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Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.

An axial vector transforms under an orthogonal matrix $Q$ asFor a proper rotation, $\det Q=1$, so this is the usual polar vector rule. Under an orientation-reversing change, the extra sign distinguishes the two. In particular, spatial inversion leaves an axial vector unchanged while reversing a polar vector.

Let $C_i=(\nabla \times u{)}_i={ϵ}_{ijk}{\partial }_j{u}_k$. The determinant identity for the Levi-Civita symbol isIt follows by multiplying the three-index determinant identity by $Q_{ir}$ and using orthogonality. The derivative transformation from part (ii) then givesHence the curl of a polar vector field is an axial vector field. The Levi-Civita determinant factor is exactly the extra sign required under reflections.

For a constant orthogonal Cartesian change of coordinates $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law for order $n$ isRepeated indices are summed. In Cartesian orthonormal coordinates the metric identifies upper and lower indices, so the same orthogonal matrix acts on every index.

An isotropic tensor has the same numerical components after every proper orthogonal change of axes: $Q^{\otimes n}T=T$ for every $Q$ with $Q^{T}Q=I$ and $\det Q=1$. For example, ${\delta }_{ij}$ is isotropic because $Q_{ik}{Q}_{jk}={\delta }_{ij}$. Some conventions require invariance under all orthogonal transformations, including reflections; this stronger convention must be distinguished for odd-order tensors, notably the Levi-Civita symbol, which is invariant under proper rotations but changes sign under its ordinary three-index transformation for a reflection.

The four displayed matrices fail closure: for example ${\sigma }_x{\sigma }_y=i{\sigma }_z$ is not one of them. Thus they do not form a group by themselves.

Let ${\sigma }_0=I$ and ${\mu }_4=\{1,-1,i,-i\}$. The enlarged set is the single-qubit Pauli groupThere are 16 distinct matrices: $I,{\sigma }_x,{\sigma }_y,{\sigma }_z$ are linearly independent, so two different ones cannot be scalar multiples, and the four phases are distinct for each nonzero matrix.

The Pauli matrices satisfyHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. Thus any product of two Pauli matrices is another Pauli matrix or the identity times a phase in ${\mu }_4$. Multiplying by the two original phases preserves membership in ${\mu }_4$, proving closure of all 16 matrices.

Associativity is inherited from matrix multiplication, the identity is $I$, andThis proves every group axiom. The resulting group is non-Abelian, since reversing ${\sigma }_x{\sigma }_y$ changes its sign.

Take the centre of the ball as the origin. An orthogonal change of variables leaves both this domain and its volume measure invariant. Consequently its Cartesian moment integrals are isotropic tensors. Equivalently, inversion $x\mapsto -x$ immediately makes every odd moment zero.

There is no nonzero invariant rank-one vector, soAn isotropic tensor of rank two is a scalar multiple of the Kronecker delta, giving ${\int }_V{x}_i{x}_{j}dV=\alpha {\delta }_{ij}$. Contract the indices and use spherical coordinates:Thus the Cartesian moments of a three-dimensional ball giveFor the last identity, one may also use the general rotation-invariant rank-three tensor $c{ϵ}_{ijk}$: the moment is fully symmetric in its indices, whereas the Levi-Civita symbol is antisymmetric, forcing $c=0$.

The origin convention is necessary. For a ball centred at a nonzero vector $c$, its first moment would instead be $(4\pi a^3/3)c_i$, so the displayed claims implicitly refer to coordinates measured from its centre.

Apply the product rule in Einstein notation, using the Levi-Civita symbol to express the cross product:The third line uses a cyclic permutation of indices in the first term and an interchange of two indices in the second. Rearranging proves the divergence of a cross product identity:

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

An order-$n$ Cartesian tensor has $3^n$ components whose values in two orthonormal Cartesian frames satisfywith repeated indices summed as in Einstein notation. It is an isotropic tensor when its components are unchanged by every proper rotation, $R\in SO(3)$.

For rank two, a half-turn about any coordinate axis changes the sign of each component having exactly one index along that axis. Invariance therefore makes every off-diagonal component zero. Quarter-turns interchanging two coordinate axes then require $T_{11}=T_{22}=T_{33}$. Thus, without assuming symmetry in advance,Conversely, the Kronecker delta is invariant because $R_{ik}{R}_{jk}={\delta }_{ij}$, so every tensor of this form is isotropic.

For rank three the general proper-rotation invariant form iswhere ${ϵ}_{ijk}$ is the Levi-Civita symbol. The convention matters: the isotropic rank-three tensor and reflections distinction says that invariance of a polar tensor under all $R\in O(3)$ would force $b=0$. A nonzero ${ϵ}_{ijk}$ is invariant under proper rotations, and is a pseudotensor under the full orthogonal group. The shell integrals below are invariant under reflections as well, so their odd moments vanish.

Under inversion $\mathbf{x}\mapsto -\mathbf{x}$, the spherical shell and its volume element are unchanged, while the product of three scalar products changes sign. Its integral is therefore its own negative:Equivalently, its rank-three moment tensor is completely symmetric and cannot have a nonzero multiple of the antisymmetric Levi-Civita symbol.

In Cartesian coordinates, write ${\partial }_i=\partial /\partial x_i$ and use the Levi-Civita symbol. The contraction of two Levi-Civita symbols givesHere the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity:

A pseudotensor has an additional determinant factor under an orthogonal transformation. Thus its rank-three transformation law isThe contracted expression $L_{ia}{L}_{jb}{L}_{kc}{ϵ}_{abc}$ is fully antisymmetric in $i,j,k$, so it equals a scalar multiple of ${ϵ}_{ijk}$. For $(i,j,k)=(1,2,3)$ it is precisely the determinant expansion of $L$, fixing that scalar to $\det L$. ThereforeThe Levi-Civita symbol consequently has the same numerical entries in every orthonormal frame when interpreted as a pseudotensor. With an ordinary rank-three tensor transformation law alone it would instead acquire a minus sign under an orientation-reversing transformation.

