The Dirac Equation

It must satisfy two conditions: cancel $[H,L_i]$ and obey the angular-momentum commutation relations. The first condition is

$$[H,S_i]=-[H,L_i]=i\hslash (\mathbf{\alpha }\times \hat{\mathbf{p}}{)}_i.$$

The Pauli matrices provide a candidate. In Relativistic QM §4, ${\alpha }_j=\left(\begin{array}{cc}0& {\sigma }_j\\ {\sigma }_j& 0\end{array}\right)$, so ${\alpha }_j{\alpha }_k=\mathrm{diag}({\sigma }_j{\sigma }_k,{\sigma }_j{\sigma }_k)$. Using $[{\sigma }_j,{\sigma }_k]=2i{\epsilon }_{jkl}{\sigma }_l$ gives

$$[{\alpha }_j,{\alpha }_k]=2i{\epsilon }_{jkl}\left(\begin{array}{cc}{\sigma }_l& 0\\ 0& {\sigma }_l\end{array}\right),$$

Define the repeated Pauli blocks as

$$\mathbf{\Sigma }=\left(\begin{array}{cc}\mathbf{\sigma }& 0\\ 0& \mathbf{\sigma }\end{array}\right).$$

Use $[{\alpha }_j,{\Sigma }_i]=-2i{\epsilon }_{ijk}{\alpha }_k$ to compute its Hamiltonian commutator:

$$[H,{\Sigma }_i]=2i(\mathbf{\alpha }\times \hat{\mathbf{p}}{)}_i.$$

Comparing with the required $i\hslash (\mathbf{\alpha }\times \hat{\mathbf{p}}{)}_i$, multiply $\mathbf{\Sigma }$ by $\hslash /2$. This gives

$$\mathbf{S}=\frac{\hslash }{2}\mathbf{\Sigma }.$$

The factor $\hslash$ originates in $[x_j,p_k]=i\hslash {\delta }_{jk}$; the factor $2$ comes from the Pauli commutator. Check the cancellation directly:

$$[H,S_i]=\frac{\hslash }{2}[H,{\Sigma }_i]=\frac{\hslash }{2}\cdot 2i(\mathbf{\alpha }\times \hat{\mathbf{p}}{)}_i=i\hslash (\mathbf{\alpha }\times \hat{\mathbf{p}}{)}_i=-[H,L_i],$$

Therefore $[H,\mathbf{J}]=0$ for $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Spin and orbital angular momentum are generally not separately conserved, but their sum is.

We must also check that $\mathbf{S}$ has the algebra of angular momentum. The Pauli commutators give

$$[S_i,S_j]=i\hslash {\epsilon }_{ijk}{S}_k,$$

These are the angular-momentum commutation relations. For example, $[S_x,S_y]=i\hslash S_z$, with cyclic versions for the other components. They describe the infinitesimal effect of rotations about different axes.

The factor $g=2$ quoted above follows from taking the Dirac equation at low momentum. Split the spinor into upper and lower pairs, $\psi =(u,v)$, and use minimal substitution $\hat{\mathbf{p}}\to \hat{\mathbf{p}}-q\mathbf{A}$ in the coupled equations of Relativistic QM §4:

$$(E-m)u=\mathbf{\sigma }\cdot (\hat{\mathbf{p}}-q\mathbf{A})v,(E+m)v=\mathbf{\sigma }\cdot (\hat{\mathbf{p}}-q\mathbf{A})u,$$

with $c=1$ and a static magnetic field, choosing zero scalar potential. For a positive-energy solution at low momentum, $E\approx m$ and the lower pair $v$ is the small component. Solve the second equation to leading order,

$$v\approx \frac{\mathbf{\sigma }\cdot (\hat{\mathbf{p}}-q\mathbf{A})}{2m}u,$$

and substitute it into the first equation:

$$(E-m)u\approx \frac{{\left[\mathbf{\sigma }\cdot (\hat{\mathbf{p}}-q\mathbf{A})\right]}^2}{2m}u.$$

