Suppose I have two extended bodies that are entangled to each other. Are the thermal properties of the objects affected in some way by entanglement?

For example, Imagine that one of the entangled objects is at some finite temperature in the vacuum of space, does the entanglement affect in any way the power lost via radiation?

If the entangled objects are considered as entangled Black Bodies,

  • Do each body radiate proportional to the surface area of each individual object $A_1$ and $A_2$, independent from the area of the other?
  • Do both objects radiate as a single black body with total area equal to $A_1 + A_2$?

Maybe if heat does not carry any information, can it be trasferred in both directions between holographic screens, or horizons? it would seem that heat should not violate causality constraints that affect information transfer


Remembering that entanglement is a property of the global quantum state, the statement "one of the entangled objects is at some finite temperature" makes no sense without some additional information. Either the local (marginal) states of $A_1$ and $A_2$ are thermal, or the global state is thermal, but not both.

If the two objects are called $A$ and $B$, their Hamiltonian can be written

$$H = H_A + H_B + H_{AB},$$ where $H_{AB}$ denotes their interaction, and $H_{A}$ ($H_B$) only affects $A$ ($B$). If the entire system is at thermal equilibrium, the quantum state is the Gibbs ensemble $$ \tau_{AB} = \frac{\mathrm{e}^{-\beta (H_A + H_B + H_{AB})}}{\mathcal{Z}_{AB}},$$ where $\beta$ is the inverse temperature and $\mathcal{Z}_{AB} = \mathrm{Tr}[\mathrm{e}^{-\beta (H_A + H_B + H_{AB})}]$ is the partition function. However, if the subsystems are locally at equilibrium, the state of (say) $A$ is $$ \tau_A = \frac{\mathrm{e}^{-\beta H_A}}{\mathcal{Z}_{A}}.$$ It is easy to see that, in general, $\tau_A \neq \mathrm{Tr}_B [\tau_{AB}]$, unless $H_{AB} = 0$. However, in this case $\tau_{AB}$ is not entangled (it is a product state).

If system is globally in an entangled thermal state $\tau_{AB}$, then the local states of $A$ and $B$ will not be at thermal equilibrium at inverse temperature $\beta$. Conversely, if the subsystems are at equilibrium, the system as a whole will not be. Note that this argument makes no reference to entanglement specifically, but rather just correlation in general. The same is true for composite classical systems.

On the other hand, one could consider entangled global states which lead to local equilibrium states, but where the subsystems are at different apparent temperatures. In this case one can see really weird effects, such as heat flowing from the "colder" subsystem to the "hot" one, apparently violating the thermodynamic arrow of time. See Partovi, Phys. Rev. E 77, 021110 (2008) (arXiv version here).

  • $\begingroup$ interesting. So in a sense, thermal state operators do not commute with entanglement state operators. In other words, if the subsystems are in well defined thermal states, they will not be highly entangled. And if the subsystem are in a well defined entanglement state, they will not be in equilibrium at a well-defined temperature $\endgroup$ – lurscher Nov 7 '15 at 22:44
  • $\begingroup$ @lurscher I am not sure that quantum non-commutativity is directly relevant. The issue is that the presence of correlations (classical or quantum) make the distinct situations of local and global thermal equilibrium incompatible with each other. $\endgroup$ – Mark Mitchison Nov 8 '15 at 18:11
  • $\begingroup$ I'm not sure either, I just thought it was curious that the non-alignment between equilibrium states and entanglement states seem to behave like mutually non-commuting observables. But probably it doesn't mean anything $\endgroup$ – lurscher Nov 8 '15 at 20:49

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