The Dirac Equation

Introduction

Relativistic spin- $\tfrac12$ fields are not built from one ordinary two-component spinor. Relativity distinguishes two different kinds of Lorentz 2-spinors:

• a left-handed spinor $\eta$, transforming as $(s_L,s_R)=(\tfrac12,0)$,
• a right-handed spinor $\xi$, transforming as $(s_L,s_R)=(0,\tfrac12)$.

Under ordinary spatial rotations, they behave the same way. Under Lorentz boosts, they transform oppositely. This difference is the core reason why the relativistic spin- $\tfrac12$ equation has to connect left-handed and right-handed spinors.

The goal is to construct a Lorentz-covariant first-order field equation for a charged spinor field. The result is the Dirac equation. In 2-spinor notation it appears as two coupled equations,

$$i\hbar(\partial_0+\vec\sigma\cdot\vec\nabla)\eta=Mc\xi, \qquad i\hbar(\partial_0-\vec\sigma\cdot\vec\nabla)\xi=Mc\eta,$$

where $\partial_0=c^{-1}\partial_t$. In 4-spinor notation the same equation becomes

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

The construction has three main steps. First, we review how left- and right-handed 2-spinors transform under rotations and boosts. Second, we show how to map between the two chiralities using either Majorana conjugation or first-order differential operators. Third, we use those maps to build the Majorana equation for a neutral spinor field and then the Dirac equation for a charged spinor field.

The final point is important: the Dirac equation is not just a relativistic Schrödinger equation for one electron. In quantum field theory, $\psi$ becomes an operator-valued field. Its quantized version destroys electrons and creates positrons. The one-particle interpretation is not enough.

Learning Objectives

• Write the transformation laws of left-handed and right-handed Lorentz 2-spinors under rotations and boosts.
• Explain why both chiralities transform identically under spatial rotations but oppositely under boosts.
• Check which spinor bilinears are invariant under rotations and boosts.
• Construct the Majorana conjugate $\mu=\sigma_2\eta^*$ and show that it flips handedness.
• Use Pauli matrix identities to understand chirality-changing spinor maps.
• Construct right-handed spinors from left-handed spinors using $i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)$.
• Construct left-handed spinors from right-handed spinors using $i\hbar(\partial_t-c\vec\sigma\cdot\vec\nabla)$.
• Derive the Majorana equation as the neutral spinor case.
• Show that the Majorana equation implies the Klein-Gordon equation.
• Explain why a charged spinor field requires two identical Majorana spinors combined into a complex field.
• Derive the Dirac equation in 2-spinor notation.
• Rewrite the 2-spinor equation using a 4-spinor and gamma matrices.
• Explain why the Dirac equation is a field equation, not a complete one-particle quantum theory by itself.

Prerequisite Knowledge

• Lorentz transformations and Lorentz covariance
• Left-handed and right-handed Lorentz 2-spinors
• Pauli matrices
• Complex conjugation and Hermitian conjugation
• Klein-Gordon equation
• Basic field-theory language: fields, particles, and operator-valued fields
• Einstein summation convention

1. Left-handed Lorentz 2-spinors

A left-handed Lorentz 2-spinor is a two-component complex field

$$\eta(\vec r,t)= \begin{pmatrix} \eta_1(\vec r,t)\\ \eta_2(\vec r,t) \end{pmatrix}.$$

It transforms in the Lorentz representation

$$(s_L,s_R)=\left(\frac12,0\right).$$

The Lorentz generators are decomposed into left- and right-handed parts:

$$\hat{\vec R}_L=\frac{\hat{\vec S}+i\hat{\vec T}}{2}, \qquad \hat{\vec R}_R=\frac{\hat{\vec S}-i\hat{\vec T}}{2}.$$

For a left-handed 2-spinor,

$$\hat{\vec R}_L=\frac{\hbar}{2}\vec\sigma, \qquad \hat{\vec R}_R=0.$$

Therefore

$$\hat{\vec S}=\hat{\vec R}_L+\hat{\vec R}_R =\frac{\hbar}{2}\vec\sigma,$$

and

$$\hat{\vec T}=-i(\hat{\vec R}_L-\hat{\vec R}_R) =-i\frac{\hbar}{2}\vec\sigma.$$

Here $\hat{\vec S}$ generates ordinary spatial rotations of spinor components, while $\hat{\vec T}$ generates boosts.