Evaluate the square with the Pauli identity $(\mathbf{\sigma }\cdot \mathbf{a})(\mathbf{\sigma }\cdot \mathbf{b})=\mathbf{a}\cdot \mathbf{b}+i\mathbf{\sigma }\cdot (\mathbf{a}\times \mathbf{b})$, taking $\mathbf{a}=\mathbf{b}=\hat{\mathbf{p}}-q\mathbf{A}$. The cross product does not vanish, because the momentum operator does not commute with $\mathbf{A}$: $(\hat{\mathbf{p}}-q\mathbf{A})\times (\hat{\mathbf{p}}-q\mathbf{A})=i\hslash q\mathbf{B}$. The square is therefore

$${\left[\mathbf{\sigma }\cdot (\hat{\mathbf{p}}-q\mathbf{A})\right]}^2=(\hat{\mathbf{p}}-q\mathbf{A}{)}^2-\hslash q\mathbf{\sigma }\cdot \mathbf{B}.$$

Inserting this and using the spin operator $\mathbf{S}=\frac{\hslash }{2}\mathbf{\sigma }$ from §1, the two-component equation for the upper pair becomes

$$(E-m)u=\left[\frac{(\hat{\mathbf{p}}-q\mathbf{A}{)}^2}{2m}-\frac{q}{m}\mathbf{S}\cdot \mathbf{B}\right]u.$$

The first term is the ordinary kinetic energy of a charged particle; the second is a magnetic interaction of the form $-\mathbf{\mu }\cdot \mathbf{B}$ with

$$\mathbf{\mu }=\frac{q}{m}\mathbf{S}.$$

For an electron, $q=-e$, so $\mathbf{\mu }=-\frac{e}{m}\mathbf{S}$. Writing the magnetic moment in the standard form $\mathbf{\mu }=g\frac{q}{2m}\mathbf{S}$ fixes $g=2$. The factor of $2$ comes from the Pauli identity, which in turn reflects the anticommutation of the Pauli matrices.

For a time-independent Hamiltonian, define the Heisenberg operator by moving the state-evolution factors onto it:

$${\mathbf{A}}_H(t)=e^{iHt/\hslash }\mathbf{A}{e}^{-iHt/\hslash },$$

At $t=0$, ${\mathbf{A}}_H(0)=\mathbf{A}$. Differentiate both exponential factors using $\frac{d}{dt}{e}^{\pm iHt/\hslash }=\pm \frac{i}{\hslash }H{e}^{\pm iHt/\hslash }$:

$$\frac{d{\mathbf{A}}_H}{dt}=\frac{i}{\hslash }\left(H{e}^{iHt/\hslash }\mathbf{A}{e}^{-iHt/\hslash }-e^{iHt/\hslash }\mathbf{A}{e}^{-iHt/\hslash }H\right)=\frac{i}{\hslash }[H,{\mathbf{A}}_H].$$

At rest ($\mathbf{p}=0$). The Hamiltonian reduces to $H=\beta m$. Substitute $\psi =w{e}^{-iEt/\hslash }$, where $w$ is a constant four-component spinor:

$$Ew=\beta mw.$$

In the representation of Relativistic QM §4, $\beta =\mathrm{diag}(1,1,-1,-1)$. Writing $w=(w_1,w_2,w_3,w_4)$ gives four equations:

$$E{w}_1=m{w}_1,E{w}_2=m{w}_2,E{w}_3=-m{w}_3,E{w}_4=-m{w}_4.$$

For a nonzero state, at least one component must survive. An upper component requires $E=m$; a lower component requires $E=-m$. Assuming $m>0$, the possibilities are:

1. $E\ne \pm m$: all four components vanish, so there is no nonzero state.
2. $E=m$: $w_3=w_4=0$, while the two upper components are free.
3. $E=-m$: $w_1=w_2=0$, while the two lower components are free.

These are the two signs in $E^2={\mathbf{p}}^2+m^2$ from Relativistic QM §2, evaluated at rest. Squaring the Dirac Hamiltonian gives this dispersion relation because its matrices anticommute.

