modified fock-schwinger method - kyoto u

IABLOKOV S.N. [1,2] and KUZNETSOV A.V. [1]

19.01.2021

Modified Fock-Schwinger method

[1] P.G. Demidov Yaroslavl State University, Yaroslavl, Russia

[2] A.A. Kharkevich Institute for Information Transmission Problems, Moscow, Russia

simplifies calculation of charged particle propagators in a constant magnetic field

2 approaches to find propagators

Canonical quantization

“Sum over solutions”

2

2 approaches to find propagators

Canonical quantization Path integral formalism

“Sum over solutions” Propagator equation

3

Sum over solutions: main features

Obtain general form

4

Orthogonalization & normalization

Sum over polarizations

Find polarizations vectors

Propagator equation: main features

Obtain general form

Orthogonalization & normalization

Sum over polariations

Find polarizations vectors

5

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿4(𝑥 − 𝑥′)

𝑆 𝑥, 𝑥′ = 𝑆 𝑥 − 𝑥′ = න𝑑4𝑝

2𝜋 4𝑒−𝑖 𝑝 𝑥−𝑥′

𝑆(𝑝)

Translational invariance is assumed

Propagator equation

Add external field

6

𝑖𝜕𝜇 → 𝑖𝐷𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)

Propagator equation

Translational invariance is lostAdd external field

7

𝑖𝜕𝜇 → 𝑖𝐷𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥) 𝑆 𝑥, 𝑥′ ≠ 𝑆 𝑥 − 𝑥′

≠ න𝑑4𝑝

2𝜋 4𝑒−𝑖 𝑝(𝑥−𝑥′) 𝑆(𝑝)

Fock-Schwinger (FS) method

Let’s solve this equation for S(x,x’):

8

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′)


Let’s solve this equation for S(x,x’): Choose a parametrization:

9

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′) 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝑈(𝑥, 𝑥′, 𝜏)

Ref: J.SchwingerPhys. Rev. 82, 664Published 1 June 1951

See also: FS method ina book on QFT byC. Itzykson, J.-B. Suber


Let’s solve this equation for S(x,x’): Choose a parametrization:

10

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′) 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝑈(𝑥, 𝑥′, 𝜏)

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝐻 𝑥, 𝜕𝑥 𝑈(𝑥, 𝑥′, 𝜏) = 𝛿(𝑥 − 𝑥′)

Ref: J.SchwingerPhys. Rev. 82, 664Published 1 June 1951

See also: FS method ina book on QFT byC. Itzykson, J.-B. Suber


In the FS method we demand the following:

11

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕

𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏

𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0



12

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:



13

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:



14

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝑖𝜕


𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:



15

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 𝑖𝜕

𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏 = 𝑈 𝑥, 𝑥′, 0 − 𝑈 𝑥, 𝑥′, −∞

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:



16

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝑈 𝑥, 𝑥′, 0 − 𝑈 𝑥, 𝑥′, −∞

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:



17

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝑈 𝑥, 𝑥′, 0 − 𝑈 𝑥, 𝑥′, −∞ = 𝛿(𝑥 − 𝑥′)

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:



18

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′)

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

Let’s check:


Solving this Schroedinger-type equation…

19

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0


Solving this Schroedinger-type equation…

20

𝑈 𝑥, 𝑥′, 𝜏 = e−i𝜏𝐻 𝑥,𝜕𝑥 + 𝜏 𝛿(𝑥 − 𝑥′)

𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕


𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0

…one obtains the following result:


21

Finally, the solution of

is the following expression:

𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 e−i𝜏𝐻+ 𝜏 𝛿(𝑥 − 𝑥′)

𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′)


22

What’s next?

𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0



23

What’s next?

𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0


In the original FS method we actually usethis result to further “massage” theoriginal differential equation:

𝑯 𝒙, 𝝏𝒙 𝑼 𝒙, 𝒙′, 𝝉 = 𝒊𝝏

𝝏𝝉𝑼 𝒙, 𝒙′, 𝝉


24

What’s next?

