1.7 Plane-wave Solutions of the Dirac Equation

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1 0 Version of February 7, 005 CHAPTER. DIRAC EQUATION It is evident that W µ is translationally invariant, [P µ, W ν ] 0. W is a Lorentz scalar, [J µν, W ], as you will explicitly show in homework. Here ǫ µνλσ is the totally antisymmetric invariant tensor, ǫ µνλσ (g µµ δω µµ )(g νν δω νν )(g λλ δω λλ )(g σσ δω σσ )ǫ µ ν λ σ ǫ µνλσ, (.67) (because ǫ µνλσ vanishes if any two indices are the same), where ǫ In the rest frame of a particle, and so P 0, P 0 E m, (.68) W 0 ǫ0ijk J ij P k 0, W i ǫijk0 J ij m m ǫijk J jk mj i, (.69a) (.69b) where the latter is the spin. Thus, the eigenvalues of W are W m s(s + ). (.70) This means for a particle with nonzero rest mass, m > 0, the irreducible representations belong to the values s 0, /,,.... For a given s, the possible value of J 3 are s 3 s, s +, s +,...,s, s. The massless limit has to be taken carefully (see homework): m 0 : W µ λp µ, λ P S P 0. (.7) λ is called the helicity, which is the spin projected along the direction of motion. There are other representations of the Poincaré group, such as tachyons, where m < 0, but they seem not to be realized in nature..7 Plane-wave Solutions of the Dirac Equation If ψ e ipx u p, where px p µ x µ p x Et, the Dirac equation becomes For a particle at rest, p 0 m, p 0, this is (γp + m)u p 0. (.7) ( γ 0 )v 0, (.73) where v u p0 is the rest-frame spinor. This means that v is an eigenvector of γ 0 with eigenvalue +. Because [Σ 3, γ 0 ] 0, (.74)

2 .7. PLANE-WAVE SOLUTIONS OF THE DIRAC EQUATION Version of February 7, 005 we can also take v to be an eigenvector of Σ 3 : Σ 3 v σ σv σ, σ ±. (.75) We can obtain u pσ from the rest-frame spinor v σ by a boost: [ u pσ exp ] α φe v σ (cosh φ + α e sinh ) φ v σ, (.76) because (α e). Here, the direction of the boost is given by that of the momentum, e p p. (.77) Because we have cosh φ v E m, (.78) and therefore u pσ cosh φ coshφ + sinh φ coshφ E + m m, E m m, ( ) E + m m + α p E m v σ E m m (E + m + α p)v σ (m + γ 0 E γ p)v σ (.79a) (.79b) (m γp)v σ. (.80) This evidently satisfies the Dirac equation (.7): (m + γp)u pσ (m + γp)(m γp)v σ (m + p )v σ 0, (.8) because (γp) p. We also note that since {v σ }, σ ±, span the twodimensional space for which γ 0, we must have the projection-operator statement ( + γ0 ) v σ v σ. (.8) σ

3 Version of February 7, 005 CHAPTER. DIRAC EQUATION If we multiply both sides of this equation by v σ, v σ σ v σ (v σ v σ ), (.83) which implies that the rest-frame spinors are orthonormal, v σv σ δ σσ. (.84) A Lorentz-invariant way of writing these two results, which you will explicitly prove in Homework, is m γp m σ u pσ u pσ γ0, (.85a) u pσγ 0 γ µ u pσ δ σσ p µ m. (.85b) We also note that v σ is an eigenvector of γ 0 with eigenvalue : γ 0 v σ v σ, Σ 3 v σ v σ, (.86) because γ 0 and Σ 3 are imaginary. The corresponding projection operator statement is ( γ0 ) vσv σ T. (.87) σ These spinors correspond to negative-energy solutions, which satisfies u pσ implying a plane wave solution of the form (m + γp)v σ, (.88) (m γp)u pσ 0, (.89) ψ e ipx u pσ. (.90) It is the appearance of these negative-energy solutions that destroys all hope of a wavefunction interpretation of ψ. (An example is given by the Klein paradox, see, e.g., Bjorken and Drell.) A partial resolution of the difficulty is the Dirac hole theory, in which all negative-energy states are filled. A vacancy (hole) in the sea of negative energy states appears as a positive-energy antiparticle; for electrons, the hole is a positron. However, we will not pursue this line of thought, for a more thoroughgoing reconstruction of the theory is necessary. A final note. Recall γ 5 γ 0 γ γ γ 3 discussed earlier. Note that we can write Σ σ 3 i [γ, γ 3 ] iγ γ 3, (.9) so iγ 5 Σ iγ 0 γ γ γ 3 iγ γ 3 γ 0 γ, (.9)

