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This question arose in the context of another discussion here: https://physics.stackexchange.com/questions/241247/semi-classical-calculation-gives-wrong-answer-for-emission I wanted to analyze the time-varying charge density of a heated tungsten filament, and Pisanty said I couldn't do it because the filament wasn't in a pure state, it was in a mixed state (a thermal state actually) so the chaarge density would be stationary. (Feel free to correct me if I am misrepresenting Pisanty's argument).

That discussion was shut down by the moderators, but it was somewhat continued here: Is there a charge density in quantum mechanics? where Pisanty elaborates on his reasons why you can't analyze the filament as a pure state.

I think he must be wrong. Isn't the whole universe in a pure state? Isn't a mixed state just something we use when we don't have (or don't want) all the detailed information? Pisanty defines the tungsten filament as a mixed state...specifically, a thermal state...and then concludes that the charge density must be stationary in time. I think he's wrong...I think he just threw away the information about the time-varying charge density when he chose to analyze the filament as a thermal state.

Isn't the filament actually in a pure state, with an enormously complicated time varying charge distribution? And if so, what is to stop us (in principle) from using Maxwell's Equations to analyze the resulting emission of radiation? Other than the fact that it might give us the wrong answer.

EDIT: It doesn't look like anyone wants to answer this: a few comments have ridiculed the notion that something as chaotic as a glowing filament could be in a "pure" state. Look, I don't like the terminology either, but I adopted it because it has a precise technical meaning as used by Pisanty in the previous discussions. A "pure" state (correct me if I'm wrong) is simply one for which the wave function is completely specified; that is, it is not a statistical mixture of various pure states. So a beam of silver atoms coming out of an aperture is, for practical purposes, in a "mixed" state of spins...half up and half down. This is different from a single atom being 1/2|up> + 1/2|down>, which is actually pure sideways.

So my question really is: isn't every system in the universe, including my tungsten filament, actually in a "pure" state...except for the fact that we simply aren't in a position to know the precise details of that state? So in cases where it doesn't make a difference, we treat it as a statistical mixture. Even the beam of silver atoms coming out of an aperture...yes, for our purposes it's a statistical mixture, but in fact isn't that beam itself in a pure state?

If no one wants to answer this I'm going to have to answer it myself.

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    $\begingroup$ the tungston filament is a thermal state. If the density matrix is calculated it will be asy to show that it is not a pure state: multiply it by itself, and if it yields the same density matrix, it is pure. Otherwise it's mixed. $\endgroup$ – Peter Diehr Mar 5 '16 at 14:04
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    $\begingroup$ it is not clear that there is a state function for the universe; if it exists it certainly cannot be written out. Coming down to earth, it is clear that there are both pure and impure states, just as there are both sums and products. $\endgroup$ – Peter Diehr Mar 5 '16 at 14:09
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    $\begingroup$ Given all the phonons and electrons bouncing around and interacting with each other, including making or destroying phonons, it is clear that that the filament is not a 'pure' state in the sense of a singular wave function. I'm not sure why you would think that it was, really. And we won't even touch the population of point defects in the W... $\endgroup$ – Jon Custer Mar 5 '16 at 15:31
  • $\begingroup$ For clarity: yes, you are misrepresenting the argument I presented here. I'm unsure why you think it's OK to ignore half of a post and claim to be representing it accurately, when that half explicitly contradicts your conclusions. The current answers by Tom and ACuriousMind do a reasonable job at explaining the parts you seem to have misunderstood; for myself I can do little beyond pointing you back to that post and asking you to read it carefully before making further claims about its contents. $\endgroup$ – Emilio Pisanty Mar 6 '16 at 15:37
  • $\begingroup$ @emilio I have no idea what you're trying to tell me. But exactly one year ago we were in a similar discussion and you said: "black-body radiation can be explained equally well with discrete energy levels on the emitters rather than the radiation". How is this different from what I'm saying about solving the filament mechanically in QM and then using Maxwell to get the radiation? $\endgroup$ – Marty Green Mar 6 '16 at 16:37
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Ok, I think the problem you're having is you can't agree on a good definition for a "pure state". Let's call our tungsten filament "system A", and the rest of the universe "system B". We say that A is in a pure state if A is not entangled with B. In other words, we can represent the state of the universe as $|\Psi_U\rangle=|\Psi_A\rangle|\Psi_B\rangle$.

In the case of a tungsten filament, that is unlikely to be the case! The filament is interacting strongly with the environment; it's releasing photons into the environment, it's interacting with air molecules, etc. The net effect of these interactions is to correlate the state of the tungsten with the environment, thus entangling A and B.

Now, of course, the entire universe is always in a pure state (unless you believe in multiverse theory), since system B is then trivial.

To elaborate a little: The entire universe is described by some wavefunction $|\Psi_U\rangle$ which depends on the coordinates of every particle in the universe. If you want to look at a particular subsystem, you want to consider the density matrix, $\rho_A$ of that subsystem. You get $\rho_A$ by taking the density matrix of the entire universe, and tracing out all the coordinates that are not part of the system you want to study.

$$ \rho_A = Tr_B(|\Psi_U\rangle\langle\Psi_U|) $$

Sometimes, you will find that $\rho_A$ is a trivial density matrix, describing a pure state $|\Psi_A\rangle$, in which case you know that A was in a pure state--a state which was not entangled with the rest of the universe! In most cases, you will find that $\rho_A$ describes an ensemble of states. In the case of the tungsten filament, the ensemble should be the ensemble described by the Boltzmann distribution. This is not because we somehow lack information about the filament; this is because the filament IS in a ensemble of different states.

