Deriving the QED Interaction from U(1) Gauge Invariance

Hello! Welcome to the next lesson in our exploration of Quantum Field Theory.

In our previous lessons, we've developed a complete quantum theory for free fermions, culminating in the derivation of the Feynman propagator for the Dirac field. We now have all the tools to describe a free electron or positron propagating through spacetime. However, the universe is not made of free particles; they interact. Our goal now is to introduce interactions in a physically principled way.

Today, we will take a monumental step by deriving the interaction between electrons and photons directly from a fundamental symmetry principle. We will start with the free Dirac Lagrangian and its simple global phase symmetry. We will then demand that this symmetry hold locally—that is, independently at every point in spacetime. You will see that this seemingly simple requirement forces the existence of a new field that mediates an interaction. This new field turns out to be the electromagnetic field, and the interaction is precisely that of Quantum Electrodynamics (QED). This is the principle of local gauge invariance, a cornerstone of the Standard Model of particle physics.

By the end of this lesson, you will understand how to apply this principle to the Dirac Lagrangian to derive the QED interaction term, $L_{int} = - e ψ ˉ γ^{μ} ψ A_{μ}$ .

1. Global U(1) Symmetry of the Dirac Lagrangian

Let's start with the Lagrangian for a free Dirac field that we've been working with:

L_{Dirac} = ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ

This Lagrangian has a continuous global symmetry. If we transform the field $ψ (x)$ by a constant phase factor, the Lagrangian remains unchanged. Let's perform the transformation:

ψ (x) \to ψ^{'} (x) = e^{- i q α} ψ (x)

where $α$ is a real, constant parameter and $q$ is a charge. The Dirac adjoint field transforms as:

ψ ˉ (x) \to ψ ˉ^{'} (x) = (e^{- i q α} ψ)^{†} γ^{0} = ψ^{†} e^{i q α} γ^{0} = e^{i q α} ψ ˉ (x)

Since $α$ is a constant, the derivative $\partial_{μ}$ doesn't act on the phase factor. The Lagrangian becomes:

L^{'} = ψ ˉ^{'} (i γ^{μ} \partial_{μ} - m) ψ^{'} = (e^{i q α} ψ ˉ) (i γ^{μ} \partial_{μ} - m) (e^{- i q α} ψ) = e^{i q α} e^{- i q α} ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ = L

The phase factors cancel, and the Lagrangian is invariant. This transformation is an element of the U(1) group (the group of complex numbers with modulus 1), and since $α$ is the same everywhere in spacetime, this is a global symmetry. As we know from Noether's theorem, this symmetry leads to a conserved current, $j^{μ} = ψ ˉ γ^{μ} ψ$ , which we identify with the electric charge-current density.

2. The Gauge Principle: Demanding Local Symmetry

The "global" nature of this symmetry implies that we must rotate the phase of the field $ψ$ by the same amount everywhere in the universe at the same time. This seems physically unnatural. What if we demand a more powerful and local principle? What if the laws of physics should be invariant under independent phase rotations at every spacetime point?

This is the gauge principle. We demand that the Lagrangian be invariant under the local U(1) gauge transformation:

ψ (x) \to ψ^{'} (x) = e^{- i q α (x)} ψ (x)

Now, the phase rotation angle $α (x)$ is a function of the spacetime coordinate $x$ .

Let's see what happens to our free Dirac Lagrangian under this local transformation. The mass term is still fine:

- m ψ ˉ ψ \to - m (e^{i q α (x)} ψ ˉ) (e^{- i q α (x)} ψ) = - m ψ ˉ ψ

However, the kinetic term now has a problem because the derivative $\partial_{μ}$ acts on $α (x)$ :

\partial_{μ} ψ (x) \to \partial_{μ} (e^{- i q α (x)} ψ (x)) = e^{- i q α (x)} \partial_{μ} ψ (x) - i q (\partial_{μ} α (x)) e^{- i q α (x)} ψ (x)

The kinetic term in the Lagrangian becomes:

ψ ˉ (i γ^{μ} \partial_{μ}) ψ \to (e^{i q α (x)} ψ ˉ) i γ^{μ} [e^{- i q α (x)} \partial_{μ} ψ - i q (\partial_{μ} α (x)) e^{- i q α (x)} ψ]

= ψ ˉ (i γ^{μ} \partial_{μ}) ψ + q (\partial_{μ} α (x)) ψ ˉ γ^{μ} ψ

The Lagrangian is no longer invariant! An extra term, $q (\partial_{μ} α (x)) ψ ˉ γ^{μ} ψ$ , has appeared.

The following video provides a clear walkthrough of this breakdown of invariance.

Deriving The QED Lagrangian | Quantum Electrodynamics

To begin, let's explore why promoting a global symmetry to a local one is a non-trivial step. This video from Dietterich Labs clearly demonstrates how the free Dirac Lagrangian, while invariant under a global U(1) phase rotation, fails to be invariant when the phase becomes dependent on the spacetime position.

Please watch from the beginning until 06:05. The video first reviews the global U(1) invariance of the Dirac Lagrangian and then shows exactly how the derivative term breaks this invariance when the transformation is made local.

