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For the Dirac representation of the gamma matrices, the unitary matrixsends ${\tilde{\gamma }}^0$ to $\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ while commuting with every spatial gamma matrix. This is the standard Weyl representation of the gamma matrices with spatial blocks $(0,{\sigma }^i;-{\sigma }^i,0)$. The similarity transformation preserves the Clifford algebra relations. Changing block-order or chirality conventions can change the displayed transformation, so its direction must be specified.

If a $k$-dimensional invariant subspace is mapped to the span of the first $k$ coordinate vectors by a similarity transformation, the transformed matrix has block formIts characteristic polynomial factors as $\det (\lambda I-B)\det (\lambda I-D)$, so its eigenvalues are the eigenvalues of the two diagonal blocks, counted with algebraic multiplicity.

In the standard Dirac representation of the gamma matrices, $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$, so ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$. The transpose signs in that representation are positive for $a=0,2$ and negative for $a=1,3$. The charge-conjugation matrix expresses the basis-independent intertwining relation $C^{-1}{\gamma }^{a}C=-({\gamma }^a{)}^T$. Under an arbitrary nonunitary similarity transformation, the Hermitizing matrix changes by inverse Hermitian congruence; it need not equal the transformed temporal gamma matrix.

The virtual similarity transformation $A^i\mapsto X{A}^i{X}^{-1}$ leaves every periodic MPS amplitude invariant by cyclicity of the matrix trace. The fundamental theorem of matrix product states says that, up to an overall phase, this is the only ambiguity for injective uniform tensors.

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

A Dirac-to-Weyl gamma-matrix similarity transform with the required signs isIt is a unitary matrix, with $U^{-1}=U^{†}=2^{-1/2}\left(\begin{array}{cc}{I}_2& I_2\\ -I_2& I_2\end{array}\right)$. Multiplication givesThe matrix $U$ commutes with every ${\tilde{\gamma }}^i$: their common block structure uses the same antisymmetric $2\times 2$ block matrix, and the scalar blocks of $U$ commute with each Pauli matrix. Hence $U{\tilde{\gamma }}^i{U}^{-1}={\tilde{\gamma }}^i$, exactly the spatial matrices specified in the Weyl representation of the gamma matrices. A similarity transformation preserves every Clifford algebra relation, so no new consistency check is needed for the transformed representation.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.

For a nonzero vector $v\in {\mathbb{R}}^n$, a Householder reflection isIt is immediately symmetric. If $P=v{v}^T/(v^{T}v)$, then $P^2=P$, soHence $H$ is also an orthogonal matrix, with $H^{-1}=H$.

The similarity transformation can be expanded asComputing $v^{T}A$, $Av$, and the scalar $v^{T}Av$ costs $O(n^2)$ arithmetic operations, after which the remaining updates are outer products and scalar multiples, also costing $O(n^2)$. Thus $HA{H}^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ matrix multiplications.

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.