1
$\begingroup$

In the usual approach to QFT presented e.g., in Weinberg's book, the state of a system is dependent on the observer. Quoting this book, in page 109 we have:

Notice how this definition is framed. To maintain manifest Lorentz invariance, in the formalism we are using here, state-vectors do not change with time - a state-vector $\Psi$ describes the whole spacetime history of a system of particles. (This is known as the Heisenberg picture, in distinction with the Schrödinger picture, where the operators are constant and the states change with time.). Thus we do not say that $\Psi_{\alpha}^\pm$ are the limits at $t\to \mp\infty$ of a time-dependent state-vector $\Psi(t)$

However, implicit in the definition of the states is a choice of the inertial frame from which the observer views the system; different observers see equivalent state-vectors, but no the same state-vector.

So Weinberg makes clear two things: (1) first to make Lorentz invariance easier to deal with, we work in the Heisenberg picture - states don't evolve in time, observables do; (2) if we have two observers $\mathcal{O}$ and $\mathcal{O}'$ they will in general describe the system with different $\Psi$ and $\Psi'$.

If then $\mathcal{O}$ and $\mathcal{O}'$ are observers in inertial frames of reference, the postulates of special relativity implies the Physics seem must be the same.

Now let's think of AQFT which seems to be as the abstract version of the Heisenberg picture.

We have one algebra $\mathfrak{A}$ of observables. This is defined entirely in terms of the canonical commutation relations. It can be defined as an algebra generated by the symbols $\phi(f)$ for $f\in C^\infty_0(M)$ and imposing suitable relations (linearity, field equations, etc).

The algebra of observables is defined without respect to any observer.

Next we define states. A state is a positive normalized functional $\omega : \mathfrak{A}\to \mathbb{C}$.

Again, by the way everything is framed, there is no mention to an observer up to here. So my question is: is this $\omega$ implicitly observer dependent (and if so, how do we see it), or is this $\omega$ one "observer agnostic state" encoding what every observer sees, which in turn can then be "resolved with respect to observers" to yield which what each of them sees?

The relation state-observer in the AQFT framework is being confusing to me, and that is what I'm trying to clarify here.

$\endgroup$
2
$\begingroup$

See how Weinberg is talking about state-vectors. He claims that different observers will see different state-vectors, albeit yielding equivalent states. Here equivalent means that they produce the same result when applied to an observable.

In more concrete terms, if $\psi$ and $\psi'$ are the state-vectors for the same state measured by two different observers, then for every observable $O$ one has

$$(\psi,O\psi) = (\psi',O\psi').$$

In particular, one deduces that the two state-vectors induce the same state $\omega$ on the algebra of observables.

$\endgroup$
  • $\begingroup$ So the point is: the definition of two state-vectors $\Psi,\Psi'$ being equivalent is that they yield the same mean values $$\langle \Psi, O\Psi\rangle = \langle \Psi',O\Psi'\rangle,$$ for any observable $O$ and so since $\omega$ gives directly this mean value, it is the "intrinsic state" instead of the description for an specific observer? $\endgroup$ – user1620696 Jul 15 '18 at 23:48
  • $\begingroup$ Yes. A state gives you an observer-independent description of the physical state of a system. Is when you fix a representation is which the state has a state-vector that observers come into play. As an analogy, think of tensors, that are genetically objects that observers-independent. When you fix a coordinate system, the representation of the tensor now depends on the frame of reference, hence the observer. $\endgroup$ – Phoenix87 Jul 16 '18 at 6:38
  • $\begingroup$ I understand the point of view. But this raises a following doubt: how does one resolve $\omega$ with respect to an observer? In the analogy with tensors, $F$ is the electromagnetic field, an abstract two-form irrespective of any observations. We then know how to resolve this into electric and magnetic fields seem by observers. So given $\gamma: \mathbb{R}\to M$ an observer, is there a known procedure to "resolve $\omega$ wrt to $\gamma$?" $\endgroup$ – user1620696 Jul 16 '18 at 13:18
  • $\begingroup$ I'm not aware of a general procedure. Consider that a state is not always representable as a state-vector in an arbitrary representation. More generally you will have to deal with a density operator, hence a density matrix. For example, if your algebra is that of compact operators, then states correspond to density operators. In this case you can fix a basis for your Hilbert space and represent the matrix wrt to the given basis. $\endgroup$ – Phoenix87 Jul 17 '18 at 12:13
  • $\begingroup$ Notice that $\langle \psi ' \vert O \vert \psi ' \rangle \neq \langle \psi \vert O \vert \psi \rangle $ if $O$ is not a Poincarè-invariant observable. $\endgroup$ – pppqqq Jan 9 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.