By Alex Bellos
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Experience has taught us that this concept is completely successful. The way to achieve our aim is to construct a four-vector since (1) can then be fulfilled. The simplest four-vector we can make from the spin and momentum operators is Χμ = ΜμνΡν s £ ΜμνΡν. _ _ T^ Λ Λ _ . e. space-time generators—these do not satisfy condition (3). 1) A = ^ W V A ε σμνλ = + 1 even permutations of 1234, — 1 odd permutations of 1234, 0 two indices equal. 13)] we can drop the Σμν component in Γσ as each term cancels because of the antisymmetry of ε ε Thus σμνλ£μνΡλ = ε Χ σμνλ\ μΡν ~~ Χν^μ) ^λ Γσ = —8σμνλΞμνΡλ 2ι = 0.
2) as a probability amplitude at a space-time point x. It is evident therefore that transformation x -* x' must reflect back into the interpretation of |y>>. 3) 38 RELATIVISTIC WAVE EQUATIONS AND FIELDS [Ch. 3 then the transformation implies that the value of the wave function of x is taken to Λχ and at the same time the value of the wave function at Λ~χχ moves to x. 4) ψ(Λ~1χ) -► ψ'(χ) —► x but if the value of the wave function remains invariant under the trans formation, then rA \ \ ,, \ /» ^ ^ f(A-xx) = ψ'(χ).
Dirac explained the negative energy states as energy levels which were normally filled, but when a particle was removed from this system, the hole created had the same properties as a particle with opposite charge and positive energy. The argument is probably more easily seen in modern language. Consider a particle (say the electron) and antiparticle (positron) with opposite properties particle — e — p — E —s antiparticle +e +p +E +s Now apply this situation to muon decay and a leptonic interaction c - μ+ -» e+vv β~μ+ -> vv If the electron has all the opposite properties to the positron then the kinematic conditions of the vv pair are unaltered.