[PPT] - New Dirac Delta function based methods - Kempf, Jackson, Morales PowerPoint Presentation

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New Dirac Delta function based methods - Kempf, Jackson, Morales

Marco Knipfer

Goethe Universitaet Frankfurt knipfer@fias.uni-frankfurt.de

July 30, 2014

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 1 / 23

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The paper I am talking about

Achim Kempf, David M. Jackson, Alejandro H. Morales New Dirac Delta function based methods with applications to perturbative expansions in quantum field theory and beyond http://arxiv.org/abs/1404.0747

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 2 / 23

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Overview

1

Introduction New way of writing the Dirac Delta New way of Fourier transforming New integration method

2

Application Quantum Field Theory Application to our work?

3

New delta, blurring, deblurring and QFT

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 3 / 23

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Fourier Transform Convention

g(y) :=

1 √ 2π

g(x)eixy dx

(1)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 4 / 23

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Fourier Transform Convention

g(y) :=

1 √ 2π

g(x)eixy dx

(1) g(x) := 1 √ 2π

g(y)e−ixy dx

(2)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 4 / 23

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New Dirac Delta

δ(x) = 1 2π

eixy dy

(3) = 1 2π

1

g(y)g(y)eixy dy (4) = 1 √ 2π 1 g(−i∂x) 1 √ 2π g(y)eixy dy

g(x)

(5)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 5 / 23

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New Dirac Delta

δ(x) = 1 2π

eixy dy

(3) = 1 2π

1

g(y)g(y)eixy dy (4) = 1 √ 2π 1 g(−i∂x) 1 √ 2π g(y)eixy dy

g(x)

(5) δ(x) = 1 √ 2π 1 g(−i∂x) g(x), (6)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 5 / 23

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New Dirac Delta

δ(x) = 1 2π

eixy dy

(3) = 1 2π

1

g(y)g(y)eixy dy (4) = 1 √ 2π 1 g(−i∂x) 1 √ 2π g(y)eixy dy

g(x)

(5) δ(x) = 1 √ 2π 1 g(−i∂x) g(x), (6) where g must be a “sufficiently well-behaved” function, e.g. a Gaussian, then

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 5 / 23

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New Dirac Delta

Example: g is a Gaussian, then δ(x) = 1 √ 2πσ

∞

n=0

( −σ

2 ∂2 x)n

n! e−x2/2σ, (7)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 6 / 23

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New Dirac Delta

Example: g is a Gaussian, then δ(x) = 1 √ 2πσ

∞

n=0

( −σ

2 ∂2 x)n

n! e−x2/2σ, (7) can be truncated → approximation of δ(x).

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New way of Fourier transforming

δ(x) = 1 √ 2π 1 g(−i∂x) g(x) , (8)

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New way of Fourier transforming

δ(x) = 1 √ 2π 1 g(−i∂x) g(x) , (8) can be inverted

g(x) =

√ 2πg(−i∂x)δ(x) . (9)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 7 / 23

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New way of Fourier transforming

δ(x) = 1 √ 2π 1 g(−i∂x) g(x) , (8) can be inverted

g(x) =

√ 2πg(−i∂x)δ(x) . (9)

Benefits?

Maybe easier to compute

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 7 / 23

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New way of Fourier transforming

δ(x) = 1 √ 2π 1 g(−i∂x) g(x) , (8) can be inverted

g(x) =

√ 2πg(−i∂x)δ(x) . (9)

Benefits?

Maybe easier to compute Put in approximation of δ(x) → approximative FT

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 7 / 23

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Second version of the FT

First one:

g(x) =

√ 2πg(−i∂x)δ(x) (10)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 8 / 23

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Second version of the FT

First one:

g(x) =

√ 2πg(−i∂x)δ(x) (10) Second one (proof easy, see paper):

g(y) =

√ 2πeixyδ(i∂x − y)g(x) (11)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 8 / 23

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Second version of the FT

First one:

g(x) =

√ 2πg(−i∂x)δ(x) (10) Second one (proof easy, see paper):

g(y) =

√ 2πeixyδ(i∂x − y)g(x) (11) How can this be used to integrate?

