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**Example text**

For the action in Eq. 73), the Hamiltonian is H = −p2 /m and the energy conjugate to proper time will be E = −m. If you make the substitutions E → −m, p2 /2m → −p2 /m in Eq. 66), you essentially get the result in Eq. 92) showing the origin of the “energy denominator”. 93) −iμ(m2 −p2 ) We next note that the factor e can be thought of as a matrix 2 element of the operator e−iμ(m + ) . Hence we can write G(x2 ; x1 ) = ∞ dμ x2 |e−iμ( +m2 −i dμ ) |x 1 0 X X Re p0 X Re p0 Im p0 X prescription and the Euclidean prescription in the momentum space.

In the second equality we have used the standard path integral prescription. Exactly as before, the sum over paths is now to be evaluated limiting ourselves to paths xi (τ ) which only go forward in the proper time τ (see Fig. 4; just as the paths in Eq. 8) were limited to those which go forward in the Newtonian absolute time t). However, we have to now allow paths like the one shown in Fig. 2 which go back and forth in the coordinate time t just as we allowed in Eq. 8) the paths which went back and forth in the y coordinate, say.

14: Construct a single particle Lorentz invariant relativistic quantum theory with a consistent interpretation. (Hint: Of course, you cannot! ) 30 This is closely related to something called the Newton-Wigner position operator in the literature. It is known that this idea has serious problems. 116) (2π)D 2ωp This shows that x|y is in general non-zero for x = y. (In fact, from our expression in Eq. ) Physically, this indicates the impossibility of localizing particles to a region smaller than (1/m) and still maintain a single particle description.