Qualitative measurement of Klauder coherent states usingBohmian Mechanics

Sanjib Dey

City University London

December 03, 2013

Based on Phys. Rev. A 88, 022116 (2013), with Prof. Andreas Fring

Sanjib Dey (City University London) Bohmian trajectories from coherent states 1 / 23

What is a coherent state?

Superposition of large no of quantum states⇒ Classical particle.For example, Glauber coherent state :

|α〉= N (α)∞

∑n=0

αn√

n!|n〉, N (α)⇒ e−

|α|22

Sometimes called the minimum uncertainty wavepacket ∆x∆p≈ }/2

Canonical coherent states : ∆x = ∆p = }/√

2a|α〉= α|α〉|α〉= eαa†−α?a|0〉= D(α)|0〉,〈β|α〉 6= δ(α−β)

Squeezed coherent state : ∆x∆p = }/2

Applications : Quantum Optics, Quantum information, Laser Physics,Mathematical Physics etc.

Sanjib Dey (City University London) Bohmian trajectories from coherent states 2 / 23

Generalised Klauder coherent state

For Hermitian H ; bounded below, nondegenerate eigenspectrum En = }ωen

and orthonormal eigenstate |φn〉 :

ψJ(x,γ) :=1

N (J)

∞

∑n=0

Jn/2e−iγen

√ρn

φn(x), J ∈ R+0

ρn := ∏nk=1 ek, N 2(J) := ∑

∞

k=0 Jk/ρk, ρ0 = 1

Properties1 Continuous in time and J.2

∫|ψJ〉〈ψJ| dµ = 1

3 Temporarily stable : e−iH tψJ(x,γ) = ψJ(x,γ+ωt), ω = Constant4 Satisfies action angle identity : 〈ψJ|H |ψJ〉= }ωJ

Sanjib Dey (City University London) Bohmian trajectories from coherent states 3 / 23

Procedure

One can analyse all the properties mathematically. Which is notsufficient to realise the quality precisely.

How would you measure the precise quality?

Draw the classical trajectories by solving :

x =∂H∂p

, p =−∂H∂x

(1)

Draw the dynamics of the coherent states of the particle and compare.

How would you draw the trajectories of the coherent states?

Sanjib Dey (City University London) Bohmian trajectories from coherent states 4 / 23

Bohmian mechanics

Quantum theory⇒ Solution of Schrodinger equation : ψ⇒ Probabilitiesof actual result.

Is it possible to find some other interpretation?

David Bohm(1952)⇒ Alternative trajectory based interpretation.

Undoubtedly successful : photodissociation problems, tunnellingprocess, atom diffraction by surfaces, high harmonic generation etc.

Bohmian mechanics =⇒ Still ongoing and controversial.Keeping interpretational issues aside =⇒ Apply it.

Sanjib Dey (City University London) Bohmian trajectories from coherent states 5 / 23

Bohmian mechanics (real case)

Time dependent Schrodinger equation :

ih∂ψ(x, t)

∂t=− h2

2m∂2ψ(x, t)

∂x2 +V(x)ψ(x, t)

WKB polar decomposition :

ψ(x, t) = R(x, t)eih S(x,t), R(x, t),S(x, t) ∈ R

Substitute ψ(x, t) into Schrodinger equation and separate real and imaginarypart :

St +(Sx)

2

2m+V(x)− h2

2mRxx

R= 0 ⇐ Quantum Hamilton-Jacobi equation

mRt +RxSx +12

RSxx = 0 ⇐ Continuity equation

Sanjib Dey (City University London) Bohmian trajectories from coherent states 6 / 23

Real Bohmian

∗ Velocity :

mv(x, t) = Sx =h2i

[ψ∗ψx−ψψ∗x

ψ∗ψ

]∗ Quantum potential :

Q(x, t) =− h2

2mRxx

R=

h2

4m

[(ψ∗ψ)2

x

2(ψ∗ψ)2 −(ψ∗ψ)xx

ψ∗ψ

]

∗ Effective potential Veff(x, t) = V(x)+Q(x, t).∗ Two options to compute quantum trajectories :

1 Solve⇒ v(x, t)2 Solve⇒ mx =−∂Veff/∂x

Sanjib Dey (City University London) Bohmian trajectories from coherent states 7 / 23

Bohmian mechanics (complex case)

∗ Decompose :ψ(x, t) = e

ih S(x,t), S(x, t) ∈ C

∗ Substitute ψ(x, t)⇒ time dependent Schrodinger equation :

St +(Sx)

2

2m+V(x)− ih

2mSxx = 0

∗ Velocity :

mv(x, t) = Sx =hi

ψx

ψ

∗ Quantum potential :

Q(x, t) =− ih2m

Sxx =−h2

2m

[ψxx

ψ− ψ2

x

ψ2

]

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Summarize

Solve canonical equations =⇒ Classical trajectoryCoherent state =⇒ Bohmian scheme =⇒ Trajectoriesof coherent stateCompare these two =⇒ Quality of coherent states

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Application : Poschl-Teller model (real case)

φn(x) =1√Nn

cosλ

( x2a

)sinκ

( x2a

)2F1

[−n,n+κ+λ;k+

12

;sin2( x

2a

)]Stationary state Bohmian :

v(t) = 0 ⇐ Not the behaviour of a classical particle.