2. Left-handed spinor under an infinitesimal rotation

Consider an infinitesimal 3D rotation by angle $\epsilon$ around the axis $\vec n$. Define

$$\vec\epsilon:=\epsilon\vec n.$$

The spatial argument changes as

$$\vec r\to \vec r' = \vec r-\vec\epsilon\times\vec r, \qquad t\to t.$$

The spinor components transform with the spin generator $\hat{\vec S}$. Hence

$$\eta'(\vec r,t) = \left(1+\frac{i}{\hbar}\vec\epsilon\cdot\hat{\vec J}\right)\eta(\vec r,t),$$

where $\hat{\vec J}$ contains both orbital and spin contributions. Written as a field transformation, this becomes

$$\eta' = \left(1+\frac{i}{\hbar}\vec\epsilon\cdot\hat{\vec S}\right) \eta(\vec r',t).$$

Using $\hat{\vec S}=\frac{\hbar}{2}\vec\sigma$,

$$\eta' = \left(1+\frac{i}{2}\vec\epsilon\cdot\vec\sigma\right) \eta(\vec r-\vec\epsilon\times\vec r,t).$$

So under rotations, a left-handed Lorentz 2-spinor behaves like the familiar spin- $\tfrac12$ two-spinor.

3. Left-handed spinor under an infinitesimal boost

For an infinitesimal boost of parameter $\epsilon$ in the $\vec n$ direction, again define

$$\vec\epsilon:=\epsilon\vec n.$$

The spacetime argument transforms as

$$\vec r\to \vec r'=\vec r-\vec\epsilon\,ct, \qquad t\to t'=t+c^{-1}\vec\epsilon\cdot\vec r.$$

For a left-handed spinor,

$$\hat{\vec T}=-i\frac{\hbar}{2}\vec\sigma.$$

Therefore

$$\eta' = \left(1+\frac{i}{\hbar}\vec\epsilon\cdot\hat{\vec T}\right) \eta(\vec r',t') = \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right) \eta(\vec r-\vec\epsilon ct,\,t+c^{-1}\vec\epsilon\cdot\vec r).$$

This boost factor is not unitary:

$$\left(1+\frac12\vec\epsilon\cdot\vec\sigma\right)^ \dagger \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right) \neq I.$$

That is not a problem. Lorentz boosts here act on the finite-dimensional space of spinor field components, not as time-evolution operators in Hilbert space.

4. Right-handed Lorentz 2-spinors

A right-handed Lorentz 2-spinor is another two-component complex field

$$\xi(\vec r,t)= \begin{pmatrix} \xi_1(\vec r,t)\\ \xi_2(\vec r,t) \end{pmatrix},$$

transforming as

$$(s_L,s_R)=\left(0,\frac12\right).$$

Now the generators are

$$\hat{\vec R}_R=\frac{\hbar}{2}\vec\sigma, \qquad \hat{\vec R}_L=0.$$

Therefore

$$\hat{\vec S}=\frac{\hbar}{2}\vec\sigma, \qquad \hat{\vec T}=+i\frac{\hbar}{2}\vec\sigma.$$

So right-handed and left-handed spinors rotate in the same way, but boost in opposite ways.

Under an infinitesimal boost,

$$\xi' = \left(1-\frac12\vec\epsilon\cdot\vec\sigma\right) \xi(\vec r-\vec\epsilon ct,\,t+c^{-1}\vec\epsilon\cdot\vec r).$$

Thus:

$$\boxed{\text{Rotations are the same for }\eta\text{ and }\xi.}$$ $$\boxed{\text{Boosts act with opposite signs on }\eta\text{ and }\xi.}$$

This is the basic chiral structure behind the Dirac equation.

5. Rotation-invariant and Lorentz-invariant spinor products

The simplest scalar-looking combinations are

$$\eta^\dagger\eta, \qquad \xi^\dagger\xi, \qquad \xi^\dagger\eta.$$

Under a rotation,

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

Since the Pauli matrices are Hermitian,

$$\left(1+\frac{i}{2}\vec\epsilon\cdot\vec\sigma\right)^\dagger = \left(1-\frac{i}{2}\vec\epsilon\cdot\vec\sigma\right),$$

and to first order in $\epsilon$,

$$\eta'^\dagger\eta' = \eta^\dagger \left(1-\frac{i}{2}\vec\epsilon\cdot\vec\sigma\right) \left(1+\frac{i}{2}\vec\epsilon\cdot\vec\sigma\right) \eta = \eta^\dagger\eta.$$

Similarly,

$$\xi'^\dagger\xi'=\xi^\dagger\xi.$$

So the ordinary norms are invariant under spatial rotations.

