Mixed up on mixed states

  • #1
nomadreid
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A pure state can be interpreted as belonging to a system, but it can also be interpreted as belonging to a single particle (although the resulting probability is in respect to the system), and as I understand it, this is now the preferred interpretation. But in https://en.wikipedia.org/wiki/Quantum_state#Mixed_states, it is stated that "A mixed quantum state is a statistical ensemble of pure states" which makes it sound as if a mixed state is only a quality of systems of equations, and cannot be considered a property of a single particle as a pure state can. Is this correct?
 

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  • #2
vanhees71
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The general quantum state is just described by a statistical operator, and it has its statistical meaning given by the foundations of the theory (for me it's Born's rule, although some people in these forums think Born's rule refers to pure states only, but I think that's making the affair just more complicated), and nothing else (many people still think after over 90 years having time to get used to QT, there should be something more, but there's no hint for something more whatsoever from a scientific point of view, and I try to stay out of philosophy as much as I can and stick to science).

The pure states are special in that they describe the situation that a complete set of compatible observables has determined values due to the preparation of the system. I think that's the right way to look at states rather than thinking pure states are the usual case. In fact they are rarely realized, and we can do so only for rather simple systems consisting of a few particles (or photons, for which it is most easily done from a technical point of view).
 
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  • #3
PeroK
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A pure state can be interpreted as belonging to a system, but it can also be interpreted as belonging to a single particle (although the resulting probability is in respect to the system), and as I understand it, this is now the preferred interpretation. But in https://en.wikipedia.org/wiki/Quantum_state#Mixed_states, it is stated that "A mixed quantum state is a statistical ensemble of pure states" which makes it sound as if a mixed state is only a quality of systems of equations, and cannot be considered a property of a single particle as a pure state can. Is this correct?
A pure state is one where you know the state of the particle or every particle in the ensemble. A mixed state is one where you know only the probability that each particle is in a given state.

For a pure state, every particle is in the same state. Let's call it ##\psi_A##.

For a mixed state, every particle is in one of several possible states. For example, you might know that 90% of the particles are in state ##\psi_A## and 10% of particles are in state ##\psi_B##.

(This knowledge, as mentioned above, comes from how you know the particle or ensemble has been prepared.)

Now, a mixed state is very different from a superposition of states. For example, in the above, both ##\psi_A## and ##\psi_B## can be described as a superposition of other states - this is completely analogous to being able to describe any vector a linear combination of any basis vectors.

Note, however, that the factors in a superposition are probability amplitudes, whereas the factors in a mixed state are probabilities. Before you go any further in QM you need to grasp this completety. This is one of the most important thing to understand in the fundamentals of QM.

To write this out. Assume we have our states as above and a set of basis states ##\{\psi_n\}##. For our pure state then we have:

##\psi_{pure} = 1 \times \psi_A = 1 \times \sum a_n \psi_n##

A particle in this state has a probability of ##1## of being in state ##A##. I've included the ##1## to emphasise that fact. Whereas, the factors ##a_n## are probability amplitudes.

In a mixed state we have:

##\psi_{mixed} = (p_A \times \psi_A) \& (p_B \times \psi_B) = (p_A \times \sum a_n \psi_n) \& (p_B \times \sum b_n \psi_n)##

Where ##p_A, p_B## are the probabilities that a particle is in state ##A, B## respectively. Note that I've used ##\&## instead of ##+## here to emphasise that you cannot directly mix probabilities and probability amplitudes in one equation.

The next step is to learn about the "density" operator, which does allow you to mix probabilities ("weighting" or "density") and probability amplitudes in the same equation.
 
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  • #4
nomadreid
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Thanks, vanhees71 and PeroK.
First, vanhees71, a quick search seems to say that the idea that the Born rule applies also to mixed states (or, as some say, generalized to them) seems to be pretty standard.
Secondly, PeroK. Thank you, I understand the mathematics that you pointed out, but my question is more a matter of interpretation (which perhaps is just the philosophy that vanhees71 avoids). I give an analogy. An earlier interpretation of experiments like the two-slit one was that the pattern on the screen was purely statistical among a bunch of electrons, but not that, as presently usually taken to be the case, that each electron was interfering with itself. In a similar fashion, a pure state refers to a single particle, although one confirms this by the measurement of many identical particles. So, does a mixed state refer to a single particle, or must one say that each particle is in a pure state but that one does not have enough information to determine which one?
 