Each energy eigenspace has dimension two. A basis of the four independent rest solutions is

$$E=+m:w=\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right);E=-m:w=\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right).$$

The four-component size follows from the matrix algebra in Relativistic QM §4. The Hamiltonian needs four mutually anticommuting matrices. The three Pauli matrices cannot accommodate a fourth in two dimensions; the smallest suitable complex representation is four-dimensional.

Choose the basis to diagonalize $S_z=\frac{\hslash }{2}{\Sigma }_z$. Label the rest solutions $w_{+}^{(1)},w_{+}^{(2)},w_{-}^{(1)},w_{-}^{(2)}$. Since ${\Sigma }_z=\mathrm{diag}(1,-1,1,-1)$,

$$S_z{w}_{\pm }^{(1)}=+\frac{\hslash }{2}{w}_{\pm }^{(1)},S_z{w}_{\pm }^{(2)}=-\frac{\hslash }{2}{w}_{\pm }^{(2)},$$

Each energy sign therefore has one spin-up and one spin-down basis state. The choice of $z$ is a basis choice; we could diagonalize spin along any axis.

Moving ($\mathbf{p}\ne 0$). Substituting a plane wave $\psi =w(p)e^{-ip\cdot x/\hslash }$ turns the differential equation into

$$(E-\mathbf{\alpha }\cdot \mathbf{p}-\beta m)w=0,$$

Write $w=(\varphi ,\chi )$ for the upper and lower two-component blocks, as in Relativistic QM §4. Then

$$(E-m)\varphi =\mathbf{\sigma }\cdot \mathbf{p}\chi ,(E+m)\chi =\mathbf{\sigma }\cdot \mathbf{p}\varphi .$$

For $E=E_p=\sqrt{{\mathbf{p}}^2+m^2}$, the second equation gives $\chi =\frac{\mathbf{\sigma }\cdot \mathbf{p}}{E_p+m}\varphi$. Choose either upper spin state, ${\chi }_{\uparrow }=(1,0{)}^T$ or ${\chi }_{\downarrow }=(0,1{)}^T$. With standard normalization, the two positive-energy spinors are

$$u_s(p)=\left(\begin{array}{c}\sqrt{E_p+m}{\chi }_s\\ \frac{\mathbf{\sigma }\cdot \mathbf{p}}{\sqrt{E_p+m}}{\chi }_s\end{array}\right)(E=+E_p),$$

At $\mathbf{p}=0$, these reduce to the positive-energy rest states up to normalization. For $E=-E_p$, the first equation instead gives $\varphi =-\frac{\mathbf{\sigma }\cdot \mathbf{p}}{E_p+m}\chi$. At this same spatial momentum, the negative-energy spinors are

$$v_s(p)=\left(\begin{array}{c}-\frac{\mathbf{\sigma }\cdot \mathbf{p}}{\sqrt{E_p+m}}{\chi }_s\\ \sqrt{E_p+m}{\chi }_s\end{array}\right)(E=-E_p),$$

Their lower blocks dominate at low momentum, as in Relativistic QM §4.

To see the map in coordinates, use the two-component form. A momentum-space plane wave $w=(\varphi ,\chi )$ solves the coupled equations

$$(E-m)\varphi =\mathbf{\sigma }\cdot \mathbf{p}\chi ,(E+m)\chi =\mathbf{\sigma }\cdot \mathbf{p}\varphi .$$

For a positive-energy solution, $\varphi$ is the large upper pair and $\chi$ the small lower pair. Now conjugate both equations. The Pauli matrices obey ${\sigma }_1^{\ast }={\sigma }_1$, ${\sigma }_3^{\ast }={\sigma }_3$, and ${\sigma }_2^{\ast }=-{\sigma }_2$, so ${\mathbf{\sigma }}^{\ast }=({\sigma }_1,-{\sigma }_2,{\sigma }_3)$, and

$$(E-m){\varphi }^{\ast }={\mathbf{\sigma }}^{\ast }\cdot \mathbf{p}{\chi }^{\ast },(E+m){\chi }^{\ast }={\mathbf{\sigma }}^{\ast }\cdot \mathbf{p}{\varphi }^{\ast }.$$

