3
$\begingroup$

We know that the Minkowski vacuum corresponds to the thermal state in a Rindler wedge at Unruh temp. But does the thermal state in one Rindler wedge at Unruh temperature uniquely map to the Minkowski vacuum? Or could there be other states in the Minkowski space field theory which also corresponds to thermal state in one of the wedges? In other words, If I am doing field theory in say the left Rindler wedge and the state is specified to be thermal at Unruh temp, does it mean that the state in the Minkowski space field theory is the vacuum, or could it be something else?

One class of states it would map to are the following: $$U_R|0\rangle$$ where $U_R$ denotes an unitary operator in the right Rindler wedge. But it's not clear that they correspond to nice Minkowski states. Are there other states in Minkowski space field theory that, on writing in the Rindler basis and tracing out degrees of freedom from one wedge, would yield the thermal state?

$\endgroup$
1
$\begingroup$

OON's idea can be made more precise to obtain an example of state which is identical to Minkowski vacuum in the left Rindler wedge but is different in the right wedge.

It is just matter of a correct choice of the local operator in OON's idea. The idea works only with certain local operators: isometries localized in the right wedge.

Take a real smooth function $g$ whose support is included in the (open) right wedge. Consider the unitary operator (though an isometry would be sufficient) $$U(g) := e^{i \phi(g)}\tag{1},$$ where I assumed that the field operator $\phi(g)$ is selfadjoint (it is essentially selfadjoint so actually $\phi(g)$ in the exponent is the closure of it). From Weyl commutation relations and Stone theorem, $$U(g) \phi(f) =\phi(f)U(g)\tag{2}$$ if $f$ and $g$ have spatially separated supports as it happens in particular if $f$ has support in the left wedge.

Define $$\Omega_g := U_g\Omega$$ where $\Omega$ is Minkowski vacuum. It is clear that $\Omega_g$ is still a unit vector and is different from $\Omega$ if $g\neq 0$. Regarding Wightman functions of the new state, (2) implies that $$\langle \Omega_g| \phi(f_1)\cdots \phi(f_n)\Omega_g \rangle= \langle \Omega| U_g^*\phi(f_1)\cdots \phi(f_n)U_g\Omega \rangle = \langle \Omega| \phi(f_1)\cdots \phi(f_n)U_g^*U_g\Omega \rangle = \langle \Omega| \phi(f_1)\cdots \phi(f_n)\Omega \rangle$$ if each $f_k$ has support in the left wedge. There $\Omega$ and $\Omega_g$ are not distinguishable. In particular $\Omega_g$ has the same thermal properties as $\Omega$ with respect to the Killing boost (it satisfies the KMS condition due to Bisognano-Wickmann theorem).

In the right wedge, $\Omega_g$ is a coherent state in view of its definition.

$\endgroup$
0
$\begingroup$

Of course there are infinite number of states with reduced density for the left wedge being in thermal state while the total state is not the Minkowski vacuum. To obtain the latter the reduced density matrix in the right state should alsobbe in the thermal state AND there also should be a specific entanglement between wedges.

Among nice Minkowski states you may consider e.g. $$ |\psi\rangle= \int_{\Omega_R} d^4x \mathcal{O}(x)|0\rangle $$ where $\mathcal{O}$ is a local operator and $\Omega_R$ is restricted to the right wedge. Any state of this type will have the same thermal state for the reduced density matrix of the left wedge.

Note that the same reduced density matrix means that observables restricted to the left wedge can't distinguish those states.

The meaning of all this is very simple. No signal passes from one wedge to another. Thus having information only about one wedge is not enough to reconstruct the state of the quantun field in the whole spacetime.

$\endgroup$
  • $\begingroup$ Thanks. That's what I had thought, but how does one explicitly show that the reduced density matrix for that state is thermal? $\endgroup$ – Nirmalya Kajuri Nov 6 '18 at 10:47
  • $\begingroup$ Reduced density matrix is $$\sum_{E^{"} }\phantom{E} _R\langle E"|\psi\rangle \langle \psi| E"\rangle_R= \sum_{E",E',E} e^{-\beta (E+E')/2)}\phantom{E} _R\langle E"|\mathcal{O}(x)|E\rangle_R \phantom{E}_R \langle E'|O(x)|E"\rangle_R |E\rangle_L \phantom{E}_L \langle E| $$ How do I proceed from here? $\endgroup$ – Nirmalya Kajuri Nov 6 '18 at 10:59
  • $\begingroup$ I think I got it: by taking the spectral decomposition of the operator $\mathcal{O}(x)$ and then tracing over its eigenstates instead of the energy eigenstates. Are there other ways to show this? $\endgroup$ – Nirmalya Kajuri Nov 6 '18 at 12:02
  • $\begingroup$ I do not think your answer is correct. You are saying in particular that $\langle \Omega| \phi(f)\phi(f) \phi(g)\phi(g) \Omega \rangle/ \langle \Omega|\phi(g)\phi(g) \Omega \rangle = \langle \Omega| \phi(f)\phi(f) \Omega \rangle$ if $f$ and $g$ have supports spacelike-separated. $f$ supported in the left wedge and $g$ in the right wedge. $\Omega$ is Minkowski vacuum. $\endgroup$ – Valter Moretti Nov 6 '18 at 14:52
  • $\begingroup$ Well, Wick theorem proves that the left-hand side of the identity above is the right hand side with the further addend $\langle \Omega| \phi(f)\phi(g) \Omega \rangle/ \langle \Omega|\phi(g)\phi(g) \Omega \rangle$. So what we should to prove is that $ \langle \Omega| \phi(f)\phi(g) \Omega \rangle=0$ $\endgroup$ – Valter Moretti Nov 6 '18 at 14:56

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.