# I Mixed states vs pure states - physical POV

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1. Jul 6, 2017

### SchroedingersLion

Hey guys,

I am having issues with understanding the physical nature of pure and mixed states. Maybe you can help me out?

1) A pure state - superposition is a state that consists of different states at the same time. It's like having several waves, each one belonging to an Eigenstate of the Hamilton operator, with known phase and amplitude, that interfere with each other, right?

Does that mean, when I have a pure superposition of |0> and |1>, I can find times and positions, where the wave of |0> has minimal deflection and the wave of |1> is at its peak, so the state looks more like a |1> state?

2) Mixed states are statistical mixtures.
A mixture of |1> and |0> means that the particle is DEFINITELY in one of these two states, but I do only know the corresponding probability.
For an ensemble, that means that for a maximal incoherent mixture, half of the particles will be in |1> and the other half in |0>, right?

Does that mean the description of mixed states is only a result of the observer's lack of knowledge, for example: " I know that physically, the particle must be represented by a wave, that corresponds either to |1> or to |0>, but I only know the amplitudes, not the phases, so I have to describe my system as a statistical mixture"

That's it for the beginning. Would be nice, if someone could clear me up on this. And please don't talk too much about density matrices, I know about the mathematical differences in their respective descriptions.

Best wishes
SchroedingersLion

2. Jul 6, 2017

### Khashishi

Energy eigenstates aren't the only possible choice of eigenstates. A pure state that is a superposition in one choice of basis states can be written as an eigenstate in another choice of basis states. (That's not true for mixed states.) For example, a particle in the spin right state is written as a linear combination of spin up and spin down states, but there isn't really a fundamental difference between spin right and spin up other than choice of coordinate system -- one is not more "mixed" than the other.

A mixed state will be mixed in any complete basis. Mixed state can arise due to incomplete knowledge or due to taking a subsystem of an entangled system. It is not "DEFINITELY" in one of the eigenstates, and a mixed state cannot be unambiguously represented as a statistical combination of pure states. For example, a 50/50 mixture of a spin up and spin down particle is the same as a 50/50 mixture of a spin left and spin right particle. Mathematically, they are the same state, and cannot be represented as a pure state in any basis.

3. Jul 6, 2017

### Khashishi

Not sure exactly what you are asking. It depends on what you mean by "looks more like a |1> state". If |0> and |1> are eigenstates of the Hamiltonian, then the absolute values of the amplitudes aren't changing in time. So, the whole wavefunction is going to look like a superposition of |0> and |1>. But some particular positions can be influenced by one eigenstate more than the other. For example, for a particle in a box superposition state,
$\psi = a_1 \sin\left(\frac{x \pi}{L}\right) + a_2 \sin\left(\frac{2x \pi}{L}\right)$
The point at L/2 is a node for the "2" state, so it looks like state "1". On the other hand, the point at L/4 is somewhat more strongly influenced by the "2" state than the "1" state, though both contribute.

The complex phases will be rotating with respect to each other.

4. Jul 6, 2017

### Staff: Mentor

Not really, no, unless you want to say that a vector which is the sum of two other vectors "consists of different vectors at the same time". Quantum states form a vector space, so you can always add any number of states together to form another state; it's just adding vectors. And you can choose the basis of the vector space however you like, so you can always make any given state a superposition, or not a superposition, depending on how you choose the basis (either choose the basis so one basis vector is pointing in the same direction as your chosen state vector, or not). This is all just basic linear algebra and is not particular to quantum mechanics.

5. Jul 6, 2017

### SchroedingersLion

Hey guys, thanks for your replies!

Ok, I remember that one has to be careful in interpreting the coefficients in the mixed states sum. In general, you can not say that the system is definitely in one of the eigenstates, BUT: If I have a source that produces known pure states with a known probability, then I have to describe my system as a mixed state, and then I can write it as a sum of these pure states, where the prefactors are the known probabilities, right?

I think you understood me correctly. Of course, the absolute values do not depend on time, my mistake. What happens if I were to measure the energy at L/2 (in the energy eigenbasis). Would I see the energy of state |1>, since the state "2" wave has a node there?