The matrix $i{\sigma }^2=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ satisfies $(i{\sigma }^2){\mathbf{\sigma }}^{\ast }(i{\sigma }^2{)}^{-1}=-\mathbf{\sigma }$, the property that will flip the frequency sign. In the standard representation $\eta =i{\gamma }^2$ has the block form $\eta =\left(\begin{array}{cc}0& i{\sigma }^2\\ -i{\sigma }^2& 0\end{array}\right)$, so applying it to $w^{\ast }$ gives

$${\psi }_c=\eta w^{\ast }=\left(\begin{array}{c}i{\sigma }^2{\chi }^{\ast }\\ -i{\sigma }^2{\varphi }^{\ast }\end{array}\right)\equiv \left(\begin{array}{c}{\varphi }_c\\ {\chi }_c\end{array}\right).$$

Multiply the conjugated equations by $i{\sigma }^2$. Because $(i{\sigma }^2)({\mathbf{\sigma }}^{\ast }\cdot \mathbf{p})=-(\mathbf{\sigma }\cdot \mathbf{p})(i{\sigma }^2)$, they become

$$(E+m){\varphi }_c=\mathbf{\sigma }\cdot \mathbf{p}{\chi }_c,(E-m){\chi }_c=\mathbf{\sigma }\cdot \mathbf{p}{\varphi }_c.$$

Multiplying both equations by $-1$ puts them in the original form with energy $-E$ and momentum $-\mathbf{p}$. Complex conjugation reverses both signs in the plane-wave phase as well. Thus the transformed spinor is proportional to $v_{s^{\prime }}(-p)$ in the fixed-momentum convention of §3, with the spin label interchanged. The reversal of momentum is essential to this identification.

To restore charge, reinsert the coupling ${\partial }_{\mu }\to {\partial }_{\mu }+\frac{iq}{\hslash }{A}_{\mu }$ from §1. Conjugation reverses the sign of every $i$, so ${\psi }_c$ obeys the equation with $q\to -q$: the antiparticle, with opposite charge and the same mass.

Quantize the amplitudes. The QFT page replaces the coefficients of §6 by operators. Before reordering, one mode contributes $E_p({\hat{a}}^{†}\hat{a}-\hat{b}{\hat{b}}^{†})$ to the Hamiltonian.

Fermion operators obey $\hat{b}{\hat{b}}^{†}=1-{\hat{b}}^{†}\hat{b}$. Substitution gives $E_p({\hat{a}}^{†}\hat{a}+{\hat{b}}^{†}\hat{b})$ plus a vacuum constant. After subtracting that constant, electrons and positrons both contribute positive energy $E_p$.

Thus ${\hat{b}}^{†}$ multiplies a negative-frequency solution but creates a positive-energy positron. This implements the reinterpretation in Relativistic QM §3.

This optional derivation checks that the spinor transformation preserves the Dirac equation. Later sections use the result without requiring these calculations.

Impose covariance. If $\psi$ solves $(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, require ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ to solve the primed equation. The derivative transforms as ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. Substitute and multiply on the left by $S^{-1}$:

$$(i\hslash S^{-1}{\gamma }^{\mu }S({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }-m)\psi =0,$$

To recover the original equation, require

$$S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }.$$

First consider a small transformation, ${\Lambda }^{\mu }{}_{\nu }={\delta }_{\nu }^{\mu }+{\omega }^{\mu }{}_{\nu }$, with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. Write $S=1+\Omega$ and keep first-order terms. Since $S^{-1}{\gamma }^{\mu }S={\gamma }^{\mu }-[\Omega ,{\gamma }^{\mu }]$, covariance becomes

$$-[\Omega ,{\gamma }^{\mu }]={\omega }^{\mu }{}_{\nu }{\gamma }^{\nu }.$$

Test $\Omega =-\frac{i}{4}{\omega }_{\rho \sigma }{\sigma }^{\rho \sigma }$. Use the matrix identity $[AB,C]=A\{B,C\}-\{A,C\}B$ and the gamma-matrix anticommutator $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$:

$$[{\gamma }^{\rho }{\gamma }^{\sigma },{\gamma }^{\mu }]={\gamma }^{\rho }\{{\gamma }^{\sigma },{\gamma }^{\mu }\}-\{{\gamma }^{\rho },{\gamma }^{\mu }\}{\gamma }^{\sigma }=2(g^{\mu \sigma }{\gamma }^{\rho }-g^{\mu \rho }{\gamma }^{\sigma }).$$

