Transcript
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
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Sum over solutions: main features
Obtain general form
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Orthogonalization & normalization
Sum over polarizations
Find polarizations vectors
Propagator equation: main features
Obtain general form
Orthogonalization & normalization
Sum over polariations
Find polarizations vectors
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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
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𝑖𝜕𝜇 → 𝑖𝐷𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)
Propagator equation
Translational invariance is lostAdd external field
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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’):
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𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′)
Fock-Schwinger (FS) method
Let’s solve this equation for S(x,x’): Choose a parametrization:
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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
Fock-Schwinger (FS) method
Let’s solve this equation for S(x,x’): Choose a parametrization:
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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
Fock-Schwinger (FS) method
In the FS method we demand the following:
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𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Fock-Schwinger (FS) method
In the FS method we demand the following:
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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𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
In the FS method we demand the following:
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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𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
In the FS method we demand the following:
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𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
In the FS method we demand the following:
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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𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
In the FS method we demand the following:
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𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝑈 𝑥, 𝑥′, 0 − 𝑈 𝑥, 𝑥′, −∞
𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
In the FS method we demand the following:
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𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝑈 𝑥, 𝑥′, 0 − 𝑈 𝑥, 𝑥′, −∞ = 𝛿(𝑥 − 𝑥′)
𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
In the FS method we demand the following:
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𝐻 𝑥, 𝜕𝑥 𝑆 𝑥, 𝑥′ = 𝛿(𝑥 − 𝑥′)
𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Let’s check:
Fock-Schwinger (FS) method
Solving this Schroedinger-type equation…
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𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
Fock-Schwinger (FS) method
Solving this Schroedinger-type equation…
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𝑈 𝑥, 𝑥′, 𝜏 = e−i𝜏𝐻 𝑥,𝜕𝑥 + 𝜏 𝛿(𝑥 − 𝑥′)
𝐻 𝑥, 𝜕𝑥 𝑈 𝑥, 𝑥′, 𝜏 = 𝑖𝜕
𝜕𝜏𝑈 𝑥, 𝑥′, 𝜏
𝑈 𝑥, 𝑥′, 0 = 𝛿(𝑥 − 𝑥′) 𝑈 𝑥, 𝑥′, −∞ = 0
…one obtains the following result:
Fock-Schwinger (FS) method
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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′)
Fock-Schwinger (FS) method
22
What’s next?
𝑆 𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏𝐻+ 𝜏 𝛿(𝑥 − 𝑥′)
Fock-Schwinger (FS) method
23
What’s next?
𝑆 𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏𝐻+ 𝜏 𝛿(𝑥 − 𝑥′)
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
Fock-Schwinger (FS) method
24
What’s next?
𝑆 𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏𝐻+ 𝜏 𝛿(𝑥 − 𝑥′)
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:
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Modified Fock-Schwinger (MFS) method
26
Modified Fock-Schwinger (MFS) method
27
Modified Fock-Schwinger (MFS) method
