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I Entangled system pure state or not?

  1. Oct 8, 2016 #1
    According to atty.. entangled system are pure (can be in superposition)
    while according to Bill.. entangled system are not pure (not in superposition)

    Here's the prove of what they stated:
    atty said:
    https://www.physicsforums.com/threa...-states-in-laymens-terms.734987/#post-4642601 message number 11:

    "In decoherence, the system consisting of environment + experiment is in a pure state and does not collapse. Here the experiment is a subsystem. Because we can only examine the experiment and not the whole system, the experiment through getting entangled with the environment will evolve from a pure state into an improper mixed state. Since the improper mixed state looks like a proper mixed state that results from collapse as long as we don't look at the whole system, decoherence is said to be apparent collapse."

    In a thread Bill said:
    "Just to reiterate - because its entangled it is not in a pure state hence not in a superposition which only applies to pure states."

    Can someone settle this clearly?
    So in an entangled system, is it in pure state (in superposition) or not?
    How can one of them (both expert) be wrong in something this basic?
     
  2. jcsd
  3. Oct 8, 2016 #2

    Simon Bridge

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    They are talking about different situations.
    What is it you are trying to understand? A fine point in semantics?
     
  4. Oct 8, 2016 #3
    But atty's "environment + experiment" is example of entanglement. Atty said it was in pure state. But Bill said entanglement can't be in pure state. Hence the conflict. It's not semantics but real descriptions.
     
  5. Oct 8, 2016 #4

    Simon Bridge

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    One referred to the complete setup and another to a subset.
    @atyy @bhobba
     
  6. Oct 8, 2016 #5

    bhobba

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    You have to ask Atty what he means.

    Here are the facts.

    A pure state is written |a>. Suppose a system can be in pure state |a> and |b>. Suppose system 1 is in pure state |a> and system 2 in pure state |b>. This is written as |a>|b>. Conversely suppose system 1 is in state |b> and system 2 in state |a>. This is written as |b>|a>. All systems are in pure states - no problem. Now for the twist with one of the main principles of QM - the superposition principle. It can be in a superposition of |a>|b> and |b>|a>. This is peculiar to QM - you can have a system like 1/root 2 |a>|b> + 1/root 2 |b>|a>. The systems are no longer in a pure state - but something new and weird. Its called entangled. It turns out if you chug through the math its in a mixed state 1/2 |a>b><b|<a| + 1/2 |b>a><a|<b|. It's the essence of and a very simple example of decoherence as well as Bell. The entangled system acts more classically (not entirely though because of Bell) but 50% of the time you will get |a>|b> and 50% of the time |b>|a>. Observe system 1 and you immediately know system 2. So you can consider system 1 to be a sort of 'measurement' device of system 2. You can even 'fool' yourself into thinking system 1 is in that state and system 2 in that other state. Things are a lot simpler which is the essence of the ignorance ensemble interpretation. Note however the Bell subtlety so it is possible to tell the difference - for 'real' decoherence there would be no way to do that - but this is just a simple illustration.

    Thanks
    Bill
     
  7. Oct 8, 2016 #6
    Thanks for the elaborate explanations above. But you stated "Just to reiterate - because its entangled it is not in a pure state hence not in a superposition which only applies to pure states." You seemed to reserve the word "superposition" only to pure state. But in the above paragraph you mentioned "Now for the twist with one of the main principles of QM - the superposition principle. It can be in a superposition of |a>|b> and |b>|a>. This is peculiar to QM - you can have a system like 1/root 2 |a>|b> + 1/root 2 |b>|a>. The systems are no longer in a pure state - but something new and weird. Its called entangled."

    So the superposition principle can still described mixed state. But why did you say the word superposition is only reserved for pure state?
    Please be clear or kindly explain because your words are laws. Thank you.
     
  8. Oct 8, 2016 #7
    Edit. I mean above the superposition principle can still described entangled systems (as you mentioned in the quotes of your statement I gave above). Although I think the density matrix is about operators hence the mixed state is not in superposition because the density matrix is a device used by humans to produce classical statistics.. no problem about that.
     
  9. Oct 8, 2016 #8

    Simon Bridge

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    ... it follows that a superposition can end up with a state that is not pure. ie. not all combinations of pure states are, themselves, pure states. Am I following you? Being a superposition is not a defining characteristic of a pure state.