𝑆 𝑥, 𝑥′ = −𝑖 න−∞

0


In the original FS method we actually usethis result to further “massage” theoriginal differential equation:

𝑯 𝒙, 𝝏𝒙 𝑼 𝒙, 𝒙′, 𝝉 = 𝒊𝝏

𝝏𝝉𝑼 𝒙, 𝒙′, 𝝉

In the modified Fock-Schwinger (MFS)method one directly evaluates the actionof exponential operator on 𝛿-function:

𝒆−𝒊𝝉𝑯+𝜺𝝉 𝜹 𝒙 − 𝒙′ = …

Ref: S. N. IABLOKOV & A. V.KUZNETSOV, 2019 J. Phys.:Conf. Ser. 1390 012078

Ref: S. N. IABLOKOV & A. V.KUZNETSOV, Phys. Rev. D102, 096015 – Published 12November 2020

Modified Fock-Schwinger (MFS) method

Assume the following decomposition of 𝛿-function:

such that:

25


26


27


28


29

Application of MFS

30

Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇− 2𝑖𝑒𝑄𝐹 𝜌

𝜇− 1 − ൗ1 𝜉 Π𝜇Π𝜌 𝐺 𝜈

𝜌𝑥, 𝑥′ = 𝛿4 𝑥 − 𝑥′ 𝛿 𝜈

𝜇

𝐹 𝜌𝜇=

0 0 0 00 0 𝐵 00 −𝐵 0 00 0 0 0 𝜌

𝜇

Π𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)

𝐴𝜇 = (0,0,−𝐵𝑥, 0)

Vector charged boson in a constant magnetic field:

Application of MFS

31



𝜌𝑥, 𝑥′ = 𝛿4 𝑥 − 𝑥′ 𝛿 𝜈

𝜇

A B C𝐹 𝜌𝜇=

0 0 0 00 0 𝐵 00 −𝐵 0 00 0 0 0 𝜌

𝜇


𝐴𝜇 = (0,0,−𝐵𝑥, 0)

A remarkable fact:

𝐴, 𝐵 = 0


Application of MFS

32



𝜌𝑥, 𝑥′ = 𝛿4 𝑥 − 𝑥′ 𝛿 𝜈

𝜇

A B C

A remarkable fact:

𝐹 𝜌𝜇=

0 0 0 00 0 𝐵 00 −𝐵 0 00 0 0 0 𝜌

𝜇


𝐴𝜇 = (0,0,−𝐵𝑥, 0)

𝐴 + 𝐵, 𝐶 = 0

𝐴, 𝐵 = 0


According to the MFS method:

