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There are some things I don't get in QFT.

I will take the scalar free field :

$$\phi(x)=\int d\tilde k (a(\vec k) e^{-ikx}+a^\dagger(\vec k) e^{ikx})$$

My first question is:

In QFT, do we consider the variable $x=(x,\vec{x})$ as a parameter in the operator $\phi$ or $\phi(x)$ would be the projection of an "abstract" hilbert space operator $\widehat{\phi}$ on the $|x \rangle$ basis ?

To illustrate what I mean: in classical quantum theory, if we take a potential $\widehat{V}$ that depends on time, we will have:

$\widehat{V}(t)$ : t is just a parameter we didn't do any projection on a basis.

And now, if I want its expression depending on $x$, I need to do a projection:

$\langle x | \widehat{V}(t) | \phi (t) \rangle= V(x,t) \phi(x,t)$

And $V(x,t)$ is the projection of the abstract hilbert space in the $x$ basis. Or to be more precise : $V(x,t)=\langle x | \widehat{V}(t) | x \rangle$

So: is $\phi(t,\vec{x})$ something like $\langle x | \widehat{\phi}(t) | x \rangle$ or $\vec{x}$ is just a parameter like $t$ in classical quantum?


Next question:

In quantum theory, when we have a state $| \phi \rangle$. And we say "well, I want to measure the observable $A$ (position or momentum for example)".

So we need to know the eigen values of $A$, the projection of $| \phi \rangle$ on the eigenspaces of $A$ to be able to predict the probability of measurement of the quantity $A$ on this state.

In QFT, we are dealing with a field. And the questions we are asking finally is "what is the amplitude in a given point of the space-time of my field?"

To answer this we would need the eigenvalues/eigenspaces of the field $\phi(x)$. But in my QFT book as in the course of my teacher we never look for things like this. For me it would be very important to compute those.

Why in practice we are not interested in those values? Don't we want to be able to predict what amplitude of the field an experimentalist would find? I don't get why we skip this part.

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  1. A quantum field is an operator-valued function (or distribution, if you want to be slightly more formally correct) on spacetime. That is, to each $x$ we associate an operator $\phi(x)$. There is no "operator" $\phi$, $\phi$ is a function that assigns an operator $\phi(x)$ to each event in spacetime.

  2. There is no such thing as $\lvert x\rangle$ in relativistic quantum field theory because there is no such thing as a relativistic position operator - the closest are the so-called Newton-Wigner operators, but they are not as useful.

  3. You certainly can look for the eigenspaces of each $\phi(x)$, but it has likewise turned out not to be a useful way to do things in QFT, in particular since we generically don't know the Hilbert space of the theory if it is not a free theory. You can compute the expectation values of the field and its powers - called correlation or n-point functions, which in turn allow us to compute all sorts of experimentally accessible amplitudes and expectation values - without doing that. Just proceed to learn QFT and you will see, you're not "skipping" anything here.

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  • $\begingroup$ re. 3: to be somewhat pedantic, sometimes the eigenstates of $\phi$ are useful. They are required, for example, in order to set up the bosonic path integral (cf. Weinberg, chapter 9). $\endgroup$ – AccidentalFourierTransform Sep 16 '17 at 13:01
  • $\begingroup$ @AccidentalFourierTransform Okay, they're not as useful as in ordinary QM, not not useful. $\endgroup$ – ACuriousMind Sep 16 '17 at 13:04

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