The definition ${\sigma }^{\rho \sigma }=\frac{i}{2}[{\gamma }^{\rho },{\gamma }^{\sigma }]$ contains both orderings. Subtract them to obtain

$$[{\sigma }^{\rho \sigma },{\gamma }^{\mu }]=\frac{i}{2}([{\gamma }^{\rho }{\gamma }^{\sigma },{\gamma }^{\mu }]-[{\gamma }^{\sigma }{\gamma }^{\rho },{\gamma }^{\mu }])=2i(g^{\mu \sigma }{\gamma }^{\rho }-g^{\mu \rho }{\gamma }^{\sigma }).$$

Insert this into the covariance condition:

$$-[\Omega ,{\gamma }^{\mu }]=\frac{i}{4}{\omega }_{\rho \sigma }\cdot 2i(g^{\mu \sigma }{\gamma }^{\rho }-g^{\mu \rho }{\gamma }^{\sigma })=-\frac{1}{2}{\omega }_{\rho \sigma }(g^{\mu \sigma }{\gamma }^{\rho }-g^{\mu \rho }{\gamma }^{\sigma })={\omega }^{\mu }{}_{\nu }{\gamma }^{\nu },$$

In the last step, use $g^{\mu \sigma }{\omega }_{\rho \sigma }=-{\omega }^{\mu }{}_{\rho }$. The two contributions add, giving the required sign.

This verifies the infinitesimal transformation. Repeating a transformation along a fixed generator gives its finite exponential. Different generators generally do not commute, so composing arbitrary rotations and boosts requires matrix multiplication rather than adding their parameters.

The following calculations apply the formula to a rotation and a boost. These transformations generate the Lorentz transformations connected to the identity.

Matrix exponentials. Write $S=e^{\Omega }$, where $\Omega =-\frac{i}{4}{\sum }_{\mu ,\nu }{\omega }_{\mu \nu }{\sigma }^{\mu \nu }$. Define the exponential by $e^M=1+M+M^2/2!+\cdots$, with $1$ the identity matrix.

We will use two properties. Taking an adjoint gives $(e^M{)}^{†}=e^{M^{†}}$ because $(M^n{)}^{†}=(M^{†}{)}^n$. Also, $e^M{e}^{-M}=1$, so $e^{-M}$ is the inverse.

Parameters and generators. The ${\omega }_{\mu \nu }$ are numbers specifying the transformation. Near the identity, $\Lambda =1+\omega$. Lower the first index using ${\omega }_{\mu \nu }=g_{\mu \rho }{\omega }^{\rho }{}_{\nu }$, with metric $g=\mathrm{diag}(1,-1,-1,-1)$.

Expanding the Lorentz condition ${\Lambda }^{T}g\Lambda =g$ to first order gives ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. There are six independent parameters. The pairs $(0,1),(0,2),(0,3)$ mix time and space and describe boosts. The spatial pairs $(1,2),(1,3),(2,3)$ describe rotations.

The ${\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }]$ are fixed $4\times 4$ matrices, called generators because they determine the infinitesimal change of the spinor. Multiplying them by the parameters and exponentiating gives a finite transformation.

They are also antisymmetric in their indices, so only six are independent. Their values do not depend on which transformation we choose.

In the double sum, reversing $(\mu ,\nu )$ changes both signs, so each pair contributes twice. The diagonal terms vanish. Therefore

$$\Omega =-\frac{i}{4}\sum _{\mu ,\nu }{\omega }_{\mu \nu }{\sigma }^{\mu \nu }=-\frac{i}{2}\sum _{\mu <\nu }{\omega }_{\mu \nu }{\sigma }^{\mu \nu }.$$

For each example, find the nonzero parameters, compute their generators, and exponentiate the resulting matrix.

