I Entangled system pure state or not?

[bluecap]

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?

 

[Simon Bridge]

They are talking about different situations.
What is it you are trying to understand? A fine point in semantics?

 

[bluecap]

They are talking about different situations.
What is it you are trying to understand? A fine point in semantics?

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.

 

[Simon Bridge]

One referred to the complete setup and another to a subset.
@atyy @bhobba

 

[bhobba]

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."

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

 

[bluecap]

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

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.

 

[bluecap]

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.

 

[Simon Bridge]

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."

... 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?

 

[bluecap]

... 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?

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 can't you? but Bill in my original post seems to say no. hence my confusion.

 

[Simon Phoenix]

but Bill in my original post seems to say no

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.

 

[Simon Bridge]

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>...

 

[vanhees71]

... 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?

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.

 

[stevendaryl]

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|.

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 |\psi\rangle + |\phi\rangle, then the corresponding density matrix is |\psi\rangle \langle \psi| + |\psi\rangle \langle \phi| + |\phi\rangle \langle \psi| + |\phi\rangle \langle \phi|. Letting |\psi\rangle = \frac{1}{\sqrt{2}} |a\rangle |b\rangle and letting |\phi\rangle = \frac{1}{\sqrt{2}} |b\rangle |a\rangle gives 4 terms in the corresponding density matrix:

\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|

 

[stevendaryl]

If you have a pure state |\psi\rangle + |\phi\rangle, then the corresponding density matrix is |\psi\rangle \langle \psi| + |\psi\rangle \langle \phi| + |\phi\rangle \langle \psi| + |\phi\rangle \langle \phi|.

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, |\psi\rangle and |\phi\rangle 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 |\psi\rangle and |\phi\rangle making the phases explicit:

|\psi\rangle = |\psi_0\rangle e^{i \theta_1}
|\phi\rangle = |\phi_0\rangle e^{i \theta_2}

Then the density matrix is modified as follows:

\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|

where \delta \theta = \theta_2 - \theta_1.

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

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

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

\rho_{eff} = |\psi_0\rangle \langle \psi_0| + |\phi_0\rangle \langle \phi_0|

 

[bhobba]

Aren't there cross-terms?

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

 

[Simon Phoenix]

Second: Is what you're saying true? Aren't there cross-terms?

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 c_i |a_i, b_i>

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.

 

[zonde]

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.

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).

 

[Simon Phoenix]

about entangled state as it is clearly mixture but with interference effects

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.

 

[zonde]

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'>

"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?

 

[vanhees71]

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

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 let's you write this down without having even to think!

 

[Simon Phoenix]

"2 spin-1/2 particles in a pure state" means 2 particles with opposite spin, right?

No it means that 2 spin-1/2 particles are prepared in a pure state - this could be an entangled state or it might not be (if either a or b are zero then we don't have an entangled state). The Schmidt expansion tells us that there's always some spin observables we can find such we can write this in the form I gave. And remember that we can only say for an entangled state that the spins are correlated upon measurement - we can't say that they have correlated spins (or opposite spin) before measurement (which would be assigning hidden variables) - so the phrase "2 particles with opposite spin" is not correct for entangled states.

 

[vanhees71]

I'd rather state it as follows: The spins of the single particles in a spin-entangled don't have determined values before measurement but very strong correlations (see my posting #12). For spin 1/2 it's exactly analogous to my example with the polarization-entangled photons.

 

[zonde]

No it means that 2 spin-1/2 particles are prepared in a pure state - this could be an entangled state or it might not be (if either a or b are zero then we don't have an entangled state).

We can say something about the state from the way we prepare it and from the measurements we make. As I understand you say that 2 entangled particles are considered in pure state because of the way we prepare them.
Polarization entangled photons usually are prepared using PDC process. In this case we have to coherently overlap photons from two downconversion processes in two outputs i.e. we mix photons from say HH pairs with photons from VV pairs.

Here is more detailed description: https://arxiv.org/abs/quant-ph/0205171

 

[stevendaryl]

$$\hat{\rho}_{\Psi}=|\Psi \rangle \langle \Psi|=\frac{1}{2} (|ab \rangle \langle ab| + |ab \rangle\langle ba| + |ba \rangle \langle ab| + |ba \rangle \langle ba|).$$
The Dirac notation let's you write this down without having even to think!

My complaint is that when one uses |\psi\rangle |\phi\rangle to mean the two-particle system in which the first particle is in state |\psi\rangle and the second particle is in state |\phi\rangle. Then when you form density matrices, it's visually confusing to understand which bras and kets correspond to which particle: |\psi\rangle |\phi \rangle\langle \psi| \langle \phi |

 

[stevendaryl]

No it means that 2 spin-1/2 particles are prepared in a pure state - this could be an entangled state or it might not be (if either a or b are zero then we don't have an entangled state).

