# State amplitude and field operator covariance in QFT

I'm studying QFT on Bogoliubov-Shirkov's "Introduction to the theory of quantized fields" (3d edition). In $§9.3$ they discuss transformation properties of quantum states and operators in QFT. Given the classical transformations of the coordinates $x$ and the set of fields $u(x)$ (the authors make a general discussion where $u(x)$ could be any set of fields, scalar, vector and so on), $$x \rightarrow x'=L(\omega)x \qquad u(x)\rightarrow u'(x')=\Lambda(\omega)u(x)$$ where $L(\omega)$ and $\Lambda(\omega)$ are appropriate representations of Poincaré group identified by the set of parameters $\omega$, we can say, making a comparison with Heisenberg and Schrödinger pictures in QM, that is completely analogous to consider the transformation of quantum states $\Phi$ by unitary operators $U(\omega)$, i.e. $\Phi'=U(\omega)\Phi$. In this way the expectation value of an operator $\hat{O}$ can be expressed in two different and equivalent ways $$\langle\Phi'|B|\Phi'\rangle= \langle\Phi|B'|\Phi\rangle$$ Now taking $B=u(x)$, the field operator, and using $\Phi'$ definition the book ends up with this formula (no. $9.15$)$$u'(x)=U^{-1}(\omega)u(x)U(\omega)$$ The point is that looking at initial transformation i would expected that $B'=u'(x')$ and not $u'(x)$. Why only the functional form of the field operator is taken in account when I consider its transformation?

I suggest you follow this link. But in short: if you transform both the coordinates and the field at the same time, you end up with the same field $u'(x') = u(x)$. If you want to study the effect of a transformation on the dynamics of the system you should consider either changing the coordinates (passive transformation) or the field (active transformation). In this case, the book is choosing the latter.

• Sorry but I don't get the point. In every case I should end up with a relation of the type (considering scalar fields) $\phi'(x)=\phi(\Lambda^{-1} x)$ if i'm using active point of view, or $\phi'(x)=\phi(\Lambda x)$ for the passive one: both have a change of coordinates. That means that $U^{-1}\phi(x)U$ should be equal with one of the two previous expressions. In every case i don't see the connection $B'\rightarrow u'(x)$ and not $B' \not\rightarrow u'(x')$. – pier94 Sep 8 '17 at 13:05
• In the active case you are transforming the field not the coordinates – caverac Sep 8 '17 at 13:09
• yes but you must end up in an equality where the other member must be the old field evaluated at the coordinates transformed with the inverse transformation $\Lambda^{-1}$. So i suppose that $U^{-1} u(x) U= u(\Lambda^{-1} x)$. – pier94 Sep 8 '17 at 13:48

\require{AMScd}

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: $$B(\omega): F \rightarrow F, u \mapsto A(\omega) \circ u \circ \Lambda(\omega)^{-1}$$

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)).

• Somehow the diagram does not work. But it works in the preview. Maybe somebody could help? – warpfel Sep 8 '17 at 13:52