3. Restoring Invariance: The Covariant Derivative

The failure of local invariance comes from the derivative term $\partial_{μ} ψ$ . The derivative of the transformed field, $(\partial_{μ} ψ)^{'}$ , picks up an extra piece proportional to $\partial_{μ} α$ . The core idea of gauge theory is to "fix" this by introducing a new field that is designed to absorb this unwanted term.

We replace the ordinary derivative $\partial_{μ}$ with a gauge covariant derivative $D_{μ}$ , defined as:

D_{μ} \equiv \partial_{μ} + i q A_{μ} (x)

Here, $A_{μ} (x)$ is a new vector field, which we call the gauge field. We now have to find how $A_{μ}$ must transform under the gauge transformation such that the new kinetic term $ψ ˉ (i γ^{μ} D_{μ}) ψ$ is invariant.

For the whole term to be invariant, the component $D_{μ} ψ$ must transform just like $ψ$ itself. That is, we require:

(D_{μ} ψ)^{'} \to e^{- i q α (x)} (D_{μ} ψ)

Let's enforce this transformation rule and see what it implies for $A_{μ}$ . The transformed covariant derivative acting on the transformed field is:

(D_{μ} ψ)^{'} = (\partial_{μ} + i q A_{μ}^{'}) (e^{- i q α (x)} ψ) = e^{- i q α (x)} [\partial_{μ} - i q (\partial_{μ} α) + i q A_{μ}^{'}] ψ

We want this to be equal to:

e^{- i q α (x)} (D_{μ} ψ) = e^{- i q α (x)} (\partial_{μ} + i q A_{μ}) ψ

Comparing the two expressions, we see that we must have:

\partial_{μ} - i q (\partial_{μ} α) + i q A_{μ}^{'} = \partial_{μ} + i q A_{μ}

This simplifies to:

A_{μ}^{'} = A_{μ} + \partial_{μ} α (x)

This is the transformation law for the gauge field $A_{μ}$ . It is exactly the well-known gauge transformation for the 4-vector potential in classical electromagnetism! This powerful result suggests that the field we were forced to introduce to preserve local symmetry is none other than the photon field. We will from now on set $q = e$ , the elementary charge.

The following segment of the video shows this derivation.

Deriving The QED Lagrangian | Quantum Electrodynamics

The solution to this problem is to replace the ordinary derivative with a 'gauge covariant derivative', which introduces a new vector field. This video explains how to define this derivative and what transformation law the new field must obey to restore local invariance.

Continue watching from 06:05 to 07:30. Pay close attention to the definition of the covariant derivative and the required transformation property of the new gauge field, Aμ.

4. The QED Lagrangian

We have achieved our goal. By replacing $\partial_{μ}$ with $D_{μ} = \partial_{μ} + i e A_{μ}$ in the Dirac Lagrangian, we construct a new Lagrangian that is invariant under local U(1) gauge transformations:

L = ψ ˉ (i γ^{μ} D_{μ} - m) ψ = ψ ˉ (i γ^{μ} (\partial_{μ} + i e A_{μ}) - m) ψ

Let's expand this out:

L = ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ - e ψ ˉ γ^{μ} ψ A_{μ}

This expression contains two parts:

$L_{Dirac} = ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ$ : The original free Dirac Lagrangian.
$L_{int} = - e (ψ ˉ γ^{μ} ψ) A_{μ} = - e j^{μ} A_{μ}$ : A new interaction term that couples the fermion current $j^{μ} = ψ ˉ γ^{μ} ψ$ to the gauge field $A_{μ}$ . This is the vertex interaction of QED.

To complete the theory, the gauge field $A_{μ}$ must also have its own dynamics. We know from classical electromagnetism that the kinetic term for the electromagnetic field is given by the gauge-invariant Lagrangian:

L_{EM} = - \frac{1}{4} F_{μν} F^{μν} where F_{μν} = \partial_{μ} A_{ν} - \partial_{ν} A_{μ}

Putting everything together, we arrive at the full Lagrangian for Quantum Electrodynamics (QED):

L_{QED} = ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ - \frac{1}{4} F_{μν} F^{μν} - e ψ ˉ γ^{μ} ψ A_{μ}

This can be written more compactly using the covariant derivative:

L_{QED} = ψ ˉ (i \neq D - m) ψ - \frac{1}{4} F_{μν} F^{μν}

This is a remarkable result. The requirement of local gauge invariance has not only dictated the existence of the photon but has also precisely fixed the form of its interaction with the electron. This principle of "gauging a symmetry" is one of the most fundamental concepts in modern physics.

The conclusion of the video you've been watching ties all these pieces together.

Deriving The QED Lagrangian | Quantum Electrodynamics

The final step is to recognize the new gauge field and combine everything into a complete, gauge-invariant Lagrangian. This video concludes the derivation by identifying the gauge field with the electromagnetic potential and assembling the full Lagrangian for Quantum Electrodynamics (QED).

Watch from 07:30 to the end (10:49). This part connects the gauge field to electromagnetism and presents the full QED Lagrangian, identifying the free field parts and the crucial interaction term.

