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50 The Dirac Equation for Relativistic Fermions In order to avoid negative probability densities, Dirac understood that one had to search for an equation that is linear in E, i.e. of first order in the time derivative ∂t. For dimensional reasons, the only possible ansatz for such an equation is H =βmc2+cα·p → ih ̄[∂t +cα·∇]ψ(r,t)=βmc2ψ(r,t), (3.13) where β and1 α = (α1, ..., αD) are dimensionless objects the mathemati- cal properties of which are fixed by the requirement that the square of the operator equation (3.13) must satisfy the relativistic dispersion relation E2 = m2c4 + c2p2 = β2m2c4 + c2(α · p)2 + mc3(βα · p + α · pβ) . In order to satisfy this equation, the introduced mathematical objects must anti-commute, β2 = 1, {αi,β} = 0, and {αi,αj} = 2δij, (3.14) in terms of the anti-commutator {A, B} ≡ AB + BA. Eq. (3.14) defines the so-called Clifford algebra, and one immediately realises that the objects β and αi must be N × N matrices, with N > 1. We summarise their main properties. (i) Because the Hamiltonian (3.13) is Hermitian, the matrices must be Hermitian, α i = α i† a n d β = β † . (ii) Furthermore, because of αi2 = β2 = 1 [see Eq. (3.14)], the matrices must be unitary αi =α† =α−1, and β =β† =β−1, ii and their eigenvalues are, therefore, ±1. (iii) If one applies the determinant to the anti-commutator, αiαj = −αjαi, one obtains with the help of the elementary properties of the determi- nant,2 det(αiαj) = det(−αjαi) = (−1)N det(αjαi) = (−1)N det(αiαj). In order to satisfy this equation, one needs to choose N to be even. The lowest possible value is, therefore, N = 2. 1One needs to introduce one αi for each space dimension. 2We use the fact that if one multiplies an N × N matrix A by a scalar λ, one has det(λA) = λN det(A), and that the determinant of a product of two matrices A and B is the product of their determinants, det(AB) = det(A) det(B).PDF Image | Physical Properties of Graphene
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