Tue Mar 17 2026 14:02:34 GMT+0000 (Coordinated Universal Time)

Construction of the QED Lagrangian

Explains how to construct the Lagrangian of QED.

Reference: (Peskin & Schroeder, 1995)

Let us demand that the Lagrangian be invariant under the local gauge transformation of the Dirac field $\psi(x)$ :

\psi(x) \mapsto e^{i\alpha(x)} \psi(x)

and try to explicitly construct such a Lagrangian.

Motivation

As an interacting theory of the Dirac field $\psi(x)$ and the Maxwell field $A_\mu(x)$ , one can naively consider:

\begin{align*} \mathcal{L}_{\text{QED}} &= \mathcal{L}_{\text{Dirac}} + \mathcal{L}_{\text{Maxwell}} + \mathcal{L}_{\text{interaction}} \\ &= \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi - \frac{1}{4}(F_{\mu\nu})^2 - e\bar{\psi}\gamma^\mu A_\mu \psi \end{align*}

Regardless of this specific theory, theories involving an electromagnetic field have the feature that there is no kinetic term for $A_0$ , and consequently, its conjugate momentum cannot be defined. Related to this feature, states with negative norms appear in the theory. However, because the Lagrangian $\mathcal{L}_{\text{QED}}$ is locally gauge invariant, such physically meaningless states do not contribute to observables. Thus, a natural question arises: are there any locally gauge-invariant theories other than QED? As shown here, the answer is No.

Terms without derivatives can be easily constructed, for example:

m\bar{\psi}(x)\psi(x)

However, the derivative of $\psi(x)$ in the direction of $n_\mu$ is

n^\mu\partial_\mu\psi = \lim_{\epsilon \to 0} \frac{\psi(x+\epsilon n) - \psi(x)}{\epsilon}

so $\psi(x+\epsilon n)$ and $\psi(x)$ transform differently, and $\partial_\mu\psi(x)$ does not transform simply like $\psi(x)$ does.

Thus, we introduce a function $U(y,x)$ that obeys the following transformation rule under local gauge transformations:

U(y,x) \mapsto e^{i\alpha(y)} U(y,x) e^{-i\alpha(x)} \tag{1}

Then, it can be seen that $U(x+\epsilon n,x)\psi(x)$ transforms in the exact same way as $\psi(x)$ under local gauge transformations.

Therefore, we define the covariant derivative as follows:

n^\mu D_\mu \psi(x) \coloneqq \lim_{\epsilon \to 0} \frac{\psi(x+\epsilon n)-U(x+\epsilon n,x)\psi(x)}{\epsilon}

To make its action more explicit, expanding $U(x+\epsilon n,x)$ gives

U(x+\epsilon n,x) = 1 - ie\epsilon n^\mu A_\mu(x) + \mathcal{O}(\epsilon^2) \tag{2}

Here $e$ is a constant introduced for convenience, and $A_\mu(x)$ can be regarded as a new vector field called a gauge field. Substituting this into the definition of the covariant derivative yields

D_\mu \psi(x) = \partial_\mu \psi(x) + ie A_\mu(x)\psi(x)

Furthermore, substituting equation (2) into equation (1) shows that the gauge field $A_\mu(x)$ transforms under local gauge transformations as

A_\mu(x) \mapsto A_\mu(x) - \frac{1}{e}\partial_\mu \alpha(x)

From this transformation rule, we can verify that $D_\mu\psi(x)$ transforms in the same way as $\psi(x)$ under local gauge transformations.

Therefore, to construct a locally gauge-invariant Lagrangian, it is necessary to:

Replace the derivative with the covariant derivative.

To interpret $A_\mu(x)$ as a physically meaningful vector field, a kinetic term for $A_\mu(x)$ is also needed. To construct this term, it is useful to look at the following transformation rule:

[D_\mu,D_\nu]\psi(x) \mapsto e^{i\alpha(x)}[D_\mu,D_\nu]\psi(x)

From this transformation rule, it can be seen that

F_{\mu\nu} \coloneqq [D_\mu,D_\nu] = ie(\partial_\mu A_\nu(x) - \partial_\nu A_\mu(x))

is invariant under local gauge transformations. This quantity is called the field strength. Therefore, to construct a locally gauge-invariant Lagrangian, it is necessary to:

Add the kinetic term $(F_{\mu\nu})^2$ for the gauge field $A_\mu(x)$ .

From the above, the renormalizable, locally gauge-invariant Lagrangian is

\mathcal{L} = \bar{\psi}(i\cancel{D}_\mu - m)\psi - \frac{1}{4}(F_{\mu\nu})^2 + c\epsilon^{\alpha\beta\mu\nu}F_{\alpha\beta}F_{\mu\nu}

The last term breaks parity and time-reversal symmetries, so if these discrete symmetries are required, it becomes

\mathcal{L}_{\text{QED}} = \bar{\psi}(i\cancel{D}_\mu - m)\psi - \frac{1}{4}(F_{\mu\nu})^2

The two free parameters are the mass $m$ and the coupling constant $e$ to the gauge field $A_\mu$ . Such a theory is called Quantum Electrodynamics (QED).

References

Peskin, M. E., & Schroeder, D. V. (1995). An Introduction to quantum field theory. Addison-Wesley.