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It's known that the Minkowski vacuum is observed as a thermal bath for Rindler observers, in paticular:

$\langle0_{M}|N_{M}|0_{M}\rangle=0 \space\space\space\space\space\space\space $ (1)

$\langle0_{M}|N_{R}|0_{M}\rangle \neq 0\space\space\space\space\space\space\space\space$ (2).

However, from eq(1) to eq(2) we change from inertial observers to Rindler observers, so why the $|0_{M}\rangle$ remains the same in these two equations, since we know that quantum state vectors transform according to different observers?

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    $\begingroup$ Where did you get the idea that "quantum state vectors transform according to different observers"? The state vector is what it is, regardless of who is observing it. $\endgroup$
    – Buzz
    Nov 27, 2020 at 3:03
  • $\begingroup$ @Buzz It's stated in many QFT books. Indeed different observers see equivalent states, but not the same states. For instance, in Weinberg's QFT book volume 1 the transformation rule of states under Lorentz transformation is given. $\endgroup$
    – DEDS
    Nov 27, 2020 at 3:18
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    $\begingroup$ Please edit the question title to be more descriptive of what you are asking. $\endgroup$ Nov 27, 2020 at 3:22

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The fact that in your two equations there's $|0_M \rangle$ comes from what you want to measure. The first one is trivial to interpret: It is the number of particles in the Minkowski vacuum for the Minkowski observer. But the second one is a bit tricky: It is the number of particles in the Minkowski vacuum from the point of view of the Rindler observer. You could have for example $\langle 0_R | N_R | 0_R \rangle$ but it is identically null. The Unruh effect is precisely the measure of the number of particles in the Minkowski vacuum from the point of view of the Rindler observer.

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  • $\begingroup$ Yes the object is the same for all observers (in our case, it's the Minkowski vacuum). However, this just implies that the corresponding state vectors are equivalent, but not necessarily the same. For instance in QFT there are transformation rules for one-particle states under Lorentz transformation. $\endgroup$
    – DEDS
    Nov 27, 2020 at 16:20
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    $\begingroup$ There's no transformation applied to the vacuum state in the case of the Unruh effect. And even if there was a transformation, as you said, the vacuum states would be equivalent, so creation and annihilation operators would give an equivalent result. Do not forget that the ladder operators for the Rindler observer are expressed in terms of the ladder operators for the Minkowski observer. $\endgroup$ Nov 27, 2020 at 16:39

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