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Invariance of integration measure under shifts in field

  1. Oct 24, 2015 #1
    I've been trying to teach myself the path integral formulation of quantum field theory and there's a point that's really bugging me: why is the integration measure ##\mathcal{D}\phi(x)## invariant under shifts in the field of the form $$\phi(x)\rightarrow\tilde{\phi}(x)=\phi(x)+\int d^{4}y\Delta(x-y)J(y),\qquad\mathcal{D}\phi(x)\rightarrow\mathcal{D}\tilde{\phi}(x)$$ (where ##\Delta(x-y)## is a Green's function corresponding to the differential operator ##\Box + m^{2}## and ##J(x)## is a source).
    Is it simply because the shift term ##\int d^{4}y\Delta(x-y)J(y)## is independent of the field configuration ##\phi(x)## at each spacetime point, or is there more to it than that?
     
  2. jcsd
  3. Oct 24, 2015 #2

    vanhees71

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    Yes, that's it! If you are in doubt about some operation in path integration, just think about it in the "lattice version", i.e., consider space and time to be a finite 4D grid of discrete points. Then you have just a usual multidimensional integral. Then it's indeed clear that the integration measure does not change by just shifting all integration variables by arbitrary constants.
     
  4. Oct 24, 2015 #3
    Ah OK ,great, I thought that was probably the case, but just wanted to make sure. Thanks for your help!
     
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