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If we consider a free (massless scalar) field $\phi$ in Minkowski space and look at it in Rindler coordinates (which correspond to what an accelerated observer sees), we find that the action of the field can be described by another free field $\tilde\phi$ related to the former one by a Bogolyubov transform.

The Unruh effect then says that the vacuum state of $\phi$ contains particles of the field $\tilde \phi$. Specifically

$$<0_M| b^\dagger_kb^{\mathstrut}_k|0_M> = (\exp(2\pi \Omega_k/a)+1)^{-1}$$

Where $b^\dagger_k$ are the ladder operators of the free field $\tilde \phi$, $\Omega_k = |k|$, and $a$ is the acceleration described by the Rindler coordinates. Since this expression is the same as the expectation value of the particle number for a thermal bath of temperature $T=a/2\pi$, the conclusion seems to be that an accelerated observer sees the Minkowski vacuum as a thermal state.

My problem/question is if this is really the case. The vacuum state is after all a pure state and the thermal state is definitely a mixed state. I don't see how changing what we call particles could ever result in creating a statistical ensemble where before there was none.

Or is something else meant here by temperature? I've seen mentions of a thermal radiation created by this temperature, which really seems to imply an understanding as a heat bath!

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  • $\begingroup$ A particle in QFT is the result of a measurement. Which and how many particles one finds depends on the experiment that one performs. An accelerated free particle count is a different experiment from a non-accelerated one. I would also note that a physical vacuum ground state is always a mixed thermal state. It can't be anything else according to the third law of thermodynamics. The case T=0K exists only in your physics books as a convenient phantasy. $\endgroup$ – CuriousOne Jun 17 '15 at 19:16
  • $\begingroup$ @CuriousOne naive question alert first. Above question and your comment is a bit beyond me at this stage, but just bear with me on one point. What is the Minkowski vacuum compared to other normal vacuums (vacuua??) Is it vacuum in flat space, i.e. why the distinction? I can ask this as a question later if its a pain answering in comments section. $\endgroup$ – user81619 Jun 17 '15 at 20:15
  • $\begingroup$ @AcidJazz: My answer is also on the naive side. The way I understand it is that the Minkowski vacuum is a state in which an inertial observer does not see any particles. That, of course, does not exist because of thermodynamics, so you can only say it's the physical vacuum in which the particle density does not matter "for our purposes". This would also imply no gravity, i.e. no curvature. I think it's pretty much the case that s.harp has in mind. $\endgroup$ – CuriousOne Jun 17 '15 at 21:03
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    $\begingroup$ @CuriousOne Saying the vacuum state does not exist because temperature is not zero is like saying the ground state of a Hamiltonian does not exist because the temperature is not zero. $\endgroup$ – s.harp Jun 18 '15 at 15:25
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    $\begingroup$ @CuriousOne The problem was of the form: out of x follows y, where x and y seem to be incompatible. Saying "x is not physically realisable, you need to take $\tilde x$ and then its not a problem anymore" is to me sidetracking the question. $\endgroup$ – s.harp Jun 18 '15 at 18:33
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The Minkowski vacuum is defined on the whole Minkowski space. The Rindler wedge covers only half of a given spacelike surface, the other half being covered by a different Rindler wegde. So that the $\tilde{\phi}$ field defined gives only half od the degrees of freedom corresponding to $\phi%$. Ehen you trace over the the DOF from the other Rindler wedge the Minkowski vacuum is a mixed thermal state for accelerated observers. The key here is that since there is a horizon you have no choice but to trace over causally inaccessible degrees of freedom, and thus comes the mixed state.

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  • $\begingroup$ "Hwen you trace over the the DOF from the other Rindler wedge the Minkowski vacuum is a mixed thermal state for accelerated observers." - If you have a pure state, that is a density matrix in the form of projection operator, and trace out a subspace you will still have a pure state. If you have a mixed state, tracing out a subspace may result in a pure state but not the other way around! $\endgroup$ – s.harp Jun 18 '15 at 15:15
  • $\begingroup$ "The Minkowski vacuum is defined on the whole Minkowski space." - This does not make sense to me, the vacuum state is not defined anywhere "on" space-time. It is defined on a Hilbert space. Although you are right that $\tilde \phi$ (as a classical field) does not fully determine $\phi$. $\endgroup$ – s.harp Jun 18 '15 at 15:18
  • $\begingroup$ @s.harp, concerning how to define de Minkowski vacuum. In the usual QFT construction one Fourier decomposes the fields and promotes the coefficients to ladder operators. The vacuum is thus defined as the state which is annihilated by all $a_k$. If you repeat this in Rindler coordinates then since your coordinates do not cover the whole Minkowski then the Fourier coefficients do not have support on the whole Minkowski space. The state lives in the Hilbert space, but it is defined via the ladder operators, and this ones in the accelerated frame do not exist for the whole Minkoswki space. (...) $\endgroup$ – cesaruliana Jun 18 '15 at 17:35
  • $\begingroup$ (...), and this is what I mean by the vaccum being defined on Minkowski. Secondly, if you have a pure state and take the partial trace with respect to some degrees of freedom then you're left with a mixed state in the reduced space if the pure state is entangled in the respective subspaces (en.wikipedia.org/wiki/Quantum_state#Mixed_states). The Minkowski Vacuum has just the sort of entanglement between the two Rindler wedges that when you perform the partial trace with respect to one of them you're left with a thermal state in the other. $\endgroup$ – cesaruliana Jun 18 '15 at 17:38
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    $\begingroup$ @s.harp, concerning the remark, it is true that two Rindler wedges do not cover the whole Minkowski spacetime, but they cover all of the cauchy surfaces. In other terms, the eigenfunctions of the klein-gordon equation in each Rindler wedge when taken together do form a complete basis for functions on the Minkowski spacetime (with soem suitable analytic conditions). Therefore you just need the Rindler wedges to get a complete equivalent Hilbert space in relation to the usual inertial one. $\endgroup$ – cesaruliana Jun 18 '15 at 20:40

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