Exercise: Plane-Wave Solutions of the Biquaternion Dirac Equation

Introduction

This is one of the worked exercises on the biquaternion Dirac equation. It exists to test its parent article, The Biquaternion Dirac Equation — Solutions and Non-Relativistic Limit, which supplies the field, the two natural spinor bases, the plane-wave solution structure, the covariant normalisations, the spin sums, and the biquaternionic mass-shell condition.

We assume the parent article throughout and do not re-derive the equation. We take over from it: the spinor-module form of the equation, $(i\gamma^\mu\partial_\mu-m)\psi=0$; the two bases, chiral and Dirac; the plane-wave branches $\psi=u(\mathbf p)e^{-i(Et-\mathbf p\cdot\mathbf x)}$ and $\psi=v(\mathbf p)e^{+i(Et-\mathbf p\cdot\mathbf x)}$ with $E=+\sqrt{\mathbf p^2+m^2}$; the momentum-space equations $(\not p-m)u=0$ and $(\not p+m)v=0$; the wave biquaternion and the mass-shell condition. The value of an exercise of this kind lies in the worked solutions, so each problem is carried to a definite answer.

The six problems are: (1) write out $u^{(r)}(\mathbf p)$ and $v^{(r)}(\mathbf p)$ explicitly at rest and at general momentum; (2) verify the covariant normalisations $\bar u u=2m$, $\bar v v=-2m$ and the orthogonality $\bar u v=0$; (3) verify the spin sums; (4) take the massless limit and exhibit the helicity–chirality locking; (5) verify the gamma-matrix algebra in the two representations and determine how the Lorentz generators change basis; (6) verify the biquaternionic mass-shell condition.

Conventions. These are inherited exactly from the parent article. The Clifford metric carried by the generators is

$$ g=\mathrm{diag}(+1,-1,-1,-1),\qquad \{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4, $$

so that with the four-momentum $p^\mu=(E,\mathbf p)$ and $p_\mu=(E,-\mathbf p)$, the slash is $\not p=\gamma^\mu p_\mu=\gamma^0E-\boldsymbol\gamma\cdot\mathbf p$. This $g$ is not the spacetime metric $\eta=\mathrm{diag}(-1,+1,+1,+1)$ carried by the $ict$ gradient; the two are negatives, $g=-\eta$, and they yield the same $\Box$. The Dirac adjoint is $\bar\psi=\psi^\dagger\gamma^0$. In the solution sections we use natural units $\hbar=c=1$, restoring $\hbar$ and $c$ in Problem 6. The two-spinors are $\xi^{(1)}=(1,0)^T$, $\xi^{(2)}=(0,1)^T$, with $\xi^{(r){}^{*}}\xi^{(s)}=\delta^{rs}$ and the same for $\eta^{(r)}$, and $\chi_\pm$ are the helicity eigenspinors $\boldsymbol\sigma\cdot\hat{\mathbf p}\,\chi_\pm=\pm\chi_\pm$.

Problem 1: The spinors at rest and at general momentum

Statement. Starting from $(\not p-m)u=0$ and $(\not p+m)v=0$, write out $u^{(r)}(\mathbf p)$ and $v^{(r)}(\mathbf p)$ explicitly at rest and at general momentum.

Solution. In the Dirac basis, $\gamma^0=\mathrm{diag}(I_2,-I_2)$ and $\gamma^k=\begin{pmatrix}0&\sigma_k\\-\sigma_k&0\end{pmatrix}$, so

$$ \not p-m=\begin{pmatrix}(E-m)I_2 & -\boldsymbol\sigma\cdot\mathbf p\\[2pt] \boldsymbol\sigma\cdot\mathbf p & -(E+m)I_2\end{pmatrix}. $$

Writing $u=(u_A,u_B)$, the equation $(\not p-m)u=0$ becomes the pair

$$ (E-m)u_A=\boldsymbol\sigma\cdot\mathbf p\,u_B,\qquad (E+m)u_B=\boldsymbol\sigma\cdot\mathbf p\,u_A . $$

