View Full Version : Minkowski vacuum: Poincare invariant, quasi-free state
Minkowski vaccum is Poincare invariant and quasi-free state.
I wonder if these two conditions fully define it or there are more
states which fulfill these conditions (or maybe Poincare
invariance alone is sufficinet). Thanks for answers.
strangerep
Oct11-10, 07:36 PM
Minkowski vaccum is Poincare invariant and quasi-free state.
I wonder if these two conditions fully define it or there are more
states which fulfill these conditions (or maybe Poincare
invariance alone is sufficient).
I'm not entirely sure what you're asking. In advanced QFT, one attempts to
construct something called an "interacting representation" of the Poincare
group. (See Weinberg vol 1). The vacuum is still defined as the lowest-energy
eigenstate of the (full, interacting) Hamiltonian -- which often does not coincide
with the lowest-energy eigenstate of the free Hamiltonian.
My question is whether a state which fulfills two condition:
(1) its two point function is invariant under action of Poincare
group and
(2) all n-point function can be express in terms of two point function by
the sum over all parings (quasi-free state)
has to be Minkowski vacuum.
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