    Lets see if I understand what you are asking...

    |a> and |b> are pure states of single particle systems.
    A 2-particle system is set up so it could be |a,b> or |b,a> ... the 2-particle system may be in state |S> = (1/√2)|a,b> + (1/√2)|b,a> (for instance).

    You want to know what it is that distinguishes |S> from |a> and |b> in terms of being a pure or a mixed state.
    Is |a,b> a pure state?

    That about right?
     
  10. Oct 9, 2016 #9
    yes.. |a,b> is pure state.
    what distinguish it is that |S> can hav choices which can be put in density matrix as mixed state.
    if true no problem about that my point is that the entangled system can be superposition, non classical state being superposition but Bill seems to be saying that there is no superposition in entanglement. For mixed state i know there is no superposition bec the density matrix is a tool for classical statistics.. no problem with that. But entangled system are unclassical and there you can use the term superposition for it cant you? but Bill in my original post seems to say no. hence my confusion.
     
  11. Oct 9, 2016 #10

    Simon Phoenix

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    Bill will correct me if I've misread things, but what I think he's saying is that if we have 2 objects a and b in the state |01> + |10>, for example, then if we write down the state of object a alone, it is not a pure state, but must be described by the mixture |0><0| + |1><1|.

    The total system is in a pure (superposition of |01> and |10>) state - but the individual subsystems a and b, considered independently (i.e. by ignoring the fact that they are part of a larger system) are in mixed states.
     
  12. Oct 9, 2016 #11

    Simon Bridge

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    Yah. Technically, it is a and b that are entangled ... |a> and |b> are pure states only of the single-particle system ... they are not pure states in the entangled system. |S> is an entangled state only in the sense that it describes entangled particles.
    It is not unusual to construct multi-particle states out of single-particle states. ie. the electron "shells" are actually single particle states but the atom usually has more than one electron.

    |S> is not a linear superposition of the states |a> and |b>, though it is a linear superposition of |ab> and |ba> - which may be the confusion over the language about pure states being a superposition and mixed states not being a superposition ... hmmm... I don't think we can represent |a> in terms of a superpositon of possible |S>...
     
  13. Oct 9, 2016 #12

    vanhees71

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    I think this becomes a bit confused now. By definition a pure state is described by an arbitrary vector in the Hilbert space (more correct is to say it's described by a vector in Hilbert space modulo a factor, i.e., a ray in Hilbert space, but that's not so important at this level of discussion). Since the Hilbert space is a vector space, any superposition of vectors is again a vector providing the description of another pure state. It's also arbitrary whether you say a state is in superosition of other states of not. You can always choose a basis (complete set of orthonormal vectors) where a given state is a member of this basis (Schmidt decomposition theorem).

    Entanglement also doesn't so much refer to states but to observables, although of course both concepts are closely related. So quantum theory offers the possibility to prepare systems in pure states where certain observables are entangled. One example is the Stern-Gerlach experiment. For simplicity I refer to non-relativistic QT. You have a (electrically neutral) particle or composite particle-like system (in the original case it was a silver atom) with a magnetic moment and let it run through an inhomogeneous magnetic field. Then the particle is deflected due to the corresponding dipole force and the deflection is proportional to the spin component in direction of the magnetic field. This leads to a spatial separation for different spin states, which makes the position and spin-##z## component of the particle (assuming the magnetic field being directed in ##z##-direction) entangled.

    In non-relativistic physics, the position ##\vec{x}## and the spin-##z## component ##s_z## form a complete set compatible observables, thus you can consider the position-spin basis which formally is a Kronecker product (as introduced in bhobbas posting above) of position and spin eigenstates, ##|\vec{x},\sigma_z \rangle##. Now from an arbitrary not too badly localized state (think of a Gaussian wave packet as an example) after running through a Stern-Gerlach apparatus you end up in a state like
    $$|\Psi \rangle=\sum_{i=-s}^s c_i |\phi_{\vec{x}_i},\sigma_z=i \rangle.$$
    here ##|\phi_{\vec{x}_i} \rangle## are the position part of the wave function, peaked around a position ##\vec{x}_i##, where a particle most probably ends up after running through the Stern-Gerlach apparatus and having a spin-##z## component with ##\sigma_z=i##. This is entanglement: If the wave packets are not too much overlapping (in practice you can build SG apparati that lead to practically non-overlapping partial beams) you have a one-to-one correspondence between the position of the particle and the value of its spin-##z## component, i.e., you have entanglement between position and spin-##z## component.

