I Entangled system pure state or not?

[bluecap]

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>"

What is this term a|0,1> exactly called?
Is this a|0,1> the infamous branch or Worlds (of Many Worlds) that we Laymen are mostly familiar about. If it is not. then what is the mathematical expression of the branches or Worlds?

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.

Your message finally makes things very clear. Thanks a lot.

 

[zonde]

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.

Let's say that relative phase between |a_1\rangle and |b_2\rangle is not fixed. As I understand relative phase between separate states becomes phase of combined state: |a_1\rangle \otimes e^{i\theta}|b_2\rangle = e^{i\theta}|a_1,b_2\rangle
then if we say that $\theta$ is not fixed it is something like pure but incoherent state. But I'm not sure if it's meaningful. Say is it possible to have ensemble that is (classical) combination of $|a_1\rangle$ and $i|a_1\rangle$? Are they the same ray mathematically, or different rays?

But certainly relative phase matters when we talk about state:
\frac{1}{\sqrt{2}}(|a_1,b_2\rangle+e^{i\theta}|b_1,a_2\rangle)
because it changes predicted result. So I don't see how phase can be considered unphysical as bhobba stated.

And you don't have to answer. I'm just explaining my confusion.

 

[vanhees71]

I don't understand where your problem is. For sure if the phase is not determined, then you don't have a superposition, but you have a mixed state. For your example it's something like the following. First we define the pure states,
$$|\psi(\theta) \rangle=\frac{1}{\sqrt{2}} (|a_1,b_2 \rangle + \exp(\mathrm{i} \theta) |b_1,a_2 \rangle).$$
Then with a probability distribution $\Theta(\theta)$ for the phase you have
$$\hat{\rho}=\int_0^{2 \pi} \mathrm{d} \theta \Theta(\theta) |\psi(\theta) \rangle \langle \psi(\theta)|.$$
If your phase is totally indetermined, you have $\Theta(\theta)=1/(2 \pi)=\text{const}$.

 

[Simon Phoenix]

What is this term a|0,1> exactly called?

OK, forgive me if I'm wrong, but it looks to me like you're trying to run before you can walk. You're interested in all the funky stuff like entanglement (and quite rightly so, because it is fascinating) but it would seem that you don't have the basic framework in place on which to hang these funky ideas.

In classical physics we describe things in terms of concepts like position, momentum, field strength, etc and each of these ideas has some intuitive appeal. If an apple falls on my head I get an immediately intuitive feel for where it is and what its momentum is

So the notion of 'state' - the state of an object - is to some extent an intuitive thing classically. We write down a list of properties and call that a state because it contains all the things we need to describe the object. Then the laws of physics tell us how that state will change. So if we apply a force to something we'll change its state of motion (i.e. its state) and that change and its subsequent motion can be figured out by solving those laws of motion.

In quantum mechanics we kind of lose this immediately intuitive feel for what is being described. It's not that we can't build up an intuition about what happens or how things evolve, but developing that intuition requires a bit of re-wiring of our mental pictures. A large part of this failure of our 'classical' intuition is because in QM we have to describe the 'state' of an object in a rather abstract way - as a vector in a complex Hilbert space (there's a bit more to it than that but that's a good enough place to start).

You'll see on here that there is still lots of heated debate about what this QM state actually means. Is it describing some objective thing or is it a mathematical device that just allows us to predict the right things? Right now we can't really say one way or another - neither of these ways of viewing the quantum state is wholly satisfactory to everybody.

So the first thing you need to do, if you want to understand more advanced things like entanglement, is to get a good feel for the basic mathematical machinery of QM. You can go quite a long way with just a few ideas from complex numbers and linear algebra and it doesn't need to involve really difficult maths.

When a state in QM is written as |a> for example, this is just notation for one of these vectors - it's a very nice notation invented by Dirac. I find it amazingly useful. But essentially when you see it think 'vector' to begin with. So like any vector it 'lives' in a vector space and it has certain properties. Adding any 2 vectors gives us another vector - or put it another way we can 'decompose' or 'expand' any given vector into a sum of other vectors. This decomposition or expansion is not unique and for a given vector there will generally be more than one way to expand it (and often there are an infinite number of different ways).

So if we walk from A to B we can label our journey as AB, but suppose we walk from A to B via C (all in straight lines), then we can see that AB is 'made up' of the journey AC followed by CB. This is nothing more than the principle of superposition so that the vector AB is equal to the sum of the vectors AC and CB. On another day we might decide to go via D so that our journey is AD plus DB. In all these cases we go from A to B - but this single journey can be 'expanded' in lots of different ways.