The second equation gives $u_B=\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}u_A$, and the first is then satisfied automatically on the mass shell $E^2=\mathbf p^2+m^2$; the upper two-spinor $u_A$ is free. Choosing $u_A=\sqrt{E+m}\,\xi^{(r)}$ and using $\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}=\dfrac{\boldsymbol\sigma\cdot\mathbf p}{\sqrt{E+m}\sqrt{E+m}}$ together with $\sqrt{E^2-m^2}=|\mathbf p|$ gives the parent's positive-frequency spinors,

$$ \boxed{\;u^{(r)}(\mathbf p)=\begin{pmatrix}\sqrt{E+m}\,\xi^{(r)}\\[2pt] \sqrt{E-m}\,(\boldsymbol\sigma\cdot\hat{\mathbf p})\,\xi^{(r)}\end{pmatrix},\qquad \hat{\mathbf p}=\frac{\mathbf p}{|\mathbf p|}.} $$

For the negative-frequency branch, $(\not p+m)v=0$ reads, in the same basis,

$$ (E+m)v_A=\boldsymbol\sigma\cdot\mathbf p\,v_B,\qquad (E-m)v_B=\boldsymbol\sigma\cdot\mathbf p\,v_A, $$

so $v_A=\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}v_B$, and with $v_B=\sqrt{E+m}\,\eta^{(r)}$,

$$ \boxed{\;v^{(r)}(\mathbf p)=\begin{pmatrix}\sqrt{E-m}\,(\boldsymbol\sigma\cdot\hat{\mathbf p})\,\eta^{(r)}\\[2pt] \sqrt{E+m}\,\eta^{(r)}\end{pmatrix}.} $$

At rest. Setting $\mathbf p=0$, $E=m$, the general formulas collapse to

$$ u^{(r)}(0)=\sqrt{2m}\begin{pmatrix}\xi^{(r)}\\0\end{pmatrix},\qquad v^{(r)}(0)=\sqrt{2m}\begin{pmatrix}0\\\eta^{(r)}\end{pmatrix}, $$

that is,

$$ u^{(1)}(0)=\sqrt{2m}\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\quad u^{(2)}(0)=\sqrt{2m}\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\quad v^{(1)}(0)=\sqrt{2m}\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\quad v^{(2)}(0)=\sqrt{2m}\begin{pmatrix}0\\0\\0\\1\end{pmatrix}. $$

At rest the positive-frequency spinors live entirely in the upper pair of components and the negative-frequency spinors entirely in the lower pair.

At general momentum. Writing $\hat{\mathbf p}=(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)$,

$$ \boldsymbol\sigma\cdot\hat{\mathbf p}=\begin{pmatrix}\cos\theta & e^{-i\phi}\sin\theta\\[2pt] e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix}. $$

Substituting into the boxed formulas gives the explicit components

$$ u^{(1)}=\begin{pmatrix}\sqrt{E+m}\\0\\ \sqrt{E-m}\cos\theta\\ \sqrt{E-m}\,e^{i\phi}\sin\theta\end{pmatrix},\qquad u^{(2)}=\begin{pmatrix}0\\ \sqrt{E+m}\\ \sqrt{E-m}\,e^{-i\phi}\sin\theta\\ -\sqrt{E-m}\cos\theta\end{pmatrix}, $$

$$ v^{(1)}=\begin{pmatrix}\sqrt{E-m}\cos\theta\\ \sqrt{E-m}\,e^{i\phi}\sin\theta\\ \sqrt{E+m}\\0\end{pmatrix},\qquad v^{(2)}=\begin{pmatrix}\sqrt{E-m}\,e^{-i\phi}\sin\theta\\ -\sqrt{E-m}\cos\theta\\0\\ \sqrt{E+m}\end{pmatrix}. $$

At $\mathbf p=0$ these reduce to the rest forms, and they solve the momentum-space equations identically for all $\mathbf p$. The two free two-spinors give the two spin states of the particle (for $u$) and of the antiparticle (for $v$), so the positive- and negative-frequency solution spaces are each two-dimensional over $\mathbb C$, and together they span the four-dimensional amplitude space.

