State amplitude and field operator covariance in QFT

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I don't really like the notation using $u'(x')$ because it leads to much confusion. Let's lay back a little bit and write everything in a different way.

We have a spacetime manifold $M$ a vectorspace $V$ and a field which is basically a map $u: M rightarrow V$. I will now employ the active point of view. We have now $M = mathbb{R}^4$, but I write $M$ since I am to lazy to write all the time this bold $mathbb{R}$.

Our fields are representations of the Lorentz group which are induced by representations of the Lorentz Group on $M$ and on $V$. We want to understand this now better.

Let $omega$ be an element of the Lorentz-Group. Then we obtain a map $Lambda(omega): M rightarrow M, x mapsto Lambda(omega)x$ where $Lambda(omega)$ denotes just the matrix in the fundamental representation. On the vector-space we have also a representation, e.g. a Spinor representation or the Vector representation, which we denote $A(omega): V rightarrow V, v mapsto A(omega)v$. Now let's call for a moment $F = {u: M rightarrow V}$ the fieldspace. We then get a representation of the Lorentz group on $F$ by: begin{equation} B(omega): F rightarrow F, u mapsto A(omega) circ u circ Lambda(omega)^{-1} end{equation}

This can be made a little bit more clear in a diagram:

begin{CD} M @>{u}>> V; @VV{Lambda(omega)}V @VV{A(omega)}V M @>{u' = B(omega)u}>> V; end{CD}

where we see directly, that $u' = A(omega) circ u circ Lambda(omega)^{-1}$.

At this stage $u'$ is a function on $M$ and if we call the element om $M$ where $u'$ is evaluated at $x in M$, we write $u'(x)$. I think the problem is clearified by thinking on $u'$ in this way.

We now have, following the lines of the book by Bogoliubov $u'(x) = U^{-1}(omega) u(x) U(x)$ with $x in M$ and anything as defined above.

Now as a little test, that this argumentation is reasonable let us calculate the infinitesimal generator generating the transformations on $F$. We restrict now to translations (okey, they are Poincare and I wrote Lorentz till now, but this does not matter), since this is more easy.

Translations are $mathbb{R}^4$ acting on $mathbb{R}$ for $a in mathbb{R}^4$ as $Lambda(a): mathbb{R}^4 rightarrow mathbb{R}^4, x mapsto x+a$. They are acting trivially on $V$, i.e. $A(a) = id.$. Hence we have $B(a): F rightarrow F, u mapsto u cong Lambda(a)^{-1} = u(cdot - a)$. Hence for $x in M$ we have in above nomenclature $u'(x) = u(x-a) = e^{-i p a}$ with $p a = a^mu partial_mu$. This reproduces the result in formula (9.19) and the formula directly above (modulo a sign since I have chosen the active point of view).

Now what is with this $x'$ floating around? Normally if a physicist writes in a book $x'$ he always means the map $x' = Lambda(omega)$ i.e. $x'(x) = Lambda(omega)^{-1} x$. If a physicist writes $u'(x')$ normally the first prime denotes the map $A(omega)$ acting on $u(x'(x))$ and the $x'$ denotes the map as I explained before. If a physicist writes $u'(x)$ the prime means, that $u' = A(omega) circ u circ x'$ i.e. $u'(x) = A(omega) u(x'(x))$.