Klauder coherent state :

ψJ(x,γ) :=1

N (J)

∞

∑n=0

Jn/2e−iγen

√ρn

φn(x)

ρn = n!(n+κ+λ)n, N 2(J) = 0F1 (1+κ+λ;J)

Classical solution :

x(t) = a arccos

[α−β

2+√

γcos

(√2Em

ta

)], α, β, γ constant

Sanjib Dey (City University London) Bohmian trajectories from coherent states 10 / 23

0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 02 . 0 0

2 . 0 1

2 . 0 2

2 . 0 3

2 . 0 4

2 . 0 5

x ( t )

t

( a )

0 5 10 15 20 252

3

4

5

6

(c)

J = 20 J = 10 J = 2 J = 20.2846

x(t)

t

Qualitatively not identical with classical trajectories !!

Look at the uncertainty of X & P

Look at the behaviour of |ψ(x, t)|2 with time too.

Sanjib Dey (City University London) Bohmian trajectories from coherent states 11 / 23

0 5 1 0 1 5 2 0 2 50

1

2

3

4

5

6

7

Q = - 0 . 3 0 7 5 9 3 Q = - 0 . 1 4 9 5 2 3 Q = - 0 . 0 4 2 5 5 5

∆x ∆p

t

( a )

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

t = 0t = 1t = 10t = 20t = 30

|(x

,t)|2

x

(b)

Not a squeezed coherent state, ∆x∆p ≫ }/2 !!Shape of the wave packet changes with time, i.e. not a classical particle!!

Need to localise the wavepacket !!How can we do that??

Sanjib Dey (City University London) Bohmian trajectories from coherent states 12 / 23

Mandel parameter

ψJ(x,γ) := 1N (J)

∞

∑n=0

Jn/2e−iγen√ρn

φn(x)

ψJ(x,γ) :=∞

∑n=0

cn(J)e−iγen |φn〉, cn =Jn/2

N (J)√

ρn⇐ weighting function

ψJ(x,γ) needs to be well localised.

To examine : check weighting probability, |cn|2⇒ Poissonian.

Deviation of |cn|2 from Poissonian is captured by Mandel parameter, Q .

If ψJ is strongly weighted around 〈n〉, Q = ∆n2

〈n〉 −1 = J ddJ ln d

dJ lnN 2

Q = 0 ⇒ Pure Poissonian, Q > 0 ⇒ Super-Poissonian.Q < 0 ⇒ Sub-Poissonian, |Q | � 1 ⇒ Quasi-Poissonian.

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Sub-Poissonian regime

0 5 1 0 1 5 2 0 2 50

1

2

3

4

5

6

7

Q = - 0 . 3 0 7 5 9 3 Q = - 0 . 1 4 9 5 2 3 Q = - 0 . 0 4 2 5 5 5

∆x ∆p

t

( a )

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

t = 0t = 1t = 10t = 20t = 30

|(x

,t)|2

x

(b)

Q =−0.307593,−0.149523,−0.042555

We are in sub-Poissonian regime !!!What happens in the quasi-Poissonian, Q→ 0 regime??

Sanjib Dey (City University London) Bohmian trajectories from coherent states 14 / 23

Q(J,κ+λ) = J2+κ+λ

0F1(3+κ+λ;J)0F1(2+κ+λ;J) −

J1+κ+λ

0F1(2+κ+λ;J)0F1(1+κ+λ;J)

One can control κ, λ and J, so that Q→ 0

0 5 10 15 20 25

0.5100

0.5103

0.5106

0.5109

0.5112

x p

t

Q= -0.000054529 Q= -0.000013634 Q= -0.000002726

(a)

0.0 0.4 0.8 1.2

0.5000055

0.5000070

0 1 2 3 4 5 60

1

2

3

t = 0 , J = 0 . 0 0 2 2 9 0 6 t = 0 . 6 5 , J = 0 . 0 0 2 2 9 0 6 t = 0 , J = 2 t = 4 , J = 2|Ψ

(x,t)|2

x

( b )

Two sets : κ = 90, λ = 100, J = 2,0.5,0.1 andκ = 2, λ = 3, J = 2,0.5,0.1

Sanjib Dey (City University London) Bohmian trajectories from coherent states 15 / 23

Quasi-Poissonian regime

0.0 0.2 0.4 0.6 0.8 1.02.00

2.01

2.02

2.03

2.04

2.05 (a)

J = 2.0 J = 0.5 J = 0.1

x(t)

t 0 5 10 15 20 25 302.00

2.01

2.02

2.03

2.04

2.05

2.06 (b)

J = 0.0022906 J = 0.00057265 J = 0.000114531

x(t)

t

Sanjib Dey (City University London) Bohmian trajectories from coherent states 16 / 23

