Quantizing the Free Dirac Field

Hello! Welcome back to our study of fermionic fields.

In our last lesson, we derived the Dirac equation, $(i γ^{μ} \partial_{μ} - m) ψ = 0$ , from first principles. We found that its solutions, the spinors $ψ (x)$ , must be four-component objects that describe both particles ( $u$ -spinors) and antiparticles ( $v$ -spinors), and intrinsically possess spin-1/2. This was a remarkable success, giving us the correct classical field equation for electrons and other fermions.

Today, we take the crucial next step from a classical field theory to a quantum one. Our learning outcome is to formulate the Lagrangian for the free Dirac field and quantize it by imposing equal-time anti-commutation relations.

This process will mirror the canonical quantization we performed for the scalar field, but with a critical twist. As we'll see, a naive application of commutation relations leads to a nonsensical theory. The resolution lies in embracing the fermionic nature of these particles, which requires a new kind of quantization rule. This will not only fix our theory but also provide a deep, fundamental explanation for the Pauli exclusion principle.

1. The Dirac Lagrangian and Hamiltonian Formalism

Let's begin by recalling the Lagrangian for the free Dirac field, which you derived in the previous lesson. Its Lorentz-invariant form is the starting point for the entire canonical procedure.

Dirac Matrices and Lorentz Spinors

To formalize the starting point, let's briefly review the Dirac Lagrangian and the derivation of its Hamiltonian. The document "Dirac Matrices and Lorentz Spinors" presents this concisely.

Please read the following two sections: "Dirac Lagrangian" (pages 9-11): Quickly review this section to refresh your memory on why the Lagrangian takes the form \mathcal{L} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi and how the Dirac conjugate \bar{\psi} = \psi^\dagger\gamma^0 ensures Lorentz invariance. "Hamiltonian for the Dirac Field" (pages 12-13): Read this section carefully. It derives the canonical momentum and the classical Hamiltonian. Pay close attention to the result for the canonical momentum \Pi_\alpha, as it differs structurally from the scalar field case.

As you've just read, the canonical procedure unfolds as follows:

Lagrangian: We start with $L = ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ$ . Note that this is first-order in the time derivative $ψ ˙ \equiv \partial_{0} ψ$ .
Canonical Momentum: The momentum conjugate to the field $ψ_{α}$ is:
$Π_{α} = \frac{\partial L}{\partial ψ ˙ _{α}} = \frac{\partial}{\partial ψ ˙ _{α}} (ψ^{†} γ^{0} (i γ^{0} \partial_{0} + i γ \cdot \nabla - m) ψ) = i (ψ^{†} γ^{0} γ^{0})_{α} = i ψ_{α}^{†}$
This is a key result: the momentum conjugate to $ψ$ is, up to a factor of $i$ , its Hermitian conjugate $ψ^{†}$ . Unlike the scalar field case, $Π$ does not involve any time derivatives.
Hamiltonian: The Hamiltonian density is found via the Legendre transform, $H = Π ψ ˙ - L$ .
$H = (i ψ^{†}) ψ ˙ - ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ = i ψ^{†} ψ ˙ - ψ^{†} γ^{0} (i γ^{0} ψ ˙ + i γ \cdot \nabla - m) ψ$
The terms involving $ψ ˙$ cancel out, leaving:
$H = ψ^{†} (- i α \cdot \nabla + β m) ψ = ψ ˉ (- i γ \cdot \nabla + m) ψ$
where we've used $α = γ^{0} γ$ and $β = γ^{0}$ . The total Hamiltonian is $H = \int d^{3} x H$ .

2. A Failed Attempt: Quantization with Commutators

Now we are ready to quantize. The standard recipe we learned for the scalar field is to promote the fields to operators and impose equal-time commutation relations (ETCRs). Let's see what happens when we try this for the Dirac field.

5. Quantizing the Dirac Field

This next reading from David Tong's QFT notes is crucial. It deliberately follows the 'wrong' path of using commutation relations to quantize the Dirac field, leading to a catastrophic failure. Understanding why this fails is the best motivation for the correct approach.

Please read Section 5.1 'A Glimpse at the Spin-Statistics Theorem' and 5.1.1 'The Hamiltonian' (pages 106-108). Focus on: The initial postulate of canonical commutation relations in Eq. (5.3). The resulting commutation relations for the creation/annihilation operators, especially the minus sign in Eq. (5.5). The final form of the Hamiltonian in Eq. (5.12) and the explanation of why it is 'a disaster'.