Actually, any two identical spin-1/2 particles are always entangled, because they must always be in an antisymmetric state under particle exchange.

 

[zonde]

I'd rather state it as follows: The spins of the single particles in a spin-entangled don't have determined values before measurement but very strong correlations (see my posting #12).

Correlation is mathematical tool that we can use to discover physical relationship behind the data we analyze. But correlation by itself is not a type of physical relationship. Maybe you mean that we have to be agnostic about possible nature of physical relationship behind correlations of entangled particles?

 

[Simon Phoenix]

Actually, any two identical spin-1/2 particles are always entangled, because they must always be in an antisymmetric state under particle exchange.

Yes that's true - I guess I was using spin-1/2 as a short hand for 2, 2 level systems. However, isn't this entanglement to do with symmetry under position exchange of indistinguishable particles? So If I had 2 labs each performing a spin measurement on a spin 1/2 particle and they both find the spins to be up - then the combined state of the 2 particles is going to be |up>|up>. In other words we won't magically get an entanglement between the 2 labs.

 

[bluecap]

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

So in summary. In an entangled system, prior to tracing out being done and prior to it being in mixed state, the entangled system is in superposition and yet it is never in pure state. Is this correct? So Superposition can occur to both pure state and pure entangled state (before the entangled system is reduced by the density matrix or tracing out (collapse in disguised) done. Correct? (kindly answer by yes or no and not just explanations that don't use yes or no so I won't be confused. Thank you.)

 

[stevendaryl]

So in summary. In an entangled system, prior to tracing out being done and prior to it being in mixed state, the entangled system is in superposition and yet it is never in pure state. Is this correct?

No. This is confusing, because if you have a two-particle system where the particles is entangled, then neither particle separately has a pure state. The pair has a pure state, but the components do not.

 

[bluecap]

No. This is confusing, because if you have a two-particle system where the particles is entangled, then neither particle separately has a pure state. The pair has a pure state, but the components do not.

Oh. How do you define Superposition then? (that is agreed by at least 99% of physicists). Because two particle entangled state are non-classical, it's not spin up or spin down entirely but they are somehow merged together (hence I thought is natural to call it superposition)..

 

[bhobba]

So in summary. In an entangled system, prior to tracing out being done and prior to it being in mixed state, the entangled system is in superposition and yet it is never in pure state. Is this correct? So Superposition can occur to both pure state and pure entangled state (before the entangled system is reduced by the density matrix or tracing out (collapse in disguised) done. Correct? (kindly answer by yes or no and not just explanations that don't use yes or no so I won't be confused. Thank you.)

No. There is no prior to being in a mixed state - it is in one. The system is in superposition, but each part acts like its in a mixed state.

Look - it is all explained in the reference I often give:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3

Its all there. Stop trying to translate it into English - you will fail.

Now let's get this over with onece and for all - exactly what part of that section is unclear and why? Not in English - but the math.

Thanks
Bill

 

[bhobba]

Oh. How do you define Superposition then?

Its agreed by 100% of physicists - superposition is simply pure states form a vector space.

Thanks
Bill

 

[stevendaryl]

Oh. How do you define Superposition then? (that is agreed by at least 99% of physicists). Because two particle entangled state are non-classical, it's not spin up or spin down entirely but they are somehow merged together (hence I thought is natural to call it superposition)..

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state. It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

 

[bluecap]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state. It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

Thanks. You explain much better. Are you a physicist or mathematician?
But isn't the words "Now, supposed that we have two particles" is similar meaning to "If the two-particle system is in state |\Psi\rangle?

 

[stevendaryl]

Thanks. You explain much better. Are you a physicist or mathematician?

I'm nothing. I've studied physics, and I've studied mathematics, but I don't do either for a living.

But isn't the words "Now, supposed that we have two particles" is similar meaning to "If the two-particle system is in state |\Psi\rangle?

Well, one way to have two particles is to have a two-particle system in a pure state. I could also have the two particles in a mixed state, or they could be entangled with yet a third particle (or even more particles). So saying that you have a two-particle system is more specific than just saying that you have two particles.

 

[zonde]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

There is an error in your reasoning.

A superposition just means a sum of pure states.

Yes of course.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle.

