Sunday, 10 December 2017

quantum field theory - Propagator and probability amplitude that a particle propagates


My QFT knowledge has very much rusted and i got confused by these few lines from Peskin and Schroeder:



p.27: " [..] the amplitude for a particle to propagate from $y$ to $x$ is $\langle 0| \phi(x) \phi(y) |0\rangle $. We will call this quantity $D(x — y)$."




(The relation with the commutator is explicitly calculated at (2.53) p.28, + bottom of p.29: $$ [\phi(x) , \phi(y)] = \cdots = D(x-y)- D(y-x) = \langle 0|[\phi(x) , \phi(y)]|0\rangle$$ the r.h.s. are implicitly understood as being proportional to the $\mathbb{1}$ operator)


Finally the expressions of the retarded and Feynman propagators are given (2.55) p.30



$$D_R := \theta (x^0 -y^0) \langle 0|[\phi(x) , \phi(y)]|0\rangle $$



and (2.60) p.31 (without commutators)



$$D_F := \theta (x^0 -y^0) \langle 0|\phi(x) \cdot \phi(y)|0\rangle + \theta (y^0 -x^0) \langle 0|\phi(y) \cdot \phi(x)|0\rangle $$



which by definition of "propagator" or "Green's function" satisfy $(\square +m^2) G(x,y)= -i\delta^4(x-y)$.





Now my confusion comes from the fact that I remember that propagators had the interpretation of the amplitude of propagation, cf. e.g. wikipedia or Peskin last 2 lines p.82, but this is wrong? (three different functions obviously cannot have the exact same interpretation!)


Remark: I am aware that from the defining relation of a Green's function, one can express the "evolution" of a solution of the (Klein-Gordon) equation, so that in some sense propagators expresses an idea of propagation




The first question is too easy so here is a second: Are propagators always combinations of the $D(x-y)$?



  • for interacting fields

  • for more general PDE?


Remark: I'm not trying to relate interacting theories to the free one, so $D(x-y)$ stands for the amplitude of propagation in each theory not the free one. The underlying idea is that $D(x-y)$ has a clear interpretation while the propagators would not have an easy interpretation on their own unless they are just simple functions of these $D(x-y)$.





No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...