Boosts are different. For a left-handed spinor,

$$\eta\to \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right)\eta.$$

Then

$$\eta'^\dagger\eta' = \eta^\dagger \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right)^\dagger \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right) \eta \neq \eta^\dagger\eta.$$

For a right-handed spinor,

$$\xi\to \left(1-\frac12\vec\epsilon\cdot\vec\sigma\right)\xi,$$

so

$$\xi'^\dagger\xi' = \xi^\dagger \left(1-\frac12\vec\epsilon\cdot\vec\sigma\right)^\dagger \left(1-\frac12\vec\epsilon\cdot\vec\sigma\right) \xi \neq \xi^\dagger\xi.$$

The mixed product behaves better:

$$\xi'^\dagger\eta' = \xi^\dagger \left(1-\frac12\vec\epsilon\cdot\vec\sigma\right)^\dagger \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right) \eta.$$

To first order,

$$\left(1-\frac12\vec\epsilon\cdot\vec\sigma\right)^\dagger \left(1+\frac12\vec\epsilon\cdot\vec\sigma\right) = 1+O(\epsilon^2),$$

therefore

$$\xi'^\dagger\eta'=\xi^\dagger\eta.$$

So $\eta^\dagger\eta$ and $\xi^\dagger\xi$ are rotation-invariant but not Lorentz-invariant. The mixed left-right bilinear $\xi^\dagger\eta$ is invariant under both infinitesimal rotations and boosts. This already hints that relativistic spinor equations naturally couple left-handed and right-handed spinors.

6. Majorana conjugation: making a new spinor from an old one

Start with a left-handed spinor $\eta$. Define

$$\mu:=\sigma_2\eta^*.$$

Using

$$\sigma_2= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix},$$

we get explicitly

$$\begin{pmatrix} \mu_1\\ \mu_2 \end{pmatrix} = \sigma_2 \begin{pmatrix} \eta_1^*\\ \eta_2^* \end{pmatrix} = \begin{pmatrix} -i\eta_2^*\\ i\eta_1^* \end{pmatrix} = -i \begin{pmatrix} \eta_2^*\\ -\eta_1^* \end{pmatrix}.$$

For a quantum field, complex conjugation is replaced by Hermitian conjugation of the field components:

$$\begin{pmatrix} \hat\mu_1\\ \hat\mu_2 \end{pmatrix} = -i \begin{pmatrix} \hat\eta_2^\dagger\\ -\hat\eta_1^\dagger \end{pmatrix}.$$

The key Pauli matrix identities are

$$\sigma_2\sigma_1^*=\sigma_2\sigma_1=-\sigma_1\sigma_2,$$ $$\sigma_2\sigma_2^*=-\sigma_2\sigma_2,$$ $$\sigma_2\sigma_3^*=\sigma_2\sigma_3=-\sigma_3\sigma_2.$$

Together,

$$\boxed{\sigma_2\sigma_k^*=-\sigma_k\sigma_2, \qquad k=1,2,3.}$$

Under a rotation,

$$\eta\to \left(1+\frac{i}{2}\epsilon_n\sigma_n\right)\eta.$$

Then

$$\mu\to \mu' = \sigma_2 \left(1+\frac{i}{2}\epsilon_n\sigma_n\right)^* \eta^*.$$

Using the identity above,

$$\mu' = \left(1+\frac{i}{2}\epsilon_n\sigma_n\right)\mu.$$

So $\mu$ is a valid 2-spinor under rotations.

Under a boost, a left-handed spinor transforms as

$$\eta\to \left(1+\frac12\epsilon_n\sigma_n\right)\eta.$$

Therefore

$$\mu\to \mu' = \sigma_2 \left(1+\frac12\epsilon_n\sigma_n\right)^* \eta^* = \left(1-\frac12\epsilon_n\sigma_n\right)\mu.$$

This is the right-handed boost rule. Therefore

$$\boxed{\mu=\sigma_2\eta^*\text{ is right-handed if }\eta\text{ is left-handed.}}$$

Similarly, if $\xi$ is right-handed, then

$$\mu:=\sigma_2\xi^*$$

is left-handed. This operation is called Majorana conjugation. It is especially important for neutral spinor fields.

