Path integral formulation: Difference between revisions

Revision as of 12:08, 26 November 2024

Revision as of 02:57, 3 December 2024

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: <math>K(x, y; T) = \langle y; T \mid x; 0 \rangle = \int_{x(0)=x}^{x(T)=y} e^{i S[x]} \,Dx.</math>

This is called the [[propagator]]. SuperposingTo differentobtain valuesthe offinal thestate initialat position$y$ {{mvar|x}}we begin with an arbitrarythe initial state {{math|''ψ''₀$\psi_0(''x'')}}$ constructsand theapply final$K(x,y; stateT)$ resulting in:

: <math>\psi_T(y) = \int_x \psi_0(x) K(x, y; T) \,dx = \int^{x(T)=y} \psi_0(x(0)) e^{i S[x]} \,Dx.</math>

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For a nonrelativistic theory, the time as measured along the path of a moving particle and the time as measured by an outside observer are the same. In relativity, this is no longer true. For a relativistic theory the propagator should be defined as the sum over all paths that travel between two points in a fixed proper time, as measured along the path (these paths describe the trajectory of a particle in space and in time):

: <math>K(x - y, \Tau) = \int_{x(0)=x}^{x(\Tau)=y} e^{i \int_0^\Tau \sqrt{{\dot {x}^2} - \alpha} \,d\tau}.</math>

The integral above is not trivial to interpret because of the square root. Fortunately, there is a heuristic trick. The sum is over the relativistic arc length of the path of an oscillating quantity, and like the nonrelativistic path integral should be interpreted as slightly rotated into imaginary time. The function {{math|''K''(''x'' − ''y'', ''τ'')}} can be evaluated when the sum is over paths in Euclidean space: