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


Analytical transparent boundary conditions for the Schrödinger

equation

OUTLINE 2-A

OUTLINE



equation

Derivation of discrete transparent boundary conditions (DTBC) for theSchrödinger equation for a higher order compact 9-point scheme

OUTLINE 2-B

OUTLINE



equation


Approximation of DTBC

OUTLINE 2-C

OUTLINE



equation



Simulation of quantum wave guides

OUTLINE 2-D

OUTLINE



equation




Extension of the DTBC to (nearly) arbitrary potentials

OUTLINE 2-E

OUTLINE



equation





more realistic simulations

OUTLINE 2-G

OUTLINE



equation





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


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.


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.


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.


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


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).


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)


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))


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


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


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


*

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


*

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


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


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.


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]


|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]


|ψ(x

,y,T

)|

T = 19500∆t


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)]


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


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


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


|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


−1 −0.5 0 0.5 1

−1

−0.5

0

0.5

1



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


numerical solution of the schrödinger equation on...

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