Poschl-Teller potential (complex case)

H =p2

2m+

V0

2

[λ(λ−1)

cos2(x/2a)+

κ(κ−1)sin2(x/2a)

]− V0

2(λ+κ)2 for 0≤ x≤ aπ

Complexify : x⇒ xr + ixi, p⇒ pr + ipi

Real and imaginary part

Hr =p2

r −p2i

2m− V0

2(λ+κ)2

+V0

[(λ2−λ)

[cosh

( xia

)cos( xr

a

)+1][

cosh( xi

a

)+ cos

( xra

)]2

−(κ2−κ)

[cosh

( xia

)cos( xr

a

)−1][

cos( xr

a

)− cosh

( xia

)]2

]

Hi =pipr

m+V0

[(λ2−λ)sinh

( xia

)sin( xr

a

)[cosh

( xia

)+ cos

( xra

)]2 − (κ2−κ)sinh( xi

a

)sin( xr

a

)[cos( xr

a

)− cosh

( xia

)]2]

Sanjib Dey (City University London) Bohmian trajectories from coherent states 17 / 23

PT-symmetry and non-hermitian Hamiltonian

Hamiltonian→ non-hermitian 6= real eigenvalues.Bender et al [Phys. Rev. Lett. 80, 5243-5246 (1998)]

P T symmetric non-hermitian Hamiltonian⇒ Real eigenvalues.P → Parity transformation, T → Time reversal

In our case P T : xr→−xr, xi→ xi, pr→ pr, pi→−pi, i→−i

Solve canonical equations of motion :

xr =12

(∂Hr

∂pr+

∂Hi

∂pi

), xi =

12

(∂Hi

∂pr− ∂Hr

∂pi

),

pr = −12

(∂Hr

∂xr+

∂Hi

∂xi

), pi =

12

(∂Hr

∂xi− ∂Hi

∂xr

)

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Classical trajectory : Poschl-Teller potential

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0

00

500

500500

(a)xi

xr

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0

1

2

3

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00

0

0 0

0

0

500

500

500

500

(b)xi

xr

- 6 - 4 - 2 0 2 4 6

- 3

- 2

- 1

0

1

2

3

Blue : x0 = 4.5, p0 = 41.8376i, E =−31.7564Black : x0 = 3+1.5i, p0 =−30.1922+0.385121i, E =−6.55991−13.5182i

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Stationary state : complex case

ψn(x) =1√Nn

cosλ

( x2a

)sinκ

( x2a

)2F1

[−n,n+κ+λ;k+

12

;sin2( x

2a

)]

-6 -4 -2 0 2 4 6

-2

-1

0

1

2x

0= ±0.1

x0= ±1.5

x0= ±2.0

x0= ±2.45

x0= 5.0

x0= 5.5

x0(t)

t

(a)

-6 -4 -2 0 2 4 6-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

x0= ±0.1x0= ±0.3x0= ±0.9x0= ±1.5x0= ±2.7x0= ±3.6x0= ±4.5x0= ±5.0x0= ±5.5

x5(t)

t

(b)

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Classical and Klauder state

- 0.9

- 0.9 - 0.9

- 0.9 - 0.9

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- 0.8 - 0.8

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- 0.3- 0.2- 0.2

- 0.2- 0.2

- 0.2 - 0.1- 0.1- 0.1- 0.1 - 0.1

xi

xr

(a)

- 20 - 15 - 10 - 5 0 5

- 3

- 2

- 1

0

1

2

3

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- 0.4 - 0.4

- 0.4

- 0.4- 0.4

- 0.4

- 0.2

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- 0.2 - 0.2

- 0.2- 0.2

- 0.2 - 0.2

- 0.2

00

00

0

0

00 0

0

0

0

0

0

0.2

0.20.2

0.2

0.20.2

0.2 0.2

0.2

0.4

0.4

0.4

0.4

0.4 0.4

0.4

0.4

0.4

0.6

0.60.6

0.6

0.6

0.6

0.6

0.6

0.6

0.80.80.8

0.8

0.8

0.8

0.8

0.8

0.8

xr

xi (b)

- 20 - 15 - 10 - 5 0 5

- 3

- 2

- 1

0

1

2

3

Sub-Poissonian regime, Q < 0

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Classical and Klauder state

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00

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(a)xi

xr

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0

1

2

3

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00

0

0 0

0

0

500

500

500

500

(b)xi

xr

- 6 - 4 - 2 0 2 4 6

- 3

- 2

- 1

0

1

2

3

Quasi-Poissonian regime, Q → 0Perfect matching : Classical⇐⇒ Klauder coherent state

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ConclusionOne must draw the trajectories of classical case and coherent states andcompare them to study the behaviour of the coherent states.

We have found an extra parameter Q which governs the behaviour of thecoherent states.

Q→ 0, Klauder state is a perfect coherent state for both real andcomplex cases.

Must take Klauder state for generalised models, instead of Glauber state.

Thank you for your attention

Sanjib Dey (City University London) Bohmian trajectories from coherent states 23 / 23