Charge and Statistics

Introduction

Quantum field theory treats particles as excitations of fields. This immediately changes how we think about particle properties. A property such as spin, charge, or statistics is not just attached to a particle as a separate label. It is tied to the structure of the field and to the algebra used to quantize that field.

This lesson focuses on two central particle properties:

1. Charge: a conserved quantity generated by a continuous internal symmetry of the field theory.
2. Statistics: the symmetry or antisymmetry of many-particle states under exchange of identical particles.

The charge part begins with two real Klein–Gordon fields of equal mass. Because the two fields have the same dynamics, the theory is invariant under rotations in the internal two-dimensional field space. Noether's theorem then gives a conserved charge. After quantization, this charge becomes an operator. In the original real-field basis, the charge operator is not diagonal. Complex linear combinations of the fields and operators diagonalize it, producing two particle sectors with opposite charges.

The statistics part starts from the fact that particles created from the same field are fundamentally indistinguishable. Ordinary canonical commutation relations lead to symmetric many-particle wavefunctions, so the quanta are bosons. Fermions require anti-commutation relations for their creation and annihilation operators. These anti-commutation relations give antisymmetric wavefunctions and Pauli exclusion.

A key warning runs through the lesson: fermions do not mean replacing every commutator by an anti-commutator. Fermionic creation and annihilation operators obey anti-commutation relations, but the general dynamical laws of quantum mechanics still use ordinary commutators.

Learning Objectives

By the end of this lesson, you should be able to:

• Explain how particle properties can arise from field properties.
• Relate spin to the transformation properties of a field.
• Derive the conserved Noether charge of two equal-mass real Klein–Gordon fields.
• Quantize the charge and express it using creation and annihilation operators.
• Show how complex linear combinations diagonalize the charge operator.
• Interpret the complex scalar field as the natural field for charged scalar particles.
• Explain why particles and antiparticles appear with opposite charges.
• Derive why canonical commutators imply Bose–Einstein statistics.
• Explain why fermions require anti-commutators.
• Derive Pauli exclusion from fermionic anti-commutation relations.
• Distinguish canonical anti-commutation relations from ordinary quantum-mechanical commutators.

Prerequisites

You should already know:

• The Klein–Gordon field.
• Canonical quantization.
• Creation and annihilation operators.
• Fock space.
• Basic Noether theorem logic.
• Basic operator algebra.
• Symmetric and antisymmetric wavefunctions.

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1. Spin as a field property

A useful starting point is spin. In QFT, spin is not an independent decoration added after quantization. It is determined by how the field transforms under rotations.

Consider a three-component vector field. Its components may be written in the spatial basis

$$\hat{\phi}_x,\qquad \hat{\phi}_y,\qquad \hat{\phi}_z.$$

The same field can also be written in a basis adapted to spin projection along the $z$-axis:

$$\hat{A}_{+},\qquad \hat{A}_0,\qquad \hat{A}_{-}.$$

The relation between the two bases is

$$\hat{\phi}_z = \hat{A}_0,$$ $$\hat{\phi}_x = \frac{\hat{A}_{+}+i\hat{A}_{-}}{\sqrt{2}},$$ $$\hat{\phi}_y = \frac{\hat{A}_{-}+i\hat{A}_{+}}{\sqrt{2}}.$$

The inverse relations are

$$\hat{A}_0 = \hat{\phi}_z,$$ $$\hat{A}_{+}=\frac{\hat{\phi}_x-i\hat{\phi}_y}{\sqrt{2}},$$ $$\hat{A}_{-}=\frac{\hat{\phi}_y-i\hat{\phi}_x}{\sqrt{2}}.$$

Particles created by the creation operators associated with $\hat{A}_s$ have spin component

$$s_z=s, \qquad s=-1,0,+1.$$

This example illustrates a general QFT principle: changing field basis is allowed when the new fields preserve the canonical commutation relations. The new basis is not a new physical theory. It is a more convenient representation of the same degrees of freedom.

This idea will be reused for charge. Two real fields will be recombined into one complex field, and that new basis will make charge diagonal.

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2. What makes a vector field a vector field?

A multi-component field is not automatically a vector field in the physical sense. A true spatial vector field has three components corresponding to the three directions of space, and it transforms under rotations in two simultaneous ways:

1. The spatial argument is rotated.
2. The components are rotated among themselves.

The generator of rotations therefore has two parts:

$$\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}},$$

where

$$\hat{\mathbf{L}}$$

generates rotations of the spatial argument, and

$$\hat{\mathbf{S}}$$

generates rotations among the field components.