We are not interested if $|a_1\rangle$ and $|b_2\rangle$ are pure states. We need $|a_1,b_2\rangle$ to be pure state in order to talk about superposition of
$$\alpha|a_1,b_2\rangle+\beta|b_1,a_2\rangle$$

So can you say that $|a_1,b_2\rangle$ is a pure state? It does not have fixed relative phase between $|a_1\rangle$ and $|b_2\rangle$. It's sum of their phases that is fixed (or something like that).

 

[bluecap]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state.

The above statements have logical error. Here's the logic or illogic.
you said |\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle
and you said "If the two-particle system is in state |\Psi\rangle,"
then its like saying "If the two-particle system is in state \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle then the first particle does not have a pure state."

but \alpha |\psi, \phi\rangle is a pure state

But you explained or follow up that "Well, one way to have two particles is to have a two-particle system in a pure state. I could also have the two particles in a mixed state, or they could be entangled with yet a third particle (or even more particles). So saying that you have a two-particle system is more specific than just saying that you have two particles."

Or better yet. Can you give example of a two-particle system in a pure state, a two-particle system in a mixed state (possible)?
Or you seemed to be saying that two-particle system is different to two particles.. please give actual example to avoid confusion. Thanks a lot!

It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

 

[bluecap]

No. There is no prior to being in a mixed state - it is in one. The system is in superposition, but each part acts like its in a mixed state.

Look - it is all explained in the reference I often give:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3

Its all there. Stop trying to translate it into English - you will fail.

Now let's get this over with onece and for all - exactly what part of that section is unclear and why? Not in English - but the math.

Thanks
Bill

Yes. My favorite is section 1.2.3. but I don't know how to type the symbols and I can't take scan the page so I'll first learn how to type the symbols here. Wait half a day. I have some critical questions about it.

 

[bhobba]

So can you say that $|a_1,b_2\rangle$ is a pure state? It does not have fixed relative phase between $|a_1\rangle$ and $|b_2\rangle$. It's sum of their phases that is fixed (or something like that).

Its a pure state - it simply means system 1 is in pure state |a> and system 2 is in pure state |b>. The vector space is the joining of the vector spaces of the system. Phases have nothing to do with it. In fact for pure states phases are meaningless because states really are operators and the operator of a pure state is |a><a| which is easily seen to be the same regardless of phase. The real writing of |a>|b> is |a>|b><b|<a| where again phase means nothing.

Thanks
Bill

 

[Simon Phoenix]

The above statements have logical error. Here's the logic or illogic

There's no error that I can see.

Let's keep things very simple and think of qubits - which are just 2 level systems. We're going to imagine that Clive prepares 2 qubits in the pure state that is described (up to a normalization constant) as |01> + |10>.

Clive puts one particle in a box and gives it to Alice and puts the other particle in a box and gives it to Bob.

So in Alice's lab there is this box with one 2 level system in it, and in Bob's lab there is a box with one 2 level system in it. Now Alice and Bob are set to work and asked to devise experiments to figure out the properties of their particles - let's assume they've been locked in their labs and can't talk to one another or otherwise communicate.

The question is can we describe the properties of the particle in Alice's lab using the mathematical object we've called a pure state? More operationally we can ask whether experiments conducted by Alice on her particle, independently of anything done by Bob, are best described mathematically by a pure state or by something else?

It's probably easier to talk about ensembles of identically prepared systems - so we imagine that Clive has furnished Alice and Bob N boxes each - all derived from N copies of the 2 qubits prepared in the pure state |01> + |10>.

So Alice has N boxes to work with. She can perform any measurement, or set of measurements, she likes on her N boxes, but she has no access whatsoever to Bob's boxes, or any information about those boxes, or what Bob is doing to his boxes.

Alice's measurement results (whatever she does) are not consistent with assuming that her qubits are in pure states. They are consistent with assuming they are in a statistical mixture of states.

Another way of seeing what's going on is to imagine that Clive fools Alice and only tells her that her particles have come from an original pure state |01> + |10>. What he's actually done is to choose, at random for each box, whether to give her a |0> qubit or a |1> qubit. The question is whether there is any set of experiments Alice can do on the qubits she has been given that will tell her whether Clive is trying to fool her, or whether he really has given her qubits that come from entangled pairs?

There are no such experiments that Alice can perform. In other words - when we look at the properties of just one qubit from an entangled pair, it's exactly the same as looking at a single qubit prepared in a statistical mixture - which is not a pure state.

So for this entangled 2 qubit system we have
total system : pure state
either qubit looked at on their own : mixed state

I'm not sure whether that helps or just confuses things further - apologies if it's the latter.

