The $E$, the $A$, the Dirac equation and the propagator

• URL: http://arxiv.org/abs/1801.08393v9
• Date: Tue, 3 Oct 2023 09:26:26 GMT
• Title: The $E$, the $A$, the Dirac equation and the propagator
• Authors: Navin Khaneja
• Abstract summary: Dirac equation in the presence of electromagnetic field is equation of Dirac spinor, $psi$, satisfying $ihbar fracpartial psipartial t. We say that Dirac equation in the presence of electromagnetic field is equation of Dirac spinor, $psi$, satisfying $ihbar fracpartial psipartial t.
• Score: 0.0
• License: http://arxiv.org/licenses/nonexclusive-distrib/1.0/
• Abstract: Consider M\o{}ller scattering. Electrons with momentum $p$ and $-p$ scatter by exchange of photon say in $z$ direction to $p+q$ and $-(p+q)$. The scattering amplitude is well known, given as Feynmann propogator $ M = \frac{(e \hbar c)^2}{\epsilon_0 V} \frac{\bar{u}(p+q) \gamma^{\mu} u(p) \ \bar{u}(-(p+q)) \gamma_{\mu} u(-p)}{q^2}$, where $V$ is the volume of the scattering electrons, $e$ elementary charge and $\epsilon_0$ permitivity of vacuum. But this is not completely correct. Since we exchange photon momentum in $z$ direction, we have two photon polarization $x,y$ and hence the true scattering amplitude should be $$ M_1 = \frac{(e \hbar c)^2}{\epsilon_0 V} \frac{ \bar{u}(p+q) \gamma^{x} u(p) \ \bar{u}(-(p+q)) \gamma_{x} u(-p)\ \ + \bar{u}(p+q) \gamma^{y} u(p) \ \bar{u}(-(p+q)) \gamma_{y} u(-p) \ }{q^2}. $$ But when electrons are nonrelativistic, $M_1 \sim 0$. This is disturbing, how will we ever get the coulomb potential, where $M \sim \frac{(e \hbar c)^2}{\epsilon_0 V q^2}$. Where is the problem ? The problem is with the Dirac equation, it is not all correct in presence of electromagnetic field. We say that Dirac equation in the presence of electromagnetic field is equation of Dirac spinor, $\psi$, satisfying $i\hbar \frac{\partial \psi}{\partial t} = H \psi$, where, $$H = (-i\hbar c \partial_j - e A_j)\alpha_j + e A_0 + mc^2 \beta, $$ where $\alpha_j , \beta$ are Dirac matrices and $A$ vector potential. But there is more to it. The true $$H = (-i\hbar c \partial_j - e A_j)\alpha_j + e A_0 + e E \cdot x + mc^2 \beta, $$ where $E$ is electric field. Its Lorentz invariant form is $$ H = -i\hbar c \partial_j \alpha_j - \frac{e}{2} \int_0^{x^{\mu}}F_{\mu \nu} d x^{\mu} \alpha^{\nu} + mc^2 \beta, $$ where $\alpha^0 = \mathbf{1}$. It is this $E$ term that gives non-vanishing propagator.

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