Problem 2: Covariant normalisation and orthogonality

Statement. Verify $\bar u^{(r)}u^{(s)}=2m\,\delta^{rs}$, $\bar v^{(r)}v^{(s)}=-2m\,\delta^{rs}$, and $\bar u^{(r)}v^{(s)}=0$, together with the Hermitian products.

Solution. The Dirac adjoint in the Dirac basis is $\bar u=u^\dagger\gamma^0=(u_A^\dagger,-u_B^\dagger)$, so $\bar u\,u=u_A^\dagger u_A-u_B^\dagger u_B$. The single algebraic input is the identity for unit $\hat{\mathbf p}$,

$$ (\boldsymbol\sigma\cdot\hat{\mathbf p})^2=|\hat{\mathbf p}|^2I_2=I_2. $$

Then

$$ \bar u^{(r)}u^{(s)} =(E+m)\,\xi^{(r){}^{*}}\xi^{(s)}-(E-m)\,\xi^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})^2\xi^{(s)} =(E+m)\delta^{rs}-(E-m)\delta^{rs}=2m\,\delta^{rs}. $$

For the negative-frequency branch the order of the two terms is reversed,

$$ \bar v^{(r)}v^{(s)} =(E-m)\,\eta^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})^2\eta^{(s)}-(E+m)\,\eta^{(r){}^{*}}\eta^{(s)} =(E-m)\delta^{rs}-(E+m)\delta^{rs}=-2m\,\delta^{rs}. $$

This relative minus sign is the standard signature of the negative-frequency branch. For the cross term,

$$ \bar u^{(r)}v^{(s)} =\sqrt{E^2-m^2}\,\xi^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})\eta^{(s)} -\sqrt{E^2-m^2}\,\xi^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})\eta^{(s)}=0, $$

since $\boldsymbol\sigma\cdot\hat{\mathbf p}$ is Hermitian and the two terms cancel exactly. The Hermitian products follow from the same formulas without the $\gamma^0$:

$$ u^{(r){}^{*}}u^{(s)}=(E+m)\delta^{rs}+(E-m)\delta^{rs}=2E\,\delta^{rs}, $$

$$ v^{(r){}^{*}}v^{(s)}=(E-m)\delta^{rs}+(E+m)\delta^{rs}=2E\,\delta^{rs}. $$

Finally, the mixed Hermitian product of the parent article requires opposite momenta and likewise vanishes:

$$ u^{(r){}^{*}}(\mathbf p)\,v^{(s)}(-\mathbf p) =\sqrt{E^2-m^2}\,\xi^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})\eta^{(s)} -\sqrt{E^2-m^2}\,\xi^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})\eta^{(s)}=0 . $$

The bilinear $\bar u u$ is a Lorentz scalar, and its value $2m$ is the covariant normalisation; $\bar u u$ is not the probability density, which is the positive quantity $u^\dagger u=2E$. As a numerical check, for $m=1$, $|\mathbf p|=0.6$ (hence $E=1.16619$), direct evaluation gives $\bar u u=2.0000$, $\bar v v=-2.0000$, $\bar u v=0$, and $u^\dagger u=v^\dagger v=2.3324=2E$, all to machine precision.

Problem 3: The spin sums

Statement. Verify the completeness relations (spin sums) $\sum_{r=1}^{2}u^{(r)}\bar u^{(r)}=\not p+m$ and $\sum_{r=1}^{2}v^{(r)}\bar v^{(r)}=\not p-m$.