Let's unpack the disaster you just read about.
The attempt was to impose:

[ψ_{α} (x, t), Π_{β} (y, t)] = i δ_{α β} δ^{(3)} (x - y) ⟹ [ψ_{α} (x, t), ψ_{β}^{†} (y, t)] = δ_{α β} δ^{(3)} (x - y)

When the field operator $ψ (x)$ is expanded in terms of creation and annihilation operators for particles ( $b, b^{†}$ ) and antiparticles ( $c, c^{†}$ ), these field commutation relations imply the following for the operators:

[b_{p}^{r}, b_{q}^{s †}] = (2 π)^{3} δ^{r s} δ^{(3)} (p - q) [c_{p}^{r}, c_{q}^{s †}] = - (2 π)^{3} δ^{r s} δ^{(3)} (p - q)

The commutator for the $c$ operators has a devastating minus sign. When we use these relations to compute the Hamiltonian operator, we find:

H = \int \frac{d ^{3} p}{( 2 π ) ^{3}} E_{p} s = 1, 2 \sum (b_{p}^{s †} b_{p}^{s} - c_{p}^{s †} c_{p}^{s})

(after dropping an infinite constant). The term $b^{†} b$ is fine; it's a number operator that counts particles, each contributing positive energy $E_{p}$ . However, the $c^{†} c$ term means that creating antiparticles lowers the total energy of the system. A state with $N_{c}$ antiparticles would have energy $- N_{c} E_{p}$ . This means the vacuum is unstable: the system could cascade downwards, creating an infinite number of particle-antiparticle pairs to reach an energy of $- \infty$ .

This is not a mathematical mistake; it's a profound statement from the theory. It tells us that quantizing a spin-1/2 field with commutation relations is physically inconsistent.

3. Success: Fermionic Quantization with Anti-commutators

The resolution lies in a cornerstone of modern physics: the spin-statistics theorem. This theorem states that integer-spin particles (like photons or Higgs bosons) must be quantized as bosons (using commutators), while half-integer spin particles (like electrons or quarks) must be quantized as fermions (using anti-commutators). Our failure was in trying to treat a fermion like a boson.

Let's now follow the correct procedure. We replace the ETCRs with equal-time anti-commutation relations (ETACRs).

5. Quantizing the Dirac Field

Let's return to Tong's notes to see how anti-commutators save the day. This section shows that simply swapping curly braces for square ones fixes everything.

Read Section 5.2 'Fermionic Quantization' and 5.2.1 'Fermi-Dirac Statistics' (pages 109-110). Pay attention to: The new quantization rules in Eq. (5.14). How these change the relations for creation/annihilation operators in Eq. (5.15) (the minus sign is gone!). The corrected Hamiltonian in Eq. (5.17). The connection to the Pauli exclusion principle in Eq. (5.21).

This is a beautiful result. Let's summarize the key points of this correct quantization scheme.

1. The Fundamental Anti-Commutation Relations

We postulate that the Dirac field operators obey:

{ψ_{α} (x, t), ψ_{β}^{†} (y, t)} = δ_{α β} δ^{(3)} (x - y)

and

{ψ_{α} (x, t), ψ_{β} (y, t)} = {ψ_{α}^{†} (x, t), ψ_{β}^{†} (y, t)} = 0

where the anti-commutator is defined as ${A, B} = A B + B A$ .

2. Creation and Annihilation Operators

When we expand the field operator $ψ (x)$ as before,

ψ (x) = \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} s \sum (b_{p}^{s} u^{s} (p) e^{- i p \cdot x} + c_{p}^{s †} v^{s} (p) e^{i p \cdot x})

the field ETACRs imply the following anti-commutation relations for the mode operators:

{b_{p}^{r}, b_{q}^{s †}} = (2 π)^{3} δ^{r s} δ^{(3)} (p - q) {c_{p}^{r}, c_{q}^{s †}} = (2 π)^{3} δ^{r s} δ^{(3)} (p - q)

with all other anti-commutators ( ${b, b}$ , ${c, c}$ , ${b, c}$ , etc.) being zero. Notice that the problematic minus sign is now gone.

