Dirac Equation and Spinor Solutions

The previous page constructed the spinor representations of the Lorentz group and introduced gamma matrices as the algebraic bridge between vectors and spinors. We now turn that representation theory into dynamics. The result is the Dirac equation: a first-order relativistic equation whose solutions describe spin-one-half modes.

The main lesson is that the Dirac equation is not a replacement for the mass shell $p^2=m^2$. It is a refined, spinorial way of imposing it. The Clifford algebra makes the operator $\gamma\cdot p-m$ a Lorentz-covariant square root of $p^2-m^2$, while the spinor components encode the two spin states of a massive fermion and the two chiralities that become independent in the massless limit.

For calculations, the page has a simple payoff: learn to recognize which equations define the $u_s(p)$ spinors, which equations define the $v_s(p)$ spinors, and why the spin sums $\sum_su_s\bar u_s$ and $\sum_sv_s\bar v_s$ are more useful than explicit spinor columns.

The square root of the Klein–Gordon operator

Chiral-basis conventions

Gamma matrices obey

$$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf 1.$$

The site-wide Dirac notation means

$$p\!\!\!/\equiv\gamma^\mu p_\mu.$$

When explicit two-component formulae are useful, we use the chiral basis

$$\gamma^\mu= \begin{pmatrix} 0&\sigma^\mu\\ \bar\sigma^\mu&0 \end{pmatrix}, \qquad \sigma^\mu=(\mathbf 1,\boldsymbol\sigma), \qquad \bar\sigma^\mu=(\mathbf 1,-\boldsymbol\sigma).$$

For covariant momentum $p_\mu=(E,-\mathbf p)$,

$$p\cdot\sigma=p_\mu\sigma^\mu=E\mathbf 1-\mathbf p\cdot\boldsymbol\sigma, \qquad p\cdot\bar\sigma=p_\mu\bar\sigma^\mu=E\mathbf 1+\mathbf p\cdot\boldsymbol\sigma.$$

The Dirac adjoint is $\bar\psi=\psi^\dagger\gamma^0$. In the helicity formulas below, $h=\pm1$ denotes the eigenvalue of $\boldsymbol\sigma\cdot\widehat{\mathbf p}$; the physical spin helicity is $h/2$. We use the chiral projectors

$$P_L={1-\gamma^5\over2}, \qquad P_R={1+\gamma^5\over2}.$$

With the block ordering used here, $\psi=(\psi_L,\psi_R)^T$ means $\psi_L=P_L\psi$ and $\psi_R=P_R\psi$.

A free scalar mode satisfies

$$(p^2-m^2)\phi(p)=0.$$

In position space, with $p_\mu\to i\partial_\mu$, this is the Klein–Gordon equation

$$(\partial^2+m^2)\phi(x)=0, \qquad \partial^2=\partial_\mu\partial^\mu.$$

For spinors, we can impose a stronger first-order equation. Define

$$\gamma\cdot p=\gamma^\mu p_\mu.$$

Using the Clifford algebra,

$$(\gamma\cdot p)^2 =\gamma^\mu\gamma^\nu p_\mu p_\nu ={1\over2}\{\gamma^\mu,\gamma^\nu\}p_\mu p_\nu =p^2\mathbf 1,$$

because $p_\mu p_\nu$ is symmetric and therefore kills the commutator part of $\gamma^\mu\gamma^\nu$. Hence

$$(\gamma\cdot p-m)(\gamma\cdot p+m)=(p^2-m^2)\mathbf 1.$$

The free Dirac equation in momentum space is

$$(\gamma\cdot p-m)u(p)=0.$$

In position space it is

$$(i\gamma^\mu\partial_\mu-m)\psi(x)=0.$$

Acting once more with $i\gamma^\mu\partial_\mu+m$ gives

$$(i\gamma^\mu\partial_\mu+m)(i\gamma^\nu\partial_\nu-m)\psi =-(\partial^2+m^2)\psi=0.$$

The overall minus sign is harmless because the expression is set to zero. The important content is that the Dirac equation implies the Klein–Gordon equation with the course-wide mostly-minus convention $(\Box+m^2)\psi=0$.