    It becomes even more fascinating if you have composite system, of which far distant parts of the entire system can have entangled observables. The mostly used examples are polarization entangled two-photon states, because it's nowadays pretty easy to provide such states by a process called parametric downconversion, where you shine with a laser into a birefringent crystal and get out two photons with less energy than the laser frequency that are entangled with respect to their polarization. You can, e.g., have the polarization states of the two photons in the state
    $$|\Psi \rangle=\frac{1}{\sqrt{2}}( |H V \rangle-|VH \rangle).$$
    You can determine in which state each of the single photons is. For entangled states that's usually a mixed state, and it is defined by a process called "partial tracing". The pure two-photon state can equivalently described by the statistical operator ##\hat{\rho}_{\Psi}=|\Psi \rangle \langle \Psi |##, i.e., a statistical operator describes a pure state if and only if it's a projection operator. The partial tracing leads to
    $$\hat{\rho}_A=\frac{1}{2} (|H \rangle \langle H|+|V \rangle \langle V|)=\frac{1}{2} \hat{1},$$
    and the same for the other photon, labeled with ##B##, which means that each single photon of the pair is completely unpolarized. However the total two-photon state tells us that the polarizations of the photons are nevertheless strictly correlated, i.e., if Alice finds for her photon that it is ##H## polarized, then Bob's photon must be ##V## polarized and vice versa. The entanglement persists over long times, if nothing during there propagation disturbes them, and this means the detection of the photons may take place at arbitrary far distances, and still although the photons themselves are strictly unpolarized the polarization of the two photons in the pair are strictly correlated due to the entanglement.
     
  14. Oct 9, 2016 #13

    stevendaryl

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    First, as an aside, the Dirac bra-ket notation, which is so elegant and clear when you're talking about a single system, becomes visually confusing to me when you have composite systems.

    Second: Is what you're saying true? Aren't there cross-terms? If you have a pure state [itex]|\psi\rangle + |\phi\rangle[/itex], then the corresponding density matrix is [itex]|\psi\rangle \langle \psi| + |\psi\rangle \langle \phi| + |\phi\rangle \langle \psi| + |\phi\rangle \langle \phi| [/itex]. Letting [itex]|\psi\rangle = \frac{1}{\sqrt{2}} |a\rangle |b\rangle [/itex] and letting [itex]|\phi\rangle = \frac{1}{\sqrt{2}} |b\rangle |a\rangle[/itex] gives 4 terms in the corresponding density matrix:

    [itex]\frac{1}{2} |a\rangle |b\rangle \langle a| \langle b| + \frac{1}{2} |a\rangle |b\rangle \langle b| \langle a| + \frac{1}{2} |b\rangle |a\rangle \langle a| \langle b| + \frac{1}{2} |b\rangle |a\rangle \langle b| \langle a|[/itex]
     
  15. Oct 9, 2016 #14

    stevendaryl

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    I just wanted to mention one reason that cross-terms may be ignored in certain situations in computing density matrices: Very often, the phases of the two elements of the superposition, [itex]|\psi\rangle[/itex] and [itex]|\phi\rangle[/itex] are either unknown, or vary rapidly. So in those cases, you can get an effective density matrix by averaging over the phases. Let me rewrite [itex]|\psi\rangle[/itex] and [itex]|\phi\rangle[/itex] making the phases explicit:

    [itex]|\psi\rangle = |\psi_0\rangle e^{i \theta_1}[/itex]
    [itex]|\phi\rangle = |\phi_0\rangle e^{i \theta_2}[/itex]

    Then the density matrix is modified as follows:

    [itex]\rho(\delta \theta) = |\psi_0\rangle \langle \psi_0| + e^{-i \delta \theta} |\psi_0\rangle \langle \phi_0| + e^{+i \delta \theta} |\phi_0\rangle \langle \psi_0| + |\phi_0\rangle \langle \phi_0| [/itex]

    where [itex]\delta \theta = \theta_2 - \theta_1[/itex].