In principle this is also what we have in QM. When we write a state as |s> = |r> + |t> then it means just the same thing. In fact we could go daft and use the Dirac notation to describe our walking journey. So the vector AB we could write as |s>, the vector |r> would be AC and the vector |t> would be CB - and then we have just used a Dirac notation applied to real vectors to describe AB = AC + CB.

Honestly, apart from some more technical details, this is really all that superposition in QM is. Because these quantum states are vectors - they have all the usual vector properties. So mathematically at least, if you know linear algebra, you'll understand much more about the basic maths behind some things in QM.

Where things get more tricky is in the meaning. It's easy to visualize the journey through real space described by AB - you could even do it, that is walk from A to B in a straight line. The vectors in QM don't live in this real space and so it's a bit more difficult to describe them in such simple terms. They are, nevertheless, simply vectors, albeit in a complex Hilbert space.

When we write |0,1> in the above discussion, for example, this is just another vector, but it's a single vector describing 2 things - in this case 2 qubits. Just like 'normal' vectors in 'real' 3D space we can form superpositions so that |0,1> + |1,0> is another perfectly good vector.

I may have pitched all that way too low - and I apologize if I have been inadvertently patronizing. If you need a bit more technical flesh on the bones then Susskind's stuff (either his lectures on youtube, or his book 'the theoretical minimum') goes into more detail and are a truly excellent place to begin.

 

[zonde]

I don't understand where your problem is.

Say is it possible to have ensemble that is (classical) combination of $|a_1\rangle$ and $i|a_1\rangle$? Are they the same ray mathematically, or different rays?

Say is it meaningful to write something like:
$$\hat{\rho}=\int_0^{2 \pi} \mathrm{d} \theta \Theta(\theta) e^{i\theta}|a_1 \rangle e^{-i\theta}\langle a_1|$$
Result seems the same as for pure state $|a_1 \rangle \langle a_1|$ as the phase cancels out. But there would't be any interference if we would perform double slit experiment with such a state (assuming it's meaningful).

 

[vanhees71]

As is very easy to see, what you get then is the pure state
$$\hat{\rho}=|a_1 \rangle \langle a_1|$$
since
$$\int_0^{2 \pi} \Theta(\theta)=1,$$.
and the $\exp$ factors cancel. I think, you have to learn the formalism (linear algebra) first, before you go further to quantum theory.

 

[stevendaryl]

Say is it meaningful to write something like:
$$\hat{\rho}=\int_0^{2 \pi} \mathrm{d} \theta \Theta(\theta) e^{i\theta}|a_1 \rangle e^{-i\theta}\langle a_1|$$
Result seems the same as for pure state $|a_1 \rangle \langle a_1|$ as the phase cancels out. But there would't be any interference if we would perform double slit experiment with such a state (assuming it's meaningful).

Overall phases are not important, but relative phases (between different elements of a superposition) are important.

Let's look at a specific case: the spin state \frac{1}{\sqrt{2}} (|U\rangle - e^{i \theta} |D\rangle). This corresponds to the density matrix: \rho(\theta) = \frac{1}{2} |U\rangle \langle U| + \frac{1}{2} e^{i \theta} |D\rangle \langle U| + \frac{1}{2} e^{-i \theta} |U\rangle \langle D| + \frac{1}{2} |D\rangle \langle D|. Now, if \theta is unknown, then we average that density matrix over all possible values of \theta:

\rho = \frac{1}{2\pi} \int d\theta \rho(\theta) = \frac{1}{2} |U\rangle \langle U| + \frac{1}{2} |D\rangle \langle D|

 

[bhobba]

OK, forgive me if I'm wrong, but it looks to me like you're trying to run before you can walk. You're interested in all the funky stuff like entanglement (and quite rightly so, because it is fascinating) but it would seem that you don't have the basic framework in place on which to hang these funky ideas.

Exactly.

You need to know a LOT more about QM before venturing into deep waters like this.

Thanks
Bill

 

[zonde]

I cleared my confusion. And I would say that notation is confusing.
Say we write a vector
$|\psi\rangle=|a_1,b_2\rangle$ but actually we mean a ray (all the vectors that are the same modulo phase factor)
and then we write
$|\phi\rangle=\frac{1}{\sqrt2}(|a_1,b_2\rangle+e^{i\theta}|b_1,a_2\rangle)$ and here $|a_1,b_2\rangle$ actually is a vector with particular phase as otherwise we couldn't speak about certain relative phase. But combined vector again is meant as a ray.

So in one case $|a_1,b_2\rangle$ means a ray (even so we write a vector) but in other case it is a vector.