Thus every component of a Dirac spinor obeys the Klein–Gordon equation. The converse is not true: a collection of four Klein–Gordon solutions is not automatically a Dirac spinor solution. The Dirac equation keeps only the spinor configurations whose components are related by the first-order constraint.

The Clifford algebra factorizes the relativistic mass-shell polynomial. The first-order equation $(\gamma\cdot p-m)u=0$ implies the Klein–Gordon condition $(p^2-m^2)u=0$, while preserving spinor transformation properties.

The size of the spinor is fixed by the Clifford algebra. In $d$ spacetime dimensions, an irreducible complex spinor has dimension $2^{\lfloor d/2\rfloor}$. In even dimension the complex Clifford algebra has one irreducible matrix representation up to equivalence; in odd dimension there are two choices distinguished by the sign of the highest-grade gamma product. In four dimensions this gives four complex components. On shell, the Dirac equation reduces the independent positive-energy solutions to two spin states, as required for a massive spin-one-half particle.

Lorentz covariance

The orbital Lorentz generators acting on functions of spacetime are

$$L_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu).$$

A spinor also needs intrinsic Lorentz generators acting on its components. In the global convention

$$\sigma^{\mu\nu}={i\over2}[\gamma^\mu,\gamma^\nu],$$

while the actual generator matrix in the spinor representation is

$$\Sigma^{\mu\nu}={1\over2}\sigma^{\mu\nu} ={i\over4}[\gamma^\mu,\gamma^\nu].$$

From the Clifford algebra one obtains

$$[\Sigma^{\mu\nu},\gamma^\rho] =i(\eta^{\nu\rho}\gamma^\mu-\eta^{\mu\rho}\gamma^\nu).$$

This is the algebraic statement that the gamma matrices carry a Lorentz vector index while acting on spinor indices. For a finite Lorentz transformation,

$$S(\Lambda)=\exp\left(-{i\over2}\omega_{\mu\nu}\Sigma^{\mu\nu}\right),$$

and

$$S(\Lambda)^{-1}\gamma^\rho S(\Lambda)=\Lambda^\rho{}_{\sigma}\gamma^\sigma.$$

If $x'=\Lambda x$ and the spinor transforms as

$$\psi'(x')=S(\Lambda)\psi(x),$$

then

$$(i\gamma^\mu\partial'_\mu-m)\psi'(x') =S(\Lambda)(i\gamma^\mu\partial_\mu-m)\psi(x).$$

So solutions are carried into solutions. Lorentz covariance is not automatic merely because the wavefunction has four components; it follows from the compatibility between $S(\Lambda)$ and the gamma matrices.

The full Lorentz generator acting on a spinor field is

$$J_{\mu\nu}=L_{\mu\nu}+\Sigma_{\mu\nu}.$$

The orbital part moves the argument of the field. The spin part rotates and boosts the spinor indices.

Weyl components and the mass term

Write a Dirac spinor in the chiral basis as

$$\psi= \begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix},$$

where $\psi_L=P_L\psi$ and $\psi_R=P_R\psi$. On the previous page, $A$ was fixed by the Hermitian-vector map $X'=AXA^\dagger$. With that choice, $\psi_R$ transforms with $A$ and $\psi_L$ with $(A^\dagger)^{-1}$. The important invariant statement is that the mass term couples the two inequivalent Weyl representations. The momentum-space Dirac equation becomes

$$\begin{pmatrix} -m&p\cdot\sigma\\ p\cdot\bar\sigma&-m \end{pmatrix} \begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix}=0,$$

or

$$(p\cdot\sigma)\psi_R=m\psi_L, \qquad (p\cdot\bar\sigma)\psi_L=m\psi_R.$$

Equivalently,

$$(E\mathbf 1-\mathbf p\cdot\boldsymbol\sigma)\psi_R=m\psi_L, \qquad (E\mathbf 1+\mathbf p\cdot\boldsymbol\sigma)\psi_L=m\psi_R.$$

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

$$(p\cdot\bar\sigma)\psi_L=0, \qquad (p\cdot\sigma)\psi_R=0.$$

This is the Weyl limit. A massless left-handed spinor and a massless right-handed spinor are independent Lorentz-covariant fields; a massive Dirac fermion needs both.