28
Modified Fock-Schwinger (MFS) method
29
Application of MFS
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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
Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇− 2𝑖𝑒𝑄𝐹 𝜌
𝜇− 1 − ൗ1 𝜉 Π𝜇Π𝜌 𝐺 𝜈
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
Π𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)
𝐴𝜇 = (0,0,−𝐵𝑥, 0)
A remarkable fact:
𝐴, 𝐵 = 0
Vector charged boson in a constant magnetic field:
Application of MFS
32
Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇− 2𝑖𝑒𝑄𝐹 𝜌
𝜇− 1 − ൗ1 𝜉 Π𝜇Π𝜌 𝐺 𝜈
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
Π𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)
𝐴𝜇 = (0,0,−𝐵𝑥, 0)
𝐴 + 𝐵, 𝐶 = 0
𝐴, 𝐵 = 0
Vector charged boson in a constant magnetic field:
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
According to the MFS method:
Application of MFS
34
𝐺 𝜈𝜌
𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏 𝐴+𝐵+𝐶 + 𝜏𝜈
𝜌𝛿4 𝑥 − 𝑥′
A B C= Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇
= −2𝑖𝑒𝑄𝐹 𝜌𝜇 = − 1 − ൗ1 𝜉 Π𝜇Π𝜌
𝐴 + 𝐵, 𝐶 = 0
𝐴, 𝐵 = 0
According to the MFS method:
Application of MFS
35
𝐺 𝜈𝜌
𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏 𝐴+𝐵+𝐶 + 𝜏𝜈
𝜌𝛿4 𝑥 − 𝑥′
A B C= Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇
= −2𝑖𝑒𝑄𝐹 𝜌𝜇 = − 1 − ൗ1 𝜉 Π𝜇Π𝜌
𝐴 + 𝐵, 𝐶 = 0
𝐴, 𝐵 = 0
e−i𝜏 𝐴+𝐵+𝐶
Separating the exponent…
According to the MFS method:
Application of MFS
36
𝐺 𝜈𝜌
𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏 𝐴+𝐵+𝐶 + 𝜏𝜈
𝜌𝛿4 𝑥 − 𝑥′
A B C= Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇
= −2𝑖𝑒𝑄𝐹 𝜌𝜇 = − 1 − ൗ1 𝜉 Π𝜇Π𝜌
𝐴 + 𝐵, 𝐶 = 0
𝐴, 𝐵 = 0
e−i𝜏 𝐴+𝐵+𝐶 = e−i𝜏 𝐶e−i𝜏 𝐴+𝐵
Separating the exponent…
According to the MFS method:
Application of MFS
37
𝐺 𝜈𝜌
𝑥, 𝑥′ = −𝑖 න−∞
0
𝑑𝜏 e−i𝜏 𝐴+𝐵+𝐶 + 𝜏𝜈
𝜌𝛿4 𝑥 − 𝑥′
A B C= Π𝜆Π𝜆 −𝑚2 𝛿 𝜌𝜇
= −2𝑖𝑒𝑄𝐹 𝜌𝜇 = − 1 − ൗ1 𝜉 Π𝜇Π𝜌
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
Separating the exponent…
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
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
The expression has a nested structure:
Application of MFS
40
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
Propagation of a scalar particle
The expression has a nested structure:
Application of MFS
41
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
Propagation of a scalar particle
Accounting for a spin
The expression has a nested structure:
Application of MFS
42
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
Propagation of a scalar particle
Accounting for a spin
Considering an arbitrary 𝜉-gauge
The expression has a nested structure:
Application of MFS
43
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
What about 𝛿-function?
Application of MFS
44
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
What about 𝛿-function?
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
Π𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)
𝐴𝜇 = (0,0,−𝐵𝑥, 0)
Application of MFS
45
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
What about 𝛿-function?
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
Π𝜆Π𝜆 −𝑚2 𝑒−𝑖 𝑝 𝑥−𝑥′
∥,𝑦𝑉𝑛 𝑥 = 𝑝∥2 −𝑚2 + 𝑄𝑒𝐵 2𝑛 + 1 𝑒
−𝑖 𝑝 𝑥−𝑥′∥,𝑦𝑉𝑛(𝑥)
Π𝜇 = 𝑖𝜕𝜇 + 𝑒𝑄𝐴𝜇(𝑥)
𝐴𝜇 = (0,0,−𝐵𝑥, 0)An example of simplification:
Application of MFS
46
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
What about 𝛿-function?
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
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
MFS allows to obtain the solution of the propagator equation directly (almost) in momentum space.
49
MFS simplifies calculations and provides additional representations of the propagator.
Conclusions
MFS allows to obtain the solution of the propagator equation directly (almost) in momentum space.
50
MFS simplifies calculations and provides additional representations of the propagator.
Drawback: MFS is as good as one’s ability to obtain/guess the form of the wave-equation’s solution.
Conclusions
MFS allows to obtain the solution of the propagator equation directly (almost) in momentum space.
51
MFS simplifies calculations and provides additional representations of the propagator.
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
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