7. Differential maps between left- and right-handed spinors

There is a second, more important way to build spinors of the opposite handedness.

Start with a left-handed spinor $\eta$, and define

$$\tilde\xi := i\hbar\left(\frac{\partial}{\partial t}+c\vec\sigma\cdot\vec\nabla\right)\eta.$$

Equivalently,

$$\tilde\xi=i\hbar(\partial_t+c\sigma_k\partial_k)\eta.$$

In components,

$$\begin{pmatrix} \tilde\xi_1\\ \tilde\xi_2 \end{pmatrix} = i\hbar \begin{pmatrix} (\partial_t+c\partial_z)\eta_1+c(\partial_x-i\partial_y)\eta_2\\ (\partial_t-c\partial_z)\eta_2+c(\partial_x+i\partial_y)\eta_1 \end{pmatrix}.$$

The Pauli identity

$$\boxed{\sigma_k\sigma_l=\delta_{kl}+i\epsilon_{klm}\sigma_m}$$

ensures that this object transforms as a spinor under rotations. Under boosts, the derivative terms and the spinor transformation combine so that

$$\tilde\xi\to \tilde\xi' = \left(1-\frac12\epsilon_n\sigma_n\right)\tilde\xi.$$

Thus $\tilde\xi$ is right-handed.

So

$$\boxed{ \tilde\xi:=i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)\eta \quad\text{is right-handed if }\eta\text{ is left-handed.} }$$

Similarly, if $\xi$ is right-handed, define

$$\tilde\eta:=i\hbar\left(\frac{\partial}{\partial t}-c\vec\sigma\cdot\vec\nabla\right)\xi.$$

Then $\tilde\eta$ is left-handed:

$$\boxed{ \tilde\eta:=i\hbar(\partial_t-c\vec\sigma\cdot\vec\nabla)\xi \quad\text{is left-handed if }\xi\text{ is right-handed.} }$$

The signs are essential:

$$\eta\text{ left-handed} \quad\xrightarrow{\ i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)\ } \quad \tilde\xi\text{ right-handed},$$ $$\xi\text{ right-handed} \quad\xrightarrow{\ i\hbar(\partial_t-c\vec\sigma\cdot\vec\nabla)\ } \quad \tilde\eta\text{ left-handed}.$$

8. Lorentz-covariant equations connecting $L$ and $R$ spinors

Because the derivative maps produce spinors of the opposite handedness, equations of the form

$$\lambda\xi = i\hbar\left(\frac{\partial}{\partial t}+c\vec\sigma\cdot\vec\nabla\right)\eta,$$ $$\lambda\eta = i\hbar\left(\frac{\partial}{\partial t}-c\vec\sigma\cdot\vec\nabla\right)\xi$$

are Lorentz-covariant, provided $\lambda$ is a Lorentz scalar constant. If such equations are valid in one inertial frame, they are valid in the same form in every inertial frame.

This is the structural skeleton of the Dirac equation. The rest of the construction determines what the constant $\lambda$ is and how to interpret $\eta$ and $\xi$.

9. The Majorana equation: neutral spinor field

For a neutral spinor field, one can close the equation using the Majorana conjugate. Start with a left-handed spinor $\eta$. Since $\sigma_2\eta^*$ is right-handed, we can impose

$$\boxed{ i\hbar\left(\frac{\partial}{\partial t}+c\vec\sigma\cdot\vec\nabla\right)\eta = Mc^2\sigma_2\eta^*. }$$

This is the Majorana equation in 2-spinor form.

It also implies the conjugate equation

$$\boxed{ i\hbar\left(\frac{\partial}{\partial t}-c\vec\sigma\cdot\vec\nabla\right)\sigma_2\eta^* = Mc^2\eta. }$$

So a single left-handed spinor and its Majorana conjugate form a closed first-order system. This is the neutral-particle version of the Dirac construction, analogous to how a real Klein-Gordon field is the neutral version of a complex Klein-Gordon field.