Solution. The spin sum is the outer product $u\bar u$, which is a $4\times4$ matrix. Using $\bar u=(u_A^\dagger,-u_B^\dagger)$ and the two-spinor completeness $\sum_r\xi^{(r)}\xi^{(r){}^{*}}=I_2$,

$$ u^{(r)}\bar u^{(r)} =\begin{pmatrix}u_Au_A^\dagger & -u_Au_B^\dagger\\[2pt] u_Bu_A^\dagger & -u_Bu_B^\dagger\end{pmatrix} \;\xrightarrow[\text{sum over }r]{}\; \begin{pmatrix}(E+m)I_2 & -(\boldsymbol\sigma\cdot\mathbf p)\\[2pt] \boldsymbol\sigma\cdot\mathbf p & -(E-m)I_2\end{pmatrix}, $$

where in the off-diagonal blocks we used $\sum_r\xi^{(r)}\xi^{(r){}^{*}}(\boldsymbol\sigma\cdot\hat{\mathbf p})=\boldsymbol\sigma\cdot\hat{\mathbf p}$ and its Hermitian conjugate. In the Dirac basis,

$$ \not p+m=\gamma^0E-\boldsymbol\gamma\cdot\mathbf p+m =\begin{pmatrix}(E+m)I_2 & -\boldsymbol\sigma\cdot\mathbf p\\[2pt] \boldsymbol\sigma\cdot\mathbf p & -(E-m)I_2\end{pmatrix}, $$

since $-\boldsymbol\gamma\cdot\mathbf p=\begin{pmatrix}0&-\boldsymbol\sigma\cdot\mathbf p\\ \boldsymbol\sigma\cdot\mathbf p&0\end{pmatrix}$ and $-E+m=-(E-m)$. This is exactly the matrix obtained above, so

$$ \sum_{r=1}^{2}u^{(r)}(\mathbf p)\,\bar u^{(r)}(\mathbf p)=\not p+m . $$

The negative-frequency sum is identical in structure with the roles of the two blocks interchanged. Carrying it out with $\sum_r\eta^{(r)}\eta^{(r){}^{*}}=I_2$,

$$ \sum_{r=1}^{2}v^{(r)}(\mathbf p)\,\bar v^{(r)}(\mathbf p) =\begin{pmatrix}-(E-m)I_2 & -\boldsymbol\sigma\cdot\mathbf p\\[2pt] \boldsymbol\sigma\cdot\mathbf p & -(E+m)I_2\end{pmatrix} =\not p-m . $$

The two spin sums are consistent with the normalisations of Problem 2, since taking the trace of $\sum u\bar u$ over the particle space returns $4m$, the trace of $\not p+m$, and the projectors annihilate one another on shell:

$$ (\not p+m)(\not p-m)=\not p^2-m^2=(p_\mu p_\nu\gamma^\mu\gamma^\nu)-m^2=(p^2-m^2)I_4=0, $$

using $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ and $p^2=E^2-\mathbf p^2=m^2$. This is the algebraic statement that the two branches close the amplitude space.

Check in the chiral basis. The same relations hold in the chiral basis with the helicity index $\pm$ replacing $r$, once the negative-frequency chiral spinors of Problem 4 are used: $\sum_\pm u_\pm\bar u_\pm=\not p+m$ and $\sum_\pm v_\pm\bar v_\pm=\not p-m$, where the sums use $\sum_\pm\chi_\pm\chi_\pm^\dagger=I_2$ and $\sum_\pm(\pm1)\chi_\pm\chi_\pm^\dagger=\boldsymbol\sigma\cdot\hat{\mathbf p}$. A numerical evaluation at the same kinematics confirms both equalities to machine precision.

Problem 4: The massless limit and helicity–chirality locking

Statement. In the chiral basis, exhibit the massless limit and show how helicity locks to chirality.