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 8 / 23

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Second version of the FT

First one:

g(x) =

√ 2πg(−i∂x)δ(x) (10) Second one (proof easy, see paper):

g(y) =

√ 2πeixyδ(i∂x − y)g(x) (11) How can this be used to integrate? ∞

−∞

g(x) dx = 2π g(0) (12)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 8 / 23

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New integration method using the new Dirac Delta

∞

−∞

g(x) dx = 2π g(0) , (13)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 9 / 23

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New integration method using the new Dirac Delta

∞

−∞

g(x) dx = 2π g(0) , (13) so by using the above representations of g

g(x) dx = 2πg(−i∂x)δ(x)|x=0

(14)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 9 / 23

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New integration method using the new Dirac Delta

∞

−∞

g(x) dx = 2π g(0) , (13) so by using the above representations of g

g(x) dx = 2πg(−i∂x)δ(x)|x=0

(14)

g(x) dx = 2πδ(i∂x)g(x)

(15)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 9 / 23

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What do we got so far?

Two representations of FT

g(x) =

√ 2πg(−i∂x)δ(x) (16)

g(y) =

√ 2πeixyδ(i∂x − y)g(x) (17)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 10 / 23

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What do we got so far?

Two representations of FT

g(x) =

√ 2πg(−i∂x)δ(x) (16)

g(y) =

√ 2πeixyδ(i∂x − y)g(x) (17)

Two representations of integrals from 0 to ∞

g(x) dx = 2πg(−i∂x)δ(x)|x=0

(18)

g(x) dx = 2πδ(i∂x)g(x)

(19)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 10 / 23

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Overview

1

Introduction New way of writing the Dirac Delta New way of Fourier transforming New integration method

2

Application Quantum Field Theory Application to our work?

3

New delta, blurring, deblurring and QFT

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 11 / 23

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Quantum Field Theory I

Generating functional (applying eqs. for FT) Z[J] =

exp
iS[φ] + i
Jφ dnx
Dφ ,

(20)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 12 / 23

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Quantum Field Theory I

Generating functional (applying eqs. for FT) Z[J] =

exp
iS[φ] + i
Jφ dnx
Dφ ,

(20) translates to Z[J] = NeiS[−iδ/δJ]δ[J] (21) = Nei

φJ dnxδ(iδ/δφ − J)eiS[φ]

(22)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 12 / 23

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Quantum Field Theory I

Generating functional (applying eqs. for FT) Z[J] =

exp
iS[φ] + i
Jφ dnx
Dφ ,

(20) translates to Z[J] = NeiS[−iδ/δJ]δ[J] (21) = Nei

φJ dnxδ(iδ/δφ − J)eiS[φ]

(22) Approximation by using approximative δ[J]

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 12 / 23

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Quantum Field Theory I

Generating functional (applying eqs. for FT) Z[J] =

exp
iS[φ] + i
Jφ dnx
Dφ ,

(20) translates to Z[J] = NeiS[−iδ/δJ]δ[J] (21) = Nei

φJ dnxδ(iδ/δφ − J)eiS[φ]

(22) Approximation by using approximative δ[J] “Usual pertubative expansion of Z[J] is not convergent and is at best asymptotic”

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 12 / 23

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Quantum Field Theory I

Generating functional (applying eqs. for FT) Z[J] =

exp
iS[φ] + i
Jφ dnx
Dφ ,

(20) translates to Z[J] = NeiS[−iδ/δJ]δ[J] (21) = Nei

φJ dnxδ(iδ/δφ − J)eiS[φ]

(22) Approximation by using approximative δ[J] “Usual pertubative expansion of Z[J] is not convergent and is at best asymptotic” This could lead to better convergence

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 12 / 23

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Quantum Field Theory II

Generating functional (applying eqs. for integration) Z[J] = NeiS[−iδ/δφ]+

Jδ/δφ dnxδ[φ]|φ=0

(23) = Nδ[iδ/δφ]eiS[φ]+i

Jφ dnx

(24)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 13 / 23

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Application to our work

To solve

dnp

1 + βpn eixp =

f (p)eixp dnp

(25)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 14 / 23

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Application to our work

To solve

dnp

1 + βpn eixp =

f (p)eixp dnp

(25)

The possibilities

1

g(x) =

√ 2πg(−i∂x)δ(x)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 14 / 23

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Application to our work

To solve

dnp

1 + βpn eixp =

f (p)eixp dnp

(25)

The possibilities

1

g(x) =

√ 2πg(−i∂x)δ(x)