 

[stevendaryl]

I cleared my confusion. And I would say that notation is confusing.
Say we write a vector
$|\psi\rangle=|a_1,b_2\rangle$ but actually we mean a ray (all the vectors that are the same modulo phase factor)
and then we write
$|\phi\rangle=\frac{1}{\sqrt2}(|a_1,b_2\rangle+e^{i\theta}|b_1,a_2\rangle)$ and here $|a_1,b_2\rangle$ actually is a vector with particular phase as otherwise we couldn't speak about certain relative phase. But combined vector again is meant as a ray.

So in one case $|a_1,b_2\rangle$ means a ray (even so we write a vector) but in other case it is a vector.

Right, it is a little confusing. In doing QM calculations, you start with a specific representation of basis states, and you fix the phases. It's pretty arbitrary how you assign the relative phases of the different basis states, but you need to pick some convention. Then in terms of the basis states, you can describe a pure state of a system as a ray. The basis states themselves are not rays, they are vectors. It doesn't make any sense to make a superposition of rays.

The main point of phases is to take into account interference. To illustrate, let's consider a two-state system with basis states |U\rangle and |D\rangle. Fix a time period T, and let:

\psi_{UU} = \langle U| e^{-iHT} |U\rangle
\psi_{UD} = \langle U| e^{-iHT} |D\rangle
\psi_{DU} = \langle D| e^{-iHT} |U\rangle
\psi_{DD} = \langle U| e^{-iHT} |U\rangle

\psi_{ij} is the probability amplitude that a system initially in state j will be found in state i a time T later. You square the amplitude to get the probability. But let's suppose we started instead in a state |\chi\rangle = \alpha |U\rangle + \beta |D\rangle. Then what's the probability amplitude that it will be found in state |D\rangle a time T later? Well, it's \alpha \psi_{DU} + \beta \psi_{DD}. You square that to get the probability:

P = |\alpha|^2 |\psi_{DU}|^2 + |\beta|^2 |\psi_{DD}|^2 + 2 Re(\alpha^* \psi_{DU}^* \beta \psi_{DD})

The term 2 Re(\alpha^* \psi_{DU}^* \beta \psi_{DD}) is an interference term, and it is an observable effect of relative phases. Multiplying the state \alpha \psi_{DU} + \beta \psi_{DD} by an overall phase will make no change to this term. But changing \alpha to \alpha e^{i \theta_1} and changing \beta to \beta e^{i \theta_2}, where \theta_1 \neq \theta_2, will change the interference term.

 

[stevendaryl]

Right, it is a little confusing. In doing QM calculations, you start with a specific representation of basis states, and you fix the phases. It's pretty arbitrary how you assign the relative phases of the different basis states, but you need to pick some convention.

Just a mathematical point. If you have some basis of states |J\rangle, then changing to a different basis means applying a unitary transformation U, with |J'\rangle \equiv U |J\rangle. To get the same physics, you have to also change the Hamiltonian H to H' = U H U^\dagger (with all other operators changed similarly). The specific case of changing the phases of the basis vectors, so that |J\rangle \Rightarrow e^{i \phi_J} |J\rangle is one particularly simple type of unitary transformation.

 

[bluecap]

OK, forgive me if I'm wrong, but it looks to me like you're trying to run before you can walk. You're interested in all the funky stuff like entanglement (and quite rightly so, because it is fascinating) but it would seem that you don't have the basic framework in place on which to hang these funky ideas.

In classical physics we describe things in terms of concepts like position, momentum, field strength, etc and each of these ideas has some intuitive appeal. If an apple falls on my head I get an immediately intuitive feel for where it is and what its momentum is

So the notion of 'state' - the state of an object - is to some extent an intuitive thing classically. We write down a list of properties and call that a state because it contains all the things we need to describe the object. Then the laws of physics tell us how that state will change. So if we apply a force to something we'll change its state of motion (i.e. its state) and that change and its subsequent motion can be figured out by solving those laws of motion.

In quantum mechanics we kind of lose this immediately intuitive feel for what is being described. It's not that we can't build up an intuition about what happens or how things evolve, but developing that intuition requires a bit of re-wiring of our mental pictures. A large part of this failure of our 'classical' intuition is because in QM we have to describe the 'state' of an object in a rather abstract way - as a vector in a complex Hilbert space (there's a bit more to it than that but that's a good enough place to start).

You'll see on here that there is still lots of heated debate about what this QM state actually means. Is it describing some objective thing or is it a mathematical device that just allows us to predict the right things? Right now we can't really say one way or another - neither of these ways of viewing the quantum state is wholly satisfactory to everybody.