Solution. In the chiral basis, $\gamma^0=\begin{pmatrix}0&I_2\\I_2&0\end{pmatrix}$, $\gamma^k=\begin{pmatrix}0&\sigma_k\\-\sigma_k&0\end{pmatrix}$, so

$$ \not p=\begin{pmatrix}0 & E-\boldsymbol\sigma\cdot\mathbf p\\[2pt] E+\boldsymbol\sigma\cdot\mathbf p & 0\end{pmatrix},\qquad \not p-m=\begin{pmatrix}-m & E-\boldsymbol\sigma\cdot\mathbf p\\[2pt] E+\boldsymbol\sigma\cdot\mathbf p & -m\end{pmatrix}. $$

The parent article gives the positive-frequency chiral spinors, indexed by helicity,

$$ u_\pm(\mathbf p)=\begin{pmatrix}\sqrt{E\mp|\mathbf p|}\;\chi_\pm\\[2pt] \sqrt{E\pm|\mathbf p|}\;\chi_\pm\end{pmatrix}, $$

with the upper block left-handed and the lower block right-handed. Verifying the equation is immediate: the upper component of $(\not p-m)u_\pm$ is $-m\sqrt{E\mp|\mathbf p|}+(E\mp|\mathbf p|)\sqrt{E\pm|\mathbf p|}$, which vanishes because $(E\mp|\mathbf p|)\sqrt{E\pm|\mathbf p|}=m\sqrt{E\mp|\mathbf p|}$; the lower component vanishes identically by the same identity.

The parent article displays the chiral positive-frequency spinors but does not write the chiral negative-frequency spinors; we construct them here, since the massless limit needs them. With the ansatz $v_\pm=(a\chi_\pm,b\chi_\pm)$, the equation $(\not p+m)v_\pm=0$ requires $ma=-(E\mp|\mathbf p|)b$ and $mb=-(E\pm|\mathbf p|)a$. The choice $b=\sqrt{E\pm|\mathbf p|}$ gives $a=-\sqrt{E\mp|\mathbf p|}$, so

$$ v_\pm(\mathbf p)=\begin{pmatrix}-\sqrt{E\mp|\mathbf p|}\,\chi_\pm\\[2pt] \sqrt{E\pm|\mathbf p|}\,\chi_\pm\end{pmatrix}. $$

The relative minus sign is fixed by the normalisation $\bar v_\pm v_\pm=-2m$; with it, the chiral spinors reproduce all the results of Problems 2 and 3.

Massless limit. Set $m=0$, so that $E=|\mathbf p|$. Then

$$ u_+(\mathbf p)\longrightarrow\sqrt{2E}\begin{pmatrix}0\\\chi_+\end{pmatrix},\qquad u_-(\mathbf p)\longrightarrow\sqrt{2E}\begin{pmatrix}\chi_-\\0\end{pmatrix}, $$

and likewise for $v_\pm$. Each solution becomes a single Weyl spinor of definite chirality, and the two labels are locked: positive helicity goes with right-handedness and negative helicity with left-handedness. This is the precise sense of the parent's statement that a massless fermion of definite helicity has definite chirality. The locking is what the mass term destroys: in the chiral basis the mass term is the off-diagonal entry $-m$ in $\not p-m$, and it is the only term in the equation that couples the two chiralities.

The Weyl equations. On shell, $\not p\,\psi=0$ splits into the two equations $(E-\boldsymbol\sigma\cdot\mathbf p)\psi_R=0$ and $(E+\boldsymbol\sigma\cdot\mathbf p)\psi_L=0$. With $E=|\mathbf p|$ these read

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

which is the helicity–chirality locking again. Restoring $c$ and $\hbar$ and passing to the time-dependent form gives the parent's Weyl equations $i\hbar\,\partial_t\psi_R=c\,\boldsymbol\sigma\cdot\hat{\mathbf p}\,\psi_R$ and $i\hbar\,\partial_t\psi_L=-c\,\boldsymbol\sigma\cdot\hat{\mathbf p}\,\psi_L$, where $\hat{\mathbf p}$ is the momentum operator (the parent uses the same symbol for the unit vector in the plane-wave formulas; in the operator equations it is $\hat{\mathbf p}=-i\hbar\nabla$). Each Weyl equation has a two-dimensional solution space, and the Dirac equation is recovered as the pair.