As a side note, this is as true for a piece of tungsten as it is for a single atom! However, it's easier to isolate single atoms from the environment because they are so small. So you could, in principle, have an atom in a pure state; if you want a macroscopic object to be in a pure state, all bets are off.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind♦ Nov 18 '16 at 0:49
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Suppose that $A$ and $B$ are (linearly independent) elements of the state space of your filament, and that $X$ and $Y$ are (linearly independent) elements of the state space of the surrounding environment. Suppose the filament and the environment occupy the state $A\otimes X+B\otimes Y$. In what pure state do you imagine the filament to be?

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I'm reluctant to get involved in this question, but the root issue is interesting enough that I will ignore my better judgement...

I'm going to reformulate your question slightly: suppose we have a universe that only contains your heated tungsten filament, lots of empty space, and the electromagnetic field initially in its vacuum state. The filament is initialized in some pure, excited state. The initial state is, therefore, not thermal. It is also not an energy eigenstate of the system, due to coupling with the electromagnetic field if for no other reason, and therefore will evolve in time. What happens?

Okay, I confess that having never done this experiment I can't tell you for sure, but I have an educated guess: the filament will reach a state that is, for all intents and purposes, indistinguishable from a thermal equilibrium state. At the same time, it will keep emitting blackbody radiation out into space, with the state remaining thermal but decreasing in temperature in a quasistatic way.

Now, the seeming paradox is that a thermal state, as has been repeated here ad nauseum, is a stationary state. But we know that the filament can't be in a stationary state: it was initialized in a non-stationary state, and it is emitting radiation. So what's going on here?

The answer is the following: thermalization in an isolated quantum system must be taken to mean that, for any initial conditions, the system evolves so that any local observable takes on thermal (and thus stationary) values. However, the overall state is still continually time-evolving. The evolution of the state all goes into generating complicated entanglement between distant parts of the system.

The result then is that you have a state that is not stationary, but appears to be a quasistationary thermal state for any observable that does not involve highly non-local joint measurements, which in practice are always inaccessible for a macroscopic system anyway. In this way the paradox is resolved.

Going back to the example at hand, the charge density, like any other simple observable, will look approximately thermal after a short time. Like any other local observable of any other equilibrium system, classical or quantum, the charge density will have a quasistatic probability distribution, but this probabilistic nature means that there is no conflict with the Maxwellian statement that a stationary charge configuration does not radiate.

This idea of thermalization in a closed quantum system is closely associated with the so-called Eigenstate Thermalization Hypothesis, and a better understanding of how this process occurs and the necessary conditions for such thermalization is a very active research topic at the moment. I discussed it a little more in a similar context here.

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The answer is no, not everything is the universe is in a pure state, at least not when considered individually. The reason is that apart from very rare cases that are extremely isolated, systems are constantly interacting with their environments. That means that there are extremely complex (practically unrecoverable) correlations between elements of the system and the environment; hence a pure state for the system alone, without the environment, cannot be defined. Specifically think of a system in canonical ensemble; the existence of a temperature for such a system implies a great number of interactions with a heat bath, which has a great number of degrees of freedom. Their constant and complex interactions means that their wave functions are inextricably intertwined; one cannot write any single one of them in a pure state.

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    $\begingroup$ You betray your own answer with the phrase "practically unrecoverable". My question is not about the practicality of writing down this or that wave function. It is not about whether there are a "great number of interactions". It is about whether, in the deepest sense, there is a precise form to the n-dimensional wave function which QM tells us is the true representation of a physical system. $\endgroup$ – Marty Green Mar 6 '16 at 15:25
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    $\begingroup$ I guess I'm not being clear. It doesnt matter if the correlations are recoverable or not. The very fact that there are correlations between the filament and its environment implies that - in the deepest, most precise sense - there is no wave function that can be defined for the filament alone. $\endgroup$ – Tom Mar 6 '16 at 15:29
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    $\begingroup$ To further this notion: imagine two individual atoms that are brought near to each other and interact. From the time of their interaction, one cannot write any single one of them in a pure state. $\endgroup$ – Tom Mar 6 '16 at 15:31
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    $\begingroup$ True, but it absolutely stops you from writing a pure state of only one of the atoms. Even if there is no molecule, and they just collide: you can no longer write a pure state for one of the atoms. $\endgroup$ – Tom Mar 6 '16 at 17:54
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    $\begingroup$ And the same is true for the filament. As long as interaction with the environment has taken place, you cannot write a pure state for the filament alone. $\endgroup$ – Tom Mar 6 '16 at 17:55
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No, not everything in the universe is in a pure state, although the universe as a whole may well be.

The reason is simple: Suppose there is some system in a pure state with density matrix $\rho$. The states of the subsystems are given by partially tracing out the other subsystems. If there is entanglement in the state $\rho$, the resultant density matrix for the subsystem is generally a mixed, not a pure state, cf. this answer of mine. That is, the existence of entanglement is inconsistent with the claim that everything in the universe is in a pure state.

The idea that everything is in a pure state and we just don't know it is a classical intuition that does not hold in quantum mechanics. Thermal states may be imagined as states where a heat bath that keeps them at that temperature has been traced out - and there is no reason to think that by some accident of nature a tungsten or any other filament should "really" be in a pure state and this thermal state is a bad apprxomation - it is constantly interacting with its environment, entangling parts of the filament with parts of the environment, yielding a mixed state when tracing out the environment.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind♦ Nov 18 '16 at 0:50

protected by Qmechanic♦ Mar 8 '16 at 6:55

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