So spin comes from the internal transformation of field components under spatial rotations. Charge will follow a parallel logic, except the relevant symmetry is not a spatial rotation. It is an internal rotation in field space.

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3. Charge and statistics as field-derived particle properties

Two important particle properties can be traced back to field-theoretic structure:

1. Charge comes from continuous internal symmetries of the field theory.
2. Bose/Fermi statistics comes from the operator algebra and the indistinguishability of field excitations.

In non-relativistic many-body theory, often called second quantization, canonically quantized fields have bosonic excitations when ordinary commutators are used. This raises a serious question:

If canonical quantization gives bosons, how can QFT describe fermions?

The answer is that fermionic fields are quantized using anti-commutation relations for the creation and annihilation operators. But before reaching fermions, we first derive charge.

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4. Two real Klein–Gordon fields with the same mass

Start with two different real classical Klein–Gordon fields of equal mass:

$$\phi_1,\qquad \phi_2.$$

Their Lagrangian is

$$L = \frac{1}{2}\sum_{j=1}^{2}\int d^3r \left[ \dot{\phi}_j^{\,2} - c^2|\nabla \phi_j|^2 - \left(\frac{Mc^2}{\hbar}\right)^2\phi_j^{\,2} \right].$$

Because the two fields have the same mass and the same dynamics, the theory is symmetric under internal rotations that mix $\phi_1$ and $\phi_2$:

$$\phi_1^\theta = \phi_1\cos\theta+\phi_2\sin\theta,$$ $$\phi_2^\theta = \phi_2\cos\theta-\phi_1\sin\theta.$$

This is not a rotation of physical space. It is a rotation in the two-dimensional internal field space spanned by $\phi_1$ and $\phi_2$.

The infinitesimal generators are obtained by differentiating with respect to $\theta$ and then setting $\theta=0$:

$$G_n=\left.\frac{\partial \phi_n^\theta}{\partial \theta}\right|_{\theta=0}.$$

Therefore

$$G_1=\phi_2,$$ $$G_2=-\phi_1.$$

Noether's theorem says that a continuous symmetry gives a conserved quantity. The basic Noether charge associated with this internal rotation is

$$\sum_{j=1}^{2}\int d^3r\,\pi_jG_j,$$

where $\pi_j$ is the momentum conjugate to $\phi_j$:

$$\pi_j=\dot{\phi}_j.$$

Multiplying a conserved quantity by any constant still gives a conserved quantity. Introduce a constant $q$, which sets the unit of charge, and define

$$Q = \frac{q}{\hbar}\sum_n\int d^3r\,\pi_nG_n.$$

Using $G_1=\phi_2$ and $G_2=-\phi_1$, this becomes

$$Q = \frac{q}{\hbar}\int d^3r\, (\pi_1\phi_2-\pi_2\phi_1).$$

This is the classical conserved charge. The important point is conceptual: charge is the conserved generator of an internal continuous symmetry.

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5. Quantizing the two-field system

After quantization, the classical fields become operators:

$$\phi_1,\phi_2 \quad\longrightarrow\quad \hat{\phi}_1,\hat{\phi}_2.$$

The Hamiltonian operator is

$$\hat{H} = \sum_{n=1}^{2}\frac{1}{2}\int d^3r \left[ \hat{\pi}_n^{\,2} + c^2|\nabla\hat{\phi}_n|^2 + \left(\frac{Mc^2}{\hbar}\right)^2\hat{\phi}_n^{\,2} \right].$$

Introduce creation and annihilation operators by expanding the fields as

$$\hat{\phi}_j(\mathbf r) = \int d^3k\, \sqrt{\frac{\hbar}{2(2\pi)^3\omega(k)}} \,e^{i\mathbf k\cdot\mathbf r} \left( \hat a_{j\mathbf k}+\hat a_{j,-\mathbf k}^{\dagger} \right),$$ $$\hat{\pi}_j(\mathbf r) = -i\int d^3k\, \sqrt{\frac{\hbar\omega(k)}{2(2\pi)^3}} \,e^{i\mathbf k\cdot\mathbf r} \left( \hat a_{j\mathbf k}-\hat a_{j,-\mathbf k}^{\dagger} \right).$$

The mode frequency is

$$\omega(k)=\sqrt{c^2k^2+\left(\frac{Mc^2}{\hbar}\right)^2}.$$

In terms of these operators, the Hamiltonian becomes

$$\hat H = \int d^Dk\,\hbar\omega_k \left( \hat a_{1\mathbf k}^{\dagger}\hat a_{1\mathbf k} + \hat a_{2\mathbf k}^{\dagger}\hat a_{2\mathbf k} \right),$$

and the momentum operator becomes

$$\hat{\mathbf P} = \int d^Dk\,\hbar\mathbf k \left( \hat a_{1\mathbf k}^{\dagger}\hat a_{1\mathbf k} + \hat a_{2\mathbf k}^{\dagger}\hat a_{2\mathbf k} \right).$$

The quantized Noether charge is

$$\hat Q = \frac{q}{\hbar}\int d^3r\, (\hat\pi_1\hat\phi_2-\hat\pi_2\hat\phi_1).$$

Using the mode expansion, this becomes

$$\hat Q = iq\int d^3k \left[ \hat a_1^{\dagger}(\mathbf k)\hat a_2(\mathbf k) - \hat a_2^{\dagger}(\mathbf k)\hat a_1(\mathbf k) \right].$$

This charge operator is not diagonal in the original $(1,2)$ basis. It mixes the two operator species. To find particles with definite charge, we need a new basis.

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6. Complex linear combinations diagonalize charge

Define new annihilation operators by complex linear combinations:

$$\hat a_{\mathbf k} = \frac{1}{\sqrt{2}} \left( \hat a_{1\mathbf k}+i\hat a_{2\mathbf k} \right),$$ $$\hat c_{\mathbf k} = \frac{1}{\sqrt{2}} \left( \hat a_{1\mathbf k}-i\hat a_{2\mathbf k} \right).$$

These operators obey canonical commutation relations:

$$[\hat c_{\mathbf k},\hat c_{\mathbf k'}^{\dagger}] = \delta^D(\mathbf k-\mathbf k') = [\hat a_{\mathbf k},\hat a_{\mathbf k'}^{\dagger}],$$ $$[\hat c_{\mathbf k},\hat a_{\mathbf k'}^{\dagger}] = 0 = [\hat a_{\mathbf k},\hat c_{\mathbf k'}^{\dagger}].$$

The Hamiltonian remains diagonal:

$$\hat H = \int d^Dk\,\hbar\omega_k \left( \hat a_{\mathbf k}^{\dagger}\hat a_{\mathbf k} + \hat c_{\mathbf k}^{\dagger}\hat c_{\mathbf k} \right).$$

The momentum also remains diagonal:

$$\hat{\mathbf P} = \int d^Dk\,\hbar\mathbf k \left( \hat a_{\mathbf k}^{\dagger}\hat a_{\mathbf k} + \hat c_{\mathbf k}^{\dagger}\hat c_{\mathbf k} \right).$$

But the charge operator becomes

$$\hat Q = q\int d^3k \left[ \hat a^{\dagger}(\mathbf k)\hat a(\mathbf k) - \hat c^{\dagger}(\mathbf k)\hat c(\mathbf k) \right].$$

Now charge is diagonal.

Therefore:

$$\hat a_{\mathbf k}^{\dagger}$$

creates a relativistic particle with mass $M$, momentum $\hbar\mathbf k$, and charge $+q$.

Similarly,

$$\hat c_{\mathbf k}^{\dagger}$$

creates a relativistic particle with mass $M$, momentum $\hbar\mathbf k$, and charge $-q$.

So particles and antiparticles appear as two charge sectors of the same complexified field system.

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7. The complex scalar field

The same diagonalization can be done directly at the field level. Define

$$\hat\phi = \frac{\hat\phi_1+i\hat\phi_2}{\sqrt{2}}.$$

This field is not Hermitian:

$$\hat\phi\neq\hat\phi^{\dagger}.$$

The conjugate momentum field is also complex:

$$\pi = \dot\phi^{\ast} = \frac{\pi_1-i\pi_2}{\sqrt{2}}.$$

In terms of $\phi$ and $\phi^{\ast}$, the classical Lagrangian is

$$L = \int d^3r \left[ \dot\phi^{\ast}\dot\phi - c^2\nabla\phi^{\ast}\cdot\nabla\phi - \left(\frac{Mc^2}{\hbar}\right)^2\phi^{\ast}\phi \right].$$

The internal rotation symmetry now takes the simple phase form

$$\phi\rightarrow e^{i\theta}\phi.$$

This is the same symmetry as before. The two-real-field rotation has become a complex phase rotation.

The Hamiltonian operator is

$$\hat H = \int d^3r \left[ \hat\pi^{\dagger}\hat\pi + c^2\nabla\hat\phi^{\dagger}\cdot\nabla\hat\phi + \left(\frac{Mc^2}{\hbar}\right)^2 \hat\phi^{\dagger}\hat\phi \right].$$

This is not a new Hamiltonian. It is the same Hamiltonian expressed in the complex-field basis.

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8. Field operator structure: particle annihilation and antiparticle creation

The complex scalar field operator can be expanded in terms of the $\hat a$ and $\hat c$ operators as

$$\hat\phi(\mathbf r) = \sqrt{\frac{\hbar}{(2\pi)^3}} \int\frac{d^3k}{\sqrt{2\omega(k)}} \,e^{i\mathbf k\cdot\mathbf r} \left( \hat a_{\mathbf k}+\hat c_{-\mathbf k}^{\dagger} \right),$$

and

$$\hat\phi^{\dagger}(\mathbf r) = \sqrt{\frac{\hbar}{(2\pi)^3}} \int\frac{d^3k}{\sqrt{2\omega(k)}} \,e^{i\mathbf k\cdot\mathbf r} \left( \hat c_{\mathbf k}+\hat a_{-\mathbf k}^{\dagger} \right).$$

This is the essential charged-field pattern:

$$\hat\phi \quad\text{destroys a particle or creates an antiparticle.}$$

The charge operator is

$$\hat Q = q\int d^3k \left[ \hat a^{\dagger}(\mathbf k)\hat a(\mathbf k) - \hat c^{\dagger}(\mathbf k)\hat c(\mathbf k) \right].$$

The field $\hat\phi$ is complex:

$$\phi\neq\phi^{\ast}.$$

Its phase symmetry is

$$\phi\rightarrow e^{i\theta}\phi.$$

This pattern also appears in the relativistic theory of electrons and positrons, except that scalar fields are replaced by four-component spinor fields.

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9. Indistinguishability of particles from the same field

Particles that are excitations of the same field are not merely similar. They are fundamentally indistinguishable. Exchanging two excitations of the same field does not produce a new physical configuration.

This is the field-theoretic origin of quantum statistics. A many-particle wavefunction must respond to particle exchange in a way consistent with the operator algebra used to create and destroy the particles.

Canonical quantization of a bosonic field uses equal-time commutators:

$$[\hat\phi(\mathbf r),\hat\phi(\mathbf r')]=0,$$ $$[\hat\pi(\mathbf r),\hat\pi(\mathbf r')]=0,$$ $$[\hat\phi(\mathbf r),\hat\pi(\mathbf r')] = i\hbar\delta^3(\mathbf r-\mathbf r').$$

These imply canonical commutation relations for the mode operators:

$$[\hat a_{\mathbf k},\hat a_{\mathbf k'}]=0,$$ $$[\hat a_{\mathbf k}^{\dagger},\hat a_{\mathbf k'}^{\dagger}]=0,$$ $$[\hat a_{\mathbf k},\hat a_{\mathbf k'}^{\dagger}] = \delta^3(\mathbf k-\mathbf k').$$

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10. Bose–Einstein statistics from commutators

Define an $N$-particle wavefunction in momentum space by

$$\psi(k_1,\ldots,k_n,\ldots,k_N) = \langle G| \left(\prod_n\hat a_{k_n}\right) |\Psi\rangle,$$

where $|G\rangle$ is the vacuum state.

For two particles,

$$\psi(k_1,k_2) = \langle G|\hat a_{k_1}\hat a_{k_2}|\Psi\rangle.$$

Since bosonic annihilation operators commute,

$$\hat a_{k_1}\hat a_{k_2} = \hat a_{k_2}\hat a_{k_1},$$

we get

$$\psi(k_1,k_2) = \langle G|\hat a_{k_2}\hat a_{k_1}|\Psi\rangle = \psi(k_2,k_1).$$

Thus the two-particle wavefunction is symmetric under exchange. The same argument extends to any number of particles:

$$\psi(\ldots,k_i,\ldots,k_j,\ldots) = \psi(\ldots,k_j,\ldots,k_i,\ldots).$$

This is Bose–Einstein statistics.

Ordinary canonical commutation relations therefore imply bosonic particles.

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11. Fermions require anti-commutators

Fermions have antisymmetric wavefunctions. For two fermions,

$$\psi(k_1,k_2) = - \psi(k_2,k_1).$$

Using the same definition of the two-particle wavefunction, exchange antisymmetry requires

$$\langle G|\hat a_{k_1}\hat a_{k_2}|\Psi\rangle = - \langle G|\hat a_{k_2}\hat a_{k_1}|\Psi\rangle$$

for all states $|\Psi\rangle$. Therefore the operators must satisfy

$$\hat a_{k_1}\hat a_{k_2} = - \hat a_{k_2}\hat a_{k_1} \qquad \forall k_1,k_2.$$

Define the anti-commutator by

$$\{\hat A,\hat B\} = \hat A\hat B+\hat B\hat A.$$

Then the fermionic exchange condition is

$$\{\hat a_k,\hat a_{k'}\}=0.$$

Taking the Hermitian conjugate gives

$$\{\hat a_k^{\dagger},\hat a_{k'}^{\dagger}\}=0.$$

These relations encode antisymmetry under exchange.

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12. The number operator and the full fermionic algebra

Fermionic creation and annihilation operators must still raise and lower particle number. Define the number operator

$$\hat N = \int d^3k\,\hat a_k^{\dagger}\hat a_k.$$

If a state has $N$ particles,

$$\hat N|\Psi\rangle=N|\Psi\rangle,$$

then the state

$$\hat a_k^{\dagger}|\Psi\rangle$$

should have $N+1$ particles:

$$\hat N\hat a_k^{\dagger}|\Psi\rangle = (N+1)\hat a_k^{\dagger}|\Psi\rangle.$$

Also,

$$\hat a_k^{\dagger}\hat N|\Psi\rangle = N\hat a_k^{\dagger}|\Psi\rangle.$$

Subtracting gives

$$[\hat N,\hat a_k^{\dagger}]|\Psi\rangle = \hat a_k^{\dagger}|\Psi\rangle.$$

Therefore

$$[\hat N,\hat a_k^{\dagger}] = \hat a_k^{\dagger}.$$

This is an ordinary commutator, not an anti-commutator.

Now compute the commutator explicitly:

$$[\hat N,\hat a_k^{\dagger}] = \int d^3k'\, \left( \hat a_{k'}^{\dagger}\hat a_{k'}\hat a_k^{\dagger} - \hat a_k^{\dagger}\hat a_{k'}^{\dagger}\hat a_{k'} \right).$$

Using fermionic anti-commutation among creation operators,

$$\hat a_k^{\dagger}\hat a_{k'}^{\dagger} = - \hat a_{k'}^{\dagger}\hat a_k^{\dagger},$$

this becomes

$$[\hat N,\hat a_k^{\dagger}] = \int d^3k'\, \left( \hat a_{k'}^{\dagger}\hat a_{k'}\hat a_k^{\dagger} + \hat a_{k'}^{\dagger}\hat a_k^{\dagger}\hat a_{k'} \right).$$

Factor out $\hat a_{k'}^{\dagger}$:

$$[\hat N,\hat a_k^{\dagger}] = \int d^3k'\, \hat a_{k'}^{\dagger} \left( \hat a_{k'}\hat a_k^{\dagger} + \hat a_k^{\dagger}\hat a_{k'} \right).$$

The quantity in parentheses is an anti-commutator:

$$[\hat N,\hat a_k^{\dagger}] = \int d^3k'\, \hat a_{k'}^{\dagger} \{\hat a_{k'},\hat a_k^{\dagger}\}.$$

For this to equal $\hat a_k^{\dagger}$, we need

$$\{\hat a_{k'},\hat a_k^{\dagger}\} = \delta^3(k-k').$$

So the full canonical anti-commutation relations are

$$\{\hat a_k,\hat a_{k'}\}=0,$$ $$\{\hat a_k^{\dagger},\hat a_{k'}^{\dagger}\}=0,$$ $$\{\hat a_k,\hat a_{k'}^{\dagger}\} = \delta^3(k-k').$$

These are the fermionic analogue of the bosonic canonical commutation relations.

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13. Pauli exclusion

The relation

$$\{\hat a_k^{\dagger},\hat a_k^{\dagger}\}=0$$

means

$$\hat a_k^{\dagger}\hat a_k^{\dagger} + \hat a_k^{\dagger}\hat a_k^{\dagger}=0.$$

Therefore

$$2(\hat a_k^{\dagger})^2=0,$$

and hence

$$(\hat a_k^{\dagger})^2=0.$$

So it is impossible to put two identical fermions into the same one-particle state. This is Pauli exclusion.

The same logic applies to entangled states involving momentum and species labels. Suppose $m,n$ label different fermion species. Then exclusion must also hold for a superposed single-particle creation operator:

$$\left( \hat a_{m k}^{\dagger}+\hat a_{n k'}^{\dagger} \right)^2=0.$$

Expanding,

$$0 = \left(\hat a_{m k}^{\dagger}\right)^2 + \left(\hat a_{n k'}^{\dagger}\right)^2 + \{\hat a_{m k}^{\dagger},\hat a_{n k'}^{\dagger}\}.$$

The first two terms vanish by Pauli exclusion, so

$$\{\hat a_{m k}^{\dagger},\hat a_{n k'}^{\dagger}\}=0.$$

For a set of different kinds of fermions, the canonical anti-commutation relations are

$$\{\hat a_{m k},\hat a_{n k'}\}=0,$$ $$\{\hat a_{m k}^{\dagger},\hat a_{n k'}^{\dagger}\}=0,$$ $$\{\hat a_{m k},\hat a_{n k'}^{\dagger}\} = \delta_{mn}\delta^3(k-k').$$

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14. Fermions and bosons together

Fermions and bosons are distinguishable from each other. Since they are different types of excitations, exchanging a fermion with a boson is not the same kind of exchange as exchanging two identical particles.

There are also no ordinary physical processes that create coherent superpositions of a fermion and a boson. The spin-statistics theorem ties fermions to half-integer spin and bosons to integer spin. A superposition of a fermion and a boson would mix different angular-momentum sectors.

For practical field theory calculations, one chooses bosonic operators and fermionic operators to commute with each other:

$$[\hat b,\hat a]=0,$$ $$[\hat b,\hat a^{\dagger}]=0,$$

where $\hat b$ is bosonic and $\hat a$ is fermionic.

So the rule is:

• boson with boson: use commutators,
• fermion with fermion: use anti-commutators,
• boson with fermion: use commutators.

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15. Even fermions still need commutators

For fermions, the creation and annihilation operators obey the canonical anti-commutation relations:

$$\{\hat a_{m k},\hat a_{n k'}\}=0,$$ $$\{\hat a_{m k}^{\dagger},\hat a_{n k'}^{\dagger}\}=0,$$ $$\{\hat a_{m k},\hat a_{n k'}^{\dagger}\} = \delta_{mn}\delta^3(k-k').$$

For bosons, the creation and annihilation operators obey the canonical commutation relations:

$$[\hat a_{m k},\hat a_{n k'}]=0,$$ $$[\hat a_{m k}^{\dagger},\hat a_{n k'}^{\dagger}]=0,$$ $$[\hat a_{m k},\hat a_{n k'}^{\dagger}] = \delta_{mn}\delta^3(k-k').$$

For mixed boson/fermion cases, bosonic and fermionic operators commute.

The crucial warning is this:

$$\text{For fermions, do not replace every commutator by an anti-commutator.}$$

The general rules of quantum mechanics still use ordinary commutators. For example,

$$[\hat N,\hat a_k^{\dagger}] = \hat a_k^{\dagger} \qquad \forall k,$$

and the Heisenberg equation of motion is still

$$i\hbar\dot{\hat A} = [\hat A,\hat H].$$

These remain commutators even when $\hat a_k$ or $\hat A$ is fermionic.

Anti-commutators are used to encode fermionic exchange and exclusion. They do not replace the commutator structure of quantum dynamics.

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Worked Example 1: Diagonalizing the charge operator

Start from the charge operator in the real-field basis:

$$\hat Q = iq\int d^3k \left[ \hat a_1^{\dagger}(\mathbf k)\hat a_2(\mathbf k) - \hat a_2^{\dagger}(\mathbf k)\hat a_1(\mathbf k) \right].$$

This is off-diagonal because it mixes species $1$ and $2$. Define

$$\hat a_{\mathbf k} = \frac{1}{\sqrt{2}} \left( \hat a_{1\mathbf k}+i\hat a_{2\mathbf k} \right),$$ $$\hat c_{\mathbf k} = \frac{1}{\sqrt{2}} \left( \hat a_{1\mathbf k}-i\hat a_{2\mathbf k} \right).$$

In this basis,

$$\hat Q = q\int d^3k \left[ \hat a^{\dagger}(\mathbf k)\hat a(\mathbf k) - \hat c^{\dagger}(\mathbf k)\hat c(\mathbf k) \right].$$

Thus $\hat a^{\dagger}$ creates charge $+q$, while $\hat c^{\dagger}$ creates charge $-q$.

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Worked Example 2: Symmetry of bosonic two-particle wavefunctions

For bosons,

$$[\hat a_{k_1},\hat a_{k_2}]=0.$$

Therefore

$$\hat a_{k_1}\hat a_{k_2} = \hat a_{k_2}\hat a_{k_1}.$$

The two-particle momentum-space wavefunction is

$$\psi(k_1,k_2) = \langle G|\hat a_{k_1}\hat a_{k_2}|\Psi\rangle.$$

Using the commutation relation,

$$\psi(k_1,k_2) = \langle G|\hat a_{k_2}\hat a_{k_1}|\Psi\rangle.$$

Hence

$$\psi(k_1,k_2)=\psi(k_2,k_1).$$

The wavefunction is symmetric under exchange, so the particles obey Bose–Einstein statistics.

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Worked Example 3: Antisymmetry of fermionic two-particle wavefunctions

For fermions,

$$\{\hat a_{k_1},\hat a_{k_2}\}=0.$$

Therefore

$$\hat a_{k_1}\hat a_{k_2} = - \hat a_{k_2}\hat a_{k_1}.$$

The two-particle wavefunction is

$$\psi(k_1,k_2) = \langle G|\hat a_{k_1}\hat a_{k_2}|\Psi\rangle.$$

Using the anti-commutation relation,

$$\psi(k_1,k_2) = - \langle G|\hat a_{k_2}\hat a_{k_1}|\Psi\rangle.$$

So

$$\psi(k_1,k_2) = - \psi(k_2,k_1).$$

The wavefunction is antisymmetric under exchange, so the particles obey Fermi–Dirac statistics.

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Worked Example 4: Pauli exclusion from anti-commutators

The fermionic creation operators obey

$$\{\hat a_k^{\dagger},\hat a_k^{\dagger}\}=0.$$

Expanding the anti-commutator gives

$$\hat a_k^{\dagger}\hat a_k^{\dagger} + \hat a_k^{\dagger}\hat a_k^{\dagger}=0.$$

Therefore

$$2(\hat a_k^{\dagger})^2=0,$$

so

$$(\hat a_k^{\dagger})^2=0.$$

Acting twice with the same fermionic creation operator gives zero:

$$\hat a_k^{\dagger}\hat a_k^{\dagger}|G\rangle=0.$$

So two identical fermions cannot occupy the same one-particle state.

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Intuition

The lesson has two main messages.

First, charge comes from symmetry. Two equal-mass real scalar fields have an internal rotation symmetry. Noether's theorem gives a conserved charge. When the theory is quantized, complex combinations of the operators diagonalize that charge, producing particles and antiparticles with opposite charge.

Second, statistics comes from operator algebra. Commuting creation and annihilation operators produce symmetric wavefunctions and bosons. Anti-commuting creation and annihilation operators produce antisymmetric wavefunctions and fermions. Pauli exclusion is not an extra rule; it follows directly from the anti-commutation algebra.

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Common Mistakes

Mistake 1: Treating charge as an arbitrary particle label

Charge is not inserted by hand. In this construction, it is the conserved Noether quantity associated with an internal continuous symmetry.

Mistake 2: Thinking the complex scalar field is a new theory

The complex scalar field is a change of basis from two real scalar fields. The Hamiltonian is the same theory written in a more useful form.

Mistake 3: Forgetting that charge is diagonal only in the complex basis

The charge operator mixes $\hat a_1$ and $\hat a_2$. The operators $\hat a$ and $\hat c$ are the charge eigen-operator basis.

Mistake 4: Thinking $\hat\phi$ only annihilates particles

For a charged complex scalar field,

$$\hat\phi \sim \hat a+ \hat c^{\dagger}.$$

So it annihilates a particle or creates an antiparticle.

Mistake 5: Thinking canonical quantization automatically gives fermions

Ordinary canonical commutators give bosons. Fermions require anti-commutation relations for creation and annihilation operators.

Mistake 6: Replacing all commutators by anti-commutators for fermions

This is wrong. The fermionic operator algebra uses anti-commutators, but quantum-mechanical generators and time evolution still use commutators:

$$[\hat N,\hat a_k^{\dagger}]=\hat a_k^{\dagger},$$ $$i\hbar\dot{\hat A}=[\hat A,\hat H].$$

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Short Summary

Particle properties in QFT are tied to field structure. Spin comes from how fields transform under rotations. Charge comes from continuous internal symmetries. For two equal-mass real Klein–Gordon fields, an internal rotation symmetry leads by Noether's theorem to a conserved charge

$$Q = \frac{q}{\hbar}\int d^3r\, (\pi_1\phi_2-\pi_2\phi_1).$$

After quantization, the charge operator is off-diagonal in the real-field basis:

$$\hat Q = iq\int d^3k \left[ \hat a_1^{\dagger}(\mathbf k)\hat a_2(\mathbf k) - \hat a_2^{\dagger}(\mathbf k)\hat a_1(\mathbf k) \right].$$

Complex operators

$$\hat a_{\mathbf k} = \frac{1}{\sqrt{2}} (\hat a_{1\mathbf k}+i\hat a_{2\mathbf k}), \qquad \hat c_{\mathbf k} = \frac{1}{\sqrt{2}} (\hat a_{1\mathbf k}-i\hat a_{2\mathbf k})$$

diagonalize it:

$$\hat Q = q\int d^3k \left[ \hat a^{\dagger}(\mathbf k)\hat a(\mathbf k) - \hat c^{\dagger}(\mathbf k)\hat c(\mathbf k) \right].$$

The operator $\hat a^{\dagger}$ creates particles of charge $+q$, and $\hat c^{\dagger}$ creates particles of charge $-q$. At the field level, this is described by a complex scalar field with symmetry

$$\phi\rightarrow e^{i\theta}\phi.$$

Statistics comes from operator algebra. Bosonic commutators imply symmetric wavefunctions and Bose–Einstein statistics. Fermionic anti-commutators imply antisymmetric wavefunctions and Fermi–Dirac statistics. Pauli exclusion follows from

$$(\hat a_k^{\dagger})^2=0.$$

But even for fermions, the general dynamical laws of quantum mechanics still use commutators, such as

$$i\hbar\dot{\hat A}=[\hat A,\hat H].$$

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Practice Problems

1. Explain why two equal-mass real Klein–Gordon fields have an internal rotation symmetry.

2. Starting from

$$\phi_1^\theta=\phi_1\cos\theta+\phi_2\sin\theta, \qquad \phi_2^\theta=\phi_2\cos\theta-\phi_1\sin\theta,$$

derive

$$G_1=\phi_2, \qquad G_2=-\phi_1.$$

3. Use the generators $G_1,G_2$ to derive

$$Q = \frac{q}{\hbar}\int d^3r\, (\pi_1\phi_2-\pi_2\phi_1).$$

4. Why is the charge operator off-diagonal in the $(\hat a_1,\hat a_2)$ basis?

5. Show that the complex combinations

$$\hat a_{\mathbf k} = \frac{1}{\sqrt{2}} (\hat a_{1\mathbf k}+i\hat a_{2\mathbf k}), \qquad \hat c_{\mathbf k} = \frac{1}{\sqrt{2}} (\hat a_{1\mathbf k}-i\hat a_{2\mathbf k})$$

diagonalize the charge operator.

6. Explain why $\hat a^{\dagger}$ and $\hat c^{\dagger}$ create particles of opposite charge.

7. Write the complex scalar field $\phi$ in terms of $\phi_1$ and $\phi_2$. Why is $\phi\neq\phi^{\dagger}$?

8. Explain the meaning of

$$\hat\phi\sim \hat a+\hat c^{\dagger}.$$

9. Why are particles created from the same field indistinguishable?

10. Use

$$[\hat a_{k_1},\hat a_{k_2}]=0$$

to show that bosonic two-particle wavefunctions are symmetric.

11. Use

$$\{\hat a_{k_1},\hat a_{k_2}\}=0$$

to show that fermionic two-particle wavefunctions are antisymmetric.

12. Derive Pauli exclusion from

$$\{\hat a_k^{\dagger},\hat a_k^{\dagger}\}=0.$$

13. Why do fermionic creation and annihilation operators obey anti-commutators while the Heisenberg equation still uses a commutator?

14. State the canonical commutation relations for bosons and the canonical anti-commutation relations for fermions.

15. Explain why bosonic and fermionic operators are taken to commute with each other.