nonlinear schrödinger equation with complex supersymmetric...

NonlinearSchrodingerequation withcomplex su-persymmetricpotentials

D. Nath∗

PTsymmetry

Supersymmetricpotentials

Non LinearSchrodingerEquation(NLSE)

Solutions

Examples

LinearStabilityAnalysis

Stabilitycondition

Conclusion

Nonlinear Schrodinger equation with complexsupersymmetric potentials

D. Nath∗

Department of MathematicsVivekananda College, Kolkata-700063, India.

Supersymmetries & Quantum Symmetries - SQS’2015August 6, 2015

∗in collaboration with P. Roy, ISI, Kolkata, India.

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Plan of the talk

1 PT symmetry

2 Supersymmetric potentials

3 Non Linear Schrodinger Equation (NLSE)SolutionsExamples

4 Linear Stability AnalysisStability condition

5 Conclusion

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

PT symmetry

• PT symmetric quantum mechanical system are invariantunder the simultaneous action of the P space and T timeinversion operations.

• P:p→ −p, x→ −x, T :i→ −i• These systems possess non-Hermitian Hamiltonians, still

they have some characteristics similar to Hermitianproblems.

• discrete energy spectrum, (partly real or completely real)

• basis states form an orthogonal set

• inner product 〈ψ(x, t)|ψ(x, t)〉PT = 〈ψ(x, t)|PT |ψ(x, t)〉

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

PT symmetric

PT symmetric Hamiltonian

• simplest PT symmetric Hamiltonian contains a 1DSchrodinger operator H

• PT H=HPT = p2

2m + V ∗(−x) = H

• such problems have been described by numerical andperturbational techniques and also several PT symmetricpotentials have been described analytically

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

PT symmetric potential

Linear Schrodinger Eq. with PT symmetric complex potential

−d2ψ

dx2+ V (x)ψ = Eψ

PT symmetric potentials (P:p→ −p, x→ −x, T :i→ −i)

• V ∗(−x) = V (x), PT symmetric potentials

• V = VR(x) + iVI(x), then VR(−x) = VR(x) &VI(−x) = −VI(x)

• PT symmetric systems can put into the Non LinearSchrodinger Equation(NLSE)

• C. M. Bender and S. Boettcher, Phys. Rev. Lett 80,(1998) 5243.

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Supersymmetric potentials

Linear Schrodinger Eq. with supersymmetric potential

−d2ψ

dx2+ V±(x)ψ = Eψ

• V±(x) = w2(x)± w′(x), supersymmetric potentials

• w(x) is the superpotential

• ψ(−)0 (x) = e−

∫w(x) dx is the ground state, with E

(−)0 = 0

• F. Cooper, A. Khare and U. Sukhatme, Supersymmetry inQuantum Mechanics, World Scientific (2001).

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Non Linear Schrodinger Equation (NLSE) withcomplex potentials and power law nonlinearity

NLSE

iΨt = −Ψxx + (V (x) + iW (x))Ψ + g|Ψ|2kΨ (1)

• For the optical beam dynamics t is a scaled propagationdistance, g = −1, corresponds to a self-focusingnonlinearity while g = 1 to a defocusing one

• same model as the Gross-Pitaevskii equation inBose-Einstein condensates

• same model as the propagation of laser radiation along thet axis of a medium

• The NLS equations describe physical phenomena in optics,Bose–Einstein condensates, as well as water waves.

• so our model is a prototype NLSE

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Potentials

Cubic nonlinearity k = 1

• V (x) + iW (x) = ε δ(x) + i γ δ′(x) Chaos 25, 023112(2015).

• V (x) + iW (x) = x2 − 2 i α x Phys. Rev. A 85, 043840(2012)

• V (x) + iW (x) = w2(x)− iw′(x), w(x) is real,arXiv:1408.2719v3 (2014), (non PT )

• V (x) + iW (x) = sech2x + i sech x tanh x

• k = 2 quintic nonlinearity

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Ψ(x, t) = e−iµtψ(x)

−d2ψdx2

+ (V + iW )ψ + g|ψ|2kψ = µψ (2)

• µ is the propagation constant, momentum number

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Our present work

Super potential of the form U(x) = UR(x) + i UI(x) andgeneral power law nonlinearity k

• V = V0Re(U2 − U ′) = V0 (U2

R − U ′R − U2I )