10. The Majorana equation implies Klein-Gordon

Apply the opposite differential operator to the Majorana equation. Starting from

$$i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)\eta = Mc^2\sigma_2\eta^*,$$

and using

$$i\hbar(\partial_t-c\vec\sigma\cdot\vec\nabla)\sigma_2\eta^* = Mc^2\eta,$$

we obtain

$$-\hbar^2 (\partial_t-c\vec\sigma\cdot\vec\nabla) (\partial_t+c\vec\sigma\cdot\vec\nabla) \eta = (Mc^2)^2\eta.$$

The crucial identity is

$$(i\hbar)^2 (\partial_t\pm c\sigma_k\partial_k) (\partial_t\mp c\sigma_l\partial_l) = (\hbar c)^2\Box,$$

where

$$\Box=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$

with the sign convention used here.

Therefore

$$\boxed{ \Box\eta=\left(\frac{Mc}{\hbar}\right)^2\eta. }$$

So the Majorana equation implies the Klein-Gordon equation for each component of $\eta$.

The converse is not true. Majorana is stricter than Klein-Gordon because it is a first-order equation relating the real and imaginary parts of the two spinor components. Majorana solutions form a subset of the complex two-field Klein-Gordon solutions.

11. Why charged spinors require two Majorana spinors

A single Majorana spinor is neutral. To construct a charged spinor field, we need a continuous internal phase symmetry, just as for the charged Klein-Gordon field.

Take two identical Majorana spinors,

$$\eta_A, \qquad \eta_B.$$

They can be rotated into one another by an internal $SO(2)$ transformation:

$$\eta_A\to \eta_A\cos\theta-\eta_B\sin\theta,$$ $$\eta_B\to \eta_B\cos\theta+\eta_A\sin\theta.$$

Now define the complex combination

$$\boxed{ \eta:=\frac{\eta_A+i\eta_B}{\sqrt2}. }$$

Under the internal rotation,

$$\eta\to e^{i\theta}\eta.$$

This is the same pattern as a charged scalar field: two real neutral fields combine into one complex charged field.

The complex conjugate is

$$\eta^*=\frac{\eta_A^*-i\eta_B^*}{\sqrt2}.$$

This sign will matter in the next step.

12. From two Majorana equations to the charged spinor equations

Each Majorana spinor satisfies

$$i\hbar\left(\frac{\partial}{\partial t}+c\vec\sigma\cdot\vec\nabla\right)\eta_{A,B} = Mc^2\sigma_2\eta_{A,B}^*.$$

Combining $\eta_A$ and $\eta_B$ gives

$$i\hbar\left(\frac{\partial}{\partial t}+c\vec\sigma\cdot\vec\nabla\right)\eta = Mc^2\sigma_2\frac{\eta_A^*+i\eta_B^*}{\sqrt2}.$$

This right-hand side is not $Mc^2\sigma_2\eta^*$, because

$$\sigma_2\eta^* = \sigma_2\frac{\eta_A^*-i\eta_B^*}{\sqrt2}.$$

The sign of the $i\eta_B^*$ term is different.

Define the right-handed spinor

$$\boxed{ \xi:=\sigma_2\frac{\eta_A^*+i\eta_B^*}{\sqrt2}. }$$

Then

$$\xi\neq \sigma_2\eta^*.$$

This is why the charged spinor equation is no longer the Majorana equation.

The two coupled equations become

$$i\hbar\left(\frac{\partial}{\partial t}+c\vec\sigma\cdot\vec\nabla\right)\eta = Mc^2\xi,$$ $$i\hbar\left(\frac{\partial}{\partial t}-c\vec\sigma\cdot\vec\nabla\right)\xi = Mc^2\eta.$$

Introducing

$$\partial_0:=c^{-1}\partial_t,$$

we can write them as

$$\boxed{ i\hbar(\partial_0+\vec\sigma\cdot\vec\nabla)\eta=Mc\xi, }$$ $$\boxed{ i\hbar(\partial_0-\vec\sigma\cdot\vec\nabla)\xi=Mc\eta. }$$

This pair of coupled first-order Lorentz-covariant equations is the Dirac equation in 2-spinor notation.

13. Structure of the 2-spinor Dirac equation

The 2-spinor Dirac equation has a very simple structure:

$$\eta\quad\text{is left-handed}, \qquad \xi\quad\text{is right-handed}.$$

The differential operator

$$i\hbar(\partial_0+\vec\sigma\cdot\vec\nabla)$$

maps $\eta$ into a right-handed spinor, so it can be equated to $Mc\xi$. The operator

$$i\hbar(\partial_0-\vec\sigma\cdot\vec\nabla)$$

maps $\xi$ into a left-handed spinor, so it can be equated to $Mc\eta$.

