naima said:
## U(a)^{\dagger} \phi (f) U(a) ## (is the dagger on the left or on the right?)
That depends very much on what goes on the right-hand-side. Under Poincare’ transformation T=(\Lambda , a) of the coordinates x \to \bar{x} = T x = \Lambda x + a , the finite-component field \varphi_{r}(x) transforms (like classical tensor fields do) by
finite-dimensional matrix representation of the
Lorentz group D(\Lambda):
\bar{\varphi}_{r}(\bar{x}) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(x) . This can be rewritten as
\bar{\varphi}_{r}(\bar{x}) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(T^{-1}\bar{x}) , \ \ \ \ \ \ (1) where T^{-1}=(\Lambda^{-1}, -\Lambda a) , is the inverse element of the Poincare’ transformation. Renaming the coordinates label in (1) as x, we find
\bar{\varphi}_{r}(x) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(T^{-1}x) . \ \ \ \ \ \ (2)
However, in QFT \varphi_{r} (x) “is” an “operator”. So, it must transform, like operators do, by (
infinite-dimensional)
unitary representation U(T) of the Poincare’ group
\bar{\varphi}_{r}(x) = U^{-1}(T) \ \varphi_{r}(x) \ U(T) . \ \ \ \ \ \ (3) From (2) and (3), we obtain
U^{-1}(T) \ \varphi_{r}(x) \ U(T) = D_{r}{}^{s}(\Lambda) \ \varphi_{s}(T^{-1}x) . \ \ \ \ \ (4) Since the set Poincare’ transformations \{T\} form a group, equation (4) must also be satisfied by the inverse element T^{-1} \equiv (\Lambda^{-1}, - \Lambda^{-1}a) :
U^{-1}(T^{-1}) \ \varphi_{r}(x) \ U(T^{-1}) = D_{r}{}^{s}(\Lambda^{-1}) \ \varphi_{s}(Tx) .
And, since the U(T) \equiv U(\Lambda , a) forms a representation, i.e., U^{-1}(T) = U(T^{-1}), we find
U(T) \ \varphi_{r}(x) \ U^{-1}(T) = D_{r}{}^{s}(\Lambda^{-1}) \ \varphi_{s}(Tx) . \ \ \ \ \ (5) So, you are free to use either (4) or (5).
Now, let us recall that the fields \varphi_{r}(x) are not operators but operator-valued (tempered)
distributions (on space-time) which become (in general unbounded) operators by smearing with “good” test functions f \in \mathcal{S}(\mathbb{R}^{4}) (the space of fast decreasing, functions in \mathbb{R}^{4}):
\varphi_{r}(f) = \int d^{4} x \ \varphi_{r}(x) \ f(x) . \ \ \ \ \ (6)
Before we proceed further, let me make the following remarks about (6) which are relevant for issues raised in this thread:
1) You can think of (6) as the mathematical statement of the fact that only space-time
averages of the field “operators” \varphi_{r}(x) are “observables”.
2) To reflect the possibility of making measurement in a
finite space-time region \mathcal{O} \subset \mathbb{R}^{4}, the space of test functions f(x), for which \varphi_{r}(f) are assumed defined, is taken to be \mathcal{D}(\mathcal{O}), i.e., the space of all infinitely differentiable functions of
compact support in space-time. Without going into detailed mathematical gibberish, we simply take the support of f(x) to be the (
closed + bounded = compact) set on which f does not vanish. In this case, we speak of a
smeared local operator, or
local observable for short.
3) In order for (6) to make sense, (i) the operators \varphi_{r}(f), for all f \in \mathcal{S}(\mathbb{R}^{4}) must have a common dense domain of definition \mathcal{G} in the Hilbert space \mathcal{H}, i.e., a dense subspace \mathcal{G} of \mathcal{H} which is (ii) stable under the actions of both \varphi_{r}(f) and U(\Lambda , a) \varphi_{r}(f) \mathcal{G} \subset \mathcal{G} , \ \ \ U(T) \mathcal{G} \subset \mathcal{G} , where the term “dense” means “dense in any admissible topology \tau of \mathcal{H}”, i.e., \bar{\mathcal{G}}^{\tau} = \mathcal{G}^{\perp \perp} = \mathcal{H} , where \bar{\mathcal{G}}^{\tau} is the closure of \mathcal{G} with respect to the topology \tau, and \mathcal{G}^{\perp} is defined by \mathcal{G}^{\perp} = \{ |\Psi \rangle \in \mathcal{H}; \ \langle \Psi | \Phi \rangle = 0 \ \ \forall |\Phi \rangle \in \mathcal{G} \} .
4) The polynomial algebras generated by operators of the form \int d^{4}x_{1} \cdots d^{4}x_{n} \varphi_{r_{1}}(x_{1}) \cdots \varphi_{r_{n}}(x_{n}) f(x_{1}, \cdots , x_{n}) with f \in \mathcal{S}(\mathbb{R}^{4n}) and with f \in \mathcal{D}(\mathcal{O}^{n}) (n=0,1, \cdots) are called the
field algebra \mathcal{A} and the
algebra of local observables \mathcal{A}(\mathcal{O}) respectively.
5) The linear functional which sends f to \langle \Psi | \varphi_{r}(f) | \Phi \rangle must be continuous with respect to the topology of \mathcal{S}(\mathbb{R}^{4}) for any |\Psi \rangle , \ |\Phi \rangle \in \mathcal{H}.
6) In a relativistic QFT with
positive-definite inner product, the following three conditions are known to be equivalent with one another:
a)
irreducibility of the field algebra \mathcal{A},
b)
cyclicity and
uniqueness of the vacuum, and
c)
cluster property.
7) All hell breaks loose in the indefinite-metric QFT.
8) Finally, “the good, the bad and the ugly” of the Axiomatic QFT can be found in the easy read textbook by
Jan Lopuszanski (1923-2008): “An Introduction to Symmetry and Supersymmetry in Quantum Field Theory” , World Scientific, 1991.
Okay, let us go back to Eq(4) (or Eq(5)) and smear the field with a good test function f
U^{-1}(T) \ \varphi_{r}(f) \ U(T) = D_{r}{}^{s} (\Lambda) \int d^{4}x \ \varphi_{s}(T^{-1}x) \ f(x) . In the left-hand-side, if we change the integration variables T^{-1}x \to x, we get
U^{-1}(T) \ \varphi_{r}(f) \ U(T) = D_{r}{}^{s} (\Lambda) \int d^{4}x \ \varphi_{s}(x) \ f(Tx) = D_{r}{}^{s} (\Lambda) \ \varphi_{s}(f_{T}) , where f_{T}(x) \equiv f(\Lambda x + a). As we said before, the above equation is equivalent to
U(T) \varphi_{r}(f) \ U^{-1}(T) = D_{r}{}^{s} (\Lambda^{-1}) \ \varphi_{s}(f_{T^{-1}}) , with f_{T^{-1}}(x) \equiv f\left(\Lambda^{-1}(x - a)\right) .
