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Can we heat a system as well as preserve its pureness?

+ 3 like - 0 dislike
131 views

This is the example from Polchinski's lecture, which seems to contradict with some answers and comments from

this question http://physics.stackexchange.com/questions/32830/is-it-theoretically-possible-to-reach-0-kelvin.

Here is the example:

Let us start by considering two sealed containers, each with a chunk of ice at 0K. We put
the first container A in contact with a heat bath at 400K until equilibrium is reached and the
container is full of steam, and then decouple the heat bath. We heat the second container
B with a laser until the same energy density is reached and again we have a container full
of steam. Is there any difference between the two? The first is in the usual mixed state
(ensemble): it is entangled with the bath. By itself it is described by a density matrix

$$ \rho_A= \frac{e^{-\beta H_A}} {  Tr e^{-\beta H_A }}  $$

The second is in some pure state

$$ \rho_B= \vert \psi >< \psi  \vert $$

My question is that whether the above process can be realized, i.e. heating and preserving pureness.

asked Sep 21, 2016 in Theoretical Physics by C Thone (105 points) [ revision history ]
edited Sep 22, 2016 by C Thone

As I understand it, in the second case in order to heat the chunk of ice the photons making up the radiation field of the laser will soon lose their coherence and stop being in a pure state due to interactions with the material making up the chunk of ice. So I dont see why there should be a difference in the final equilibrium state of the system in the two cases.

3 Answers

+ 3 like - 0 dislike

No, strictly speaking. Let us consider a scattering of a photon from a still electron. After scattering the electron may move somehow. Its state depends on the scattered photon state: only the whole system is in a pure state, both final particles being entangled. Now, if you average over the final photon states, you get a density matrix description for the electron. Worse, after scattering, you will have many soft photons in the final state. Taking them into account in calculation is necessary for predicting a non-zero probability of scattering. But experimentally we usually do not care about soft photons and their precise states (they carry away too few energy), so experimentally we take any output as "the same" and desirable, i.e., we add up different outputs. Here we again factually average over one subsystem and inevitably get a density matrix description for the other subsystem.

I personally cannot imagine a steam volume without electromagnetic radiation, a black body radiation, for example.

answered Sep 22, 2016 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Sep 22, 2016 by Vladimir Kalitvianski

I have some questions about the above argument. Imagine that ice is in pure state $\vert \Psi >_{ice}$ and one photon is in pure state $\vert \psi>$. Through Hamiltonian evolution, how to judge the final composite state? Is it seperable or entangled? why is this process not the so called local operation in LOCC?

It is directly visible from the total Hamiltonian. If it contains some interaction between the initial states, then we may only have a pure state for the total system. It will be, say, a superposition of the eigenstates of the total Hamiltonian with the same total energy: $$|\Psi\rangle_{E_{{\rm{in}}}}= \sum_k C_k|\Psi_{{\rm{ice}}}\rangle_{E_{{\rm{ice}}}(k)}|\psi_{{\rm{ph}}}\rangle_{E_{{\rm{ph}}}(k)}, \;E_{{\rm{ph}}}(k)+E_{{\rm{ice}}}(k)=E_{{\rm{in}}}.$$ If there is no interaction, then the initial direct product remains a direct product and no entanglement may occur. You may forward the photon towards some absorber and still have the ice in the same state, whatever happens to the photon. It means mathematically that the ice Hamiltonian variables are totally separated from the rest of the system (universe). No interaction, no change.

On the other hand, if the "ice" is some macroscopic system with many degrees of freedom, then the incoming photon excites many degrees of freedom at once and the low energy levels close to each other are excited very effectively as they take few energy, so the superposition description above becomes impractical. We average over them and get a density matrix for the ice. See the answer by Dierk Bormann.

+ 3 like - 0 dislike

Of course it is (theoretically) possible to put the system in a pure state with an energy per atom corresponding to 400 K. Usually one would rather call this process "putting the system into an excited state" than "heating the system". However, from a practical point of view there is little difference.

The density of states of a macroscopic system at the given energy will be extremely high, so there is a huge number of such states available. If the system is ergodic (in a loose sense, which essentially all real systems are) then any such state dynamically generates the microcanonical ensemble for that energy, which for all practical purposes is completely indistinguishable from the canonical ensemble described by your density matrix \(\rho_A\). That's the beauty of statistical mechanics. 

answered Sep 22, 2016 by Dierk Bormann (70 points) [ no revision ]
+ 2 like - 0 dislike

The concept of heat and heating is intrinsically thermodynamic and, like (nonzero absolute) temperature, makes sense only in local equilibrium, where the system is described by a density matrix. Thus if you can heat something to 400K, the end product will be a state in local equilibrium - else it was no heating and the designation 400K is meaningless.

For a laser to heat a system to 400K, assume for simplicity the nuclei to remain fixed. Then the fate of the ice is described by QED for the joint system consisting of the photons from the laser and the electrons from the ice. Since the laser cannot be targeted to infinite accuracy, it will have a thermal spread in momenta, hence is already described by a density operator. thus even when the ice was initially at 0K and hence in a pure state (density operator of rank 1) it will start getting mixed once the laser interacts with the ice. Since multiparticle systems are strongly mixing unless extremely dilute (which is not the case for ice), local equilibrium will very soon be established, long before the 400K are achieved.

The situation would be different if, as in your comment to one of the answers, only a single photon in a pure state interacted with ice of 0k in a pure state. Then the joint system would remain pure - though only as long as decoherence by the environment is ignored. It would be practically impossible to keep the ice at 0K (and hence in a pure state) even without the laser....

answered Sep 23, 2016 by Arnold Neumaier (12,375 points) [ no revision ]

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