• W = W0 Im(U2 − U ′) = W0 (2URUI − U ′I)• V (−x) = V (x), W (−x) = −W (x), i.e.,UR(−x) = −UR(x), UI(−x) = UI(x)

ψ = ψ0 e−

∫(UR+iAUI)dx

(V0 − 1)(U2R − U

′R) + (A2 − V0)U2

I + gψ2k0 e−2k

∫UR(x) dx = µ

(2URUI − U′I)(W0 −A) = 0 (3)

• for linear Schrodinger Eq. g = 0

• for exact SUSY V0 = W0 = A = 1, µ = 0

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PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

• (2URUI − U′I) = 0, generates real potential

• for PT symmetric complex potentials we choose W0 = A

• for NLSE V0 = W0 = A = 1 is not possible

• V0 6= 1 or (and )W0 = A 6= 1

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Characteristic of our model

The basic characteristic of our model is the following

• We are trying to use supersymmetric potential of thelinear Schrodeinger Eq. into the nonlinear one.

• However this can not be done exactly and therefore oneneeds to deform the procedure at two stages

• at the level of potential (V0,W0) and secondly at the levelof solution (A0) (which again causes a deformation of thepotential).

• We call it deformed SUSY complex potential

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PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Find UR(x), UI(x) such that they satisfy theconditions (3)

Case 1. V0 = A2, UR(x) = −f ′(x)f(x)

(V0 − 1)f′2(x) +

g0ψ2k0

k+1 f2k+2(x) = µf2(x) (4)

• this is a first order ordinary nonlinear differential equation

• for different parameter values we get different solutions

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PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Solutions for case 1

• (i) f(x) = α sech1k (βx + γ) with µ = (V0−1)β2

k2, and

g(αψ0)2k = − (k+1)(V0−1)β2

k2, PT

• (ii) f(x) = α sec1k (βx + γ) with µ = − (V0−1)β2

k2, and

g(αψ0)2k = − (k+1)(V0−1)β2

k2, PT

• (iii) f(x) = α cosech1k (βx + γ) with µ = (V0−1)β2

k2, and

g(αψ0)2k = − (k+1)(V0−1)β2

k2, non PT

• (iv) f(x) = α cosec1k (βx + γ) with µ = − (V0−1)β2

k2, and

g(αψ0)2k = − (k+1)(V0−1)β2

k2, non PT

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Find UR(x), UI(x) such that they satisfy theconditions

Case 2. V0 6= A2, UR(x) = −f ′(x)f(x)

(V0 − 1)f′′

+ (A20 − V0)U2

I (x) f +g0ψ2k

0k+1 f

2k+21 = µ (5)

• this is a NLSE with real potential we can solve Eq.(5) forsome particular UI(x)’s.

• these particular solutions can be obtained from thesolutions for case 1 by properly choosing the parameters.

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Example 1. UR = tanh xk , UI = C0sechmx

PT symmetric potential

• V = V0k2

[1− (k + 1) sech2x− k2 C2

0 sech2mx]

• W = W0 C0k (2 + km) sechmx tanh x

Ψ(x, t) = ψ0 sech1k x e−i{µt+W0 C0 2F1[

12, 1+m

2, 32,− sinh2 x] sinh x}

where

• ψ0 =[(k+1)(V0−1)

gk2

] 12k, µ = V0−1

k2, V0 = W 2

0

• 2F1[a, b, c, x] is a Hypergeometric function.

• For V0 6= W 20 the solution also exists when m = 0, 1

• M. Abramowitz and I.A. stegun, Handbook ofMathematical functions with formulas, graphs andMathematical tables, Dover, 1972.

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Example 2. UR = − tan xk , UI = C0secmx

PT symmetric periodic potential

• V = V0k2

[−1 + (k + 1) sec2x− k2 C2

0 sec2mx]

• W = −W0 C0k (2 + km) secmx tan x

Ψ(x, t) = ψ0 sec1k x e−i{µt+W0 C0 2F1[

12, 1+m

2, 32,sin2 x] sin x}

• ψ0 =[(k+1)(1−V0)

gk2

] 12k, µ = 1−V0

k2, V0 = W 2

0

• 2F1[a, b, c, x] is a Hypergeometric function.