So the first thing you need to do, if you want to understand more advanced things like entanglement, is to get a good feel for the basic mathematical machinery of QM. You can go quite a long way with just a few ideas from complex numbers and linear algebra and it doesn't need to involve really difficult maths.

When a state in QM is written as |a> for example, this is just notation for one of these vectors - it's a very nice notation invented by Dirac. I find it amazingly useful. But essentially when you see it think 'vector' to begin with. So like any vector it 'lives' in a vector space and it has certain properties. Adding any 2 vectors gives us another vector - or put it another way we can 'decompose' or 'expand' any given vector into a sum of other vectors. This decomposition or expansion is not unique and for a given vector there will generally be more than one way to expand it (and often there are an infinite number of different ways).

So if we walk from A to B we can label our journey as AB, but suppose we walk from A to B via C (all in straight lines), then we can see that AB is 'made up' of the journey AC followed by CB. This is nothing more than the principle of superposition so that the vector AB is equal to the sum of the vectors AC and CB. On another day we might decide to go via D so that our journey is AD plus DB. In all these cases we go from A to B - but this single journey can be 'expanded' in lots of different ways.

In principle this is also what we have in QM. When we write a state as |s> = |r> + |t> then it means just the same thing. In fact we could go daft and use the Dirac notation to describe our walking journey. So the vector AB we could write as |s>, the vector |r> would be AC and the vector |t> would be CB - and then we have just used a Dirac notation applied to real vectors to describe AB = AC + CB.

Honestly, apart from some more technical details, this is really all that superposition in QM is. Because these quantum states are vectors - they have all the usual vector properties. So mathematically at least, if you know linear algebra, you'll understand much more about the basic maths behind some things in QM.

Where things get more tricky is in the meaning. It's easy to visualize the journey through real space described by AB - you could even do it, that is walk from A to B in a straight line. The vectors in QM don't live in this real space and so it's a bit more difficult to describe them in such simple terms. They are, nevertheless, simply vectors, albeit in a complex Hilbert space.

When we write |0,1> in the above discussion, for example, this is just another vector, but it's a single vector describing 2 things - in this case 2 qubits. Just like 'normal' vectors in 'real' 3D space we can form superpositions so that |0,1> + |1,0> is another perfectly good vector.

I may have pitched all that way too low - and I apologize if I have been inadvertently patronizing. If you need a bit more technical flesh on the bones then Susskind's stuff (either his lectures on youtube, or his book 'the theoretical minimum') goes into more detail and are a truly excellent place to begin.

I'm familiar about Hilbert space and vectors. My first lesson in QM is imagining what it's like to be inside Hilbert Space as mentioned in the book "The Many Worlds of Schrodinger Rabbits".. I have read Feybook Little book of light and sound on vector addition. And we laymen mostly are more quantum than Newtonian.. it's because we are born and exposed to concepts like...

* Is the Cat in superposition of both alive and dead
* Is the Cat both alive and dead at same time in different Worlds
* all powerful Consciousness that can collapse wave function
* Ominipresent quantum potential and pilot wave
* Back and forward traveling Cramer waves
* Objective collapse either from GRW or Penrose gravitational source
etc.

So you see. We are anything but Newtonian. We are quantum at heart... so I can easily learn the math (basic math).. in this connection. I'd like to know something about this very interesting paper 100 years of the Quantum written by Max Tegmark

A question about page 6 of https://arxiv.org/pdf/quant-ph/0101077v1.pdf

"The second unanswered question in the Everett picture was more subtle but equally important: what physical mechanism picks out the classical states — face up and face down for the card — as special? The problem was that from a mathematical point of view, quantum states like "face up plus face down" (let’s call this "state alpha") or "face up minus face down" ("state beta", say) are just as valid as the classical states "face up" or "face down".

So just as our fallen card in state alpha can collapse into the face up or face down states, a card that is definitely face up — which equals (alpha + beta)/2 — should be able to collapse back into the alpha or beta states, or any of an infinity of other states into which "face up" can be decomposed. Why don’t we see this happen?

Decoherence answered this question as well. The calculations showed that classical states could be defined and identified as simply those states that were most robust against decoherence. In other words, decoherence does more than just make off-diagonal matrix elements go away. If fact, if the alpha and beta states of our card were taken as the fundamental basis, the density matrix for our fallen card would be diagonal to start with, of the simple form

density matrix = [1 0]
--------------------[0 0]

since the card is definitely in state alpha. However, decoherence would almost instantaneously change the state to

density matrix = [1/2 0]
--------------------[0 1/2]

so if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. In contrast, if we put the card in the state "face up", it would stay "face up" in spite of decoherence. Decoherence therefore provides what Zurek has termed a "predictability sieve", selecting out those states that display some permanence and in terms of which physics has predictive power."