Under an orthogonal matrix $R$, the Levi-Civita symbol satisfiesUse this twice, and use $R_{mc}{R}_{mf}={\delta }_{cf}$ for the contracted index. It follows thatTherefore $\boxed{T_{ijkl}^{\prime }=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}=T_{ijkl}}$: it is a fourth-rank isotropic tensor. Although a single alternating symbol is a pseudotensor under reflections, the two determinant factors multiply to one, so this product is an ordinary tensor even for improper orthogonal transformations.

For a Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$ with $R$ an orthogonal matrix, a rank-$n$ tensor obeyswith repeated indices summed according to Einstein notation. A tensor is isotropic if $T_{i_1\cdots i_n}^{\prime }=T_{i_1\cdots i_n}$ for every proper rotation matrix $R$. Under proper rotations the Levi-Civita symbol is an isotropic third-rank tensor; if reflections are included, it is a pseudotensor. The convention using proper rotations is the one relevant to the rank-three classification here.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.

The divergence theorem states that, for a continuously differentiable vector field $\mathbf{H}$ on a bounded region $V$ with piecewise smooth boundary $S$ and outward unit normal $\hat{\mathbf{n}}$,To apply it, first derive the divergence of a cross product using the Levi-Civita symbol and the product rule:In the second term the minus sign comes from interchanging the first two indices of the Levi-Civita symbol. The divergence theorem with $\mathbf{H}=\mathbf{F}\times \mathbf{G}$ now yields

The given angular momentum commutation relations imply $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. By linearity,Similarly,For the remaining commutators, the commutator product rule gives, for each $j$,The last sum vanishes because the Levi-Civita symbol is antisymmetric under $i\leftrightarrow k$, while $L_i{L}_k+L_k{L}_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves

If $L^2\psi =\lambda {\hslash }^2\psi$ and $L_3\psi =m\hslash \psi$, these identities yieldOne must also prove that this candidate eigenstate is nonzero. Since the angular momentum operators are self-adjoint operators, $L_{+}^{\ast }=L_{-}$, and expanding the product givesThereforeThis is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The angular momentum raising operator really does produce a nonzero simultaneous eigenstate with the claimed eigenvalues.

To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity givesso $q^2\le \lambda$. Thus the sequence cannot remain nonzero for every integer $r\ge 0$. At its last nonzero state $v=(L_{+}{)}^k\psi$, one has $L_{+}v=0$, and the squared norm identity givesUnder the paper's initial inequality the first step is nonzero, so in fact $k\ge 1$. An angular momentum ladder endpoint must therefore be reached: bounded projection follows from positivity, and the final raising norm fixes the total angular momentum eigenvalue. Writing $j=m+k$, the complementary product $L_{+}{L}_{-}=L^2-L_3^2+\hslash L_3$ gives $\Vert L_{-}v{\Vert }^2=2j{\hslash }^2\Vert v{\Vert }^2$, also showing $j\ge 0$.

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

Choose $A^{\mu }=(\Phi ,\mathbf{A})$, so that $F_{i0}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$, with ${ϵ}_{123}=1$. The definitions of the polarization-magnetization tensor imply $M^{0i}=-P_i$ and $M^{ij}={ϵ}_{ijk}{M}_k$. Separating temporal and spatial indices,The Levi-Civita symbol contraction equals $2{\delta }_{k\ell }$, givingThus $\mathbf{P}$ is the electric dipole moment per volume, or electric polarization, and $\mathbf{M}$ is the magnetic dipole moment per volume, or magnetization. The signs are those of dipole interactions in a Lagrangian; the associated fixed-dipole potential energies have the opposite signs. Consistently, the equation in part (a) has componentsso ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\mathbf{J}}_b={\partial }_t\mathbf{P}+\nabla \times \mathbf{M}$ are the bound charge and bound current.

With outgoing physical photon polarization vectors, the Feynman rule givesThe overall phase does not affect the decay width. Replacing either polarization vector by its own four-momentum gives zero by antisymmetry. Thus the Ward identity permits replacement of each physical photon polarization sum by $-{\eta }^{\mu \nu }$ inside this amplitude. The printed sum is not an equality of free-standing physical two-polarization tensors: for a photon moving along $z$, its physical $00$ component is zero whereas $-{\eta }_{00}=-1$. Gauge-dependent longitudinal terms cancel only after contraction with the transverse amplitude.

The contraction of the two Levi-Civita symbols givessince $q_1^2=q_2^2=0$ and $2q_1\cdot q_2=m_a^2$. No initial spin average is needed for the spin-zero axion. The result is independent of direction in its rest frame. Including the identical-particle factor explained in solution therefore givesOmitting $1/2!$ would overestimate the physical rate by two. The normalization agrees with the standard axion-photon convention.^[ref]

For a twice continuously differentiable deformation map, each row of its cofactor matrix has zero divergence in reference coordinates. In three dimensions write $(\mathrm{cof}Dy{)}_{iA}=\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{ABC}{y}_{j,B}{y}_{k,C}$. Applying ${\partial }_A$ gives terms in which a symmetric pair of second partial derivatives contracts with an antisymmetric Levi-Civita symbol, so each term vanishes. This identity is the cancellation that makes the Piola-Kirchhoff stress tensor transform force balance correctly between configurations.