I have one question about local gauge invariance of the spinor and scalar theories.

For the scalar complex field with lagrangian $L_{0}$ requirement of local gauge invariance leads us to the Lagrangian $$ L = L_{0} - J_{\mu}A^{\mu} + q^{2}\varphi \varphi^{*} A_{\mu}A^{\mu} - \frac{1}{4 }F_{\mu \nu}F^{\mu \nu} = L_{0} + L_{el.} + q^{2}\varphi \varphi^{*} A_{\mu}A^{\mu}, \qquad (.1) $$ where $J_{\mu}$ is the conserved quantity of the theory giving by $(.1)$,

$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ is tensor of electromagnetic field,

$A_{\mu}$ is the local gauge invariance electromagnetic field, $$ A_{\mu} \to A_{\mu} - iq\partial_{\mu}f \Rightarrow L_{el.} = inv. $$ So, my question: (after quantization) how the summand $J_{\mu}A^{\mu}$ describes interaction between two charges $Q, -Q$ of the scalar field?

Maybe, the spinor case is analogical, because the lagrangian of local gauge invarince spinor field is very similar to $(.1)$, except the last summand.


1 Answer 1


Reference : Zee, Quantum Theory in a nutshell, Champter I.4, first edition

You have to get a electromagnetic propagator $D_{\mu\nu}(x)$, to the desired precision level.

At first order, we do not consider the modifications to the electromagnetic propagator due to the term $q^{2}\varphi \varphi^{*} A_{\mu}A^{\mu}$

We have :

$$Z(J) = \langle 0_J|e^{-iH\int dx^o} |0_J \rangle = \int [DA]e^{i\int d^4x J.A +A \square A} \tag{1}$$

After integration of the RHS, We have :

$-H_{int}\int dx^o = -\frac{1}{2}\int d^4x d^4y J^\mu(x) D_{\mu\nu}(x-y)J^\nu(y) \tag{2}$

Take $J^0(x)= Q_1\delta(\vec x- \vec x_1)+Q_2\delta(\vec x- \vec x_2)$, as unique non-zero value coordinate for the current, and conserve only cross-terms $Q_1Q_2$. You then express the propagator $D_{\mu\nu}(x-y)$ as a Fourier transform of propagator in momentum space $D_{\mu\nu}(k)$, then integrate on $y^o$, this gives a delta function with constraint $k^o$ to be zero. This leaves a term $\int dx^o$ on the RHS which simplify with the LHS term. At the end, up to a sign and a constant factor, the interaction energy $H_{int}(\vec x)$ is the spatial Fourier transform of the electromagnetic propagator in momentum space with $k^0=0$. With $D(k)\sim \frac{1}{\vec k^2}$, you get an interaction energy in $\frac{Q_1Q_2}{r}$

  • 1
    $\begingroup$ I made an error (now corrected in the answer) : the interaction energy goes like $\frac{Q_1Q_2}{r}$, there is no minus sign, that is : 2 identical charges are repelling, so the interaction energy is positive. $\endgroup$
    – Trimok
    Sep 3, 2013 at 15:02

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