meandering1

A meandering proof of the covariance of the Dirac equation PhL 10.8.08 1. The Lie Algebra for the abstract Lorentz Group generators Jμν is derivable from the known coordinate transformations of special relativity and is this: [ Jμν, Jαβ] = -i (gνβ Jμα + gμα Jνβ- gμβ Jνα - gνα Jμβ) index = 0,1,2,3 g = diag(1,-1,-1,-1) from which it can be shown that each Jμν is antisymmetric when its labels are switched, Jμν = - Jνμ. An NxN matrix has (N2-N)/2 elements in its upper or lower triangle, which here is (42-4)/ 2 = 6. Thus, there are only 6 distinct Jμν generators, as befits the 6 parameter Lorentz Group. 2. The traditional rotation and boost generators are these. rotation Ji = Jjk (where jk completes the cyclic order begun by i) boost Ki = Ji0 and the Lie Algebra shown above becomes [ Ji, Jj] = +i εijkJk [ Ji, Kj] = +i εijk Kk [ Ki, Kj] = -i εijkJk The first two commutators just say that J and K are rotational 3-vectors, while the last says that boosts don't commute and results in effects like the Thomas Precession. 3. The Lorentz Group Lie Algebra can be "decoupled" simply by defining Li ≡ ½ (Ji + iKi) Ri ≡ ½ (Ji – iKi) with inverses Ji = (Li + Ri) Ki = i(Ri – Li) and we then have [Li, Lj] = i εijk Lk [Ri, Rj] = i εijk Rk [Li, Rj] = 0 which says that Li and Ri (left and right) are normal rotation group generators and are completely independent. It follows that the finite dimensional irreducible representations of the Lorentz Group can be written as direct products of rotation group representations. This fact is summarized by the notation jLjR where L2= jL (jL+ 1) 1 R2= jR (jR+ 1) 1 dim = (2jL+1) dim = (2jR+1) where 1 is the identity matrix in whatever space we are in. 4. Apart from the trivial representation 0 0 , the lowest two representations are ½ 0 and 0 ½ which are sometimes called the left and right Weyl spinor representations. In the ½ 0 we set Ri = 0 to get the trivial group representation DR = exp(0) = 1, and we then have Ji = (Li + Ri) = Li = σi/2 Ki = i(Ri – Li) = -iLi = -iσi/2 In the 0 ½ representation we have Li = 0 and the results are the same as above except the boost generators have the opposite sign Ji = (Li + Ri) = Ri = σi/2 Ki = i(Ri – Li) = +iRi = +iσi/2 5. A Majorana 4-spinor has the form ψ = (fL, gR) where f and g are 2-spinors which transform according to our left and right Weyl representations. For such a spinor, we would have the following 4x4 rotation and boost generator matrices, based on what has been said above: Ji = σi/2 σi/2 = ½ = ½ Σi = ½ εijk Jjk = Jjk (cyclic) Ki = –iσi/2 iσi/2 = ½ = Ji0 and such a spinor then belongs to the representation (½ 0) (0 ½) of the Lorentz Group where indicates a direct sum, as shown in the matrices above. 6. This is not, however, the official Dirac representation. We have to do a "shuffle" on the boost generators (only) as follows Ki' = - γ6 Ki where γ6 ≡ Thus "shuffle" does not affect the underlying Lie Algebra, and it is easy to show that [ Ji, Jj] = +i εijkJk [ Ji, K'j] = +i εijk K'k [ K'i, K'j] = -i εijkJk and we end up with Ji = ½ = ½ σjk Ki' = i = ½ σi0 => Jμν = ½ σμν where σμν ≡ i/2 [ γμ, γν] and this is the official Dirac representation of the generators, which is seen to be "in covariant form" with respect to the gamma matrices. The corresponding Lorentz Transformation in this 4x4 Dirac space is then given by U(ω) = exp(-i/2 ωμνJμν) = exp [– i ( rJ + bK')] = exp(-i/4 ωμν σμν) where we have the usual antisymmetric (when both indices are lower) parameter matrix ωμν = but ωμν = gμαωαν = 7. For contrast, the normal 4-vector representation of the Lorentz group has generators given by (Jμν)αβ = i ( gμαgνβ – gναgμβ) // note that Jμν = - Jνμ and the resultant 4x4 Lorentz Transformation is then Λ(ω) = exp(-i/2 ωμνJμν) = exp [– i ( rJ + bK)] = exp(ω) where the final form is so compact that the generators appear to have vanished altogether, but this is just due to their extreme "thinness" in this representation, their being made of nothing but metric tensors. The J and K generator matrices have the following typical forms rotation Ji = Jjk boost Ki = Ji0 (J1)μν = (K1)μν = which we know exponentiate into finite rotations and boosts exp(-irJ1)μν = exp(-ibK1)μν = The 4-vector representation just stated is in fact a "shuffle" of the ½ ½ irreducible representation of the Lorentz group which of course has dimensionality (2jL+ 1) (2jR+ 1) = 2*2 = 4. In this direct product one can represent the left ½ by a 2x2 matrix g, and the right ½ by g*. These 2x2 g matrices are in fact group elements of SL(2,C) [S = special => det(g) = 1] which we know is ≈ SO(3,1) = the Lorentz Group. So our representation of ½ ½ is then DL DR = g g*. The "shuffle" is then given by the following equation in which one can see the "two g's" appearing on the right hand side. Λμν = ½ tr(σμ g σν g†) where g = g(ω) and Λ = Λ(ω). This "shuffle" arises from the 2x2 transformation X' = gXg† where X = xμσμ = detX = xμxμ in which ones sees that detX' = detX so the transformation results in x'μx'μ = xμxμ. The Λμν equation above is obtained by right multiplying X' = gXg† by σν and taking the trace of both sides. 