3. The Hamiltonian and a Stable Vacuum

Let's re-calculate the Hamiltonian using these new rules. The expression before re-ordering the operators was:

H = \int \frac{d ^{3} p}{( 2 π ) ^{3}} E_{p} s \sum (b_{p}^{s †} b_{p}^{s} - c_{p}^{s} c_{p}^{s †})

Now, to normal-order the second term, we use the anti-commutator: $c_{p}^{s} c_{p}^{s †} = - c_{p}^{s †} c_{p}^{s} + {c_{p}^{s}, c_{p}^{s †}}$ . This gives:

H = \int \frac{d ^{3} p}{( 2 π ) ^{3}} E_{p} s \sum (b_{p}^{s †} b_{p}^{s} + c_{p}^{s †} c_{p}^{s} - {c_{p}^{s}, c_{p}^{s †}})

The final term is an infinite c-number (proportional to $δ (0)$ ), which we discard via normal ordering. The normal-ordered Hamiltonian is:

: H := \int \frac{d ^{3} p}{( 2 π ) ^{3}} E_{p} s = 1, 2 \sum (b_{p}^{s †} b_{p}^{s} + c_{p}^{s †} c_{p}^{s})

This Hamiltonian is positive-definite. The operators $b_{p}^{s †}$ create particles of energy $E_{p}$ , and the operators $c_{p}^{s †}$ create antiparticles, which also have positive energy $E_{p}$ . The theory is now stable and physically sensible.

Test your understanding!

Show how the anti-commutation relations ${ψ_{α} (x), ψ_{β} (y)} = 0$ naturally lead to the Pauli Exclusion Principle, which states that no two fermions can occupy the same quantum state. (Hint: Consider creating two identical particles.)

Show answer

Let's create a single-particle state by acting on the vacuum $∣0 ⟩$ with a creation operator: $∣ p, s ⟩ = b_{p}^{s †} ∣0 ⟩$ .
Now, let's try to create a second particle in the exact same state:

b_{p}^{s †} ∣ p, s ⟩ = b_{p}^{s †} b_{p}^{s †} ∣0 ⟩

The fundamental anti-commutation relations for the creation operators include ${b_{p}^{s †}, b_{p}^{s †}} = 0$ . Let's expand this:

{b_{p}^{s †}, b_{p}^{s †}} = b_{p}^{s †} b_{p}^{s †} + b_{p}^{s †} b_{p}^{s †} = 2 (b_{p}^{s †})^{2} = 0

This implies that $(b_{p}^{s †})^{2} = 0$ . Therefore, when we try to create a two-particle state with identical quantum numbers, we get:

b_{p}^{s †} b_{p}^{s †} ∣0 ⟩ = 0

The resulting state is the null vector. It is impossible to have two identical fermions in the same state. The Pauli principle is a direct mathematical consequence of fermionic quantization.

This connects back to the intuitive "Dirac Sea" picture from the last lesson. The reason all the negative-energy states could be "filled" without collapsing is precisely the Pauli exclusion principle, which forbids two electrons from occupying the same state. Our quantization formalism using anti-commutators is the rigorous, modern embodiment of that physical idea.

Dirac Sea Model — The Dirac Sea model proposed that the vacuum is a state where all negative-energy electron states are filled. Exciting an electron to a positive-energy state leaves a "hole," which is an antiparticle (positron). The stability of this sea relies on the Pauli exclusion principle, which is mathematically guaranteed by the anti-commutation relations we've just established.

Conclusion

In this lesson, we have successfully quantized the free Dirac field, completing the transition from a classical description to a quantum one.

Key Takeaways:

The classical Dirac Lagrangian is $L = ψ ˉ (i γ^{μ} \partial_{μ} - m) ψ$ , and the corresponding canonical momentum is $Π = i ψ^{†}$ .
Quantizing the Dirac field with commutation relations fails, leading to a theory with an unstable vacuum and energy that is not bounded below.
The correct procedure, in accordance with the spin-statistics theorem, is to impose equal-time anti-commutation relations (ETACRs) on the field operators.
This "fermionic quantization" resolves the instability, yielding a positive-definite Hamiltonian: $H = \int \dots \sum_{s} (b^{s †} b^{s} + c^{s †} c^{s})$ .
The operators $b^{s †}$ create particles, and $c^{s †}$ create antiparticles, both with positive energy.
A profound consequence of the anti-commutation relations is the Pauli Exclusion Principle, which is now seen not as an ad-hoc rule but as a fundamental requirement of a consistent quantum theory of fermions.

Preview of the Next Lesson:
With the creation and annihilation operators for fermions and antifermions now properly defined, we are ready to build the state space of our theory. In the next lesson, we will formally construct the Fock space for fermions, solidifying our understanding of multi-particle fermionic states and the implications of the exclusion principle.

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