Rotation. A spatial rotation leaves time unchanged, so ${\omega }_{0i}=0$. Only the three spatial pairs remain:

$${\Omega }_{\text{rot}}=-\frac{i}{2}\left({\omega }_{12}{\sigma }^{12}+{\omega }_{13}{\sigma }^{13}+{\omega }_{23}{\sigma }^{23}\right).$$

We need the generators ${\sigma }^{ij}=\frac{i}{2}({\gamma }^i{\gamma }^j-{\gamma }^j{\gamma }^i)$. Compute the gamma-matrix products first.

Use the standard representation from Relativistic QM §4: ${\gamma }^0=\beta$ and ${\gamma }^i=\beta {\alpha }_i$. The matrices from §1 have the following block form, with each entry a $2\times 2$ block and ${\sigma }_i$ the Pauli matrices:

$$\beta =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),{\alpha }_i=\left(\begin{array}{cc}0& {\sigma }_i\\ {\sigma }_i& 0\end{array}\right).$$

Multiply the blocks in row-by-column order, keeping their order within each product:

$${\gamma }^i=\beta {\alpha }_i=\left(\begin{array}{cc}1\cdot 0+0\cdot {\sigma }_i& 1\cdot {\sigma }_i+0\cdot 0\\ 0\cdot 0+(-1)\cdot {\sigma }_i& 0\cdot {\sigma }_i+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& {\sigma }_i\\ -{\sigma }_i& 0\end{array}\right).$$

For $i\ne j$, multiply two spatial gamma matrices. The off-diagonal blocks vanish:

$${\gamma }^i{\gamma }^j=\left(\begin{array}{cc}0\cdot 0+{\sigma }_i\cdot (-{\sigma }_j)& 0\cdot {\sigma }_j+{\sigma }_i\cdot 0\\ (-{\sigma }_i)\cdot 0+0\cdot (-{\sigma }_j)& (-{\sigma }_i)\cdot {\sigma }_j+0\cdot 0\end{array}\right)=\left(\begin{array}{cc}-{\sigma }_i{\sigma }_j& 0\\ 0& -{\sigma }_i{\sigma }_j\end{array}\right).$$

Subtract the reverse product. Each diagonal block becomes the negative Pauli commutator:

$$[{\gamma }^i,{\gamma }^j]=\left(\begin{array}{cc}-[{\sigma }_i,{\sigma }_j]& 0\\ 0& -[{\sigma }_i,{\sigma }_j]\end{array}\right).$$

From §1, $[{\sigma }_i,{\sigma }_j]=2i{\epsilon }_{ijk}{\sigma }_k$. The symbol ${\epsilon }_{ijk}$ is $+1$ for $(1,2,3)$, changes sign under an index exchange, and vanishes when indices repeat. Grouping the repeated blocks into the spin matrix ${\Sigma }^k=\mathrm{diag}({\sigma }_k,{\sigma }_k)$ gives

$$[{\gamma }^i,{\gamma }^j]=-2i{\epsilon }_{ijk}{\Sigma }^k.$$

Multiply by $i/2$ to obtain the generator. Since $(i/2)(-2i)=1$,

$${\sigma }^{ij}=\frac{i}{2}[{\gamma }^i,{\gamma }^j]={\epsilon }_{ijk}{\Sigma }^k.$$

Thus ${\sigma }^{12}={\Sigma }^3$, ${\sigma }^{13}=-{\Sigma }^2$, and ${\sigma }^{23}={\Sigma }^1$. Each generator corresponds to rotation about the axis perpendicular to its coordinate plane. All three are Hermitian because the Pauli matrices are Hermitian.