• For V0 6= W 20 the solution also exists when m = 0, 1

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Example 3. UR = sn(x,m) dn(x,m)k cn(x,m) , UI(x) = C0

cn(x,m)

PT symmetric periodic potential

• V (x) = ξ−1cn−2(x,m) + ξ0 + ξ1cn

2(x,m)

• W (x) =(2k − 1

)W0C0

sn(x,m)dn(x,m)cn2(x,m)

Ψ(x, t) = ψ0 cn1k (x,m) e

−i{µt+

W0C0√1−m

ln[dn(x,m)cn(x,m)

+√1−m sn(x,m)

cn(x,m)]}

• ψ0 =[m(V0−1)(k+1)

gk2

] 12k, µ = (2m−1)(V0−1)

k2

• C0 ={

(k−1)(V0−1)(1−m)k2(W 2

0−V0)

} 12

, V0 6= W 20

• ξ−1 ={

(1−k)(1−m)k2

− C20

}V0

• ξ0 ={

(1−k)(2m−1)k2

+ (V0−1)(2m−1)k

}V0

• ξ1 = −{m(1−k)k2

+ 2m(V0−1)k

}V0

• sn(x,m), cn(x,m), dn(x,m) are Jacobi Elliptic functions.18 / 29


D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Linear stability analysis

Perturbed solution

Ψ(x, t) = {ψ(x) + [v(x) + w(x)] eλt + [v∗(x)− w∗(x)] eλ∗t}eiµt

(6)

• |v|, |w| << |ψ|, neglecting 2nd and higher order of v, w.

Eigen value Eq.

i

(h0 ∇2 + h1

∇2 + h2 h3

)(vw

)= λ

(vw

)(7)

h0 = g k2

(ψ2 − ψ∗2

)|ψ|2k−2 + i W (x)

h1 = µ+ V (x) + g|ψ|2k +{|ψ|2 − 1

2

(ψ2 + ψ∗2

)}gk|ψ|2k−2

h2 = µ+ V (x) + g|ψ|2k +{|ψ|2 + 1

2

(ψ2 + ψ∗2

)}gk|ψ|2k−2

h3 = −gk2

(ψ2 − ψ∗2

)|ψ|2k−2 + i W (x) (8)

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Stability condition

λ = λR + iλI

• [v(x) + w(x)] eλt = [v(x) + w(x)] [cosλIt+ i sinλIt]eλRt

• [v∗(x)− w∗(x)] eλ∗t =

[v∗(x)− w∗(x)] [cosλIt− i sinλIt]eλRt

• If λR > 0, limt→∞

eλRt →∞, and the corresponding solution

is unstable.

• Therefore the solution Ψ(x, t) is stable if λR in nonpositive.

• To solve Eq.(7) we use Fourier Collocation Method (FCM)

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Eigen value spectra & time evolution of Ψ(x, t)stable solution (m = 1)

Stable mode of example 1

−4 −3 −2 −1 0 1 2 3 4 5

x 10−12

−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

Im[λ

]

(a) k=0.45, g=−1, C0=0.1, V

0=0.995,W

0=−0.1 HbL k=0.45, g=-1, C0=0.1, V0=0.995, W0=-0.1

-40 -20 0 20 40

X

0

20

40

60

80t

0.000

0.005

0.010

ÈYÈ

• (a) λR = Re(λ) < 5× 10−12 ≈ 0

• (b) stable solution

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Eigen value spectra & time evolution of Ψ(x, t)stable solution

Stability condition for Re(λ)

−0.05 0 0.05−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

Im[λ]

(a) k=2

−0.4 −0.2 0 0.2 0.4−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

(b) k=3

−0.4 −0.2 0 0.2 0.4−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

(c) k=4

HdL k=2, g=-1, C0=0.1, V0=0.01, W0=-0.1

-40 -20 0 20 40

X

0

20

40

60

80t

0.0

0.5

1.0

ÈYÈ

HeL k=3, g=-1, C0=0.1, V0=0.01, W0=-0.1

-40 -20 0 20 40

X

0

20

40

60

80t

0.0

0.5

1.0

ÈYÈ

HfL k=4, g=-1, C0=0.1, V0=0.01, W0=-0.1

-40 -20 0 20 40

X

0

20

40

60

80t

0.0

0.5

1.0

ÈYÈ

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Stable solution

Stable mode of example 1 for focusing nonlinearity (g = −1)