The momentum matrices $p\cdot\sigma$ and $p\cdot\bar\sigma$ map between the two Weyl representations. The mass term identifies the two chiralities inside one massive Dirac spinor.

At rest, $p=(m,0,0,0)$, the positive-frequency equation gives

$$\psi_L=\psi_R.$$

Thus the rest-frame positive-frequency spinors may be written as

$$u_s(0)=\sqrt m \begin{pmatrix} \chi_s\\ \chi_s \end{pmatrix}, \qquad s=1,2,$$

where $\chi_s$ are ordinary two-component Pauli spinors. The spin label $s$ is a rotation spin label, not a chirality label.

Plane-wave spinors

For positive-frequency solutions, take

$$\psi(x)=u_s(p)e^{-ip\cdot x}, \qquad p^0=E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}.$$

Then

$$(\gamma\cdot p-m)u_s(p)=0.$$

For negative-frequency solutions, write

$$\psi(x)=v_s(p)e^{+ip\cdot x}, \qquad p^0=E_{\mathbf p}>0,$$

which gives

$$(\gamma\cdot p+m)v_s(p)=0.$$

Here $p^0$ in the label of $v_s(p)$ is still positive. The sign change comes from the phase: $i\partial_\mu e^{+ip\cdot x}=-p_\mu e^{+ip\cdot x}$. This convention is what later lets the $v$ branch appear as antiparticle creation rather than as a negative-energy particle.

A compact chiral-basis solution uses the unique positive-semidefinite Hermitian square roots of $p\cdot\sigma$ and $p\cdot\bar\sigma$. They exist because, for $p^0=E_{\mathbf p}>0$ on shell, the eigenvalues are $E_{\mathbf p}\mp|\mathbf p|\geq0$:

$$u_s(p)= \begin{pmatrix} \sqrt{p\cdot\sigma}\,\chi_s\\ \sqrt{p\cdot\bar\sigma}\,\chi_s \end{pmatrix},$$

and

$$v_s(p)= \begin{pmatrix} \sqrt{p\cdot\sigma}\,\eta_s\\ -\sqrt{p\cdot\bar\sigma}\,\eta_s \end{pmatrix}.$$

Here $\chi_s$ and $\eta_s$ are orthonormal two-component spinor bases. On shell,

$$(p\cdot\sigma)(p\cdot\bar\sigma)=m^2\mathbf 1,$$

so

$$(p\cdot\sigma)\sqrt{p\cdot\bar\sigma}=m\sqrt{p\cdot\sigma}, \qquad (p\cdot\bar\sigma)\sqrt{p\cdot\sigma}=m\sqrt{p\cdot\bar\sigma}.$$

This verifies the two coupled Weyl equations directly.

Writing a plane wave as $e^{-iq\cdot x}$, the $u$ mode has Fourier momentum $q=p$ while $v_s(p)e^{+ip\cdot x}$ has $q=-p$. The label $p$ in both spinors nevertheless has $p^0=E_{\mathbf p}>0$; after quantization the lower-frequency branch becomes the antiparticle creation term.

For explicit calculations it is often useful to choose Pauli helicity eigenspinors

$$\boldsymbol\sigma\cdot\widehat{\mathbf p}\,\chi_\lambda=\lambda\chi_\lambda, \qquad \lambda=\pm1.$$

The physical helicity carried by the two-component spinor is $\lambda/2$. With this notation,

$$(p\cdot\sigma)\chi_\lambda=(E-\lambda|\mathbf p|)\chi_\lambda, \qquad (p\cdot\bar\sigma)\chi_\lambda=(E+\lambda|\mathbf p|)\chi_\lambda.$$

The positive-frequency spinor becomes

$$u_\lambda(p)= \begin{pmatrix} \sqrt{E-\lambda|\mathbf p|}\,\chi_\lambda\\ \sqrt{E+\lambda|\mathbf p|}\,\chi_\lambda \end{pmatrix}.$$