Hm, but isn't this the general interpretation of a superposition? You have two waves from two states, interfering each other to form a third state?
If I am in a superposition of two eigenstates of H and I measure the energy, I will get either the energy of the first or the energy of the second state, there is no in between, so it makes sense to interpret the superposition as "both states at once" before measuring. Or would it be more accurate to say "with respect to energy, this superposition consists of two energy eigenstates"?

6. Jul 6, 2017

### jfizzix

The difference between a pure, even superposition of $|0\rangle$ and $|1\rangle$:
e.g., $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$
.. and an even mixture of $|0\rangle$ and $|1\rangle$:
e.g., $\hat{\rho} = \frac{1}{2} |0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle1|$
is that when measuring the right observable of a pure state, one can get a single outcome 100 percent of the time, while the evenly mixed state will yield 50/50 probabilities no matter what observable is measured.

A quantum mixed state can indeed represent a mixture of quantum states. For example, an ensemble of imperfectly prepared particles will have a statistical distribution of quantum states, relative to the precision of the preparation.
However, mixed states also accurately represent the statistics of individual particles, when they happen to be entangled with other particles.
When the joint state of a pair of particles is maximally entangled, the state of either particle by itself is best represented as a maximally mixed state.

7. Jul 6, 2017

### Physics Footnotes

Suppose you have a system with two energy eigenstates $|a\rangle$ and $|b\rangle$ with eigenvalues $a$ and $b$, respectively. When you perform an energy measurement and get a value of $a$, say, this does not mean that the state of the system was $|a\rangle$. In fact, from an individual measurement you cannot determine what the initial state $|\psi\rangle$ was. At best you can say that the initial state had an overlap with $|a\rangle$ (i.e. $\langle a|\psi\rangle \neq 0$) or, equivalently, has a non-zero $|a\rangle$ component in an energy basis expansion.

If the initial state of the system was $|\psi\rangle=\frac {|a\rangle + |b\rangle} {\sqrt {2}}$, for example, then measurements on this state would give values of $a$ and $b$ 50% of the time each. None of these measurements in and of themselves tell you the initial state of the system.

Also, you should think of the state $|\psi\rangle$ as a state in and of itself, rather than being genuinely composed of other individual states. A spatial displacement, for example, is a vector in and of itself. It is convenient to express it in terms of other vectors, but it is still a new entity with its own identity and properties.

EDITED TO ADD: You might also find it helpful to read this post and this one for a more detailed explanation of these points.

Last edited: Jul 6, 2017
8. Aug 9, 2017

### SchroedingersLion

Good evening,

I have been busy with other courses the last weeks so I completely forgot about this thread. Now I am back and I need to understand this topic in order to learn for another examination in 2 weeks :D

@Phyics Footnotes
Thanks, I've red your posts and I understand. I always used to think about superpositions as if they were "2 states at once", and not an "unique state on itself", since if we have a superposition of energy eigenstates and I were to measure the energy, I would get EITHER the energy of state 1 or state 2, and NOT some strange mixture energy. However, this is not because the superposition is not a unique state, but because it is not an EIGENstate of H.

Now back to a few questions:
If I have a particle source that creates state 1 with a certain probability p and state 2 with 1-p, then I can describe my system as a statistical mixture: p|1><1|+(1-p)|2><2|. So this time, I know for certain that each particle is either in |1> or in |2>.
So I describe every particle in this ensemble as being in a mixed state. Just because I don't KNOW in which pure state my particle is. I know it has to be in a pure state, but I have to describe it as a mixed state, right? Feels a bit strange.

9. Aug 10, 2017

### vanhees71

Yes, you even know that it is in either the pure state $|1 \rangle \langle 1|$ or in the pure state $|2 \rangle \langle 2|$, but before measuring it, you don't know in which. The only thing you know that it's in state 1 with probability $p$# and in state 2 with probability $1-p$. That's described by the mixed state
$$\hat{\rho}=p |1 \rangle \langle 1| + (1-p) |2 \rangle \langle 2 |.$$
Any state, whose statistical operator is not a projection operator is called "mixed". A pure state is always represented by a projection operator.

10. Aug 10, 2017

### DrDu

No, generally you don't know that it is either in state 1 or 2, as the decomposition of a mixed state into pure states isn't unique.

11. Aug 10, 2017

### Staff: Mentor

In general you don't, but for some preparation procedures you can.