• g = −1, C0 = 0.1, V0 = 0.81,W0 = −0.9

−40 −20 0 20 400

20

6080

0

0.5

1

t

(d) k=3

x−40 −20 0 20 400

20

6080

0

0.5

1

t

(c) k=2

x

|ψ(x

,t)|

−1.5 −1 −0.5 0 0.5 1 1.5

x 10−11

−2000

−1000

0

1000

2000

Re[λ]

Im[λ

]

(a) k=2

−3 −2 −1 0 1 2 3 4

x 10−12

−2000

−1000

0

1000

2000

Re[λ]

(b) k=3

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Stable solution

Stable mode of example 1 for defocusing nonlinearity (g = 1)

• g = 1, C0 = 0.1, V0 = 1.01,W0 = −1.05

−5 0 5 10

x 10−12

−2000

−1000

0

1000

2000

Re[λ]

Im[λ

]

(a) k=2

−5 0 5

x 10−12

−2000

−1000

0

1000

2000

Re[λ]

(b) k=3

−5 0 5

x 10−12

−2000

−1000

0

1000

2000

Re[λ]

(c) k=4

−40 −20 0 20 40 0

20

60

80

0

0.2

0.4

t

(d) k=2

x

|ψ(x

,t)|

−40 −20 0 20 40 0

20

60

80

0

0.5

1

t

(e) k=3

x −40 −20 0 20 40 0

20

60

80

0

0.5

1

t

(f) k=4

x

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Eigen value spectra & time evolution of Ψ(x, t)unstable solution

Unstable mode of example 1

• (a) g = −1, C0 = 1, V0 = 0.25,W0 = −0.5

−0.2 −0.1 0 0.1 0.2−5

−4

−3

−2

−1

0

1

2

3

4

5

Re[λ]

Im[λ]

(a) k=2

−1 −0.5 0 0.5 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Re[λ]

(b) k=3

HcL k=2, g=-1, C0=1, V0=0.25, W0=-0.5

-40 -20 0 20 40

X

0

20

40

60

80t

0

1´ 1016

2´ 1016

3´ 1016

4´ 1016

ÈYÈ

HdL k=3, g=-1, C0=1, V0=0.25, W0=-0.5

-40 -20 0 20 40

X

0

20

40

60

80t

0

2

4

ÈYÈ

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Eigen value spectra & time evolution of Ψ(x, t)unstable solution (m = 1)


• g = 1, C0 = 0.1, V0 = 0.25,W0 = −0.5

−4 −2 0 2 4

x 104

−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

Im[λ

]

(a) k=2

−4 −2 0 2 4

x 104

−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

(b) k=3

−2 −1 0 1 2

x 104

−2000

−1500

−1000

−500

0

500

1000

1500

2000

Re[λ]

(c) k=4

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Eigen value spectra & time evolution of Ψ(x, t)unstable solution


• g = −1,m = 0.75, V0 = 0.5,W0 = −0.5

−0.08 −0.06 −0.04 −0.02 0 0.02 0.04 0.06 0.08−1.5

−1

−0.5

0

0.5

1

1.5x 10

4

Im[λ

]

(a) k=2

−4000 −3000 −2000 −1000 0 1000 2000 3000 4000−1.5

−1

−0.5

0

0.5

1

1.5x 10

4 (b) k=3

−5000 0 5000−1

−0.5

0

0.5

1x 10

4

Re[λ]

Im[λ

]

(c) k=4

−6000 −4000 −2000 0 2000 4000 6000−1

−0.5

0

0.5

1x 10

4

Re[λ]

(d) k=4.45

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Conclusion

• A method of constructing exact solutions of NLSE withpower law nonlinearity and complex PT symmetricpotentials has been suggested.

• We have analyzed Linear stability analysis of thesesolutions .

• In this case we find that for suitably chosen parametervalues stable solutions can be found for g = 1, k = 2, 3 aswell as for g = −1, k = 2, 3.

• Stable solutions for non integral values of the nonlinearityparameter k has also been found for the example 1 forfocusing (g=-1) as well as de-focusing (g=1) nonlinearity.

• Using these method we can solve NLSE with nonPT symmetric potentials as well.

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D. Nath∗

PTsymmetry



Solutions

Examples


Stabilitycondition

Conclusion

Thank You

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