8. Consider the following commutator relation which we assume exists in some unspecified space, [Jμν, pα ] = -i ( gαμpν - gανpμ) where Jμν are the Lorentz generators, and pα is some object with α = 0,1,2,3. By doing the exponentiation, the above implies that exp(+i/2 ωμνJμν) pα exp(-i/2 ωμνJμν) = U-1 pα U = exp(ω)αν pν =Λαν pν or just U-1 pα U = Λαν pν which says that pμ transforms as a 4-vector object in whatever space pμ happens to live. If pμ lives in the vector representation space, then it is a regular 4-vector such as momentum. But the equation is valid in any space in which the starting commutator relation is valid. In the Dirac space, we have Jμν = ½ σμν = i/4 [ γμ, γν] and we have the commutator relation [Jμν, γα ] = -i ( gαμγν - gανγμ) so in this example we have pα = γα = a 4-index object living in the Dirac space. It follows then that U-1 γα U = Λαν γν where U(ω) = exp(-i/2 ωμνJμν) = exp(-i/4 ωμν σμν) One implication of this fact is another fact, namely U-1' U= where ≡ aμγμ which we prove as follows: U-1' U = U-1a'μγμ U = (a'μ) (U-1γμ U) = (Λμβaβ)(Λμν γν) = (Λμβ Λμν) aβ γν = δβν aβ γν = aβ γβ = where we use our general orthogonality rule for any curvilinear coordinate transformation δbc = Tac Tab, of which the Lorentz transformation is just a linear example. 9. Armed with the above, it is trivial to show the "covariance" of the Dirac equation, which is this in "the primed frame" (where the 4x4 matrix aspect of the equation is implied), (' - m)ψ'(x') = 0 where μ ≡ i∂μ , the differential operator. Since ψ belongs to our Dirac representation, we know that ψ'(x')= Uψ(x) where U(ω) = exp(-i/2 ωμνJμν) = exp(-i/4 ωμνσμν) = 4x4 Dirac LT then we can apply U-1 from the left to get U-1(' - m) Uψ(x) = 0 => (U-1 ' U- m) ψ(x) = 0 => ( - m) ψ(x) = 0 QED Thus, the Dirac equation has the same "form" in both the primed and unprimed frames which are related by a Lorentz Transformation. Bjorken and Drell volume 1 expends an entire chapter (Chapter 2) deriving this fact. 10. Finally, as a coda, we comment more on the Left and Right spinor representations of the Lorentz Group, where we now use the symbol Λ to represent 2x2 matrices: ( means g*) spinor of the first kind (undotted indices) ξ'a = Λabξb Λab = D(1/2,0)(g)ab = g spinor of the 2nd kind (dotted indices) ξ' = Λξ Λ = D(0,1/2)(g)ab = The dots are necessary so one can tell how a given index transforms when one constructs "spinor tensors" and watches them transform. For example, here is a mixed rank-2 "spinor tensor" with one index of each kind X' a = Λaa'Λ' Xa'' where we can regard Λaa'Λ'= [D(1/2,1/2)(g)] aa'' = [D(1/2,0)(g)] aa' [D(0,1/2)(g)] ' = gaa' bb' as described earlier. By putting indices in proper matrix order, we can write the above "spinor tensor" transformation in the following more compact notation. X' = gXg† Clearly X a transforms according to the ½ ½ representation of the Lorentz Group. The Pauli matrices can be written (σμ)a and transform exactly as such rank 2 "spinor tensors" and the four σμ thus form a "basis" for this ½ ½ representation. One can then construct a linear combination X = xμσμ and this leads to the homomorphism result quoted above that Λμν = ½ tr(σμ g σν g†) which relates the 4-vector representation Λμν to the canonical ½ ½ representation. As presented above, the 2-spinors of either type are written with contravariant upper indices. For either type of spinor, one can define a covariant lower index spinor as follows, aα ≡ εαβ aβ = covariant first kind spinor where ε ≡ iσ2 = = "the spinor metric tensor" Whereas the 4-vector metric tensor gμv is diagonal and symmetric, the " spinor metric tensor " is off diagonal and antisymmetric. The compatible rule for raising a spinor index is this aβ = aτετβ εμν ≡ εμν where summed indices are always put adjacent. It is easy to show that the matrix ε acts as a similarity transformation changing g to (gT)-1, which we could call an "automorphism of SL(2,C)". So we have (gT)-1 = ε g ε-1 note: ε-1 = εT = ε† = -ε and ε2 = -1 => gT ε g = ε Suppose ψ and φ are each 2-spinors of the first kind, which means ψ' = gψ for transformation. Then from the above ψ'T ε φ' = (gψ)Tε (gφ) = ψTgTεg φ = ψT ε φ or ψ'αφ'α = ψαφα which shows that the contraction of contravariant with covariant spinors produces a world scalar, just as happens with 4-vectors in the vector representation of the Lorentz Group. If one contracts a spinor with itself, the result is 0 due to the antisymmetry of the spinor metric tensor ψαψα = ψα εαβ ψβ = εαβ ψα ψβ = 0 so all spinors have norm 0 If we define ||ψ||2 = ψαψα. Just as with regular tensors, one can combine "spinor tensors" in various complicated ways, and the only complication is that there is only one basic vector, but there are two basic spinors, so one must use the dots on indices as needed. Just as in "tensor analysis" one has "spinor analysis" in which usual zoo animals appear, such as the covariant derivative involving Christoffel symbols Γραs, ψα;s ≡ ∂sψα – Γραs ψρ