 

[bluecap]

A superposition just means a sum of pure states. If |\psi\rangle is a pure state, and |\phi\rangle is a pure state, then \alpha |\psi\rangle + \beta |\phi\rangle (where \alpha and \beta are complex numbers such that |\alpha|^2 + |\beta|^2 = 1) is another pure state that is a superposition of |\psi\rangle and |\phi\rangle.

Now, suppose that we have two particles. Then let |\psi, \phi\rangle be the state in which the first particle is in the pure state |\psi\rangle and the second particle is in the pure state |\phi\rangle. Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle

If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state. It's not |\psi\rangle, and it's not |\phi\rangle, and it's not any superposition of the two. Similarly, the second particle does not have a pure state. But the two-particle system has a pure state. That's what entanglement means: it means that the composite system has a pure state, but the components do not.

If you want to describe the first particle by itself, you would have to use a mixed state. It's in a mixture of |\psi\rangle and |\phi\rangle, but it is not in a superposition.

Simon Phoenix.. Thanks a lot for your description. It is very clear and it described the latter of Stevendaryl statement "where If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state".

But in the sentences before it. Stevendaryl mentioned :
"Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle"

Here the first particle is in the pure state |\phi\rangle. This is not obviously the copy of Alice or Bob in your example. So what example of Alice and Bob can you give where the first particle (Alice copy) is in pure state?? Thanks so much for the help guys!

 

[Simon Phoenix]

This is not obviously the copy of Alice or Bob in your example

Yes, it is. Let's rewrite what Stevendaryl correctly wrote, but in terms of qubits in the hope that this lends some clarity for you :

"Let |0,1> be the state in which the first particle is in the pure state |0> and the second particle is in the pure state |1>. Similarly we let |1,0> be the state in which the first particle is in the pure state |1> and the second particle is in the pure state |0>

Now we have two distinct pure states |0,1> and |1,0>

We can form a superposition of these two different pure states as a|0,1> + b|1,0>"

For this superposition, is the first particle in the pure state |0> or is it in the pure state |1>?

It is in neither of these pure states - but a statistical mixture of these states.

If the combined state of both particles is a pure state AND particle 1 is in a pure state then the second particle must also be in a pure state. Furthermore the total state of both particles can be described by a product state |particle1>|particle2>

If two particles are entangled - so of the form a|0,1> + b|1,0> then there is NO possible way to decompose this state into a product of 2 individual pure states like this.

The statement combined state = pure, particle 1 = pure, automatically implies we can't have the entangled (superposition) state above.

 

[vanhees71]

Simon Phoenix.. Thanks a lot for your description. It is very clear and it described the latter of Stevendaryl statement "where If the two-particle system is in state |\Psi\rangle, then the first particle does not have a pure state".

But in the sentences before it. Stevendaryl mentioned :
"Similarly, let |\phi, \psi\rangle be the state in which the first particle is in the pure state |\phi\rangle and the second particle is in the pure state |\psi\rangle.Now, we can form a superposition for the two-particle system as follows:

|\Psi\rangle = \alpha |\psi, \phi\rangle + \beta |\phi, \psi\rangle"

Here the first particle is in the pure state |\phi\rangle. This is not obviously the copy of Alice or Bob in your example. So what example of Alice and Bob can you give where the first particle (Alice copy) is in pure state?? Thanks so much for the help guys!

Stevendaryl is correct. The state of particle 1 is defined by tracing out particle 2 in the statistical operator of the two-particle system, which in this case is a projection operator, because the two-particle state is in a pure state,
$$\hat{\rho}_{12}=|\Psi \rangle \langle \Psi |=|\alpha|^2 |\psi \phi \rangle \langle \psi \phi|+\alpha^* \beta |\psi \phi \rangle \langle \phi \psi| +\alpha \beta^* |\phi \psi \rangle \langle \psi \phi| + |\beta|^2 |\phi \psi \rangle \langle \phi \psi |.$$
Now for simplicity we assume that $|\phi \rangle$ and $|\psi \rangle$ are orthonormal vectors. Then we can take them to be part of a complete orthonormal set (Schmidt decomposition theorem) and take the partial trace over this set to find
$$\hat{\rho}_{1}=\mathrm{Tr}_2 \hat{\rho}_{12} = |\alpha|^2 |\psi \rangle \langle \psi|+|\beta|^2 |\phi \rangle \langle \phi|,$$
which is not a projection operator and thus represents a mixed state.