12. Aug 10, 2017

### SchroedingersLion

Yeah I was explicitely talking about a particle source that gives me state |1> or state |2> with known probabilities.

So next step, how is it generally? What is the physical nature of a mixed state. Lets talk about a single particle.
I have red it has something to do with coherence. In a pure superposition, I also know the phases of the waves, since I know the coefficients a,b in a|1>+b|2>.
However, wenn I have a mixture, that can be represented as, lets say, |p|² |1><1| + |q|² |2><2|, I do not know anything about the phase of the complex coefficients p and q.

13. Aug 10, 2017

### Staff: Mentor

That's the sort of question that has no answer within the mathematical formalism of QM. The state tells you the probability of getting various results from various possible measurements, but that's all the physical nature that's there.

Many people are unsatisfied with this state of affairs, which is why we hear so many questions along the lines of "but what's REALLY happening under the covers?" and why there are so many interpretations of quantum mechanics.... But the universe doesn't seem to care much about whether we're satisfied with its rules.

14. Aug 11, 2017

### Staff: Mentor

Your question is both easy and hard - I will not touch the hard stuff since it is interpretive and will not lead to an answer in the terms you have posed your question.

First a state is an operator. Forget what beginner and some intermediate texts tell you - they are wrong - but we all have to start somewhere.

Specifically the are positive operators of unit trace. Since this is an I level thread you should know what that means from a course in linear algebra. If not see the following:
http://quantum.phys.cmu.edu/CQT/chaps/cqt03.pdf

Now by definition states of the form |u><u| are called pure. States of the form ∑pi |ui><ui| are called mixed, where since it is a positive operator of unit trace you can easily prove the pi are positive and ∑pi = 1. Straight away you can see they are likely to have a connection to probability and the Born rule shows exactly what that connection is. I will leave that as an exercise for you.

Now, again you can show it, its not hard, all states are either are either pure or mixed. Pure states are easily mapped to a vector space and show why phase makes no difference to states |cu><cu| = |u><u| where c is a complex number of unit length ie a phase factor.

Without telling you they are only a subset of states, beginner texts tell you they are part of a vector space. That is only pure states.

Its not hard - what does it mean physically. Well the Born Rule for all states says - the average of the outcome of an observable O, E(O), is E(O) = Trace(SO) where S is the state the system is in. In fact it to some extent can be proved via Gleason's theorem, which you can look up. Playing with the Born rule will show you all sorts of things - eg if you assume continuity the state of a system after observing it is a pure state and an eigenvector of the observable. This is the so called collapse postulate - which isn't a postulate at all and a simple consequence of the Born Rule.

That all there really is to it theory wise, but what it means - that's much much harder and the part of interpretations.

If you want to look into that - it cant be done in a post - here is THE textbook on it:
https://www.amazon.com/Decoherence-Classical-Transition-Frontiers-Collection/dp/3540357734

Thanks
Bill

15. Aug 11, 2017

### DrDu

I don't see this. All of the information we have is encoded in the density matrix. If the density matrix does not allow for a unique decomposition into eigenstates (which is basically only possible for pure states) then we also can't say that it is a mixture of specific eigenstates.

16. Aug 11, 2017

### DrDu

No, this is representation dependent. Any state, whether pure or mixed can be represented as |u><u|.
E.g. take the operators of angular momentum for a spin 1/2 particle. You can represent them (neglecting all hbar prefactors) by Pauli matrices $\sigma_i$, but you can also represent them as
$\sigma_i \otimes 1_2$. In the latter case, a mixed state with a times eigenstate 1 of s_z and (1-a) times eigenstate -1 of s_z can be represented as
$| \phi \rangle= a |1\rangle |1 \rangle + (1-a) |-1 \rangle |-1 \rangle$.

Edit: Let me dwell a little further on this:
The representation of the state $\phi$ isn't unique. Apparently, we could also write
$| \phi \rangle = a| 1 \rangle |-1 \rangle +(1-a) | -1 \rangle | 1 \rangle$.
If I call the first representation $|\phi_1\rangle$ and the second one $|\phi_2\rangle$, then any combination $|\tilde{\phi}\rangle= \lambda |\phi_1\rangle + (1-\lambda) |\phi_2 \rangle$ represents the same mixed state. Hence, in contrast to pure states, mixed states can be represented as sums of other states. This also holds for density matrices. Pure states are the extremal points of the convex set formed by all states.

