Gupta-Bleuler formalism

The Gupta-Bleuler formalism (after Suraj N. Gupta and Konrad Bleuler ) is a method for quantization of gauge theories , with which the Lorenz gauge of the classical electrodynamics in the quantum electrodynamics can be transferred. Such a calibration fixation is necessary in order to avoid the occurrence of unphysical degrees of freedom such as temporally and longitudinally polarized photons in the context of quantum electrodynamics.

In the Gupta-Bleuler formalism, the Lorenz calibration of the four-potential of classical electrodynamics ,, is replaced by the two weaker conditions and . This means that the derivative itself is not zero, but only its expected value for each state . ${\displaystyle \partial _{\mu }A^{\mu }=0}$ ${\displaystyle \partial _{\mu }A_{(+)}^{\mu }|\Psi \rangle =0}$ ${\displaystyle \langle \Psi |\partial _{\mu }A_{(-)}^{\mu }=0}$ ${\displaystyle \partial _{\mu }A^{\mu }}$ ${\displaystyle \langle \Psi |\partial _{\mu }A^{\mu }|\Psi \rangle }$ ${\displaystyle |\Psi \rangle }$

background

See also : quantum electrodynamics  and quantum field theory

The naive Lagrangian of the photon read with the field strength tensor : ${\displaystyle F_{\mu \nu }=\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }}$

${\displaystyle {\mathcal {L}}=-{\frac {1}{4}}F_{\mu \nu }F^{\mu \nu }}$

and without calibration fixation leads to various problems, since the four-potential with four degrees of freedom should adequately describe a physical object with only two degrees of freedom, the photon. The occurrence of this problem becomes clear by forming the Euler-Lagrange equation and the conjugate momentum. The following applies to the Euler-Lagrange equation

${\displaystyle \Box A^{\nu }+\partial ^{\nu }\partial ^{\mu }A_{\mu }=0}$,

which is the classical result, and for the conjugate momentum

${\displaystyle \pi ^{\mu }={\frac {\partial {\mathcal {L}}}{\partial (\partial _{0}A_{\mu })}}=F^{\mu 0}}$.

Since the field strength tensor is antisymmetric by design, the time-like component of the conjugate momentum disappears . Hence the commutator relation ${\displaystyle \pi ^{0}=0}$

${\displaystyle \left[A_{\mu }({\vec {x}},t),\pi _{\nu }({\vec {y}},t)\right]=\mathrm {i} g_{\mu \nu }\delta ^{(3)}({\vec {x}}-{\vec {y}})}$

in the event not be valid. ${\displaystyle \mu =\nu =0}$

Calibration fixation of the Lagrangian

To remedy this, it can be demanded that the Euler-Lagrange equation of the photon takes the form of a wave equation , as is the case in classical electrodynamics using the Lorenz calibration . This is done in the Gupta-Bleuler formalism by introducing an additional term in the Lagrangian:

${\displaystyle {\mathcal {L}}=-{\frac {1}{4}}F_{\mu \nu }F^{\mu \nu }-{\frac {1}{2\xi }}(\partial _{\mu }A^{\mu })^{2}}$

Feynman calibration is used in this article . This simplifies the Euler-Lagrange equation for the four-potential operator to ${\displaystyle \xi =1}$

${\displaystyle \Box A^{\nu }=0}$,

whereas the conjugate momentum contains an additional term:

${\displaystyle \pi ^{\mu }=F^{\mu 0}-\partial _{\nu }A^{\nu }\delta ^{\mu 0}}$.

Not in contradiction to the formulas of classical electrodynamics and thus not violating the principle of correspondence , the choice of is completely free until then . It would even be wrong to claim that the comparison of the Euler-Lagrange equations before and after calibration fixation inevitably means that only a zero would be added to the calibration fixation! ${\displaystyle \partial _{\mu }A^{\mu }}$ ${\displaystyle \partial _{\mu }A^{\mu }=0}$

Temporally and longitudinally polarized photons

It is known from classical electrodynamics that electromagnetic waves in a vacuum are transverse waves and photons only have two degrees of freedom, which manifest themselves in the two transverse directions of polarization. If, however, the four potential is represented in Fourier decomposition, the result is with the four degrees of freedom and four linearly independently chosen basis vectors ${\displaystyle \lambda }$ ${\displaystyle \epsilon _{\mu }(k,\lambda )}$

${\displaystyle A_{\mu }=\int {\frac {\mathrm {d} ^{3}{\vec {k}}}{(2\pi )^{3}}}{\frac {1}{\sqrt {2E_{k}}}}\sum _{\lambda =0}^{3}\left(\epsilon _{\mu }(k,\lambda )\alpha (k,\lambda )e^{-\mathrm {i} kx}+\epsilon _{\mu }^{*}(k,\lambda )\alpha ^{\dagger }(k,\lambda )e^{\mathrm {i} kx}\right)}$.

