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Ivancevic_Applied-Diff-Geom

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1178 <strong>Applied</strong> <strong>Diff</strong>erential <strong>Geom</strong>etry: A Modern IntroductionAgain, the p 0 sign can be easily found: p µ and ẋ µ are time–like andp µ ẋ µ = −γ √ (−1) p g < 0, so that sgn(p 0 ) = −sgn(ẋ 0 ).The result is similar to (6.287):p 0 + E p sgn(ẋ 0 ) = 0,the ‘Hamiltonian’ is equal to zero, and the total Hamiltonian isH T = u(p 0 + E p (x, −→ p )) + v i p µ x µ ,i .Here, p+1 momenta are un–physical ones. We assume that det(x i,k ) ≠ 0(x i,k is an p × p matrix) and exclude momenta p 1 , . . . , p p from E p . Due to(6.297) we havep i = ([x .,. ] −1 ) il (p 0 x 0,l − p > x >,l ).Here we denoted all the components of p µ with µ > p by the lower index‘ > ’, and ([x .,. ] −1 ) il ≡ d il is a matrix inverse of x l,i . Then (6.298) turns intoquadratic equationp 02 ( 1 − d il x 0,l d ik x 0,k)+ 2p0 d il ( p > x >,l )d ik x 0,k−d il (p > x >,l )d ik (p > x >,k ) − (−1) p γ 2 ζ(x) = 0.It has two real roots of opposite signs iff d il x 0,l d ik x 0,k < 1. The sufficientcondition is that the norm of x i,k as a linear operator is greater than thelength of pD vector x 0,l . If x 0,l = 0 the simple answer exists:∑E p = E p (x, p > ) = √ p (p i (x, p > )) 2 + p 2 > + (−1) p γ 2 ζ(x).i=1For the wave function, taking (6.288) into account, we get path integrals∫ψ(x) = D n+1 pD n+1˜x exp(i[p µ ∆x µ −∫=− ωu(p 0 + E p (˜x, p ⊥ )) − ωv i p µ x µ ,i ])ψ(˜x) =D n−p p > D n+1˜x exp(i[−p > ∆x > − ωuE p (˜x, p > ) + ωv i p > x >,i ])r∏× δ(∆x 0 − ωu − ωv i x ′ 0,i) δ(−∆x l + ωv i x l,i )ψ(˜x).l=1

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