, ∆θ = 2π /128
|ψ(r
,θ,t)
|
T = 0.125
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1
0
1
0
0.5
1
1.5
2
2.5
3
3.5
solution on Ω1 = [0,1]x[0,2π] at time t= 0.25, ∆t = ∆r = 1/128, ∆θ = 2π /128
|ψ(r
,θ,t)
|
T = 0.25
−1 −0.5 0 0.5 1
−1
−0.5
0
0.5
1
initial function on Ω1 = [0,1]x[0,2π], ∆r = 1/128, ∆θ = 2π /128
−1 −0.5 0 0.5 1
−1
−0.5
0
0.5
1
solution on Ω1 = [0,1]x[0,2π] at time t= 0.125, ∆t = ∆r = 1/128, ∆θ = 2π /128
−1 −0.5 0 0.5 1
−1
−0.5
0
0.5
1
solution on Ω1 = [0,1]x[0,2π] at time t= 0.25, ∆t = ∆r = 1/128, ∆θ = 2π /128
SCHRÖDINGER EQUATION ON CIRCULAR DOMAINS 32
Error due to the scheme/TBC:
L(ψ, ϕ, tn,Ω) :=||ψ(rj , θk, tn) − ϕ(rj , θk, tn)||Ω,2
||ϕ(rj , θk, tn)||Ω,2
ψ: numerical solution
ϕ: exact solution or numerical reference solution
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.510
−4
10−3
10−2
10−1
100
t
relative L2 error due to the scheme
J=K=64, ∆t=1/64J=K=128, ∆t=1/128J=K=256, ∆t=1/256
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.510
−16
10−15
10−14
10−13
10−12
10−11
10−10
t
relative L2−error due to the TBC
J=K=64, ∆t=1/64
J=K=128, ∆t=1/128
J=K=256, ∆t=1/256
SCHRÖDINGER EQUATION ON CIRCULAR DOMAINS 33