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So if there are a billion boxes in a warehouse, I would like to know conceptually how to tell if it is in a pure state.

I know that if it is in a pure state (not mixed) that the density matrix has only one nonzero diagonal element and that the trace is 1, whereas a mixed state would not be idempotent and the trace would be less than 1.

I'm thinking that the warehouse resembles a microcanonical ensemble or a canonical ensemble since the box count is fixed. I also think it wise to imagine it is well-insulated (isolated) so that there is no entanglement with the environment.

Would a good way to tell if it is in a pure state would be to measure each box's temperature? (Too many, right) Or instead measure a few times several boxes to make sure the measurement doesn't change and is the same for all in the sample? Or would one just look at the thermostate? ha ha please forget that last remark.

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Do all these billion boxes contain the same pure state? Or are they all different pure states? –  Peter Shor Mar 11 '13 at 19:54
    
Well I guess the best answer would be it is unknown. After all, if they were all the same pure state, or even different pure states, than wouldn't this coherent state be pure? (As in pure but not maybe an eigenstate) –  nate Mar 11 '13 at 20:15

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up vote 3 down vote accepted

Mathematically, a normalized (unit trace) density matrix describes a pure state only if $1,0$ are the only eigenvalues that the density matrix has.

Physically, there can't be any objective answer to the question whether a physical system is in a pure state or a mixed state. The state vector or the density matrix aren't classical degrees of freedom; instead, they are collections of complex numbers that encode subjective probabilistic knowledge about the system.

A system in a pure state – any pure state – represents the maximum knowledge one may have about the system in quantum mechanics. It is an analogy of the point on the phase space in classical physics. A system in the mixed state is always just a probabilistic mixture of pure states, $$ \rho = \sum_i p_i |i\rangle\langle i|$$ for some orthogonal ket vectors, and this mixed state is a counterpart of probabilistic distributions on phase space in classical physics. We use several ket vectors with $p_i$ strictly in between $0$ and $1$ because we don't know which one is realized. But in principle, we could know. Perhaps, there exists another observer who does know and who describes the situation by a pure state.

For example, we may talk about a photon (with a known momentum) coming from a light bulb. If we know its linear polarization, we will describe it by a pure state. If we don't know the polarization, we may describe it by a mixed state. But it may be the same photon and other people may know less or more about the photon's polarization.

So there can't exist any operational procedure that objectively settles the pure/mixed question. The answer depends on one's knowledge – an important and universal fact about any quantum mechanical theory.

Incidentally, the comments about the temperature are completely unrelated to the original question. It is not sensible to associate a temperature with a single degree of freedom. So if a box is a microscopic state and not a macroscopic piece of wood, it doesn't have any temperature. Nevertheless, multi-body systems with many particles may be found in microcanonical, canonical, or grand canonical ensembles. When we talk about many particles, it is possible to reconstruct the distribution function and it does become measurable – in this sense, the ensembles are distinguishable by a measurement. However, if we talk about a single particle, it again makes no sense to physically ask whether it obeys a microcanonical or canonical distribution. These are two possible density matrices that the particle may pick and different density matrices describe different subjective knowledge about the system. For one observer, a particular electron may be in a microcanonical ensemble, for another observer, it may be in a canonical ensemble.

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Well I don't see any comments but yours so can't explain in my mind the downvote. Your comments make a lot of sense - in particular, I know temperature was not such a good comment to make now, depending upon the situation. Also, maybe it is best to visualize the pure state by the surface of a Bloch sphere and mixed states inside the sphere... (en.wikipedia.org/wiki/Qubit) I guess then it will take a little time to see why there is no physicality associated with the matrix. Also, I guess the lack in ability to measure all the boxes (atoms or wood) is the lack of knowledge... –  nate Mar 11 '13 at 18:45
    
One last comment/question: If pure states are those that are characterized by a single wavefunction, then in the warehouse example it seems impossible to ever have a pure state, even if it were in fact in one (unless we could measure some observable quick enough so that the system did not evolve). Since we have a lack of knowledge about the system then we have to use probabilistic assumptions and therefore have a "mixed state" with its associated density matrix. Is this correct? –  nate Mar 11 '13 at 19:16
    
Dear Nate, thanks for your comments. I don't understand your situation with the warehouse accurately enough - I have the same question as Peter Shor above, and others - so it's hard to answer this additional question... Yes, the surface of the sphere etc. is a good representation of a pure state. Being inside the Bloch sphere, away from the surface, means to be unsure which point on the surface is realized. –  Luboš Motl Mar 11 '13 at 20:04
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"Physically, there can't be any objective answer to the question whether a physical system is in a pure state or a mixed state." Suppose I'm given a beam of spin 1/2 particles and a Stern-Gerlach apparatus. I measure the average spins along three axes. The 2x2 density matrix is Hermitian so it has three parameters which are fixed by the measurements. The density matrix is now known and an eigenvalue decomposition says if it's pure or mixed. All inertial observers would agree with the eigenvalues so the result is objective. –  Stephen Blake Mar 11 '13 at 20:59
    
Well in @LubošMotl photon example, one can know the polarization of a photon -> pure state, or not -> mixed state. Your example was also able to know if the state was pure or not, based on average spins along three axes. You also knew your basis was spanned by 2 vectors. But most important your example could determine, through an average, of whether the state was pure or not. Am I crazy or does this seem the opposite of what you quoted? Or is the operative word, "objective", and if so, is that the most one can get (without being omniscient) - objective? –  nate Mar 11 '13 at 21:13

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