------
What I'd like clarification is the following. It says above that if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. Whereas if it is face-up.. it will stay face up in spite of decoherence.. Does it mean that the alpha and beta state are really still there just we don't see it because it says if you measure it you could still get random outcome?

 

[bluecap]

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

So the symbol >< being related to the projection operator is simply the infamous "collapse" into eigenstate (or selection of the single outcome if Many Worlds is preferred choice), right? We quantum laymen need to know at least just the very basic and I think this ><, projection operator, collapse thing is one thing we need to keep in mind

"http://quantummechanics.ucsd.edu/ph130a/130_notes/node185.html
"Projection Operators

and Completeness
Now we move on a little with our understanding of operators. A ket vector followed by a bra vector is an example of an operator. For example the operator which projects a vector onto the

eigenstate is

"

 

[bluecap]

This thread has given me very intuitive understanding of entanglement, superposition, pure state, mixed state, density matrix etc. enough for me to understand a lof of this paper http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

Now practical applications in daily life. Entanglement between us and environment occurs all the time, the mystery is how improper mixed state turned into proper mixed state. For now let's ignore how.

But I need to know if the entanglement with the environment is in the vicinity or direct line of sight.. or beyond line of sight.. meaning is it possible my body particles are entangled with Stevendaryl bodies and all the rest of you? If you'd say no. Why not since entanglement knows no distance.. so how can you say my body is only entangled with our environment that is direct line of sight?

 

[stevendaryl]

But I need to know if the entanglement with the environment is in the vicinity or direct line of sight.. or beyond line of sight.. meaning is it possible my body particles are entangled with Stevendaryl bodies and all the rest of you? If you'd say no. Why not since entanglement knows no distance.. so how can you say my body is only entangled with our environment that is direct line of sight?

Particles become entangled through interactions, and interactions are local. However, once entangled, particles stay entangled even after they get far apart. And if A is entangled with B, and B becomes entangled with C, then A will be entangled with C, as well. So, everything becomes entangled with everything else.

In the "collapse" interpretation, the way that a particle can become unentangled from other particles is by measurments. After a measurement, the measured particle is in a pure state (well, if it was an ideal measuement), and is unentangled with anything else. Without collapse, or at least with Many-Worlds, there is never any unentangling associate with measurement; measurement is, on the contrary, a matter of the measuring device becoming entangled with the object being measured.

 

[bluecap]

Particles become entangled through interactions, and interactions are local. However, once entangled, particles stay entangled even after they get far apart. And if A is entangled with B, and B becomes entangled with C, then A will be entangled with C, as well. So, everything becomes entangled with everything else.

So if I met Trump once and you also met him once then we really are entangled.. aren't we? (seriously)

In the "collapse" interpretation, the way that a particle can become unentangled from other particles is by measurments. After a measurement, the measured particle is in a pure state (well, if it was an ideal measuement), and is unentangled with anything else. Without collapse, or at least with Many-Worlds, there is never any unentangling associate with measurement; measurement is, on the contrary, a matter of the measuring device becoming entangled with the object being measured.

In the double slit experiment, after measurement, the screen decohered the electron (or photon) producing mixed state (and the particle hits the detector).. so how can the electron (or particle) return to being in pure state. Please give an actual example of your statement. Thank you.

 

[stevendaryl]

So if I met Trump once and you also met him once then we really are entangled.. aren't we? (seriously)

Yes. Whenever some quantum-mechanical process could have gone this way or that way, and which way it went affected you, then that means that you are entangled with everyone else who was affected by it.

In the double slit experiment, after measurement, the screen decohered the electron (or photon) producing mixed state (and the particle hits the detector).. so how can the electron (or particle) return to being in pure state. Please give an actual example of your statement. Thank you.

Decoherence is entanglement---the microscopic object (electron) becomes entangled with the detector, the electromagnetic field, the observer, etc. I'm not sure what you mean by "so how can the electron return to being in [a] pure state". If you assume collapse, then measuring the electron to have spin-up in a non-destructive way causes the electron in a pure spin-up state. If the electron smashes into a screen, that's destructive, so we don't talk about the state of the electron afterward.

 

[bluecap]

Yes. Whenever some quantum-mechanical process could have gone this way or that way, and which way it went affected you, then that means that you are entangled with everyone else who was affected by it.

Really. Well. Emotions are caused by biochemistry in the body. So in the Syrian mass protests that led to the tragic civil war, the protesters emotions (or biochemistry) are entangled with one another causing it to become contagious (in negative way) and spiral out of control. Is this correct? If not.. what is wrong with what I just described?