Problem 5: The gamma-matrix algebra and the change of basis

Statement. Verify the Clifford algebra in the chiral and Dirac representations, identify $\gamma_5$, and determine how the Lorentz generators change basis.

Solution. In both representations the gamma matrices have the block forms

$$ \text{chiral:}\quad \gamma^0_c=\begin{pmatrix}0&I_2\\I_2&0\end{pmatrix},\quad \gamma^k_c=\begin{pmatrix}0&\sigma_k\\-\sigma_k&0\end{pmatrix}; $$

$$ \text{Dirac:}\quad \gamma^0_d=\begin{pmatrix}I_2&0\\0&-I_2\end{pmatrix},\quad \gamma^k_d=\begin{pmatrix}0&\sigma_k\\-\sigma_k&0\end{pmatrix}. $$

The spatial generators are the same in the two representations; only $\gamma^0$ differs. Direct multiplication gives $(\gamma^0)^2=I_4$, $(\gamma^k)^2=-I_4$, $\gamma^0\gamma^k=-\gamma^k\gamma^0$, and $\gamma^j\gamma^k=-\gamma^k\gamma^j$ for $j\neq k$ (the last because $\sigma_j\sigma_k=-\sigma_k\sigma_j$ off diagonal). Hence $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4$ with $g=\mathrm{diag}(+1,-1,-1,-1)$ in both bases, as the parent states. The chirality operator is

$$ \gamma_5=i\gamma^0\gamma^1\gamma^2\gamma^3: \qquad \gamma_5^c=\begin{pmatrix}-I_2&0\\0&I_2\end{pmatrix},\qquad \gamma_5^d=\begin{pmatrix}0&I_2\\I_2&0\end{pmatrix}. $$

In both bases $\gamma_5^2=I_4$ and $\gamma_5$ anticommutes with every generator, $\{\gamma_5,\gamma^\mu\}=0$; the projectors $P_L=\tfrac12(1-\gamma_5)$, $P_R=\tfrac12(1+\gamma_5)$ select the chiralities, and in the chiral basis they are simply the two diagonal blocks.

Change of basis. The parent states that the two bases are related by a fixed unitary transformation but does not exhibit it; we construct it here. Define

$$ U=\frac{1}{\sqrt2}\begin{pmatrix}I_2&I_2\\-I_2&I_2\end{pmatrix},\qquad U^\dagger=\frac{1}{\sqrt2}\begin{pmatrix}I_2&-I_2\\I_2&I_2\end{pmatrix}. $$

Then $U^\dagger U=I_4$, so $U$ is unitary, and a direct computation gives

$$ U\gamma^\mu_c\,U^\dagger=\gamma^\mu_d,\qquad U\gamma_5^c\,U^\dagger=\gamma_5^d . $$

The two ingredients are that $U$ commutes with each spatial $\gamma^k_c$, so the spatial generators are unchanged, and that $U\gamma^0_cU^\dagger$ produces the diagonal $\gamma^0_d$ and swaps the diagonal $\gamma_5^c$ for the off-diagonal $\gamma_5^d$. Since $\gamma^\mu_d=U\gamma^\mu_cU^\dagger$, the spinor transforms as $\psi_d=U\psi_c$.

The Lorentz generators. The generators of the spinor representation are $S^{\mu\nu}=\tfrac{i}{4}[\gamma^\mu,\gamma^\nu]$, related to the parent's tensor bilinear by $\sigma^{\mu\nu}=\tfrac i2[\gamma^\mu,\gamma^\nu]=2S^{\mu\nu}$. In the chiral basis they evaluate to

$$ S^{jk}_c=\tfrac12\epsilon^{jkl}\begin{pmatrix}\sigma_l&0\\0&\sigma_l\end{pmatrix} =\tfrac12\epsilon^{jkl}\,\sigma_l\otimes I_2, \qquad S^{0k}_c=\tfrac i2\begin{pmatrix}-\sigma_k&0\\0&\sigma_k\end{pmatrix} =\tfrac i2\,\gamma_5^c\,(\sigma_k\otimes I_2). $$