The mass term couples the two chiralities. If $M=0$, the equations decouple:

$$i\hbar(\partial_0+\vec\sigma\cdot\vec\nabla)\eta=0, \qquad i\hbar(\partial_0-\vec\sigma\cdot\vec\nabla)\xi=0.$$

For $M\neq0$, left and right components are tied together.

This is the heart of relativistic spin- $\tfrac12$ dynamics: a massive spinor field needs both handedness sectors.

14. Four-spinor notation

The two 2-spinors are packaged into one 4-spinor:

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

Define the four $4\times4$ gamma matrices

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

With

$$\partial_\mu=(\partial_0,\partial_1,\partial_2,\partial_3), \qquad \partial_0=c^{-1}\partial_t,$$

and Einstein summation over $\mu=0,1,2,3$, the coupled 2-spinor equations become

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

This is the standard compact form of the Dirac equation.

15. What the Dirac equation is and is not

The equation

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

is a relativistically covariant field equation for a charged spinor field of mass $M$.

It is not simply a relativistic Schrödinger equation for one electron. In quantum field theory, $\psi$ is promoted to an operator-valued field. The quantized Dirac field acts on Fock space, the Hilbert space containing states with any number of particles. In that setting, $\psi$ destroys electrons and creates positrons.

To fully describe relativistic electrons, one needs the Hamiltonian operator whose Heisenberg equation of motion gives the Dirac equation for the field operator:

$$i\hbar\frac{d\hat\psi}{dt}=[\hat\psi,\hat H].$$

The classical-looking Dirac equation is therefore only one part of Dirac quantum field theory.

Worked Examples

Example 1: Why $\sigma_2\eta^*$ flips handedness

Let $\eta$ be left-handed. Under a boost,

$$\eta\to \left(1+\frac12\epsilon_k\sigma_k\right)\eta.$$

Define

$$\mu=\sigma_2\eta^*.$$

Then

$$\mu\to \sigma_2\left(1+\frac12\epsilon_k\sigma_k\right)^*\eta^*.$$

Using

$$\sigma_2\sigma_k^*=-\sigma_k\sigma_2,$$

we get

$$\mu\to \left(1-\frac12\epsilon_k\sigma_k\right)\mu.$$

This is the right-handed boost rule. Therefore $\mu$ is right-handed.

Example 2: Why the derivative operator maps left to right

Start with

$$\tilde\xi=i\hbar(\partial_t+c\sigma_k\partial_k)\eta.$$

Under boosts, both the derivatives and the spinor components transform. The derivative part changes schematically as

$$\partial_t+c\sigma_k\partial_k \to \partial_t-c\epsilon_k\partial_k+c\sigma_k\partial_k-\sigma_k\epsilon_k\partial_t,$$

while the left-handed spinor transforms as

$$\eta\to\left(1+\frac12\epsilon_n\sigma_n\right)\eta.$$

Using

$$\sigma_k\sigma_l=\delta_{kl}+i\epsilon_{klm}\sigma_m,$$

the transformed $\tilde\xi$ becomes

$$\tilde\xi\to\left(1-\frac12\epsilon_n\sigma_n\right)\tilde\xi.$$

That is exactly the right-handed boost rule.

Example 3: Why Dirac is not Majorana

For a charged spinor field,

$$\eta=\frac{\eta_A+i\eta_B}{\sqrt2}.$$

But

$$\eta^*=\frac{\eta_A^*-i\eta_B^*}{\sqrt2}.$$

The right-handed field that appears in the charged equation is

$$\xi=\sigma_2\frac{\eta_A^*+i\eta_B^*}{\sqrt2},$$

not

$$\sigma_2\eta^*= \sigma_2\frac{\eta_A^*-i\eta_B^*}{\sqrt2}.$$

Therefore the charged equation does not close on $\eta$ and $\sigma_2\eta^*$. It needs an independent right-handed spinor $\xi$. That is why the charged spinor equation is Dirac, not Majorana.

Intuition

Relativity splits spin- $\tfrac12$ fields into left-handed and right-handed pieces. Ordinary rotations cannot tell the difference between them, but boosts can.

