Fermionic Fock Space and Pauli Exclusion

Hello! Welcome back to our journey into Quantum Field Theory.

In our last lesson, we successfully quantized the free Dirac field. We discovered that a naive application of commutation relations led to a physically inconsistent theory with an unstable vacuum. The solution was to impose equal-time anti-commutation relations, a choice dictated by the spin-statistics theorem for half-integer spin particles. This led us to a positive-definite Hamiltonian and properly defined creation operators ( $b_{p}^{s †}, c_{p}^{s †}$ ) and annihilation operators ( $b_{p}^{s}, c_{p}^{s}$ ) for particles and antiparticles, which obey the fundamental algebra ${b, b^{†}} \propto δ$ , ${c, c^{†}} \propto δ$ , etc.

Today, we will use these operators to formally build the space in which our quantum states live. The learning outcome for this lesson is to construct the Fock space for fermions, demonstrating the Pauli exclusion principle. We will define the vacuum, see how to build multi-particle states, and make the connection between the anti-commutation algebra and the fundamental properties of fermions explicit.

1. The Structure of Fermionic Fock Space

Just as with the scalar field, the state space for a theory with a variable number of particles is called the Fock space, $F$ . It is constructed as a direct sum of Hilbert spaces for a fixed number of particles:

F = H_{0} \oplus H_{1} \oplus H_{2} \oplus \dots

where $H_{n}$ is the space of $n$ -particle states.

The foundation of this space is the vacuum state, $∣0 ⟩$ , which is the unique state containing no particles or antiparticles. By definition, it is annihilated by all annihilation operators:

b_{p}^{s} ∣0 ⟩ = 0 and c_{p}^{s} ∣0 ⟩ = 0 for all p, s

The vacuum state forms the basis for the zero-particle space, $H_{0}$ . All other states in the Fock space are built by acting on this vacuum with creation operators.

The video below provides an excellent overview of the fermionic Fock space, how creation and annihilation operators allow us to navigate it, and how it fundamentally differs from the bosonic case.

Fermion creation and annihilation operators

To begin, please watch the following video from the 'Professor M does Science' channel. It provides a clear, conceptual introduction to the fermionic Fock space and the role of creation/annihilation operators.

Please watch these three segments: Introduction to Fermionic Fock Space (00:34 - 01:28): This part explains why the Pauli exclusion principle limits occupation numbers to 0 or 1. The Creation Operator (01:28 - 02:40): Focus on how the creation operator adds a particle and, crucially, what happens when you try to add a particle to an already occupied state. Structure of Fock Space (11:45 - 14:31): This segment beautifully illustrates how all possible states are constructed from the vacuum and directly contrasts the finite nature of fermionic states with the infinite possibilities for bosons.

As the video explained, the entire structure of the fermionic Fock space is a direct consequence of the algebraic rules obeyed by the creation and annihilation operators. Let's now explore this more formally.

2. Building States and the Pauli Exclusion Principle

We can now construct the basis states for each $n$ -particle subspace, $H_{n}$ , by repeatedly acting on the vacuum with our creation operators.

One-Particle States

The one-particle space $H_{1}$ is spanned by states created by acting on the vacuum with a single creation operator. A state with one particle of momentum $p$ and spin $s$ is:

∣ p, s ⟩_{p} \equiv b_{p}^{s †} ∣0 ⟩

Similarly, a state with one antiparticle is:

∣ p, s ⟩_{a} \equiv c_{p}^{s †} ∣0 ⟩

These states form a basis for $H_{1}$ .

Two-Particle States and Fermi-Dirac Statistics

Now, consider a two-particle state created by acting with two creation operators:

∣ p_{1}, s_{1}; p_{2}, s_{2} ⟩ \equiv b_{p_{1}}^{s_{1} †} b_{p_{2}}^{s_{2} †} ∣0 ⟩

What happens if we swap the two particles? We can use the fundamental anti-commutation relation ${b_{p_{1}}^{s_{1} †}, b_{p_{2}}^{s_{2} †}} = 0$ , which means $b_{p_{1}}^{s_{1} †} b_{p_{2}}^{s_{2} †} = - b_{p_{2}}^{s_{2} †} b_{p_{1}}^{s_{1} †}$ . Applying this gives:

∣ p_{1}, s_{1}; p_{2}, s_{2} ⟩ = - b_{p_{2}}^{s_{2} †} b_{p_{1}}^{s_{1} †} ∣0 ⟩ = - ∣ p_{2}, s_{2}; p_{1}, s_{1} ⟩

The state vector picks up a minus sign upon the interchange of two particles. This is the defining characteristic of Fermi-Dirac statistics. The anti-symmetric nature of fermionic states is not an extra postulate, but a direct result of the anti-commutation algebra we were forced to adopt.