Decoherence is entanglement---the microscopic object (electron) becomes entangled with the detector, the electromagnetic field, the observer, etc. I'm not sure what you mean by "so how can the electron return to being in [a] pure state". If you assume collapse, then measuring the electron to have spin-up in a non-destructive way causes the electron in a pure spin-up state. If the electron smashes into a screen, that's destructive, so we don't talk about the state of the electron afterward.

Ok.

 

[stevendaryl]

Really. Well. Emotions are caused by biochemistry in the body. So in the Syrian mass protests that led to the tragic civil war, the protesters emotions (or biochemistry) are entangled with one another causing it to become contagious (in negative way) and spiral out of control. Is this correct? If not.. what is wrong with what I just described?

I'm not going to get into specifics about emotions and civil war, etc. But in terms of Many-Worlds, we only share the same world because we are entangled. Entanglement just means that my state is correlated with your state. If we influence one another, or are influenced by something in common, then we will become correlated/entangled.

 

[bluecap]

I'm not going to get into specifics about emotions and civil war, etc. But in terms of Many-Worlds, we only share the same world because we are entangled. Entanglement just means that my state is correlated with your state. If we influence one another, or are influenced by something in common, then we will become correlated/entangled.

in terms of Many-Worlds, we only share the same world because we are entangled.
how about in Bohmian, Copenhagen, Objective Collapse, Cramers, Ensemble, how do you describe it in terms of the description "we only share the same world because we are entangled", can we say for example "In terms of Bohmian, we only share the same pilot wave because we are entangled"? how about others? thank you.

 

[stevendaryl]

in terms of Many-Worlds, we only share the same world because we are entangled.
how about in Bohmian, Copenhagen, Objective Collapse, Cramers, Ensemble, how do you describe it in terms of the description "we only share the same world because we are entangled", can we say for example "In terms of Bohmian, we only share the same pilot wave because we are entangled"? how about others? thank you.

So let me illustrate with Schrodinger's cat.

You have a radioactive atom that is in a superposition of a decayed atom and an undecayed atom. The decayed atom triggers the release of poison, killing a cat. The undecayed atom leaves the cat alive.

So in the Many-Worlds interpretation, the state of the universe is a superposition of two possibilities:

1. \psi_{dead}: The atom is decayed, the poison is released, the cat is dead, Schrodinger sees a dead cat, etc.
2. \psi_{alive}: The atom is not decayed, the poison is not released, the cat is alive, Schrodinger sees a live cat, etc.

So the atom, the poison, the cat and Schrodinger (and basically the rest of the universe) are all entangled. (That is, there is a state for the composite system of atom + poison + cat + Schrodinger, but the composite state doesn't "factor" into a product of the state for the atom \times the state of the poison \times the state of the cat \times the state of Schrodinger.) Note: \psi_{alive} does not describe an entangled composite system, and \psi_{dead} does not describe an entangled composite system, but the superposition \psi_{alive} + \psi_{dead} does describe an entangled composite system.

In the Bohmian interpretation, you have exactly the same state of the universe as in Many-Worlds, so there is still the same entanglement. The difference is that in the Bohmian interpretation, there is more to the state of the universe than the wave function; the world has an ACTUAL state, where all particles have definite positions at all times. So the cat is either actually dead, or actually alive, although the wave function alone doesn't tell you which is the case.

In collapse interpretations, once Schrodinger checks on his cat, the wave function collapses to one of the possibilities \psi_{alive} or \psi_{dead}. Since neither of those states is entangled, then after the collapse, nothing is entangled.

So it's really only in collapse interpretations does entanglement ever disappear.

 

[bluecap]

So let me illustrate with Schrodinger's cat.

You have a radioactive atom that is in a superposition of a decayed atom and an undecayed atom. The decayed atom triggers the release of poison, killing a cat. The undecayed atom leaves the cat alive.

So in the Many-Worlds interpretation, the state of the universe is a superposition of two possibilities:

1. \psi_{dead}: The atom is decayed, the poison is released, the cat is dead, Schrodinger sees a dead cat, etc.
2. \psi_{alive}: The atom is not decayed, the poison is not released, the cat is alive, Schrodinger sees a live cat, etc.

So the atom, the poison, the cat and Schrodinger (and basically the rest of the universe) are all entangled. (That is, there is a state for the composite system of atom + poison + cat + Schrodinger, but the composite state doesn't "factor" into a product of the state for the atom \times the state of the poison \times the state of the cat \times the state of Schrodinger.) Note: \psi_{alive} does not describe an entangled composite system, and \psi_{dead} does not describe an entangled composite system, but the superposition \psi_{alive} + \psi_{dead} does describe an entangled composite system.

