0
$\begingroup$

When deriving the existence of the photon, we start with the free Lagrangian $\mathcal L_{\text{free}} = \bar{\psi}\left( i\gamma^{\mu}\partial_{\mu}-m\right)\psi$ and require $U(1)$ local gauge invariance: $\psi(x)\rightarrow \psi^{'}(x) =e^{iq\alpha\left(x\right)}\psi(x)$, etc.

Question:

What do $q$ and $\alpha\left(x\right)$ mean? In our scriptum, we had the following paragraph: enter image description here

So is it correct to state that $\alpha(x)$ is the (running) coupling constant of QED, while $q$ is the electric charge? I don't read this out to 100%, to be honest.

I also took a look at chapter 10 of our scriptum, but it doesn't specify there what $q$ means..

$\endgroup$
2
  • 1
    $\begingroup$ No, $\alpha(x)$ here has nothing to do with the fine-structure constant, it's merely a (spacetime-dependent) parameter for the local transformation. $\endgroup$ – Nihar Karve Feb 7 at 16:22
  • $\begingroup$ Hi @NiharKarve, thanks! Could you possibly also comment on the last sentence: "However, the interpretation of $q$ is modified by [...]." $\endgroup$ – user248824 Feb 7 at 16:30
0
$\begingroup$

A generic element $U$ of a gauge group $G$ can be written in term of the exponential of the generators $T_a$ of the group:

\begin{equation} U=e^{i\alpha_a T_{a}} \end{equation}

where a sum over $a$ is intended. In gauge theory we promote $\alpha$ to $\alpha(x)$. In our case the group is $U(1)$ and we have only one generator $Q$, which is a number:

\begin{equation} U=e^{i\alpha(x)Q} \end{equation}

So $\alpha(x)$ is not the fine structure constant, it is just an arbitrary real function of $x$. The generator $Q$ is interpreted as the (total) electric charge.

Gauge invariance then becomes equivalent to the conservation of charge. In fact, if $E(\phi(x)_i)$ is some expression containing product of fields, each of those fields, under a $U(1)$ transformation, will transform as

\begin{equation} \phi(x)_i\to e^{i\alpha(x)Q_i}\phi(x), \hspace{5mm} \phi^{*}(x)_i\to e^{-i\alpha(x)Q_i}\phi^{*}(x) \end{equation}

Hence

\begin{equation} E(\phi(x)_i) \to e^{i(Q_1+Q_2+...)}E(\phi(x)_i) \end{equation}

For the expression to be invariant, we have then to require that

\begin{equation} \sum Q_i =0 \end{equation}

which is the statement of conservation of charge.

About the last statement of the page: it turns out that when you evaluate loop diagrams (representing some scattering or annihilation process), the integrals representing them are infinite. You could circumnavigate this problem by absorbing the infinities into counterterms added to the Lagrangian. This has the effect of modifying the couplings. In particular, the value of the electric charge gets some modifications. This is a pure quantum effect, it does not appear in tree level processes (which can be interpreted classically).

$\endgroup$
4
  • $\begingroup$ Hi @Ruben Campos Delgado, thanks a lot! About the last statement: Oooh, now I see! So we now that $\alpha_{\text{QED}} = \frac{e^2}{4\pi}$, and since $\alpha_{\text{QED}}$ isn't constant, the electric charge $Q$ isn't either! :) $\endgroup$ – user248824 Feb 7 at 19:48
  • $\begingroup$ I would just have a brief follow-up question, because I just learned about QCD $SU\left(3\right)$ symmetry. So my question is: In the case of QCD, we write the transformation as $\hat U = \exp\left[ i\mathbf{\alpha}\left( x\right)\cdot g_{S}\mathbf{\hat{T}} \right]$. But in this case, the $\mathbf{\hat{T}}$ are already the generators of the $SU(3)$ group, so why do we need $g_{S}$, see Thomson "Modern Particle Physics" p. 244 e. g.? :) Because in your first paragraph, you write that $\hat{U} = \exp\left[ i\alpha_{a}T_{a}\right]$, where the $T_{a}$ are the generators.. $\endgroup$ – user248824 Feb 7 at 19:54
  • $\begingroup$ The coupling is always a number. In the special case of $U(1)$ the generator is a number too. But for the other groups the generators can be viewed as matrices. So Thomson is simply renaming $\alpha(x)\to \alpha(x)g_s$. The final form is exactly the one that I wrote in my paragraph. $\endgroup$ – Ruben Campos Delgado Feb 8 at 9:18
  • $\begingroup$ Oh, I see, thanks! $\endgroup$ – user248824 Feb 8 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy