Dirac Quantum Field Theory

Introduction

The Dirac equation is a relativistically covariant first-order field equation for a charged spinor field. But by itself it is not yet the full quantum field theory of spin-$\tfrac12$ matter.

In quantum field theory, the spinor field $\psi(\mathbf r,t)$ is not the state vector of one electron. It is a field. After quantization, it becomes an operator-valued field acting on a Fock space containing states with arbitrary numbers of particles and antiparticles.

So the real task is not only to write the Dirac equation. The real task is to construct the Hamiltonian operator whose Heisenberg equation of motion gives the Dirac equation:

$$i\hbar \frac{\partial \hat\psi}{\partial t} = [\hat\psi,\hat H].$$

To build Dirac quantum field theory, we need several steps:

1. construct Lorentz scalars and four-vectors from spinors,
2. write a Lorentz-scalar Lagrangian density,
3. find the canonical momentum field,
4. derive the Hamiltonian,
5. diagonalize the Hamiltonian using normal modes,
6. decide whether the modes must be quantized bosonically or fermionically,
7. derive the momentum and charge operators.

The main result is sharp: a spin-$\tfrac12$ Dirac field cannot be quantized consistently with bosonic commutation relations. Bosonic quantization makes the Hamiltonian unbounded from below. Fermionic anti-commutation relations fix the sign problem and give a stable theory. This is one route to the spin-statistics connection for the Dirac field.

The quantized field has two kinds of excitations. The $a$-operators create particles, and the $c$-operators create antiparticles. Both have positive energy, but they carry opposite charge.

Learning Objectives

• Explain why the Dirac equation alone is not yet a full quantum field theory.
• Construct Lorentz scalars and four-vectors from left- and right-handed two-spinors.
• Understand why $\bar\psi = \psi^\dagger \gamma^0$ is needed.
• Write the Dirac Lagrangian density.
• Derive the canonical momentum field of the Dirac field.
• Derive the Hamiltonian from the Dirac Lagrangian.
• Expand the Dirac field in normal modes using eigenspinors $u_\pm(k)$ and $v_\pm(k)$.
• Understand the momentum-space Dirac Hamiltonian as a $4\times4$ matrix eigenvalue problem.
• Explain why bosonic quantization gives an unstable Hamiltonian.
• Explain why fermionic quantization gives a Hamiltonian bounded from below.
• Derive the momentum and charge operators of the free Dirac field.

Prerequisite Knowledge

• Left-handed and right-handed Lorentz two-spinors
• Dirac equation in two-spinor and four-spinor notation
• Gamma matrices
• Pauli matrices
• Lorentz covariance
• Canonical field theory
• Noether charges
• Creation and annihilation operators
• Bosonic commutators and fermionic anti-commutators

1. Scalars from two-spinors

Let $\eta$ be a left-handed Lorentz two-spinor and let $\xi$ be a right-handed Lorentz two-spinor.

Under an infinitesimal rotation, both transform with the same spin-$\tfrac12$ matrix:

$$\eta \to \eta' = \left(1+\frac{i}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right)\eta,$$ $$\xi \to \xi' = \left(1+\frac{i}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right)\xi.$$

Therefore the mixed product

$$\xi^\dagger \eta$$

is invariant under rotations:

$$\xi^\dagger\eta \to \xi'^\dagger\eta' = \xi^\dagger \left(1+\frac{i}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right)^\dagger \left(1+\frac{i}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right) \eta = \xi^\dagger\eta$$

to first order in $\boldsymbol\epsilon$.

Under boosts, left- and right-handed spinors transform oppositely:

$$\eta \to \eta'= \left(1+\frac{1}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right)\eta,$$ $$\xi \to \xi'= \left(1-\frac{1}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right)\xi.$$

So

$$\xi^\dagger\eta \to \xi'^\dagger\eta' = \xi^\dagger \left(1-\frac{1}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right)^\dagger \left(1+\frac{1}{2}\boldsymbol\epsilon\cdot\boldsymbol\sigma\right) \eta = \xi^\dagger\eta$$

again to first order. Thus

$$\boxed{\xi^\dagger\eta \text{ is a Lorentz scalar}.}$$

Since

$$\eta^\dagger\xi = (\xi^\dagger\eta)^*,$$

it is also a Lorentz scalar.

This is the first key structural point: Lorentz scalars are naturally made by pairing a left-handed spinor with a right-handed spinor.

2. Four-vectors from two-spinors

Let $v_\mu=(v_0,\mathbf v)$ be a covariant four-vector. The combinations

$$\tilde\xi := (v_0 + \boldsymbol\sigma\cdot\mathbf v)\eta$$

and

$$\tilde\eta := (v_0 - \boldsymbol\sigma\cdot\mathbf v)\xi$$

transform respectively as right-handed and left-handed spinors.

Since left-right scalar products are Lorentz scalars, the quantities

$$\eta^\dagger \tilde\xi = \eta^\dagger(v_0+\boldsymbol\sigma\cdot\mathbf v)\eta$$

and

$$\xi^\dagger \tilde\eta = \xi^\dagger(v_0-\boldsymbol\sigma\cdot\mathbf v)\xi$$

are Lorentz scalars. Expanding gives

$$\eta^\dagger(v_0+\boldsymbol\sigma\cdot\mathbf v)\eta = v_0\eta^\dagger\eta + \mathbf v\cdot(\eta^\dagger\boldsymbol\sigma\eta),$$ $$\xi^\dagger(v_0-\boldsymbol\sigma\cdot\mathbf v)\xi = v_0\xi^\dagger\xi - \mathbf v\cdot(\xi^\dagger\boldsymbol\sigma\xi).$$

If the contraction of an object with an arbitrary covariant four-vector is a scalar, then that object is a contravariant four-vector. Therefore

$$\boxed{ A^\mu=(A^0,\mathbf A) = (\eta^\dagger\eta,\eta^\dagger\boldsymbol\sigma\eta) }$$

and

$$\boxed{ B^\mu=(B^0,\mathbf B) = (\xi^\dagger\xi,-\xi^\dagger\boldsymbol\sigma\xi) }$$

are contravariant Lorentz four-vectors.

This construction will become the four-spinor current $\bar\psi\gamma^\mu\psi$.

3. Four-spinor notation

A charged Dirac spinor contains one left-handed and one right-handed two-spinor:

$$\psi(\mathbf r,t)= \begin{pmatrix} \eta\\ \xi \end{pmatrix} = \begin{pmatrix} \eta_1\\ \eta_2\\ \xi_1\\ \xi_2 \end{pmatrix}.$$

Use the gamma matrices

$$\gamma^0= \begin{pmatrix} 0_{2\times2} & I_{2\times2}\\ I_{2\times2} & 0_{2\times2} \end{pmatrix},$$ $$\gamma^k= \begin{pmatrix} 0_{2\times2} & -\sigma_k\\ \sigma_k & 0_{2\times2} \end{pmatrix}, \qquad k=1,2,3.$$

Then direct matrix multiplication gives

$$\psi^\dagger\gamma^0\psi = \xi^\dagger\eta+\eta^\dagger\xi.$$

This is a real Lorentz scalar. Similarly,

$$\psi^\dagger\gamma^0\gamma^\mu\psi = (\eta^\dagger\eta+\xi^\dagger\xi, \eta^\dagger\boldsymbol\sigma\eta-\xi^\dagger\boldsymbol\sigma\xi).$$

This is a contravariant Lorentz four-vector built symmetrically from the left- and right-handed parts.

The insertion of $\gamma^0$ is essential. For Lorentz spinors, time and space are not treated like four Euclidean directions. The object $\psi^\dagger\psi$ is not a Lorentz scalar. The correct scalar uses $\gamma^0$.

This motivates the standard Dirac adjoint:

$$\boxed{\bar\psi := \psi^\dagger\gamma^0.}$$

With this notation,

$$\boxed{\bar\psi\psi = \psi^\dagger\gamma^0\psi}$$

is a scalar, and

$$\boxed{\bar\psi\gamma^\mu\psi=\psi^\dagger\gamma^0\gamma^\mu\psi}$$

is a four-vector.

Another common abbreviation is Feynman slash notation. For any four-vector $A_\mu$,

$$\boxed{\slashed A := A_\mu\gamma^\mu.}$$

So, for example,

$$\slashed\partial := \gamma^\mu\partial_\mu.$$

4. Why the Dirac equation is not the full quantum theory

The Dirac equation in four-spinor notation is

$$i\hbar\gamma^\mu\partial_\mu\psi=Mc\psi.$$

This is a relativistically covariant field equation, but it is not yet the full quantum field theory.

In the Schrödinger picture, the Hamiltonian evolves the quantum state:

$$i\hbar\frac{d}{dt}|\Psi\rangle=\hat H|\Psi\rangle.$$

Here $|\Psi\rangle$ is a state vector in the full Hilbert space of the theory. It is not the spinor field $\psi(\mathbf r,t)$.

In the Heisenberg picture, operators evolve according to

$$i\hbar\frac{\partial \hat A}{\partial t}=[\hat A,\hat H].$$

For the Dirac field operator, this equation should reproduce the Dirac equation:

$$i\hbar\frac{\partial\hat\psi}{\partial t} = [\hat\psi,\hat H] = c\gamma^0(-i\hbar\boldsymbol\gamma\cdot\boldsymbol\nabla+Mc)\hat\psi.$$

So the central question is:

$$\boxed{\text{What is }\hat H\text{, and what is }\hat\psi\text{ as an operator?}}$$

5. The Dirac Lagrangian density

A relativistic field theory must have a Lorentz-scalar Lagrangian density. The natural scalar terms built from $\psi$ are

$$\bar\psi\gamma^\mu\partial_\mu\psi$$

and

$$\bar\psi\psi.$$

After a convenient rescaling of the field, the free Dirac Lagrangian density can be written as

$$\boxed{ \mathcal L = i\hbar c\,\bar\psi\gamma^\mu\partial_\mu\psi - Mc^2\bar\psi\psi. }$$

Using $\bar\psi=\psi^\dagger\gamma^0$, this is

$$\mathcal L = i\hbar c\,\psi^\dagger\gamma^0\gamma^\mu\partial_\mu\psi - Mc^2\psi^\dagger\gamma^0\psi.$$

The Euler-Lagrange equation of this Lagrangian is the Dirac equation.

6. Canonical momentum field

The canonical momentum conjugate to $\psi$ is

$$\Pi_\psi = \frac{\partial\mathcal L}{\partial(\partial_t\psi)}.$$

Since

$$\partial_0=\frac{1}{c}\partial_t,$$

the time-derivative part of the Lagrangian is

$$i\hbar c\,\psi^\dagger\gamma^0\gamma^0\partial_0\psi = i\hbar\psi^\dagger\partial_t\psi.$$

Therefore

$$\boxed{ \Pi_\psi=i\hbar\psi^\dagger. }$$

This first-order structure is different from scalar field theory, where the canonical momentum is proportional to a time derivative of the field. For the Dirac field, the canonical momentum is directly proportional to $\psi^\dagger$.

7. The Dirac Hamiltonian

The Hamiltonian is obtained by the Legendre transform

$$H=\int d^3r\left(\Pi_\psi\frac{\partial\psi}{\partial t}-\mathcal L\right).$$

Substituting the Dirac Lagrangian gives

$$\boxed{ H = \int d^3r \left( -i\hbar c\,\psi^\dagger\gamma^0\boldsymbol\gamma\cdot\boldsymbol\nabla\psi + Mc^2\psi^\dagger\gamma^0\psi \right). }$$

Equivalently, using $\Pi=i\hbar\psi^\dagger$,

$$\boxed{ H = \int d^3r \left( c\Pi\gamma^0\boldsymbol\gamma\cdot\boldsymbol\nabla\psi - i\frac{Mc^2}{\hbar}\Pi\gamma^0\psi \right). }$$

This Hamiltonian gives the Dirac equation as the canonical equation of motion for $\psi$.

8. Normal-mode expansion of the Dirac field

To quantize the theory, we need to diagonalize the Hamiltonian.

For a scalar Klein-Gordon field, Fourier transformation almost immediately diagonalizes the free Hamiltonian. For a Dirac field, Fourier transformation alone is not enough because $\psi$ has four spinor components.

We use a momentum-space expansion of the form

$$\boxed{ \psi(\mathbf r) = \sum_{s=\pm}\frac{1}{(2\pi)^{3/2}} \int d^3k\,e^{i\mathbf k\cdot\mathbf r} \left( u_s(\mathbf k)a_s(\mathbf k,t)+v_s(\mathbf k)c_s^*(-\mathbf k,t)\right). }$$

Many texts write the positive-energy spinors as $u_s(\mathbf k)$. Using that notation, the same expansion is

$$\psi(\mathbf r) = \sum_{s=\pm}\frac{1}{(2\pi)^{3/2}} \int d^3k\,e^{i\mathbf k\cdot\mathbf r} \left(u_s(\mathbf k)a_s(\mathbf k,t)+v_s(\mathbf k)c_s^*(-\mathbf k,t)\right).$$

Here

$$\psi= \begin{pmatrix} \psi_1\\\psi_2\\\psi_3\\\psi_4 \end{pmatrix}, \qquad u_s= \begin{pmatrix} u_{s1}\\u_{s2}\\u_{s3}\\u_{s4} \end{pmatrix}, \qquad v_s= \begin{pmatrix} v_{s1}\\v_{s2}\\v_{s3}\\v_{s4} \end{pmatrix}.$$

The goal is to choose $u_s$ and $v_s$ so that the Hamiltonian becomes diagonal in the mode amplitudes. Ideally, it would have the form

$$H=\sum_{s=\pm}\int d^3k\,E(k) \left(|a_s(\mathbf k)|^2+|c_s(\mathbf k)|^2\right).$$

The actual sign structure will be more subtle and will force fermionic quantization.

9. Momentum-space Hamiltonian

The Hamiltonian is

$$H = \int d^3r \left( -i\hbar c\psi^\dagger\gamma^0\boldsymbol\gamma\cdot\boldsymbol\nabla\psi + Mc^2\psi^\dagger\gamma^0\psi \right).$$

In Fourier space, write the spinor amplitude as $\Psi(\mathbf k)$. Then

$$H = \int d^3k \left( \hbar c\Psi^\dagger\gamma^0\boldsymbol\gamma\cdot\mathbf k\Psi + Mc^2\Psi^\dagger\gamma^0\Psi \right).$$

So

$$\boxed{ H = \int d^3k\, \Psi^\dagger\gamma^0 (\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2) \Psi. }$$

For each momentum $\mathbf k$, diagonalization reduces to a $4\times4$ matrix eigenvalue problem:

$$M_{4\times4}(\mathbf k) := \gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2).$$

The eigenspinors are defined by

$$\boxed{ \gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2)u_s(\mathbf k) = E(k)u_s(\mathbf k), }$$ $$\boxed{ \gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2)v_s(\mathbf k) = -E(k)v_s(\mathbf k). }$$

The energy is

$$\boxed{ E(k)=\sqrt{(Mc^2)^2+\hbar^2c^2k^2} = \sqrt{(Mc^2)^2+c^2p^2}, \qquad p=\hbar k. }$$

The matrix has two degenerate positive eigenvalues $+E(k)$ and two degenerate negative eigenvalues $-E(k)$.

The eigenspinors are chosen orthonormal:

$$u_{s'}^\dagger u_s=\delta_{s's}, \qquad v_{s'}^\dagger v_s=\delta_{s's},$$ $$u_{s'}^\dagger v_s=0, \qquad v_{s'}^\dagger u_s=0.$$

Equivalently,

$$u_{s'}^\dagger u_s=v_{s'}^\dagger v_s=\delta_{s's}, \qquad u_{s'}^\dagger v_s=v_{s'}^\dagger u_s=0.$$

10. Rest-frame eigenspinors

For $\mathbf k=0$, the eigenvalue problem is simple because

$$E(0)=Mc^2.$$

A convenient normalized basis is

$$\boxed{ u_+(0)=\frac{1}{\sqrt2} \begin{pmatrix} 1\\0\\1\\0 \end{pmatrix}}$$ $$\boxed{ u_-(0)=\frac{1}{\sqrt2} \begin{pmatrix} 0\\1\\0\\1 \end{pmatrix}}$$ $$\boxed{v_+(0)=\frac{1}{\sqrt2} \begin{pmatrix} 1\\0\\-1\\0 \end{pmatrix}}$$ $$\boxed{v_-(0)=\frac{1}{\sqrt2} \begin{pmatrix} 0\\1\\0\\-1 \end{pmatrix}}.$$

The $u$-spinors span the positive-energy subspace at rest, while the $v$-spinors span the negative-energy subspace at rest.

11. Eigenspinors at nonzero momentum

For $\mathbf k\neq0$, the momentum-dependent eigenspinors can be obtained by boosting the rest-frame spinors.

The result is

$$\boxed{ u_s(\mathbf k) = \frac12 \left( \sqrt{\frac{E(k)}{Mc^2}+1} - \sqrt{\frac{E(k)}{Mc^2}-1}\, \frac{\mathbf k\cdot\boldsymbol\gamma}{k} \right) u_s(0). }$$

and

$$\boxed{v_s(\mathbf k) = \frac12 \left( \sqrt{\frac{E(k)}{Mc^2}+1} + \sqrt{\frac{E(k)}{Mc^2}-1}\, \frac{\mathbf k\cdot\boldsymbol\gamma}{k} \right)v_s(0). }$$

The square roots are important. They are what make the boost normalization work correctly.

These spinors satisfy

$$\gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2)u_s(\mathbf k) = E(k)u_s(\mathbf k),$$ $$\gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2)v_s(\mathbf k) = -E(k)v_s(\mathbf k).$$

Useful identities for checking this are

$$E(k)^2=(Mc^2)^2+\hbar^2c^2k^2,$$ $$\hbar ck = Mc^2 \sqrt{\frac{E(k)}{Mc^2}+1} \sqrt{\frac{E(k)}{Mc^2}-1},$$ $$\gamma^i\gamma^j=-\delta^{ij}, \qquad \gamma^0\gamma^i=-\gamma^i\gamma^0.$$

12. The diagonalized classical Hamiltonian

Substituting the normal-mode expansion into the Hamiltonian gives

$$H = \int d^3k\, \Psi^\dagger\gamma^0 (\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2) \Psi.$$

Using the positive- and negative-energy eigenspinors gives

$$\boxed{ H = \sum_{s=\pm}\int d^3k\,E(k) \left(a_s^*(\mathbf k)a_s(\mathbf k)-c_s(\mathbf k)c_s^*(\mathbf k)\right). }$$

The minus sign in the $c$-sector is the crucial problem. It is not a harmless convention. It determines which quantization rule gives a stable Hamiltonian.

13. Why bosonic quantization fails

Suppose we quantize bosonically:

$$a_s\to \hat a_s, \qquad a_s^*\to \hat a_s^\dagger,$$ $$c_s\to \hat c_s, \qquad c_s^*\to \hat c_s^\dagger,$$

with canonical commutation relations.

Then the Hamiltonian contains

$$-E(k)\hat c_s(\mathbf k)\hat c_s^\dagger(\mathbf k).$$

For bosons,

$$\hat c\hat c^\dagger=\hat c^\dagger\hat c+1.$$

So this term becomes

$$-E(k)(\hat c^\dagger\hat c+1).$$

The energy decreases without bound as more $c$-quanta are added. Therefore bosonic quantization gives a Hamiltonian that is unbounded from below.

$$\boxed{\text{Bosonic quantization of the Dirac field is unstable.}}$$

14. Fermionic quantization fixes the Hamiltonian

Now quantize fermionically, using anti-commutation relations. For fermions,

$$\{\hat c,\hat c^\dagger\}=1,$$

so

$$\hat c\hat c^\dagger=1-\hat c^\dagger\hat c.$$

Therefore

$$-\hat c\hat c^\dagger = \hat c^\dagger\hat c-1.$$

The Hamiltonian becomes

$$\boxed{ \hat H = \sum_{s=\pm}\int d^3k\,E(k) \left( \hat a_s^\dagger(\mathbf k)\hat a_s(\mathbf k) + \hat c_s^\dagger(\mathbf k)\hat c_s(\mathbf k) \right) -E_0. }$$

The constant $E_0$ is the vacuum-energy offset. It can be removed by normal ordering when appropriate.

The important physical point is that the Hamiltonian is now bounded from below. Both $a$-quanta and $c$-quanta have positive energy.

$$\boxed{\text{Spin-}\tfrac12\text{ Dirac fields must be quantized fermionically.}}$$

This is the stability argument behind the spin-statistics connection for the Dirac field.

15. Momentum operator

The momentum comes from spatial translation symmetry:

$$\psi(\mathbf r,t)\to\psi(\mathbf r-\mathbf a,t).$$

The infinitesimal generator is

$$G=-\boldsymbol\nabla\psi.$$

The corresponding Noether momentum is

$$\mathbf P = \int d^3r\,\Pi G.$$

Using $\Pi=i\hbar\psi^\dagger$, this becomes

$$\mathbf P = \int d^3r\,i\hbar\psi^\dagger\boldsymbol\nabla\psi.$$

In momentum space,

$$\mathbf P = \int d^3k\,\hbar\mathbf k\,\Psi^\dagger\Psi.$$

After fermionic quantization,

$$\hat{\mathbf P} = \sum_{s=\pm}\int d^3k\,\hbar\mathbf k \left( \hat a_s^\dagger(\mathbf k)\hat a_s(\mathbf k) + \hat c_s(-\mathbf k)\hat c_s^\dagger(-\mathbf k) \right).$$

Reordering the fermionic operators and relabeling $\mathbf k\to-\mathbf k$ gives the standard form

$$\boxed{ \hat{\mathbf P} = \sum_{s=\pm}\int d^3k\,\hbar\mathbf k \left( \hat a_s^\dagger(\mathbf k)\hat a_s(\mathbf k) + \hat c_s^\dagger(\mathbf k)\hat c_s(\mathbf k) \right). }$$

Thus both particle and antiparticle excitations carry ordinary positive momentum $\hbar\mathbf k$.

16. Charge operator

The charge comes from the global phase symmetry

$$\psi(\mathbf r,t)\to e^{-i\theta}\psi(\mathbf r,t).$$

For an infinitesimal transformation, the generator is

$$G=-i\psi.$$

The Noether charge is

$$Q=\int d^3r\,\Pi G.$$

Using $\Pi=i\hbar\psi^\dagger$,

$$Q = \int d^3r\,(i\hbar\psi^\dagger)(-i\psi) = \hbar\int d^3r\,\psi^\dagger\psi.$$

Equivalently,

$$Q=\int d^3r\,\rho(\mathbf r,t), \qquad \rho=\hbar\psi^\dagger\psi.$$

After quantization, the charge operator becomes

$$\boxed{ \hat Q = \hbar\sum_{s=\pm}\int d^3k \left( \hat a_s^\dagger(\mathbf k)\hat a_s(\mathbf k) - \hat c_s^\dagger(\mathbf k)\hat c_s(\mathbf k) \right). }$$

The $a$-quanta and $c$-quanta therefore carry opposite charge.

This is the same structural pattern as the charged scalar field, but now for a spin-$\tfrac12$ fermionic field:

$$\text{particle sector: } a_s^\dagger, \qquad \text{antiparticle sector: } c_s^\dagger.$$

Both have positive energy, but their charges differ by a sign.

17. Interpretation of the quantized Dirac field

The quantized Dirac field has the schematic form

$$\hat\psi(\mathbf r,t) \sim \sum_s\int d^3k \left( u_s(\mathbf k)\hat a_s(\mathbf k,t) + v_s(\mathbf k)\hat c_s^\dagger(-\mathbf k,t) \right)e^{i\mathbf k\cdot\mathbf r}.$$

So the field operator has two roles:

$$\boxed{\hat\psi \text{ destroys a particle or creates an antiparticle}.}$$

For the electron field, this means that $\hat\psi$ destroys electrons and creates positrons. The adjoint field performs the opposite operations.

This is why the Dirac equation should not be interpreted as a single-particle Schrödinger equation. In quantum field theory, it is the field equation obeyed by an operator-valued spinor field.

Worked Example 1: Why $\bar\psi\psi$ is the scalar mass term

A naïve expression like

$$\psi^\dagger\psi$$

is positive-looking, but it is not a Lorentz scalar. It treats the spinor components as if boosts were unitary rotations, which they are not.

The correct scalar is

$$\bar\psi\psi=\psi^\dagger\gamma^0\psi.$$

Writing $\psi=(\eta,\xi)^T$, this becomes

$$\bar\psi\psi = \xi^\dagger\eta+\eta^\dagger\xi.$$

This couples left-handed and right-handed spinors, exactly as a Lorentz scalar should. That is why the Dirac mass term is

$$-Mc^2\bar\psi\psi.$$

Worked Example 2: Why fermionic quantization fixes the sign problem

The diagonalized classical Hamiltonian contains

$$E(k)(a^*a-cc^*).$$

Bosonic quantization gives

$$E(k)(\hat a^\dagger\hat a-\hat c\hat c^\dagger) = E(k)(\hat a^\dagger\hat a-\hat c^\dagger\hat c-1),$$

which becomes arbitrarily negative as more $c$-quanta are added.

Fermionic quantization gives

$$\hat c\hat c^\dagger=1-\hat c^\dagger\hat c,$$

so

$$E(k)(\hat a^\dagger\hat a-\hat c\hat c^\dagger) = E(k)(\hat a^\dagger\hat a+\hat c^\dagger\hat c)-E(k).$$

The remaining negative term is only a constant vacuum-energy shift, not an instability. The excitation spectrum is positive.

Intuition

The Dirac equation becomes a proper quantum field theory only after the Hamiltonian is built and quantized.

The essential logic is:

• $\psi$ is a field, not a one-particle state.
• Lorentz covariance forces the use of $\bar\psi=\psi^\dagger\gamma^0$.
• The scalar Lagrangian gives a first-order Hamiltonian.
• Fourier transformation leaves a $4\times4$ spinor matrix to diagonalize.
• The matrix has positive- and negative-energy eigenspinors.
• The negative-energy sector produces a dangerous sign.
• Fermionic anti-commutation relations turn that sign into a stable antiparticle sector.

The result is a field whose excitations are fermions with positive energy. The two operator families create particles and antiparticles of opposite charge.

Common Mistakes

• Treating $\psi(\mathbf r,t)$ as the wavefunction of one electron.
• Forgetting that the QFT state vector lives in Fock space, not in spinor-component space.
• Using $\psi^\dagger\psi$ as a Lorentz scalar mass term.
• Forgetting the definition $\bar\psi=\psi^\dagger\gamma^0$.
• Thinking Fourier transformation alone diagonalizes the Dirac Hamiltonian.
• Ignoring the spinor eigenvalue problem at each momentum.
• Missing the negative sign in the $c$-sector of the diagonalized classical Hamiltonian.
• Thinking bosonic quantization is a harmless alternative.
• Forgetting that fermionic anti-commutation relations are what make the Hamiltonian bounded from below.
• Forgetting that particles and antiparticles have opposite charge but positive energy.

Short Summary

Dirac quantum field theory starts by constructing Lorentz scalars and four-vectors from spinors. A left-right product such as $\xi^\dagger\eta$ is a scalar, while combinations like

$$(\eta^\dagger\eta,\eta^\dagger\boldsymbol\sigma\eta)$$

and

$$(\xi^\dagger\xi,-\xi^\dagger\boldsymbol\sigma\xi)$$

are four-vectors. In four-spinor notation, this becomes the standard Dirac adjoint

$$\bar\psi=\psi^\dagger\gamma^0,$$

so that $\bar\psi\psi$ is a scalar and $\bar\psi\gamma^\mu\psi$ is a four-vector.

The free Dirac Lagrangian is

$$\mathcal L=i\hbar c\bar\psi\gamma^\mu\partial_\mu\psi-Mc^2\bar\psi\psi.$$

Its canonical momentum field is

$$\Pi_\psi=i\hbar\psi^\dagger,$$

and its Hamiltonian is

$$H=\int d^3r \left( -i\hbar c\psi^\dagger\gamma^0\boldsymbol\gamma\cdot\boldsymbol\nabla\psi +Mc^2\psi^\dagger\gamma^0\psi \right).$$

In momentum space, diagonalization requires solving

$$\gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2)u_s(\mathbf k)=E(k)u_s(\mathbf k),$$ $$\gamma^0(\hbar c\boldsymbol\gamma\cdot\mathbf k+Mc^2)v_s(\mathbf k)=-E(k)v_s(\mathbf k),$$

with

$$E(k)=\sqrt{(Mc^2)^2+\hbar^2c^2k^2}.$$

The diagonalized classical Hamiltonian contains

$$H=\sum_s\int d^3k\,E(k)(a_s^*a_s-c_sc_s^*).$$

Bosonic quantization makes this Hamiltonian unbounded from below. Fermionic quantization gives

$$\hat H = \sum_s\int d^3k\,E(k) (\hat a_s^\dagger\hat a_s+ \hat c_s^\dagger\hat c_s)-E_0,$$

which is bounded from below. The momentum and charge operators become

$$\hat{\mathbf P} = \sum_s\int d^3k\,\hbar\mathbf k (\hat a_s^\dagger\hat a_s+ \hat c_s^\dagger\hat c_s),$$ $$\hat Q = \hbar\sum_s\int d^3k (\hat a_s^\dagger\hat a_s- \hat c_s^\dagger\hat c_s).$$

Thus the quantized Dirac field has stable fermionic excitations, with particle and antiparticle sectors carrying opposite charge.

Practice Problems

1. Why is $\bar\psi\psi$ a Lorentz scalar while $\psi^\dagger\psi$ is not?

2. Show explicitly that $\xi^\dagger\eta$ is invariant under infinitesimal boosts.

3. Derive the four-vector

$$(\eta^\dagger\eta,\eta^\dagger\boldsymbol\sigma\eta)$$

from the scalar contraction with $v_\mu$.

4. Derive the canonical momentum field

$$\Pi_\psi=i\hbar\psi^\dagger$$

from the Dirac Lagrangian.

5. Starting from the Dirac Lagrangian, derive the Hamiltonian density.

6. Why is the Dirac Hamiltonian not diagonalized by Fourier transformation alone?

7. What matrix must be diagonalized at each momentum $\mathbf k$?

8. Verify that the rest-frame spinors $u_\pm(0)$ have eigenvalue $+Mc^2$, while $v_\pm(0)$ have eigenvalue $-Mc^2$.

9. Explain why the $c$-sector appears with a minus sign in the diagonalized classical Hamiltonian.

10. Show explicitly why bosonic quantization makes the Hamiltonian unbounded from below.

11. Show how fermionic anti-commutation relations change

$$-\hat c\hat c^\dagger$$

into

$$\hat c^\dagger\hat c-1.$$

12. Derive the momentum operator from translation symmetry.

13. Derive the charge operator from global phase symmetry.

14. Explain why the $a$- and $c$-quanta have opposite charge.

15. Explain why $\hat\psi$ destroys particles and creates antiparticles.