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In special relativity, we are introduced with lorentz-transformations, as changes of the reference frame. One and the same physical situation, observed by two observers, moving relative to each other, is seen as a change of the coordinate system.

An event is thus represented by a point in spacetime (which in special relativity can be described as a vector space). The change in the reference frame is thus just a change in the basis of the vector space. The observables "position" and "time" that are seen by the observers can be identified with the coordinates / components. The same can be said about the components of the electromagnetic-field-tensor.

So my question is now about quantum mechanics. We have operators, states, and observables: What should I identify these with?

In case of nonrelativistic quantum mechanics: Does $\hat{x}$ already describe the components , or the fundamental object?

The same goes for QFTs: Is $\hat{A}^{\mu}(x)$ the field operator that stands for the mathematical object itself (a tensor at the space-time-point with spacetime-coordinates x), or does it already describe the components of this tensor for the chosen system of reference?

Maybe to give more information about what I find confusing: My guess is that, since $\hat{x}$ is the observable, it corresponds to the components of the point in space that I want to measure. But on the other hand, we know that $\hat{x}$ is an element form a vector space as well, that can be written aus a superposition of other operators.

Now, when I apply a "passive transformation" (for example a rotation, to stay out of relativistic treaties), I would usually expect that this passive transformation changes the components $x_i$, but leafs $x= x^i \hat{e}_i$ invariant.

However, if in QM I identify $\hat{\vec{x}}$ with the coordinates, then the passive transformation would change the whole operator.

Is this just a conventional thing, and I could formulate Quantum mechanics independent of a choosen coordinate system? This clings to another question where I asked about a rigorous underpinning of "vector operators" here.

Quantumwhisp
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  • Can you measure a vector or a tensor? – DanielC Jul 14 '21 at 20:32
  • $\hat x$ is a physical object. In the coordinate basis it looks like x. In the momentum basis, it looks like $i\hbar \partial_p $. You call these components? – Cosmas Zachos Jul 14 '21 at 21:28
  • @CosmasZachos What you describe is another instance of "objects and components", that I didn't mention in my question. Yes, I would call these components as well (as I would call $\Psi(x)$ the components of $|\Psi\rangle$ in the basis $|x\rangle$).

    That we have another instance of vectors and components makes it more difficult for me to reconcile all these things.

     – Quantumwhisp Jul 14 '21 at 23:47
  • I suspect you'd profit from studying representations of Lie groups before you get to the peculiar Heisenberg group of QM. – Cosmas Zachos Jul 15 '21 at 00:22
  • @DanielC by your question, you probably want to point me to that the answer to my question is the former, $\hat{x}$ corresponds to the components in the classical case. Is that right? – Quantumwhisp Jul 15 '21 at 09:42
  • @CosmasZachos You think the answer to my question involves the Heisenberg Group? I decided to leaf out the fact that hilbert space and operators have a vector-space-structure as well, because I thought it would just put another layer of complexity to the question. – Quantumwhisp Jul 15 '21 at 09:44

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