4 .8. IRREDUCIBLE REPRESENTATIONS OF THE LORENTZ GROUP3 Version of February 7, 005 or generally, Then the boost equation (.80) becomes u pσ γ 0 γ iγ 5 Σ. (.93) (E + m + iγ 5 Σ p)v σ. (.94) We can take p to be the quantization direction for Σ: Σ p p σ E m σ, (.95) where now σ is the helicity. Further, from the Homework, iγ 5 σv σ v σ. (.96) So then we can write u pσ ( E + m vσ + ) E m v σ. (.97) m Left- and right-handed spinors are obtained by projecting with ( iγ 5): u L,R ( iγ 5)u. (.98) Note that iγ 5 is a good quantum number, the chirality, if m 0. Consider a general Lorentz transformation, u L,R ( ( iγ 5) + iδω Σ δv ) α u ( + iδω Σ ± δv ) Σ u L,R, (.99) because α iγ 5 Σ. This indeed is the correct transformation properties for the (/, 0) and (0, /) representations, respectively. See (.3), (.4)..8 Irreducible Representations of the Lorentz Group Another way to describe Lorentz transformations is the following. Associated with any four-vector x µ is a Hermitian matrix, ( ) x x 0 + x 3 x ix x + ix x 0 x 3 x 0 + x τ, (.00) where τ are the Pauli matrices. The scalar length of x µ is given by the determinant of this matrix: x µ x µ (x 0 ) + x x detx. (.0)

5 4 Version of February 7, 005 CHAPTER. DIRAC EQUATION We can extract x µ from the matrix x as follows: x µ Tr(xτµ ), τ µ (, τ). (.0) If A is any matrix with determinant unity, det A, we can construct a new matrix ˆx by the transformation which has the same determinant as x: ˆx AxA, (.03) det ˆx detx, (.04) so the corresponding four-vector has the same length as that of x µ : ˆx µˆx µ x µ x µ. That is, A corresponds to a restricted Lorentz transformation, SO(3, ) SL(, C), (.05) the latter being the group of transformations induced by complex matrices with determinant, a special linear group. The irreducible representations of the Lorentz group are given by, as a generalization of the above transformation of a vector, ξ α...α j; β... β k A αρ A αjρ j A β σ A βk σ k ξ ρ...ρ j; σ... σ k. (.06) This belongs to the representation ( j, k), which is characterized by j undotted indices (which transform with A) and k dotted indices (which transform with A ). For more details on this way of proceeding see Gel fand, Minlos, and Shapiro, Representations of the Rotation and Lorentz Groups, Pergamon Press, 963. Tensors are not, in general, irreducible representations. For example, consider a general second rank tensor, A µν. It can be decomposed as follows: where and A µν g µν A + F µν + T µν, (.07) T µν T νµ, T µ µ 0, (symmetric and traceless) (.08) F µν F νµ, (antisymmetric). (.09) The count of independent components is consistent: (3 + 3) + (3 3). (.0) The latter count refers to the spinorial representation: (, ) (, ) (, ) (, ) (, 0) (0, ) (0, 0), (.) where the first term corresponds to T µν, the second and third to F µν, and the last to A.