What does it mean "but the composite state doesn't "factor" into a product of the state for the atom \times the state of the poison \times the state of the cat \times the state of Schrodinger"? I'm not familiar with this statement about "doesn't "factor" into a product of the state for the atom \times the state of the poison.. etc...: what the \times and "doesn't factor" mean? Thank you.

 

[stevendaryl]

What does it mean "but the composite state doesn't "factor" into a product of the state for the atom \times the state of the poison \times the state of the cat \times the state of Schrodinger"? I'm not familiar with this statement about "doesn't "factor" into a product of the state for the atom \times the state of the poison.. etc...: what the \times and "doesn't factor" mean? Thank you.

If you have two objects, A and B, then you have a joint wave function (or state) \Psi_{A,B}. Sometimes the joint wave function can be written as a product of component wave functions: \Psi_{A,B} = \psi_A \cdot \psi_B, where \psi_A only involves A, and \psi_B only involves B. In that case, we say that the composite wave function "factors". But now, suppose the joint wave function is a superposition: \Psi_{A,B} = \alpha \psi_A \cdot \psi_B + \beta \phi_A \cdot \phi_B. You can't write that as a product of component wave functions. It doesn't factor.

It's analogous to being able to factor multinomials: The expression xy + x + y + 1 can be factored into (x+1) \cdot (y+1), where the first factor refers only to x and the second factor refers only to y. In contrast, x+y can't be written as something involving only x times something involving only y.

 

[stevendaryl]

What does it mean "but the composite state doesn't "factor" into a product of the state for the atom \times the state of the poison \times the state of the cat \times the state of Schrodinger"? I'm not familiar with this statement about "doesn't "factor" into a product of the state for the atom \times the state of the poison.. etc...: what the \times and "doesn't factor" mean? Thank you.

There is an analogous notion of "entanglement" in classical probability theory: You have a joint probability P(A \wedge B) for two events A and B. If A and B are independent, then the joint probability factors as follows:

P(A \wedge B) = P(A) P(B)

But if the events are not independent, then the joint probability doesn't factor.

For example, for a random person, the probability of being male is 1/2, and the probability of being blue-eyed is 1/20 (suppose). Since these are independent, the probability of being a blue-eyed male is 1/2 \cdot 1/20 = 1/40. On the other hand, if the probability of having a beard is 1/20, then the probability of being a bearded male is 1/20, not 1/40. They aren't independent, since only males have beards (pretty much).

 

[bluecap]

There is an analogous notion of "entanglement" in classical probability theory: You have a joint probability P(A \wedge B) for two events A and B. If A and B are independent, then the joint probability factors as follows:

P(A \wedge B) = P(A) P(B)

But if the events are not independent, then the joint probability doesn't factor.

For example, for a random person, the probability of being male is 1/2, and the probability of being blue-eyed is 1/20 (suppose). Since these are independent, the probability of being a blue-eyed male is 1/2 \cdot 1/20 = 1/40. On the other hand, if the probability of having a beard is 1/20, then the probability of being a bearded male is 1/20, not 1/40. They aren't independent, since only males have beards (pretty much).

Thank you for your clear explanations. A question that keeps bugging me for days about about page 6 of https://arxiv.org/pdf/quant-ph/0101077v1.pdf

"The second unanswered question in the Everett picture was more subtle but equally important: what physical mechanism picks out the classical states — face up and face down for the card — as special? The problem was that from a mathematical point of view, quantum states like "face up plus face down" (let’s call this "state alpha") or "face up minus face down" ("state beta", say) are just as valid as the classical states "face up" or "face down".

So just as our fallen card in state alpha can collapse into the face up or face down states, a card that is definitely face up — which equals (alpha + beta)/2 — should be able to collapse back into the alpha or beta states, or any of an infinity of other states into which "face up" can be decomposed. Why don’t we see this happen?

Decoherence answered this question as well. The calculations showed that classical states could be defined and identified as simply those states that were most robust against decoherence. In other words, decoherence does more than just make off-diagonal matrix elements go away. If fact, if the alpha and beta states of our card were taken as the fundamental basis, the density matrix for our fallen card would be diagonal to start with, of the simple form

density matrix = [1 0]
--------------------[0 0]

since the card is definitely in state alpha. However, decoherence would almost instantaneously change the state to

density matrix = [1/2 0]
--------------------[0 1/2]

so if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. In contrast, if we put the card in the state "face up", it would stay "face up" in spite of decoherence. Decoherence therefore provides what Zurek has termed a "predictability sieve", selecting out those states that display some permanence and in terms of which physics has predictive power."