The square roots are exactly what makes the two first-order equations true, because

$$(E+\lambda|\mathbf p|)(E-\lambda|\mathbf p|)=E^2-\mathbf p^2=m^2.$$

Spin sums and projectors

The standard spinor normalization used in the next page is

$$\bar u_r(p)u_s(p)=2m\delta_{rs}, \qquad \bar v_r(p)v_s(p)=-2m\delta_{rs},$$

and

$$u_r(p)^\dagger u_s(p)=2E_{\mathbf p}\delta_{rs}, \qquad v_r(p)^\dagger v_s(p)=2E_{\mathbf p}\delta_{rs}.$$

The first line is Lorentz invariant. The second line is the equal-time Hilbert-space normalization used in canonical quantization.

The spin sums are

$$\sum_{s=1}^2 u_s(p)\bar u_s(p)=\gamma\cdot p+m, \qquad \sum_{s=1}^2 v_s(p)\bar v_s(p)=\gamma\cdot p-m.$$

These identities are usually more important than the explicit spinors. They convert sums over external spin states into traces of gamma matrices, and they are independent of the particular spin basis used for the two states. In practice, one often chooses helicity spinors for external particles but immediately replaces spin sums by the covariant numerators above. Whenever a calculation seems to differ by a factor of $2m$ or $2E_{\mathbf p}$, check the spinor normalization before changing the physics.

For $m>0$, the positive-frequency numerator defines a projector:

$$\Lambda_+(p)={\gamma\cdot p+m\over2m}, \qquad \Lambda_+^2=\Lambda_+.$$

Indeed,

$$(\gamma\cdot p+m)^2=p^2+2m\gamma\cdot p+m^2=2m(\gamma\cdot p+m),$$

where $p^2=m^2$ was used. Since $\operatorname{tr}\Lambda_+=2$, this projector has rank two, matching the two spin states of a massive positive-energy fermion. The corresponding projector onto solutions of $(\gamma\cdot p+m)v=0$ at the same positive-energy momentum label is

$$\Lambda_-(p)={m-\gamma\cdot p\over2m}.$$

It obeys $\Lambda_-v_s=v_s$ and $\Lambda_-^2=\Lambda_-$. Because the covariant normalization of the $v$ spinors is negative,

$$\sum_s v_s(p)\bar v_s(p) =\gamma\cdot p-m =-2m\Lambda_-(p).$$

Neither $\Lambda_+$ nor $\Lambda_-$ should be used at $m=0$, where division by $2m$ is singular; the chiral and helicity projectors are the appropriate massless objects.

Spin sums turn basis-dependent spinor products into covariant matrices. With the stated normalization, $\sum_su_s\bar u_s=2m\Lambda_+$ while $\sum_sv_s\bar v_s=-2m\Lambda_-$; the minus sign in the latter is tied to $\bar v_sv_s=-2m$.

The same numerator appears in the free Dirac propagator,

$$S_F(p)={i(\gamma\cdot p+m)\over p^2-m^2+i\epsilon},$$

which will be derived after quantizing the Dirac field.

Chirality and helicity in the massless limit

When $m=0$, the Dirac equation separates into two Weyl equations:

$$i\bar\sigma^\mu\partial_\mu\psi_L=0, \qquad i\sigma^\mu\partial_\mu\psi_R=0.$$

For a positive-energy mode with $p^0=|\mathbf p|$,

$$p\cdot\bar\sigma=|\mathbf p|(1+\boldsymbol\sigma\cdot\widehat{\mathbf p}), \qquad p\cdot\sigma=|\mathbf p|(1-\boldsymbol\sigma\cdot\widehat{\mathbf p}).$$

Therefore

$$\boldsymbol\sigma\cdot\widehat{\mathbf p}\,\psi_L=-\psi_L, \qquad \boldsymbol\sigma\cdot\widehat{\mathbf p}\,\psi_R=+\psi_R.$$

Since the spin operator is $\mathbf S=\boldsymbol\sigma/2$, a positive-energy left-handed Weyl mode has helicity $-1/2$, while a positive-energy right-handed Weyl mode has helicity $+1/2$.

For $m=0$, chirality and helicity lock together for positive-energy particle modes. Left-handed modes have spin antiparallel to momentum; right-handed modes have spin parallel to momentum.

For massive fermions the distinction matters. Chirality is a Lorentz-representation label, implemented by

$$P_L={1-\gamma^5\over2}, \qquad P_R={1+\gamma^5\over2}.$$

Helicity is the projection of spin along momentum. A massive particle can be viewed from a frame that overtakes it and reverses its momentum, so helicity is not Lorentz invariant for massive particles. A massless particle cannot be overtaken, and helicity becomes invariant. This is why weak-interaction phrases such as “left-handed fermion” must be read carefully: for a massive fermion, a chirality projection and a helicity measurement are related but not identical.

Dirac adjoint, Lagrangian, and current

The ordinary Hermitian inner product $\psi^\dagger\psi$ is invariant under spatial rotations, but not under boosts, because finite-dimensional spinor boost matrices are not unitary. The gamma matrices instead obey

$$(\gamma^0)^\dagger=\gamma^0, \qquad (\gamma^i)^\dagger=-\gamma^i,$$

or, equivalently,

$$(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0.$$

This identity implies

$$S(\Lambda)^\dagger\gamma^0S(\Lambda)=\gamma^0.$$

The invariant matrix is therefore $\gamma^0$, not the identity. This motivates the Lorentz-covariant adjoint

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

Indeed, if $\psi'=S\psi$, then

$$\bar\psi' =\psi^\dagger S^\dagger\gamma^0 =\bar\psi S^{-1}.$$

It follows immediately that $\bar\psi\psi$ is a Lorentz scalar and $\bar\psi\gamma^\mu\psi$ is a Lorentz vector.

The free Dirac Lagrangian is

$$\mathcal L=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi.$$

Varying $\bar\psi$ gives the Dirac equation,

$$(i\gamma^\mu\partial_\mu-m)\psi=0.$$

Varying $\psi$ and integrating by parts gives the adjoint equation

$$i(\partial_\mu\bar\psi)\gamma^\mu+m\bar\psi=0.$$

The Lagrangian is invariant under the global phase symmetry

$$\psi\mapsto e^{i\alpha}\psi, \qquad \bar\psi\mapsto \bar\psi e^{-i\alpha}.$$

The associated current is

$$j^\mu=\bar\psi\gamma^\mu\psi.$$

Using the Dirac equation and its adjoint,

$$\partial_\mu j^\mu =(\partial_\mu\bar\psi)\gamma^\mu\psi+\bar\psi\gamma^\mu\partial_\mu\psi =im\bar\psi\psi-im\bar\psi\psi=0.$$

The time component is

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

As a classical wave equation this gives a positive density. In quantum field theory the deeper interpretation is that $j^\mu$ is a charge current, while the negative-frequency modes become antiparticle creation modes.

The same gamma algebra organizes all local Dirac bilinears. A basis for the $4\times4$ complex matrices contains

$$1+4+6+4+1=16$$

independent elements:

Gamma structure	Bilinear	Proper-Lorentz type
$\mathbf 1$	$\bar\psi\psi$	scalar
$\gamma^\mu$	$\bar\psi\gamma^\mu\psi$	vector
$\sigma^{\mu\nu}$	$\bar\psi\sigma^{\mu\nu}\psi$	antisymmetric tensor
$\gamma^\mu\gamma^5$	$\bar\psi\gamma^\mu\gamma^5\psi$	axial vector
$\gamma^5$	$\bar\psi\gamma^5\psi$	pseudoscalar

The adjectives “axial” and “pseudo” distinguish behavior under parity; under the proper orthochronous Lorentz group they transform like a vector and scalar, respectively. This $1+4+6+4+1$ decomposition is also a useful completeness check for gamma-matrix manipulations.