Substitute these generators into the exponent:

$${\Omega }_{\text{rot}}=-\frac{i}{2}\left({\omega }_{12}{\Sigma }^3-{\omega }_{13}{\Sigma }^2+{\omega }_{23}{\Sigma }^1\right).$$

The parameters are real, so the matrix in parentheses is Hermitian. Multiplication by $-i$ makes ${\Omega }_{\mathrm{rot}}$ anti-Hermitian, meaning ${\Omega }_{\mathrm{rot}}^{†}=-{\Omega }_{\mathrm{rot}}$.

Consequently $S^{†}=e^{-{\Omega }_{\mathrm{rot}}}=S^{-1}$. The rotation matrix is unitary.

Collect the angles into $\mathbf{\theta }=({\omega }_{23},-{\omega }_{13},{\omega }_{12})$. Each entry specifies rotation about the corresponding axis. The exponent becomes a dot product:

$$S=\exp \left(-\frac{i}{2}\mathbf{\theta }\cdot \mathbf{\Sigma }\right),$$

This is the rotation formula used above.

For a rotation about $z$, set ${\omega }_{12}=\theta$ and the other angles to zero. Since ${\Sigma }^3=\mathrm{diag}(1,-1,1,-1)$, exponentiate each diagonal entry:

$$S=\mathrm{diag}\left(e^{-i\theta /2},e^{i\theta /2},e^{-i\theta /2},e^{i\theta /2}\right).$$

Spin up and spin down acquire opposite phases. At $\theta =2\pi$, every entry is $-1$, so the spinor changes sign. At $4\pi$ it returns exactly, confirming the property described in §5.

Boost along $x$. A boost mixes time with one spatial coordinate. For motion along $x$, only the pair $(0,1)$ contributes. We first express its parameter in terms of the frame's velocity.

The coordinate transformation from Special Relativity is $t^{\prime }=\gamma (t-vx)$ and $x^{\prime }=\gamma (x-vt)$, using $c=1$. Define rapidity $\phi$ by $v=\tanh \phi$. Then $\gamma =\cosh \phi$ and $\gamma v=\sinh \phi$, giving

$$t^{\prime }=t\cosh \phi -x\sinh \phi ,x^{\prime }=x\cosh \phi -t\sinh \phi .$$

For small $\phi$, these become $t^{\prime }\approx t-\phi x$ and $x^{\prime }\approx x-\phi t$. Read the off-diagonal entries of $\Lambda =1+\omega$ from those coefficients:

$${\omega }^0{}_1=-\phi ,{\omega }^1{}_0=-\phi .$$

Lowering the first index uses $g_{00}=1$, so

$${\omega }_{01}=g_{00}{\omega }^0{}_1=-\phi .$$

Only one pair remains in the reduced exponent:

$${\Omega }_{\text{boost}}=-\frac{i}{2}{\omega }_{01}{\sigma }^{01}.$$

Compute ${\sigma }^{01}=\frac{i}{2}[{\gamma }^0,{\gamma }^1]$. Use ${\gamma }^0=\beta$ and the block form of ${\gamma }^1$ found above:

$${\gamma }^0{\gamma }^1=\left(\begin{array}{cc}1\cdot 0+0\cdot (-{\sigma }_1)& 1\cdot {\sigma }_1+0\cdot 0\\ 0\cdot 0+(-1)\cdot (-{\sigma }_1)& 0\cdot {\sigma }_1+(-1)\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& {\sigma }_1\\ {\sigma }_1& 0\end{array}\right).$$

This is ${\alpha }_1$ from §1, also written ${\alpha }^1$ here. Reversing the multiplication gives

$${\gamma }^1{\gamma }^0=\left(\begin{array}{cc}0\cdot 1+{\sigma }_1\cdot 0& 0\cdot 0+{\sigma }_1\cdot (-1)\\ (-{\sigma }_1)\cdot 1+0\cdot 0& (-{\sigma }_1)\cdot 0+0\cdot (-1)\end{array}\right)=\left(\begin{array}{cc}0& -{\sigma }_1\\ -{\sigma }_1& 0\end{array}\right)=-{\alpha }^1,$$