    If [itex]\delta \theta[/itex] is unknown, or varies rapidly, then we can get rid of it by averaging over all possible values:

    [itex]\rho_{eff} = \frac{1}{2\pi} \int \rho(\delta \theta) d \delta \theta[/itex]

    When you do this averaging, the terms involving [itex]\delta \theta[/itex] average to zero, leaving:

    [itex]\rho_{eff} = |\psi_0\rangle \langle \psi_0| + |\phi_0\rangle \langle \phi_0| [/itex]
     
    Last edited: Oct 9, 2016
  16. Oct 9, 2016 #15

    bhobba

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    Yes cross terms are possible but for simplicity I chose one that didn't have them. I chose the state 1/root2 |a>|b> + 1/root 2 |b>a>

    And the Dirac notation can be a nightmare.

    Thanks
    Bill
     
  17. Oct 9, 2016 #16

    Simon Phoenix

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    Well for a bi-partite system you can always transform to the Schmidt basis for the state - so writing an entangled pure state of a bipartite system as

    i ci |ai, bi>

    is always possible. This then gives nice diagonal reduced density operators, but you don't get a diagonal density matrix for the complete state in this basis.

    In fact I think the only basis in which an entangled state |e> gives a diagonal density operator is the one of which |e> is a member - in which case the density operator is just |e><e|. Can't be otherwise, since for a pure state the entropy must be zero. A density matrix with eigenvalues 1/2, 1/2 has an entropy of ln 2.
     
    Last edited: Oct 9, 2016
  18. Oct 9, 2016 #17

    zonde

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    I think it's not only notation. Is there such a term as coherent mixture (vs incoherent mixture)?
    I suppose not. Interference in double slit is still considered interference of single particle (that goes over two paths). But this idea is falsified by experiment that produced interference between two phase locked laser beams.
    And this inconsistency shows up when we speak about entangled state as it is clearly mixture but with interference effects. But it's not so different from single particle interference that too should be considered mixture with interference effects (coherent mixture).
     
  19. Oct 9, 2016 #18

    Simon Phoenix

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    I'm starting to get a bit confused by what states people are referring to in this thread.

    Let's try to get a bit specific. If we take 2 spin-1/2 particles in a pure state then this state can always be written in some appropriate basis (the Schmidt basis) as

    |ψ> = a |01'> + b |10'>

    Note here that the bases for the individual particles do not have to represent the same spin-directions. This is a pure state with entropy zero for any a,b. The only basis in which its density matrix does not have off-diagonal elements is the orthonormal basis of which |ψ> is a member

    The reduced density operators for the particles are diagonal in their respective Schmidt bases and have non-zero entropy when either a or b have magnitude less than 1. If, for example, we had |a| = 1, then b = 0 and we no longer have an entangled state. In general (when neither |a| or |b| is equal to unity) the individual particles are in mixed states.

    I think some of the confusion is because it's not being made clear enough (at least not to me) whether it is the single-particle reduced density operator that is being referred to, or the complete density operator for the 2-particle state.
     
  20. Oct 10, 2016 #19

    zonde

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    "2 spin-1/2 particles in a pure state" means 2 particles with opposite spin, right? But opposite spin can be (UP,DOWN) or (DOWN, UP) and we make two different observation for these two cases.So should we view "opposite spin" as more basic concept than (UP,DOWN) and (DOWN, UP) observations?
     
  21. Oct 10, 2016 #20

    vanhees71

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    The Dirac notation is brilliant (using LaTeX to type it, helps a lot to make your posting readable!).

    In fact there are cross terms, and they are crucial for the fascinating properties of entangled states. In your case you have
    $$|\Psi \rangle=\frac{1}{\sqrt{2}} (|ab \rangle + |ba \rangle)$$
    and the corresponding stat. op. thus reads
    $$\hat{\rho}_{\Psi}=|\Psi \rangle \langle \Psi|=\frac{1}{2} (|ab \rangle \langle ab| + |ab \rangle\langle ba| + |ba \rangle \langle ab| + |ba \rangle \langle ba|).$$
    The Dirac notation lets you write this down without having even to think!
     
    Last edited: Oct 10, 2016
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