Summary

The Dirac equation

$$(i\gamma^\mu\partial_\mu-m)\psi=0$$

combines Lorentz covariance, the Clifford algebra, and spin-one-half representation theory. Squaring the operator gives the Klein–Gordon equation, so Dirac spinors live on the same mass shell as scalar particles, but with spinor polarizations attached to each momentum.

A massive Dirac spinor contains left- and right-handed Weyl components coupled by the mass. In the massless limit the components decouple, and chirality becomes tied to helicity for positive-energy particle modes.

The positive-frequency spinors $u_s(p)$ obey $(\gamma\cdot p-m)u_s=0$, while the negative-frequency spinors $v_s(p)$ obey $(\gamma\cdot p+m)v_s=0$. Their spin sums,

$$\sum_s u_s(p)\bar u_s(p)=\gamma\cdot p+m, \qquad \sum_s v_s(p)\bar v_s(p)=\gamma\cdot p-m,$$

are the practical heart of spinor calculations and foreshadow the numerator of the fermion propagator. In both sums $p^0$ is the positive on-shell energy.

Common pitfalls

Chirality and helicity differ at nonzero mass. They coincide for positive-energy massless particle modes, but a boost can reverse the momentum of a massive particle without changing its chirality.

Spinor boost matrices need not be unitary. The finite-dimensional matrix $S(\Lambda)$ is not generally unitary for boosts. This does not contradict unitarity of the Lorentz transformation on the Hilbert space.

The $v_s(p)$ spinors are not negative-probability states. Their label still has $p^0=+E_{\mathbf p}$. The negative Fourier frequency is carried by $e^{+ip\cdot x}$, and after quantization this mode multiplies an antiparticle creation operator.

The momentum label in the $v$ spin sum has positive energy. In $\sum_s v_s(p)\bar v_s(p)=\gamma\cdot p-m$, do not silently replace $p^0=+E_{\mathbf p}$ by $-E_{\mathbf p}$.

Spinor normalizations fix the completeness relations. The displayed spin sums assume $\bar u_su_{s'}=2m\delta_{ss'}$ and $\bar v_sv_{s'}=-2m\delta_{ss'}$. Changing normalization changes the prefactors.

The Dirac adjoint is essential. The scalar $\bar\psi\psi$ and vector $\bar\psi\gamma^\mu\psi$ are Lorentz covariant; $\psi^\dagger\psi$ alone is neither a Lorentz scalar nor boost invariant.

Explicit columns are basis dependent. Spin sums and bilinears are usually safer covariant objects. Projectors divided by $2m$ are only defined for $m>0$.

Exercises

Exercise 1: squaring the Dirac equation

Show directly that the Dirac equation implies the Klein–Gordon equation for each spinor component.

Solution

Start from

$$(i\gamma^\mu\partial_\mu-m)\psi=0.$$

Act on the left with $i\gamma^\nu\partial_\nu+m$:

$$0=(i\gamma^\nu\partial_\nu+m)(i\gamma^\mu\partial_\mu-m)\psi.$$

The cross terms cancel, leaving

$$0=-\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu\psi-m^2\psi.$$

Because $\partial_\nu\partial_\mu$ is symmetric, the commutator part of $\gamma^\nu\gamma^\mu$ drops out:

$$\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu ={1\over2}\{\gamma^\nu,\gamma^\mu\}\partial_\nu\partial_\mu =\partial^2.$$

Therefore

$$(\partial^2+m^2)\psi=0.$$

Exercise 2: chiral equations imply the mass shell

Using the chiral-basis equations

$$(E\mathbf 1-\mathbf p\cdot\boldsymbol\sigma)\psi_R=m\psi_L, \qquad (E\mathbf 1+\mathbf p\cdot\boldsymbol\sigma)\psi_L=m\psi_R,$$

verify the helicity-basis spinor

$$u_\lambda(p)= \begin{pmatrix} \sqrt{E-\lambda|\mathbf p|}\,\chi_\lambda\\ \sqrt{E+\lambda|\mathbf p|}\,\chi_\lambda \end{pmatrix}, \qquad \boldsymbol\sigma\cdot\widehat{\mathbf p}\,\chi_\lambda=\lambda\chi_\lambda, \qquad \lambda=\pm1.$$