 

[bluecap]

Yes. I understood now. Thanks to all the folks for all valuable help. Now let me apply this to atoms as this is my main motivation. In the atoms. Electrons are entangled to the nucleus. Now is the nucleus alone or electrons alone in pure states? Or in statistical mixture? But then electrons are quantum objects, nucleus are quantum objects.. by quantum objects I mean describable as ray of trace 1 or pure state. Therefore if the nucleus is in pure state, and electrons are pure state, then can we say the electrons are entangled to the nucleus or the electrons and nucleus are (or atom) in pure state? (Let's assume the atom (nucleus and electrons) are isolated and not in interaction with its environment (no photons or thermal sources to perturb the atom). Thank you.

 

[vanhees71]

Take as the most simple example the hydrogen atom first. For a thorough discussion about the state of the proton and the electron in this case (in the realm of non-relativistic QM) see

https://arxiv.org/abs/quant-ph/9709052

 

[stevendaryl]

We are not interested if $|a_1\rangle$ and $|b_2\rangle$ are pure states. We need $|a_1,b_2\rangle$ to be pure state in order to talk about superposition of
$$\alpha|a_1,b_2\rangle+\beta|b_1,a_2\rangle$$

I'm not sure I understand what you're saying. |a_1,b_2\rangle is a pure state of the composite system. |b_1,a_2\rangle is a pure state of the composite system. The sum \alpha |a_1,b_2\rangle + \beta |b_1, a_2\rangle is a pure state of the composite system.

So can you say that $|a_1,b_2\rangle$ is a pure state?

Yes. The superposition of pure states produces a pure state.

 

[bluecap]

No. There is no prior to being in a mixed state - it is in one. The system is in superposition, but each part acts like its in a mixed state.

Look - it is all explained in the reference I often give:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3

Its all there. Stop trying to translate it into English - you will fail.

Now let's get this over with onece and for all - exactly what part of that section is unclear and why? Not in English - but the math.

Thanks
Bill

Ok let me address section 1.2.3. I don't know how to type the symbols. But how is equation 1.21 derived or

p1=|psi1> < psi1| = 1/2 (|arrow upz>< arrowupz| + |arrowupupz><arrowdownz|+|arrow down z> <arrowupz|+|arrowdownz><arrowdownz|)??

and prior to that is written "We prepare the superposition |psi1> = 1/sqrt (2) (|arrowupz> + |arrowdown z>)

Why is there a square root of 2 and why are there many ><... what does the symbol >< mean and all the arrows?

Hope you can just address the above and not let me read a whole book about the math. And what kind of math are they called?

 

[bhobba]

Ok let me address section 1.2.3. I don't know how to type the symbols. But how is equation 1.21 derived or

p1=|psi1> < psi1| = 1/2 (|arrow upz>< arrowupz| + |arrowupupz><arrowdownz|+|arrow down z> <arrowupz|+|arrowdownz><arrowdownz|)??

and prior to that is written "We prepare the superposition |psi1> = 1/sqrt (2) (|arrowupz> + |arrowdown z>)

Why is there a square root of 2 and why are there many ><... what does the symbol >< mean and all the arrows?

Hope you can just address the above and not let me read a whole book about the math. And what kind of math are they called?

I can't give you a course in the math of QM here. See Susskind:
https://www.amazon.com/dp/0465062903/?tag=pfamazon01-20

But a few answers to the many you rase.

1. 1/Root2 is for normalisation.
2. The symbol |a><b| is an operator invented by Dirac. If you apply it to the bra |u> you get the bra |a><b|u> (in fact it's a projection operator.)

As a beginner you asked a question where even the terms you use (entanglement and pure state) are at least intermediate concepts, probably even advanced.

No wonder you are finding it hard.

You want more advanced answers, then you must learn the more advanced concepts.

Thanks
Bill

 

[stevendaryl]

Ok let me address section 1.2.3. I don't know how to type the symbols. But how is equation 1.21 derived or

p1=|psi1> < psi1| = 1/2 (|arrow upz>< arrowupz| + |arrowupupz><arrowdownz|+|arrow down z> <arrowupz|+|arrowdownz><arrowdownz|)??

and prior to that is written "We prepare the superposition |psi1> = 1/sqrt (2) (|arrowupz> + |arrowdown z>)

Why is there a square root of 2 and why are there many ><... what does the symbol >< mean and all the arrows?

Hope you can just address the above and not let me read a whole book about the math. And what kind of math are they called?

I think that what you're really asking for is a course in multiparticle quantum mechanics. "Why is there a square-root of 2?" is best answered as part of a course. The short answer is that states are usually normalized, so that the sum of the squares of their components (in an orthonormal basis) add up to 1. (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 = 1/2 + 1/2 = 1

 

[zonde]

why are there many ><... what does the symbol >< mean

This might help: http://www.mathpages.com/home/kmath638/kmath638.htm