------
Stevendaryl. What I'd like clarification is the following. It says above that if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. Whereas if it is face-up.. it will stay face up in spite of decoherence.. Does it mean that the alpha and beta state are really still there just we don't see it because it says if you measure it you could still get random outcome? Thanks you.

 

[Mentz114]

Particles become entangled through interactions, and interactions are local. However, once entangled, particles stay entangled even after they get far apart. And if A is entangled with B, and B becomes entangled with C, then A will be entangled with C, as well. So, everything becomes entangled with everything else.

Is it possible for A to become entangled with an earlier version of itself ?

 

[stevendaryl]

Is it possible for A to become entangled with an earlier version of itself ?

Interesting question. I'm not sure exactly what that would mean.

 

[Mentz114]

Interesting question. I'm not sure exactly what that would mean.

If A interacts with B, B with C, C with D in that order and then A interacts with D.

 

[Zafa Pi]

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?

My source is Nielsen & Chuang's "Q-Computation & Q-Information" (highly recommended and easy to read as opposed to Bill's recommendations). They say:
"A quantum system whose state |ψ⟩ is known exactly is said to be in a pure state. In this case the density operator is simply ρ = |ψ⟩⟨ψ|."
A mixed state is when the state is not completely known, is an ensemble of states with associated probabilities and
ρ = Σp_k•|ψ_k⟩⟨ψ_k|. A state is pure if tr(ρ²) = 1, and is mixed if tr(ρ²) < 1.

Using Q- info notation |0⟩ = [1,0] the 2D vector, |1⟩ = [0,1], |0⟩|0⟩ = |00⟩ is the tensor product |0⟩⊗|0⟩ which can be realized as the vector [1,0,0,0] in 4D. The superposition √½(|00⟩ + √½|11⟩) = √½[1,0,0,1] is in the 4D tensor product space, but is not a tensor product and is thus an entangled state, and further it is completely known. It is a pure state. In the "Bell basis" consisting of, √½(|00⟩ + |11⟩), ½(|00⟩ - |11⟩), √½(|10⟩ + |01⟩), √½(|01⟩ - |10⟩) the density operator for √½(|00⟩ + |11⟩) is the 4 by 4 matrix with 1 in the upper left corner and 0s elsewhere.

 

[bluecap]

I'm not going to get into specifics about emotions and civil war, etc. But in terms of Many-Worlds, we only share the same world because we are entangled. Entanglement just means that my state is correlated with your state. If we influence one another, or are influenced by something in common, then we will become correlated/entangled.

For context. Let's get back to this punchline. "But in terms of Many-Worlds, we only share the same world because we are entangled".

Let's say I met Sabine (let's be more practical instead of the more inaccessible Trump) and Stevendaryl also met Sabine. Then we 3 share the same world (o branch) because we are entangled. Our World is one mixed state component. It means if I feel sad or happy. It won't affect Sabine or Stevendaryl because we are already in single outcome (mixed state). Does it mean we are forever in that one mixed state or world. For example. Me and Sabine and Stevendaryl worlds are all happy. So I cna no longer access the other branch of Sabine where she is sad? As well as access the other branch where Stevendaryl is said. I can imagine me having different worlds or branches.. and I know I'm stuck in one world or branch (one mixed state chosen). But does it mean once I entangled with Sabine and we correlate in one world.. it is one world forever correlated and I can't access Sabine or Stevendaryl other branches forever no matter what kind of interaction? I wonder if this is called multi-body entanglement (maybe there is some papers about this).

 

[stevendaryl]

Let's say I met Sabine (let's be more practical instead of the more inaccessible Trump) and Stevendaryl also met Sabine. Then we 3 share the same world (o branch) because we are entangled. Our World is one mixed state component. It means if I feel sad or happy. It won't affect Sabine or Stevendaryl because we are already in single outcome (mixed state). Does it mean we are forever in that one mixed state or world. For example. Me and Sabine and Stevendaryl worlds are all happy. So I cna no longer access the other branch of Sabine where she is sad?

Yes, that's right. The splitting of worlds in Many-Worlds is irreversible, in the same way that if smash a glass bottle, the parts will never form together into a bottle.

As well as access the other branch where Stevendaryl is said. I can imagine me having different worlds or branches.. and I know I'm stuck in one world or branch (one mixed state chosen). But does it mean once I entangled with Sabine and we correlate in one world.. it is one world forever correlated and I can't access Sabine or Stevendaryl other branches forever no matter what kind of interaction? I wonder if this is called multi-body entanglement (maybe there is some papers about this).

For practical purposes, other branches of a macroscopic superposition are forever inaccessible.