The creation and annihilation operators introduced here or fulfill the commutator relation ${\displaystyle \alpha ^{\dagger }}$ ${\displaystyle \alpha }$

${\displaystyle \left[\alpha (k,\lambda ),\alpha ^{\dagger }(k',\lambda ')\right]=-g_{\lambda \lambda '}(2\pi )^{3}\delta ^{(3)}({\vec {k}}-{\vec {k}}')}$,

as can be shown by explicitly setting up the conjugate momentum in Fourier representation.

By selecting a coordinate system can be in direction set so that in the following time-like, transverse and describe longitudinally polarized photons. The Hamilton operator for the photon is then ${\displaystyle {\vec {k}}}$ ${\displaystyle z}$ ${\displaystyle \lambda =0}$ ${\displaystyle \lambda =1,2}$ ${\displaystyle \lambda =3}$ ${\displaystyle H}$

${\displaystyle H=\int {\frac {\mathrm {d} ^{3}{\vec {k}}}{(2\pi )^{3}}}E_{k}\left(\sum _{\lambda =1}^{3}\alpha ^{\dagger }(k,\lambda )\alpha (k,\lambda )-\alpha ^{\dagger }(k,0)\alpha (k,0)\right)}$,

whereby the result implies that photons polarized in time lead to negative energies.

Application of the formalism

If the derivative of the four potential is derived, then applies to the choice ${\displaystyle k=(E_{k},0,0,k)}$

${\displaystyle \partial _{\mu }A^{\mu }=\int {\frac {\mathrm {d} ^{3}{\vec {k}}}{(2\pi )^{3}}}\mathrm {i} {\sqrt {\frac {E_{k}}{2}}}\left((\alpha (k,3)-\alpha (k,0))e^{-\mathrm {i} kx}-(\alpha ^{\dagger }(k,3)-\alpha ^{\dagger }(k,0))e^{\mathrm {i} kx}\right)}$

In this context one defines

${\displaystyle \partial _{\mu }A_{(+)}^{\mu }=\int {\frac {\mathrm {d} ^{3}{\vec {k}}}{(2\pi )^{3}}}\mathrm {i} {\sqrt {\frac {E_{k}}{2}}}\left(\alpha (k,3)-\alpha (k,0)\right)e^{-\mathrm {i} kx}}$,

${\displaystyle \partial _{\mu }A_{(-)}^{\mu }=\int {\frac {\mathrm {d} ^{3}{\vec {k}}}{(2\pi )^{3}}}\mathrm {i} {\sqrt {\frac {E_{k}}{2}}}\left(\alpha ^{\dagger }(k,3)-\alpha ^{\dagger }(k,0)\right)e^{\mathrm {i} kx}}$.

It follows from the condition of the Gupta-Bleuler formalism mentioned at the beginning that already in the integrand

${\displaystyle \langle \Psi |\alpha ^{\dagger }(k,3)=\langle \Psi |\alpha ^{\dagger }(k,0)}$,

${\displaystyle \alpha (k,3)|\Psi \rangle =\alpha (k,0)|\Psi \rangle }$

must apply. From this relationship, the expected value of the Hamilton operator for the energy is the non-negative quantity

${\displaystyle E=\langle \Psi |H|\Psi \rangle =\langle \Psi |\int {\frac {\mathrm {d} ^{3}{\vec {k}}}{(2\pi )^{3}}}E_{k}\sum _{\lambda =1}^{2}\alpha ^{\dagger }(k,\lambda )\alpha (k,\lambda )|\Psi \rangle }$,

since the contributions of the longitudinal and temporal polarization cancel each other out.

Individual evidence

1. ↑ Suraj N. Gupta: Theory of Longitudinal Photons in Quantum Electrodynamics . In: Proceedings of the Physical Society. Section A . tape 63 , no. 7 , 1950, pp. 681 , doi : 10.1088 / 0370-1298 / 63/7/301 .
2. ↑ Konrad Bleuler: A new method for the treatment of longitudinal and scalar photons . In: Helvetica Physica Acta . tape 23 , no. 5 , 1950, pp. 567 ff ., doi : 10.5169 / seals-112124 ( digital copy ).