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Geometrical significance of gauge invariance of the QED Lagrangian

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The QED Lagrangian is invariant under $\psi(x) \to e^{i\alpha(x)} \psi (x)$, $A_{\mu} \to A_{\mu}- \frac{1}{e}\partial_{\mu}\alpha(x)$. What is the geometric significance of this result? Also why is $D_{\mu}=\partial_{\mu}+ieA_{\mu}(x)$ called the covariant derivative?

This post imported from StackExchange Physics at 2014-03-22 17:12 (UCT), posted by SE-user ramanujan_dirac
asked Jan 13, 2013 in Theoretical Physics by ramanujan_dirac (235 points) [ no revision ]

2 Answers

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The scalar QED Lagrangian in your question is for a complex scalar field $\psi(x)$ interacting with an electromagnetic field given by potential $A_{\mu}(x)$. At any point $x$, the scalar field is a complex number. We model this situation by constructing a space - a vector bundle $V$ - which is isomorphic to $M \ X \ \mathbb{C}$. In general a bundle is only locally isomorphic to the product space, since it might have twists in it, but we'll ignore this here. The bundle has a projection map onto the spacetime manifold $\pi: V \rightarrow M$. The set of points which are projected onto $x \in M$ is called the fibre over $x$, and denoted $F_x$. Each $F_x$ is isomorphic to the complex numbers $\mathbb{C}$.

Now to make explicit such an isomorphism, we effectively choose a coordinate $z$ for each fibre. So our bundle $V$ then has coordinates $\{x^{\mu}, z \}$. Our spacetime field $\psi(x)$ as a map from $M \rightarrow V$ is called a section of $V$. If we compose a section with the projection $\pi$ we get back the spacetime point we started with. We can think of the choice of fibre coordinates as a gauge choice. A gauge transformation is the choice of a new fibre coordinate, related to the old by $z \rightarrow g.z$ where $g \in U(1)$. In the case of a local gauge transformation, this new choice of coordinate becomes $z \rightarrow g(x).z$ where $g$ is now a function of $x$.

Now, given the interpretation of $\psi(x)$ as a section of $V$, in order to construct the Lagrangian, we need to be able to differentiate it i.e. we need to be able to compute a derivative which is a limit $$ \lim_{\Delta x \to 0}\frac{\psi(x+\Delta x)-\psi(x)}{\Delta x} $$ The problem is: $\psi(x)$ lives in the fibre $F_x$ over $x$, and $\psi(x+dx)$ lives in the fibre $F_{x+dx}$ over $x+dx$. These are different spaces, so we can't perform the subtraction unless we can map points in $F_{x+dx}$ to points in $F_x$. If we've chosen a gauge, this is no problem - we have an explicit mapping of both fibres to the complex numbers, so we can perform the subtraction, but we want something that makes sense when we make changes of gauge, in particular local changes of gauge.

The recipe to do this is to introduce a connection. If we start at a point $p$ in the fibre $F_x$ and infinitesimally perturb the point $x$ to $x+dx$, to specify where $p$ moves to, we need to give it in general a horizontal component (in the M coordinate direction), and a vertical component (in the fibre directions). Given a gauge, describing an infinitesimal fibrewise displacement is easy - we just apply an infinitesimal element of the gauge group. Such an infinitesimal element belongs to the Lie algebra of the group. For $U(1)$, this Lie algebra is just the real numbers, so the vertical displacements corresponding to movement of $p$ in the 4 spacetime coordinate directions are just given by four real numbers. As a function of spacetime coordinates, they become four functions $A_{\mu}(x)$. The gauge covariant derivative is then just $$ D_{\mu}\psi(x) = \partial_{\mu}\psi(x) + A_{\mu}(x)\psi(x)$$

If we perform a local gauge transformation $$\psi(x) \rightarrow \psi'(x) = g(x)\psi(x)$$ then, provided we make a corresponding transformation $$ A_{\mu}(x) \rightarrow A'_{\mu}(x) = A_{\mu}(x) + g(x)\partial_{\mu}g^{-1}(x) \ \ (1) $$ the gauge covariant derivative transforms like $$ D_{\mu}\psi'(x) = D_{\mu}(g(x)\psi(x)$$ $$ = \partial_{\mu}(g(x)\psi(x)) + A'_{\mu}(x)g(x)\psi(x)$$ $$ = \partial_{\mu}(g(x)\psi(x)) + [A_{\mu}(x)+g(x)\partial_{\mu}g^{-1}(x)]g(x)\psi(x)$$ $$ = g(x)\partial_{\mu}\psi(x) + (\partial_{\mu}g)\psi(x) + A_{\mu}(x)g(x)\psi(x) + g(x)(\partial_{\mu}g^{-1}(x))g(x)\psi(x)$$ $$ = g(x)(\partial_{\mu}\psi(x) + A_{\mu}(x)\psi(x) = g(x)D_{\mu}\psi(x)$$

Where for the last step we used $$ 0 = \partial_{\mu}1 = \partial_{\mu}(g(x)g^{-1}(x)) = (\partial_{\mu}g(x))g^{-1}(x)+g(x)(\partial_{\mu}g^{-1}(x)) $$ So $D_{\mu}\psi(x)$ transforms covariantly, in a way that ensures the Lagrangian is gauge invariant. If we write $g(x) = e^{i\alpha(x)}$ then the transformation law (1) becomes $$ A_{\mu}(x) \rightarrow A'_{\mu}(x) = A_{\mu}(x) - i\partial_{\mu}\alpha(x) $$

This post imported from StackExchange Physics at 2014-03-22 17:12 (UCT), posted by SE-user twistor59
answered Jan 13, 2013 by twistor59 (2,490 points) [ no revision ]
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You can interpret gauge invariance in terms of fiber bundles. One can think of a fiber as the space of configurations of the gauge field, connected by gauge transformations.

The covariant derivative is called "covariant" because it transforms covariantly, i.e. $\begin{equation}D'_\mu=U^{-1}(x)D_\mu U(x)\end{equation}.$

This post imported from StackExchange Physics at 2014-03-22 17:12 (UCT), posted by SE-user Frederic Brünner
answered Jan 13, 2013 by Frederic Brünner (920 points) [ no revision ]

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