2

g(y) =

√ 2πeixpδ(i∂x − y)g(x)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 14 / 23

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Application to our work

To solve

dnp

1 + βpn eixp =

f (p)eixp dnp

(25)

The possibilities

1

g(x) =

√ 2πg(−i∂x)δ(x)

2

g(y) =

√ 2πeixpδ(i∂x − y)g(x)

3

g(x) dx = 2πg(−i∂x)δ(x)|x=0

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 14 / 23

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Application to our work

To solve

dnp

1 + βpn eixp =

f (p)eixp dnp

(25)

The possibilities

1

g(x) =

√ 2πg(−i∂x)δ(x)

2

g(y) =

√ 2πeixpδ(i∂x − y)g(x)

3

g(x) dx = 2πg(−i∂x)δ(x)|x=0

4

g(x) dx = 2πδ(i∂x)g(x)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 14 / 23

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Application to our work

To solve

dnp

1 + βpn eixp =

f (p)eixp dnp

(25)

The possibilities

1

g(x) =

√ 2πg(−i∂x)δ(x)

2

g(y) =

√ 2πeixpδ(i∂x − y)g(x)

3

g(x) dx = 2πg(−i∂x)δ(x)|x=0

4

g(x) dx = 2πδ(i∂x)g(x) Since 3-4 follow from 1-2, they should not give any new method for solving eq. (25). Right?

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 14 / 23

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#1

The Fourier Transformation #1

dnp

1 + βpn eixp =

f (p)eixp dnp

(26)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 15 / 23

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#1

The Fourier Transformation #1

dnp

1 + βpn eixp =

f (p)eixp dnp

(26) The equation g(x) = √ 2πg(−i∂x)δ(x) has so be extended to n dimensions

f (x) = (2π)n/2f (−i∇x)δ(x)

(27) = (2π)n/2 1 1 + β(−i∇x)n δ(x), (28)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 15 / 23

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#1

The Fourier Transformation #1

dnp

1 + βpn eixp =

f (p)eixp dnp

(26) The equation g(x) = √ 2πg(−i∂x)δ(x) has so be extended to n dimensions

f (x) = (2π)n/2f (−i∇x)δ(x)

(27) = (2π)n/2 1 1 + β(−i∇x)n δ(x), (28) That’s where we started some time ago . . .

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 15 / 23

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#2

The Fourier Transformation #2

dnp

1 + βpn eixp =

f (p)eixp dnp

(29)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 16 / 23

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#2

The Fourier Transformation #2

dnp

1 + βpn eixp =

f (p)eixp dnp

(29)

f (x) = (2π)n/2eixpδn(i∇p − x)f (p)

(30) = (2π)n/2eixpδn(i∇p − x) 1 1 + βpn (31) = (2π)n/2eixp ∞ δn (i(nβs) − x) e−(1−βpn) ds , (32)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 16 / 23

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#2

The Fourier Transformation #2

dnp

1 + βpn eixp =

f (p)eixp dnp

(29)

f (x) = (2π)n/2eixpδn(i∇p − x)f (p)

(30) = (2π)n/2eixpδn(i∇p − x) 1 1 + βpn (31) = (2π)n/2eixp ∞ δn (i(nβs) − x) e−(1−βpn) ds , (32) and now?

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 16 / 23

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#2

The Fourier Transformation #2

dnp

1 + βpn eixp =

f (p)eixp dnp

(29)

f (x) = (2π)n/2eixpδn(i∇p − x)f (p)

(30) = (2π)n/2eixpδn(i∇p − x) 1 1 + βpn (31) = (2π)n/2eixp ∞ δn (i(nβs) − x) e−(1−βpn) ds , (32) and now? In eq. (31) use approximation for the δ-function → approx. solution

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 16 / 23

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#2

The Fourier Transformation #2

dnp

1 + βpn eixp =

f (p)eixp dnp

(29)

f (x) = (2π)n/2eixpδn(i∇p − x)f (p)

(30) = (2π)n/2eixpδn(i∇p − x) 1 1 + βpn (31) = (2π)n/2eixp ∞ δn (i(nβs) − x) e−(1−βpn) ds , (32) and now? In eq. (31) use approximation for the δ-function → approx. solution Maybe in eq. (32) something more can be done. Reminds me of that Fadeev Popov trick.

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 16 / 23

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Another idea

Take the equation for the Dirac Delta δ(x) = 1 √ 2π 1 g(−i∂x) g(x) , (33) and use g =

1 1+βpn

δn(x) = 1 (2π)n/2 1 g(−i∇x) g(x) (34) = 1 (2π)n/2 (1 + β(−i∇x)n)

1

(2π)n/2 eixp 1 1 + βpn dnp . (35) Rearranging the last line gives

eixp

1 + βpn dnp = 1 (2π)n δn(x) − (−i)nβ

∇n

x

(ip)n

eixp 1 1 + βpn dnp (36)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 17 / 23

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Another idea

eixp

1 + βpn dnp = 1 (2π)n δn(x) − (−i)nβ

(ip)neixp

1 1 + βpn dnp (37)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 18 / 23

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Another idea

eixp

1 + βpn dnp = 1 (2π)n δn(x) − (−i)nβ

(ip)neixp

1 1 + βpn dnp (37) The integral only gives the energy density, has to be integrated again from 0 to r (spher. coords.).

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 18 / 23

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Another idea

eixp

1 + βpn dnp = 1 (2π)n δn(x) − (−i)nβ

(ip)neixp

1 1 + βpn dnp (37) The integral only gives the energy density, has to be integrated again from 0 to r (spher. coords.). Maybe the last integral is somehow easier to solve, but I didn’t go further.

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 18 / 23

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Overview

1

Introduction New way of writing the Dirac Delta New way of Fourier transforming New integration method

2

Application Quantum Field Theory Application to our work?

3

New delta, blurring, deblurring and QFT

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 19 / 23

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Blurring a signal

Blurred signal fB fB(y) :=

f (x)g(x − y) dx ,

(38) e.g. Gaussian blurring.

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 20 / 23

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Blurring a signal

Blurred signal fB fB(y) :=

f (x)g(x − y) dx ,

(38) e.g. Gaussian blurring. Also possible: Deblurring the signal. Deblurring operator Dg DgfB(y) = 1 √ 2π g(i∂y) fB(y) = f (y) , (39) proof see paper. For Gaussian:

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 20 / 23

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Blurring a signal

Blurred signal fB fB(y) :=

f (x)g(x − y) dx ,

(38) e.g. Gaussian blurring. Also possible: Deblurring the signal. Deblurring operator Dg DgfB(y) = 1 √ 2π g(i∂y) fB(y) = f (y) , (39) proof see paper. For Gaussian: DGauss = e− 1

2a ∂2 y =

∞

n=0

1 n!

−∂2

y

2a n , (40) can be truncated to get an approximation.

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 20 / 23

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Euclidean φ4 theory

Generating functional: Z[J] =

e
dnx(− 1

2 φ(x)(−∆+m2)φ(x)−λφ4(x)+J(x)φ(x))Dφ

(41) in quadratic term substitute φ by derivatives. Z[J] = e− 1

2

dnx

δ δJ(x) (−∆+m2) δ δJ(x)

Gaussian deblurring operator
e
dnx(−λφ4(x)+J(x)φ(x))Dφ

(42)

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 21 / 23

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Euclidean φ4 theory

Generating functional: Z[J] =

e
dnx(− 1

2 φ(x)(−∆+m2)φ(x)−λφ4(x)+J(x)φ(x))Dφ

(41) in quadratic term substitute φ by derivatives. Z[J] = e− 1

2

dnx

δ δJ(x) (−∆+m2) δ δJ(x)

Gaussian deblurring operator
e
dnx(−λφ4(x)+J(x)φ(x))Dφ

(42) Can be expanded in powers of

1 √ λ (strong coupling).

Marco Knipfer (FIAS) New Dirac Delta July 30, 2014 21 / 23

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Euclidean φ4 theory

Generating functional: Z[J] =

e
dnx(− 1

2 φ(x)(−∆+m2)φ(x)−λφ4(x)+J(x)φ(x))Dφ

(41) in quadratic term substitute φ by derivatives. Z[J] = e− 1

2

dnx

δ δJ(x) (−∆+m2) δ δJ(x)

Gaussian deblurring operator
e
dnx(−λφ4(x)+J(x)φ(x))Dφ

(42) Can be expanded in powers of

1 √ λ (strong coupling).

The typical way of pulling φ4 in front of the integral is analytically not justified (see ref [10,11] in paper).

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The End

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