A Lorentz-covariant massive equation must connect these two pieces. Majorana conjugation gives one way to relate them for a neutral field. Differential operators give another way to map one handedness into the other. Combining these ideas leads first to the Majorana equation and then, after making a complex charged field from two identical neutral spinors, to the Dirac equation.

The Dirac equation is therefore not magic. It is the simplest Lorentz-covariant first-order equation for a massive charged spinor field.

Common Mistakes

• Thinking left-handed and right-handed spinors differ under ordinary rotations. They do not.
• Forgetting that the difference appears under boosts.
• Treating Lorentz boosts on spinor components as unitary Hilbert-space time evolution.
• Assuming $\eta^\dagger\eta$ is Lorentz-invariant just because it is rotation-invariant.
• Forgetting that mixed left-right bilinears are the natural Lorentz scalar structures.
• Confusing $\sigma_2\eta^*$ with ordinary complex conjugation.
• Thinking the Majorana equation and Dirac equation are the same.
• Forgetting that Majorana implies Klein-Gordon, but Klein-Gordon does not imply Majorana.
• Missing why a charged spinor field needs two Majorana spinors.
• Treating the Dirac equation as a complete one-particle theory of the electron.
• Forgetting that in QFT the Dirac field destroys electrons and creates positrons.

Short Summary

Left-handed and right-handed Lorentz 2-spinors transform identically under spatial rotations but oppositely under boosts. Because of this, the ordinary spinor norms $\eta^\dagger\eta$ and $\xi^\dagger\xi$ are not Lorentz-invariant, while mixed left-right structures behave correctly. The Majorana conjugate $\sigma_2\eta^*$ turns a left-handed spinor into a right-handed one, and the differential operator $i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)$ does the same. Similarly, $i\hbar(\partial_t-c\vec\sigma\cdot\vec\nabla)$ maps right-handed spinors to left-handed spinors.

Closing the system with the Majorana conjugate gives the neutral Majorana equation,

$$i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)\eta=Mc^2\sigma_2\eta^*.$$

Iterating this equation gives the Klein-Gordon equation for each component. To obtain a charged spinor field, two identical Majorana spinors are combined into a complex field $\eta=(\eta_A+i\eta_B)/\sqrt2$. The corresponding right-handed spinor is $\xi=\sigma_2(\eta_A^*+i\eta_B^*)/\sqrt2$, which is not $\sigma_2\eta^*$. This gives the coupled first-order Dirac equations

$$i\hbar(\partial_0+\vec\sigma\cdot\vec\nabla)\eta=Mc\xi, \qquad i\hbar(\partial_0-\vec\sigma\cdot\vec\nabla)\xi=Mc\eta.$$

Packaging $\eta$ and $\xi$ into a 4-spinor $\psi$ gives the standard form

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

This is a Lorentz-covariant field equation for a charged spinor field. In QFT, $\psi$ becomes an operator-valued field, not a single-particle wavefunction.

Practice Problems

1. Why do left-handed and right-handed Lorentz 2-spinors transform the same way under rotations but differently under boosts?

2. Show that $\eta^\dagger\eta$ is invariant under infinitesimal rotations but not under infinitesimal boosts.

3. Show that $\xi^\dagger\eta$ is invariant under infinitesimal boosts to first order in $\epsilon$.

4. Use $\sigma_2\sigma_k^*=-\sigma_k\sigma_2$ to show that $\mu=\sigma_2\eta^*$ has the opposite handedness from $\eta$.

5. Why is Majorana conjugation especially natural for neutral spinor fields?

6. Expand the components of

$$\tilde\xi=i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)\eta.$$

7. Why does $i\hbar(\partial_t+c\vec\sigma\cdot\vec\nabla)$ map a left-handed spinor into a right-handed spinor?

8. Derive the Majorana equation from the condition that the right-handed spinor is proportional to $\sigma_2\eta^*$.

9. Show that the Majorana equation implies the Klein-Gordon equation.

10. Why is one Majorana spinor not enough to describe a charged spinor field?

11. Explain why $\xi\neq\sigma_2\eta^*$ in the charged spinor construction.

12. Derive the two coupled 2-spinor equations of the Dirac equation.

13. Write the gamma matrices $\gamma^0$ and $\gamma^k$ used in the 4-spinor form.

14. Explain why the Dirac equation is not, by itself, the full quantum theory of the electron.