  • #5
PeroK
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Thanks, vanhees71 and PeroK.
First, vanhees71, a quick search seems to say that the idea that the Born rule applies also to mixed states (or, as some say, generalized to them) seems to be pretty standard.
Secondly, PeroK. Thank you, I understand the mathematics that you pointed out, but my question is more a matter of interpretation (which perhaps is just the philosophy that vanhees71 avoids). I give an analogy. An earlier interpretation of experiments like the two-slit one was that the pattern on the screen was purely statistical among a bunch of electrons, but not that, as presently usually taken to be the case, that each electron was interfering with itself. In a similar fashion, a pure state refers to a single particle, although one confirms this by the measurement of many identical particles. So, does a mixed state refer to a single particle, or must one say that each particle is in a pure state but that one does not have enough information to determine which one?
If you are asking these questions you can't possibly have understood what I posted. A pure state refers to a pure state. That may apply to a single particle or a whole ensemble of particles. A mixed state referes to a mixed state. That may apply to a single particle or a whole ensemble.

For example, suppose you have a preparation process that produces particles in state ##\psi_A## 90% of the time and in state ##\psi_B## 10% of the time. If a single particle emerges from that preparation process then that particle has a mixed state.

Note that you can talk about the state of that particle: it has a definite state (A or B), you just don't know which one.

But, with a superposition of basis states, the particle is not in any of the basis states. Its state is the superposition.

PS: this has nothing to do with an electron "interfering with itself".
 
  • #6
kith
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So, does a mixed state refer to a single particle, or must one say that each particle is in a pure state but that one does not have enough information to determine which one?
Your phrasing seems odd to me because even classically, you are never "forced" to take the second option. If you flip a coin and don't record the result, you cannot distinguish the first from the second option experimentally. (But of course, the first corresponds to a rather strange ontology.)

QM introduces something new the other way round: if you want to assign a quantum state to a single particle under all circumstances, you have to assign a mixed state in certain situations (namely in the case that the particle is entangled with another system).
 
  • #7
nomadreid
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Thank you for your replies, PeroK and kith.
First, PeroK's postscript. I understand that my question had nothing to do with the electron interfering with itself. This was merely a (an imperfect) analogy to try to distinguish a property belonging to a single versus an ensemble of particles. Imperfect analogies have the annoying property that some of the non-analogous parts can be misinterpreted as being part of the analogy. Sorry for the poor phrasing. Anyway, you answered my question well in your
a mixed state. That may apply to a single particle or a whole ensemble.
(I wouldn't go so far as to say I had understood "nothing" of what you wrote, but I will admit that I apparently missed that point in your first post. The reason I pose questions which are going out on a limb is to get them possibly shot down, which is my best way of learning. That is, rather than hide my possible misunderstandings, I flaunt them, risking o:) for which I am grateful.)
So, to continue flaunting my uncertainty:
kith: I am a little confused by your statement
you have to assign a mixed state in certain situations (namely in the case that the particle is entangled with another system)
I thought that entanglement meant that the system was now in a pure state; that is, I had thought that an interaction between two systems which could be expressed as a single tensor product was not considered entangled, so that your statement sounds a system is both a pure and a mixed state simultaneously (other than a pure state considered as a trivial case of a mixed state). Or did you mean that the system is in a pure state but due to only part of the system being "visible", it appears as a mixed state? Am I confused as to the terminology, or as to my interpretation of your phrase?
I am grateful for continued criticism and clarification.
 
  • #8
PeroK
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Thank you for your replies, PeroK and kith.
First, PeroK's postscript. I understand that my question had nothing to do with the electron interfering with itself. This was merely a (an imperfect) analogy to try to distinguish a property belonging to a single versus an ensemble of particles. Imperfect analogies have the annoying property that some of the non-analogous parts can be misinterpreted as being part of the analogy. Sorry for the poor phrasing. Anyway, you answered my question well in your

(I wouldn't go so far as to say I had understood "nothing" of what you wrote, but I will admit that I apparently missed that point in your first post. The reason I pose questions which are going out on a limb is to get them possibly shot down, which is my best way of learning. That is, rather than hide my possible misunderstandings, I flaunt them, risking o:) for which I am grateful.)
So, to continue flaunting my uncertainty:
One issue I think is that you are trying to identify mixed and pure states with other properties and concepts. Let's go back to a simple, and not entirely satisfactory, classical analogy. But, it might do to make a point. If you have a box of apples:

The apples are in a "pure" state if all the apples are the same colour, say.