The Pauli Exclusion Principle

Now for the crucial moment. What if we try to create two particles in the exact same quantum state? That is, we set $(p_{1}, s_{1}) = (p_{2}, s_{2}) = (p, s)$ . The two-particle state becomes:

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

From the anti-commutation relation, we know that for any operator $A$ , ${A, A} = AA + AA = 2 A^{2}$ . Since ${b_{p}^{s †}, b_{p}^{s †}} = 0$ , it immediately follows that:

(b_{p}^{s †})^{2} = 0

The creation operator for a given state, when applied twice, gives zero. Therefore, our attempt to create a two-particle state with identical quantum numbers results in:

∣ p, s; p, s ⟩ = (b_{p}^{s †})^{2} ∣0 ⟩ = 0

The state is the null vector; it does not exist in the Hilbert space. This is the Pauli exclusion principle: no two identical fermions can occupy the same quantum state. Like Fermi-Dirac statistics, it is a direct and profound consequence of quantizing the field with anti-commutators.

The following reading from David Tong's notes, which we used in the last lesson, formalizes these points concisely.

5. Quantizing the Dirac Field

To see this argument in a written, rigorous format, please review the following short section from David Tong's QFT notes.

Please read the subsection 5.2.1 'Fermi-Dirac Statistics' (page 110). This section explicitly shows how the anti-commutation relations lead to the anti-symmetric nature of two-particle states (Eq. 5.21) and the Pauli exclusion principle.

3. The Occupation Number Representation

We can now formalize the structure of the Fock space using the occupation number representation. A general basis state is uniquely specified by listing the number of particles $n_{p, s}$ for each possible single-particle state $(p, s)$ .

A general $N$ -particle state can be written as:

∣ n_{p_{1}, s_{1}}, n_{p_{2}, s_{2}}, \dots ⟩ = (b_{p_{1}}^{s_{1} †})^{n_{p_{1}, s_{1}}} (b_{p_{2}}^{s_{2} †})^{n_{p_{2}, s_{2}}} \dots ∣0 ⟩

(A similar expression holds for antiparticles with $c^{†}$ operators).

Because $(b_{p}^{s †})^{2} = 0$ , the occupation number $n_{p, s}$ for any fermionic state can only take two possible values:

$n_{p, s} = 0$ (the state is empty)
$n_{p, s} = 1$ (the state is occupied by one particle)

This stands in stark contrast to bosons, where the commutation relation $[a^{†}, a^{†}] = 0$ allows $a^{†}$ to be applied any number of times, leading to occupation numbers $n = 0, 1, 2, \dots, \infty$ . For fermions, each "slot" in the quantum state space can either be empty or full, but no more. This simple binary property, rooted in the anti-commutation algebra, underpins the structure of atoms, the stability of matter, and the behavior of electrons in the metals and semiconductors you have studied in your condensed matter courses.

Test your understanding!

Consider a normalized two-particle state $∣Ψ ⟩ = N b_{p_{1}}^{s_{1} †} b_{p_{2}}^{s_{2} †} ∣0 ⟩$ , where the states are distinct: $(p_{1}, s_{1}) \neq = (p_{2}, s_{2})$ . Calculate the normalization constant $N$ by computing the inner product $⟨ Ψ∣Ψ ⟩ = 1$ .

(Assume the vacuum is normalized, $⟨ 0∣0 ⟩ = 1$ , and recall that the bra corresponding to $∣Ψ ⟩$ is $⟨ Ψ∣ = N^{*} ⟨ 0∣ b_{p_{2}}^{s_{2}} b_{p_{1}}^{s_{1}}$ .)