Both are block diagonal, so both preserve chirality, $[S^{\mu\nu},\gamma_5]=0$; but they differ in their action on the two chiralities. The rotation generators are even: they act with the same sign on the left- and right-handed blocks. The boost generators are odd: they act with opposite signs, which is the algebraic reason a boost rotates the two chiralities in opposite senses while leaving each in its own chirality. For a general change of basis the generators transform by conjugation, $S^{\mu\nu}_d=U S^{\mu\nu}_c U^\dagger$, and here explicitly

$$ S^{jk}_d=S^{jk}_c,\qquad S^{0k}_d=\tfrac i2\,\gamma_5^d\,(\sigma_k\otimes I_2) =\tfrac i2\begin{pmatrix}0&\sigma_k\\\sigma_k&0\end{pmatrix}. $$

In the Dirac basis the rotation generators are again block diagonal, while the boost generators are block off-diagonal; this is the representation in which the non-relativistic reduction separates large and small components (the next exercise). Finally, the structure matches the geometric-algebra dictionary of the companion articles: the chirality-even generators are built from $\sigma_l$ alone and correspond to the real-quaternion rotation generators in $\mathbb H_{\mathbb B}$, whereas the chirality-odd boost generators carry the extra factor $\gamma_5$ and correspond to the Hermitian (boost) generators in $\mathbb M_+$.

Problem 6: The biquaternionic mass-shell condition

Statement. Verify the biquaternionic mass-shell condition of the parent article and recover the relativistic energy–momentum relation.

Solution. The parent's plane-wave ansatz carries the wave biquaternion $\tilde k$, which the parent identifies with the four-wavevector

$$ \tilde K=\frac{i\omega}{c}\,e_0+\mathbf k,\qquad \mathbf k=k_1e_1+k_2e_2+k_3e_3 . $$

This is an element of the material sector $\mathbb M_-$: imaginary scalar part, real vector part. Since the biquaternion norm of $\mathbb B$ is $N(\tilde Q)=\tilde Q\tilde Q^{\natural}=\sum_\mu Q_\mu^2$,

$$ N(\tilde K)=\tilde K\tilde K^{\natural}=\left(\frac{i\omega}{c}\right)^2+k_1^2+k_2^2+k_3^2=-\frac{\omega^2}{c^2}+\mathbf k^2 . $$

The relativistic dispersion relation is $\omega^2=c^2\mathbf k^2+\dfrac{m^2c^4}{\hbar^2}$, so

$$ N(\tilde K)=-\frac{1}{c^2}\left(c^2\mathbf k^2+\frac{m^2c^4}{\hbar^2}\right)+\mathbf k^2=-\frac{m^2c^2}{\hbar^2}, $$

which is exactly the parent's condition

$$ \boxed{\;\tilde k\tilde k^{\natural}=-\frac{m^2c^2}{\hbar^2}.\;} $$

Equivalently, with $E=\hbar\omega$ and $\mathbf p=\hbar\mathbf k$, the wave biquaternion is the four-momentum biquaternion divided by $\hbar$, $\tilde K=\tilde P/\hbar$; the biquaternion norm being quadratic, $N(\tilde K)=N(\tilde P)/\hbar^2=-m^2c^2/\hbar^2$, exactly as $N(\tilde P)=\tilde P\tilde P^{\natural}=-m^2c^2$ in the companion kinematics. The mass-shell condition is therefore not a new postulate: it is the statement that the four-wavevector is a timelike vector of $\mathbb M_-$ with the same biquaternion norm as the four-momentum.

The two roots of the dispersion relation, $\omega=\pm\sqrt{c^2\mathbf k^2+m^2c^4/\hbar^2}$, are the two frequency branches, and they are precisely the positive- and negative-frequency spinors $u$ and $v$ constructed above; the two spin labels within each branch give the four-dimensional complex solution space that the parent article quotes for the massive equation. As a numerical check, for an electron ($mc^2=0.510998950\ \mathrm{MeV}$) at $|\mathbf p|=1\ \mathrm{MeV}/c$, the dispersion relation gives $E=1.122996\ \mathrm{MeV}$ and $E^2-\mathbf p^2c^2=0.261120\ \mathrm{MeV}^2=(mc^2)^2$; the corresponding wave biquaternion has $N(\tilde K)=-6.706054\times10^{24}\ \mathrm{m}^{-2}$, agreeing with $-m^2c^2/\hbar^2$ to one part in $10^{15}$.

Summary

The six problems test the parent article's plane-wave solutions and confirm them.

1. The spinors. In the Dirac basis the solutions are $u^{(r)}=\big(\sqrt{E+m}\,\xi^{(r)},\ \sqrt{E-m}\,(\boldsymbol\sigma\cdot\hat{\mathbf p})\xi^{(r)}\big)^T$ and $v^{(r)}=\big(\sqrt{E-m}\,(\boldsymbol\sigma\cdot\hat{\mathbf p})\eta^{(r)},\ \sqrt{E+m}\,\eta^{(r)}\big)^T$, with explicit rest and general-momentum forms.
2. Normalisation. The covariant normalisations $\bar u^{(r)}u^{(s)}=2m\delta^{rs}$, $\bar v^{(r)}v^{(s)}=-2m\delta^{rs}$, and the orthogonality $\bar u^{(r)}v^{(s)}=0$ all follow from the single identity $(\boldsymbol\sigma\cdot\hat{\mathbf p})^2=I_2$; the Hermitian products are $2E\delta^{rs}$.
3. Spin sums. The completeness relations $\sum_r u^{(r)}\bar u^{(r)}=\not p+m$ and $\sum_r v^{(r)}\bar v^{(r)}=\not p-m$ follow from the two-spinor completeness relation, and satisfy $(\not p+m)(\not p-m)=0$ on shell.
4. Massless limit. In the chiral basis each helicity eigenstate becomes a single Weyl spinor, with positive helicity locking to right-handedness and negative helicity to left-handedness; the mass term is the only term coupling the two chiralities.
5. Basis change. The chiral and Dirac representations obey the same Clifford algebra with $g=\mathrm{diag}(+1,-1,-1,-1)$; the unitary $U=\tfrac{1}{\sqrt2}\left(\begin{smallmatrix}I_2&I_2\\-I_2&I_2\end{smallmatrix}\right)$ interpolates between them, and the Lorentz generators transform by conjugation, rotations acting chirality-even and boosts chirality-odd.
6. Mass shell. The biquaternionic condition $\tilde k\tilde k^{\natural}=-m^2c^2/\hbar^2$ is the biquaternion-norm statement $\tilde K=\tilde P/\hbar$, and reproduces $E^2=\mathbf p^2c^2+m^2c^4$.

Two items needed in this exercise are not displayed in the parent article: the chiral negative-frequency spinors $v_\pm$ (constructed in Problem 4) and the explicit change-of-basis matrix $U$ (constructed in Problem 5). The parent states the existence of both but does not exhibit them; the constructions above are the ones that reproduce its stated results.

Summary of Notation

Further Reading

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics (McGraw-Hill, 1964), for the plane-wave spinors, the normalisations, and the spin sums in the standard notation.
• C. Itzykson and J.-B. Zuber, Quantum Field Theory (McGraw-Hill, 1980), for the spin sums, the chiral and Dirac representations, and the Lorentz generators.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory (Addison-Wesley, 1995), for the two representations, the spinor normalisation, and the completeness relations.
• Pertti Lounesto, Clifford Algebras and Spinors (Cambridge, 2001), for the Clifford algebra and the spinor representations.
• The parent article of this exercise: The Biquaternion Dirac Equation — Solutions and Non-Relativistic Limit, and its companion The Dirac Equation in Biquaternionic Form.