Application of MFS

33

𝐺 𝜈𝜌

𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 e−i𝜏𝐻+ 𝜏𝜈

𝜌𝛿4 𝑥 − 𝑥′

A B C= Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇

= −2𝑖𝑒𝑄𝐹 𝜌𝜇 = − 1 − ൗ1 𝜉 Π𝜇Π𝜌


Application of MFS

34

𝐺 𝜈𝜌

𝑥, 𝑥′ = −𝑖 න−∞

0

𝑑𝜏 e−i𝜏 𝐴+𝐵+𝐶 + 𝜏𝜈




𝐴 + 𝐵, 𝐶 = 0

𝐴, 𝐵 = 0


Application of MFS

35

𝐺 𝜈𝜌

𝑥, 𝑥′ = −𝑖 න−∞

0





𝐴 + 𝐵, 𝐶 = 0

𝐴, 𝐵 = 0

e−i𝜏 𝐴+𝐵+𝐶

Separating the exponent…


Application of MFS

36

𝐺 𝜈𝜌

𝑥, 𝑥′ = −𝑖 න−∞

0





𝐴 + 𝐵, 𝐶 = 0

𝐴, 𝐵 = 0

e−i𝜏 𝐴+𝐵+𝐶 = e−i𝜏 𝐶e−i𝜏 𝐴+𝐵



Application of MFS

37

𝐺 𝜈𝜌

𝑥, 𝑥′ = −𝑖 න−∞

0





𝐴 + 𝐵, 𝐶 = 0

𝐴, 𝐵 = 0

e−i𝜏 𝐴+𝐵+𝐶 = e−i𝜏 𝐶e−i𝜏 𝐴+𝐵 = e−i𝜏 𝐶e−i𝜏 𝐵e−i𝜏 𝐴


Application of MFS

38

𝐺 𝜈𝜇= −𝑖 න

−∞

0

𝑑𝜏 𝛿4 𝑥 − 𝑥′e+i𝜏 1− ൗ1 𝜉 Π𝜇Π𝜌 𝑒−2𝜏𝑒𝑄𝐹 𝜌

𝜇

e−i𝜏 Π𝜆Π𝜆−𝑚2 𝛿 𝜈

𝜇+ 𝜏

Application of MFS

39


−∞

0


𝜇


𝜇+ 𝜏

The expression has a nested structure:

Application of MFS

40


−∞

0


𝜇


𝜇+ 𝜏

Propagation of a scalar particle


Application of MFS

41


−∞

0


𝜇


𝜇+ 𝜏


Accounting for a spin


Application of MFS

42


−∞

0


𝜇


𝜇+ 𝜏


Accounting for a spin

Considering an arbitrary 𝜉-gauge


Application of MFS

43


−∞

0


𝜇


𝜇+ 𝜏

What about 𝛿-function?

Application of MFS

44


−∞

0


𝜇


𝜇+ 𝜏


𝛿4 𝑋 − 𝑋′ =

𝑛=0

∞

න𝑑3𝑝∥,𝑦2𝜋 3

𝑒−𝑖 𝑝 𝑋−𝑋′

∥,𝑦𝑉𝑛 𝑥 𝑉𝑛(𝑥′)

here, 𝑉𝑛 𝑥 are simple harmonic oscillator eigenfunctions


𝐴𝜇 = (0,0,−𝐵𝑥, 0)

Application of MFS

45


−∞

0


𝜇


𝜇+ 𝜏


𝛿4 𝑥 − 𝑥′ =

𝑛=0

∞


𝑒−𝑖 𝑝 𝑥−𝑥′



Π𝜆Π𝜆 −𝑚2 𝑒−𝑖 𝑝 𝑥−𝑥′

∥,𝑦𝑉𝑛 𝑥 = 𝑝∥2 −𝑚2 + 𝑄𝑒𝐵 2𝑛 + 1 𝑒

−𝑖 𝑝 𝑥−𝑥′∥,𝑦𝑉𝑛(𝑥)


𝐴𝜇 = (0,0,−𝐵𝑥, 0)An example of simplification:

Application of MFS

46


−∞

0


𝜇


𝜇+ 𝜏


𝛿4 𝑥 − 𝑥′ =

𝑛=0

∞


𝑒−𝑖 𝑝 𝑥−𝑥′



The rest of the calculations are boring straightforward…

Charged massive vector boson propagator in a constant magnetic field in arbitrary ξ-gaugeobtained using the modified Fock-Schwinger method; S. N. Iablokov and A. V. KuznetsovPhys. Rev. D 102, 096015 – Published 12 November 2020

Application of MFS

47

Charged massive vector boson propagator in a constant magnetic field in arbitrary ξ-gaugeobtained using the modified Fock-Schwinger method; S. N. Iablokov and A. V. KuznetsovPhys. Rev. D 102, 096015 – Published 12 November 2020

Conclusions

MFS allows to obtain the solution of the propagator equation directly (almost) in momentum space.

48

Conclusions


49

MFS simplifies calculations and provides additional representations of the propagator.

Conclusions


50


Drawback: MFS is as good as one’s ability to obtain/guess the form of the wave-equation’s solution.

Conclusions


51


Drawback: MFS is as good as one’s ability to obtain/guess the form of the wave-equation’s solution.

Can be applied for the constant electric field configuration, which is relevant for the Schwinger effect.

Thank you for your attention

52

modified fock-schwinger method - kyoto u

Documents