Last edited: Aug 11, 2017
17. Aug 11, 2017

### vanhees71

Of course, but I understood the original question such as if somebody is preparing a system either in state $|1 \rangle \langle 1|$ or $|2 \rangle \langle 2 |$ with probabilities $p$ and $1-p$ respectively.

18. Aug 11, 2017

### DrDu

The OP said: "A mixture of |1> and |0> means that the particle is DEFINITELY in one of these two states, but I do only know the corresponding probability."

I can't say why, but with the EPR paradoxon, Bells inequalties and all that stuff in mind, I am reluctant to make statements about a quantum system being in some definite state assuming is only unknown to me.

19. Aug 11, 2017

### Staff: Mentor

As much as I hate contradicting my fellow science adviser and someone who has a doctorate - I much prefer to listen to such - this is wrong.

For example how does it fit with Gleason? It proves, providing you have non contextuality, all states are positive operators of unit trace - assuming the strong principle of superposition any positive operator is a legit state. Mixed states are different, by their very definition, to pure states. The state 1/2 |a><a| + 1/2 |b><b| can never be put in the form |u><u| - its simply not possible by the axioms of linear algebra and what a positive operator of unit trace is. The decomposition is not unique - one can find other mixed states that observationaly are the same - but not a pure state.

See page 53 Ballentine - the following is equivalent to being a pure state. If p is the state it is pure iff trace (p^2) =1 - it is mixed otherwise ie if trace (p^2) <1. Nothing can change that - if trace (p^2) <1 it cant be made to be equal to one. Think of the Born rule and take the state as the observable - Trace (p^2) < 1 will give observational outcomes different statistically to trace (p^2) = 1.

Thanks
Bill

20. Aug 11, 2017

### vanhees71

But that's the operational definition of mixed states, leading to the formal description of the states as a statistical operator.

Where are your specific quibbles concerning the EPR paradoxon, Bell's inequalities and the like?

21. Aug 11, 2017

### vanhees71

But that's the operational definition of mixed states, leading to the formal description of the states a

But that's the operational definition of mixed states.

Where are your specific quibbles concerning the EPR paradoxon, Bell's inequalities and the like?

22. Aug 11, 2017

### DrDu

I think it is operationally ok to define a statistical operator as the outcome of a measurement where the result is not observed. I am only concerned whether we can really say that the system is really in a definite eigenstate but we do not know which one. This may be a philosophical question, but I think it is the original question of the OP.

23. Aug 11, 2017

### Staff: Mentor

Indeed.

But if that was the only way to get mixed states all our interpretation issues would be over.

Its of zero importance of course to the Ensemble interpretation - it does't care how the state was formed, but the fact is states formed by that operational definition are not formed the same way as mixed states from decoherence. One in such cases cant say for sure it was in some state from the mixture prior to observation and therein lies the whole argument about if decoherence solves the observation problem or not.

Thanks
Bill

24. Aug 11, 2017

### vanhees71

I don't understand your problems with this interpretation of the mixed state. Usually it's quoted as "the measurement problem", i.e., if you have given a system in some pure state $\hat{R}_0=|\psi \rangle \langle \psi|$ and you measure an observable (for which $|\psi \rangle$ is not an eigenstate), then (assuming you have made an ideal von Neumann filter measurement), before taking note of the outcome, you have to associate the statistical operator
$$\hat{R}_1=\sum_{n} |\langle n \rangle \rangle n| \hat{R}_0|n \rangle \langle n|,$$
where $|n \rangle$ denote a complete orthonormal set of eigenstates of the measured observable, and this is a mixed state. There cannot be a unitary time evolution from $\hat{R}_0$ to $\hat{R}_1$, and that's taken as the "measurement problem" by proponents of the collapse conjecture, where they take this "transition" as a physical process rather than an adaption of the probabilistic description after gaining information (i.e., in this case the knowledge that somebody has done a von Neumann filter measurement on the system, but not taking note of the measured result).

25. Aug 11, 2017

### vanhees71

Yes, and that's why I always say that the ensemble interpretation solves all the apparent paradoxa concerning the socalled "measurement problem".