Solution

Let

$$\psi_L=A_\lambda\chi_\lambda, \qquad \psi_R=B_\lambda\chi_\lambda,$$

with

$$A_\lambda=\sqrt{E-\lambda|\mathbf p|}, \qquad B_\lambda=\sqrt{E+\lambda|\mathbf p|}.$$

The first Weyl equation gives

$$(E\mathbf 1-\mathbf p\cdot\boldsymbol\sigma)\psi_R =(E-\lambda|\mathbf p|)B_\lambda\chi_\lambda.$$

Using $E^2-\mathbf p^2=m^2$,

$$(E-\lambda|\mathbf p|)B_\lambda =(E-\lambda|\mathbf p|)\sqrt{E+\lambda|\mathbf p|} =m\sqrt{E-\lambda|\mathbf p|}=mA_\lambda.$$

So the first equation holds. Similarly,

$$(E\mathbf 1+\mathbf p\cdot\boldsymbol\sigma)\psi_L =(E+\lambda|\mathbf p|)A_\lambda\chi_\lambda=mB_\lambda\chi_\lambda,$$

so the second equation also holds.

Exercise 3: the positive-energy spin-sum projector

Show that

$$\Lambda_+(p)={\gamma\cdot p+m\over2m}$$

is a projector on shell and acts as the identity on any $u_s(p)$.

Solution

Since $(\gamma\cdot p-m)u_s(p)=0$,

$$\gamma\cdot p\,u_s(p)=m u_s(p).$$

Thus

$$\Lambda_+(p)u_s(p) ={\gamma\cdot p+m\over2m}u_s(p)=u_s(p).$$

On shell, $p^2=m^2$, so

$$(\gamma\cdot p+m)^2 =(\gamma\cdot p)^2+2m\gamma\cdot p+m^2 =p^2+2m\gamma\cdot p+m^2 =2m(\gamma\cdot p+m).$$

Therefore

$$\Lambda_+^2={\gamma\cdot p+m\over2m}=\Lambda_+.$$

Exercise 4: chirality versus helicity at zero mass

For a massless positive-energy spinor, show that $\psi_L$ has helicity $-1/2$ and $\psi_R$ has helicity $+1/2$.

Solution

For $m=0$,

$$(p\cdot\bar\sigma)\psi_L=0, \qquad (p\cdot\sigma)\psi_R=0.$$

With $p^0=|\mathbf p|$,

$$p\cdot\bar\sigma=|\mathbf p|(1+\boldsymbol\sigma\cdot\widehat{\mathbf p}), \qquad p\cdot\sigma=|\mathbf p|(1-\boldsymbol\sigma\cdot\widehat{\mathbf p}).$$

Hence

$$\boldsymbol\sigma\cdot\widehat{\mathbf p}\,\psi_L=-\psi_L, \qquad \boldsymbol\sigma\cdot\widehat{\mathbf p}\,\psi_R=+\psi_R.$$

The spin operator is $\mathbf S=\boldsymbol\sigma/2$, so the helicities are $-1/2$ and $+1/2$.

Exercise 5: current conservation

Use the Dirac equation and its adjoint to prove current conservation for $j^\mu=\bar\psi\gamma^\mu\psi$.

Solution

The Dirac equation gives

$$\gamma^\mu\partial_\mu\psi=-im\psi.$$

The adjoint equation gives

$$(\partial_\mu\bar\psi)\gamma^\mu=im\bar\psi.$$

Therefore

$$\partial_\mu j^\mu =(\partial_\mu\bar\psi)\gamma^\mu\psi +\bar\psi\gamma^\mu\partial_\mu\psi =im\bar\psi\psi-im\bar\psi\psi=0.$$

Further reading

• Coleman, Sidney. Lectures on Quantum Field Theory. World Scientific, 2019, chapters 19–21.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007, sections 34–38.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, sections 5.4–5.6 and 14.1.
• Zee, A. Quantum Field Theory in a Nutshell. 2nd ed., Princeton University Press, 2010, chapters II.1–II.2.