The apples are in a mixed state if the apples have different colours with certain probabilities.

Now, "pure state" and "mixed state" refer to nothing else. Any question that tries to associate this concept of mixed and pure with any other apple property or concept is fundamentally flawed. And, yes, of course, "pure" is just a special case of "mixed". Perhaps this is a good idea for you to adopt: all states are mixed, and a "pure" state is just a special case of a mixed state where the weighing is ##100\%## for a certain state.

In this respect "pure" and "mixed" refer to the classical concept of weighting or probabilities. If you try to see more in these concepts than this, then you are going wrong, I believe.

The difference between apples and quantum particles lies not in different concepts of "pure" and "mixed", but in the quantumness of the objects themselves.
 
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  • #9
nomadreid
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Thanks, PeroK.
you are trying to identify mixed and pure states with other properties
If you try to see more in these concepts than this, then you are going wrong
I suppose you are right that I should just concentrate on the mathematics as vanhees71 stated, echoing Mermin's famous "Shut up and calculate." It was not just myself who tries to look at other properties, but some books which I use as sources; for example, after the characterization about the trace being equal or less than one, the characterization most often repeated seems to be that a pure system is one that cannot be "factored" (whereby "product" would be "tensor product"), and that a non-trivial mixed system can be. However, I can see that trying to say whether we assign it to a single particle or not becomes metaphysics, given that we don't get the probabilities from measurement of a single particle.
 
  • #10
PeroK
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Thanks, PeroK.


I suppose you are right that I should just concentrate on the mathematics as vanhees71 stated, echoing Mermin's famous "Shut up and calculate." It was not just myself who tries to look at other properties, but some books which I use as sources; for example, after the characterization about the trace being equal or less than one, the characterization most often repeated seems to be that a pure system is one that cannot be "factored" (whereby "product" would be "tensor product"), and that a non-trivial mixed system can be. However, I can see that trying to say whether we assign it to a single particle or not becomes metaphysics, given that we don't get the probabilities from measurement of a single particle.
Mathematically, you have to be careful as you have classical probabilities and quantum probability amplitudes to deal with. The consequences of this are clearly significant. What I'm saying is not to let the mathematical complexities to which this leads obscure what is essentially a simple concept.

It seems to me that this is the opposite of "shut up and calculate". Which perhaps we are using out of context anyway.
 
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  • #11
Demystifier
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A pure state can be interpreted as belonging to a system, but it can also be interpreted as belonging to a single particle (although the resulting probability is in respect to the system), and as I understand it, this is now the preferred interpretation. But in https://en.wikipedia.org/wiki/Quantum_state#Mixed_states, it is stated that "A mixed quantum state is a statistical ensemble of pure states" which makes it sound as if a mixed state is only a quality of systems of equations, and cannot be considered a property of a single particle as a pure state can. Is this correct?
A full answer to your question depends on the interpretation of QM. A minimal answer that does not depend on interpretation is that both pure and mixed states are mathematical tools by which one can calculate probabilities of measurement outcomes in different situations.
 
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  • #12
nomadreid
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Thanks, PeroK and Demystifier. As Demystifier says, it depends on the interpretation, and that gets into metaphysics which is always iffy (albeit irresistibly tantalizing). And PeroK, you are right, I was using that quote out of context, which lent to it a positive, rather than the original negative, connotation: It originated in an article by
Mermin in Physics Today (April, 1989) as a dismissive stance toward the Copenhagen interpretation (even though later he was not quite as categorical): "If I were forced to sum up in one sentence what the Copenhagen interpretation says to me, it would be 'Shut up and calculate!'"
 
  • #13
kith
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I thought that entanglement meant that the system was now in a pure state
Again your terminology seems slightly odd to my: why "now"? If we have two separated systems, we describe them by a product state. Introducing an interaction between the systems leads to an entangled state. Both states are pure.

The states of the individual subsystems on the other hand are pure only in the case of the product state. In the case of the entangled state, if we want to assign a state to only one of the subsystems, we get a mixed state (google "reduced density matrix").
 
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  • #14
nomadreid
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Thanks, kith. That makes sense, and I looked up the reduced density matrix, which jives with what you wrote.
 
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