For a more formal and text-based account of this derivation, the following reading is excellent. It concisely covers the same steps we've just discussed.

6. Quantum Electrodynamics

This section from David Tong's lecture notes provides a rigorous, written derivation of the coupling of fermions to electromagnetism via the gauge principle.

Please read Section 6.3.1, "Coupling to Fermions" (pages 136-137). Focus on how the Lagrangian is rewritten with the covariant derivative in equation (6.67) and how the gauge transformations (6.68) ensure its invariance.

Test your understanding!

Consider a complex scalar field $ϕ$ with the Lagrangian $L = (\partial_{μ} ϕ^{*}) (\partial^{μ} ϕ) - m^{2} ϕ^{*} ϕ$ . This theory also has a global U(1) symmetry $ϕ \to e^{- i α} ϕ$ . Following the same logic as for the Dirac field, derive the full gauge-invariant Lagrangian for this scalar field interacting with a U(1) gauge field $A_{μ}$ . This theory is known as "Scalar QED".

Show answer

Promote the symmetry to local: We require invariance under $ϕ (x) \to e^{- i e α (x)} ϕ (x)$ .
Introduce the covariant derivative: The kinetic term breaks the local symmetry. We replace the ordinary derivative $\partial_{μ}$ with the covariant derivative $D_{μ} = \partial_{μ} + i e A_{μ}$ .
Construct the invariant Lagrangian: We replace all derivatives in the original Lagrangian with covariant derivatives.
$L = (D_{μ} ϕ)^{*} (D^{μ} ϕ) - m^{2} ϕ^{*} ϕ$
The term $ϕ^{*} ϕ$ is already locally invariant. Let's check the kinetic term. We need to use the fact that $(D_{μ} ϕ) \to e^{- i e α (x)} (D_{μ} ϕ)$ . Therefore, its complex conjugate transforms as $(D_{μ} ϕ)^{*} \to e^{+ i e α (x)} (D_{μ} ϕ)^{*}$ . The product is then invariant:
$(D_{μ} ϕ)^{*} (D^{μ} ϕ) \to [e^{+ i e α (x)} (D_{μ} ϕ)^{*}] [e^{- i e α (x)} (D^{μ} ϕ)] = (D_{μ} ϕ)^{*} (D^{μ} ϕ)$
Add the gauge field dynamics: To get the full theory, we must include the kinetic term for the gauge field itself, $- \frac{1}{4} F_{μν} F^{μν}$ .
The full Lagrangian for Scalar QED is:
$L_{Scalar QED} = (D_{μ} ϕ)^{*} (D^{μ} ϕ) - m^{2} ϕ^{*} ϕ - \frac{1}{4} F_{μν} F^{μν}$
If you expand the kinetic term, $((\partial_{μ} - i e A_{μ}) ϕ^{*}) ((\partial^{μ} + i e A^{μ}) ϕ)$ , you will find the interaction terms between the scalar field and the photon.

Conclusion

In this lesson, we have seen one of the most elegant and powerful ideas in theoretical physics. By demanding that a simple global phase symmetry of the free Dirac Lagrangian be elevated to a local gauge symmetry, we were forced to introduce an interaction mediated by a vector gauge field, which we identified as the photon. This procedure, known as the gauge principle or minimal coupling, not only predicted the existence of the electromagnetic field but also precisely determined the form of the QED interaction vertex.

Key Takeaways:

The free Dirac Lagrangian has a global U(1) symmetry, leading to the conservation of electric charge.
Promoting this to a local U(1) gauge symmetry (where the phase rotation depends on spacetime position) breaks the invariance of the kinetic term.
Invariance is restored by replacing the ordinary derivative $\partial_{μ}$ with the gauge covariant derivative $D_{μ} = \partial_{μ} + i e A_{μ}$ .
This procedure introduces a new vector field $A_{μ}$ , the gauge field, which must transform as $A_{μ} \to A_{μ} + \partial_{μ} α$ to cancel unwanted terms. This is precisely the gauge transformation of the electromagnetic potential.
The full QED Lagrangian is $L_{QED} = ψ ˉ (i \neq D - m) ψ - \frac{1}{4} F_{μν} F^{μν}$ , which contains the free Dirac Lagrangian, the free Maxwell (photon) Lagrangian, and the crucial interaction term $L_{int} = - e ψ ˉ γ^{μ} ψ A_{μ}$ .

Preview of the Next Lesson:
Now that we have the full QED Lagrangian, we must properly quantize the new field we've introduced: the electromagnetic field $A_{μ}$ . This presents some unique challenges compared to the scalar and Dirac fields due to the gauge redundancy. In the next lesson, we will tackle the quantization of the electromagnetic field in a covariant gauge and derive the photon propagator, another essential building block for our Feynman diagram toolkit.

Recommended resources

1. Global U(1) Symmetry of the Dirac Lagrangian

2. The Gauge Principle: Demanding Local Symmetry

3. Restoring Invariance: The Covariant Derivative

4. The QED Lagrangian

Conclusion