State amplitude and field operator covariance in QFT

Question

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?

Answer 1

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.

Answer 2

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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. $u'(x) = A(\omega) u(x'(x))$.