The two gamma matrices anticommute, so

$$[{\gamma }^0,{\gamma }^1]={\gamma }^0{\gamma }^1-{\gamma }^1{\gamma }^0={\alpha }^1-(-{\alpha }^1)=2{\alpha }^1,$$

Substituting into the generator definition gives

$${\sigma }^{01}=\frac{i}{2}[{\gamma }^0,{\gamma }^1]=i{\alpha }^1.$$

The matrix ${\alpha }^1$ is Hermitian. The factor $i$ therefore makes ${\sigma }^{01}$ anti-Hermitian:

$$({\sigma }^{01}{)}^{†}=(i{\alpha }^1{)}^{†}=-i{\alpha }^1=-{\sigma }^{01}.$$

This differs from the Hermitian rotation generators.

Now insert the parameter and generator into the exponent:

$${\Omega }_{\text{boost}}=-\frac{i}{2}{\omega }_{01}{\sigma }^{01}=-\frac{i}{2}(-\phi )(i{\alpha }^1).$$

The scalar coefficient is $(-i/2)(-\phi )i=-\phi /2$. The two factors of $i$ produce a real exponent:

$$S=\exp \left(-\frac{\phi }{2}{\alpha }^1\right).$$

Here the exponent is Hermitian, so $S^{†}=S$. A nonzero boost is not unitary as a matrix acting on these four components: $S^{†}S=e^{-\phi {\alpha }^1}\ne 1$. This does not imply that Lorentz transformations fail to preserve the physical Hilbert-space norm.

To evaluate the exponential, use $({\alpha }^1{)}^2=1$. Even powers contribute the identity and odd powers contribute ${\alpha }^1$:

$$e^{-\phi {\alpha }^1/2}=\underset{\cosh (\phi /2)}{\underbrace{[1+\frac{(\phi /2{)}^2}{2!}+\cdots ]}}-{\alpha }^1\underset{\sinh (\phi /2)}{\underbrace{[\frac{\phi }{2}+\frac{(\phi /2{)}^3}{3!}+\cdots ]}},$$

Thus $S=\cosh (\phi /2)-{\alpha }^1\sinh (\phi /2)$.

At small velocity, $S\approx 1-\frac{\phi }{2}{\alpha }^1$. Its off-diagonal blocks mix the upper and lower spinor components with relative size $\phi /2\approx v/2$. Since $|\mathbf{p}|\approx m|v|$, this gives the small-component size $|\mathbf{p}|/(2m)$ obtained in §3. It provides an independent check of those moving solutions.

For a boost along a unit vector $\hat{\mathbf{n}}$, use ${\omega }_{0k}=-\phi {\hat{n}}_k$ and ${\sigma }^{0k}=i{\alpha }^k$. Then $S=\exp [-\frac{\phi }{2}\mathbf{\alpha }\cdot \hat{\mathbf{n}}]$.

In this representation, pure rotations give unitary component matrices and pure boosts give Hermitian component matrices. Their different factors of $i$ account for the different exponential forms.

For the negative-frequency phase written here, the spatial wave is $e^{-i\mathbf{p}\cdot \mathbf{x}/\hslash }$. Its classical momentum is therefore $-\mathbf{p}$. Replace $\mathbf{p}$ by $-\mathbf{p}$ in the negative-energy spinor of §3. With ${\chi }_{\uparrow }=(1,0{)}^T$ and ${\chi }_{\downarrow }=(0,1{)}^T$, the spinors in this expansion are

$$u_s(p)=\left(\begin{array}{c}\sqrt{E_p+m}{\chi }_s\\ \frac{\mathbf{\sigma }\cdot \mathbf{p}}{\sqrt{E_p+m}}{\chi }_s\end{array}\right),v_s(p)=\left(\begin{array}{c}\frac{\mathbf{\sigma }\cdot \mathbf{p}}{\sqrt{E_p+m}}{\chi }_s\\ \sqrt{E_p+m}{\chi }_s\end{array}\right),$$

The relabeling changes the sign of the upper block of $v$. The low-momentum structure remains the one found in Relativistic QM §4: $u$ has a large upper block, while $v$ has a large lower block.