Show answer

We start with the expression for the inner product:

⟨ Ψ∣Ψ ⟩ = ∣ N ∣^{2} ⟨ 0∣ b_{p_{2}}^{s_{2}} b_{p_{1}}^{s_{1}} b_{p_{1}}^{s_{1} †} b_{p_{2}}^{s_{2} †} ∣0 ⟩ = 1

Our goal is to move the annihilation operators to the right until they act on $∣0 ⟩$ , which gives zero. We use the anti-commutation relations ${b_{p}^{r}, b_{q}^{s †}} = (2 π)^{3} δ^{r s} δ^{(3)} (p - q)$ .

First, let's move $b_{p_{1}}^{s_{1}}$ past $b_{p_{1}}^{s_{1} †}$ .

b_{p_{1}}^{s_{1}} b_{p_{1}}^{s_{1} †} = - b_{p_{1}}^{s_{1} †} b_{p_{1}}^{s_{1}} + {b_{p_{1}}^{s_{1}}, b_{p_{1}}^{s_{1} †}} = - b_{p_{1}}^{s_{1} †} b_{p_{1}}^{s_{1}} + (2 π)^{3} δ^{(3)} (0)

Let's plug this into the main expression:

⟨ Ψ∣Ψ ⟩ = ∣ N ∣^{2} ⟨ 0∣ b_{p_{2}}^{s_{2}} (- b_{p_{1}}^{s_{1} †} b_{p_{1}}^{s_{1}} + (2 π)^{3} δ^{(3)} (0)) b_{p_{2}}^{s_{2} †} ∣0 ⟩

The term with $b_{p_{1}}^{s_{1}}$ on the right can be moved past $b_{p_{2}}^{s_{2} †}$ (picking up a minus sign since the states are different) until it hits $∣0 ⟩$ and vanishes. This leaves:

⟨ Ψ∣Ψ ⟩ = ∣ N ∣^{2} ⟨ 0∣ b_{p_{2}}^{s_{2}} ((2 π)^{3} δ^{(3)} (0)) b_{p_{2}}^{s_{2} †} ∣0 ⟩

Now we commute $b_{p_{2}}^{s_{2}}$ past $b_{p_{2}}^{s_{2} †}$ :

⟨ Ψ∣Ψ ⟩ = ∣ N ∣^{2} ((2 π)^{3} δ^{(3)} (0)) ⟨ 0∣ (- b_{p_{2}}^{s_{2} †} b_{p_{2}}^{s_{2}} + (2 π)^{3} δ^{(3)} (0)) ∣0 ⟩

Again, the term with $b_{p_{2}}^{s_{2}}$ on the right annihilates the vacuum. We are left with:

⟨ Ψ∣Ψ ⟩ = ∣ N ∣^{2} ((2 π)^{3} δ^{(3)} (0))^{2} ⟨ 0∣0 ⟩ = 1

For a discrete set of states with Kronecker delta normalization ${b_{i}, b_{j}^{†}} = δ_{ij}$ , the calculation is simpler and gives $⟨ Ψ∣Ψ ⟩ = ∣ N ∣^{2} = 1$ , so $N = 1$ . For our continuum normalization, we see that the normalization factor must absorb the infinite factors of $(2 π)^{3} δ^{(3)} (0)$ . This is a common feature in QFT, where states are often normalized "per unit volume". In practice, we often work with unnormalized states and account for normalization factors when calculating physical quantities like cross-sections.

Conclusion

In this lesson, we have formally constructed the state space for the free Dirac field. By building up from the vacuum, we have seen in concrete terms how the anti-commutation relations for fermionic operators govern the properties of the theory.

Key Takeaways:

The Hilbert space for fermions is a Fock space, whose basis states are constructed by acting on the vacuum state $∣0 ⟩$ with creation operators.
The anti-commutation relation ${b^{†}, b^{†}} = 0$ directly implies that multi-fermion states are anti-symmetric under particle exchange, which is the origin of Fermi-Dirac statistics.
The fact that $(b^{†})^{2} = 0$ is the mathematical embodiment of the Pauli exclusion principle: the occupation number of any given fermionic state can only be 0 or 1.
The fermionic Fock space for a given mode is fundamentally simple—it's just a two-dimensional space spanned by $∣0 ⟩$ and $∣1 ⟩$ .

Preview of the Next Lesson:
We now have the free Dirac field operator and the space of states it acts upon. The next logical step is to describe how a free fermion propagates through spacetime. This will lead us to one of the most important tools in QFT: the Feynman propagator. In the next lesson, we will derive the propagator for the Dirac field, which will be an essential building block when we finally introduce interactions.

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