Math of environment-particle entanglement

[Edward Wij]

I'd like to understand the math of how something specific in the environment (like photons) can be said to be entangled with a system (like dust particles).

Im reading a newbie friendly intro to superposition and entanglement "Home of the Wave Function"... I'd like to visualize entanglement using purely Schroedinger wave equations. I read that in entanglement (is what follows right)?

"We have already discussed the linear superposition of different quantum waves for the same system. But supposed we have two or more systems. Two or more quantum systems can share the same quantum wave. When this happens, we say that the systems are entangled with each other. The separate systems no longer have quantum waves of their own. The process "quantum connects" the two systems no matter how far apart they get from each other, provided that they are protected against collitions with other particles.

Getting to Hilbert space. I know the Schroedinger Wave is a vector in Hilbert space.. but in the following passage where vanesh mentioned:

"In the case that a quantum state of a multi-system is NOT one of those very special states which are "factorisable" (psi(p,q) = f(p) g(q)), then we say that the quantum state is an ENTANGLED state of the subsystems."

What is meant by factorisable and what does this mean psi(p,q) = f(p) g(q)? what is f and g? I know vectors and components axis in Hilbert space... What is the equivalent of factorisable in pure Schroedinger wave equation?

Can you give an actual example of how an actual specific thing in the environment can be entangled with a system in decoherence? like photon with dust? How would this form entanglement where it is not factorisable?

 

[jfizzix]

The quantum state of a pair of particles is said to be entangled if you can't express it as a mixture of product states (as opposed to a superposition)

If you have a single wavefunction for two particles that is non-factorizeable, that means you cannot express it as the product of two wavefunctions, one for each particle.

if a photon scatters off an atom, the photon and atom interact in the scattering process. The quantum state describing the two of them together becomes entangled. The way that this can create entanglement can be seen in terms of conservation laws.

The collision between atom and photon must conserve total momentum, but the uncertainties in the momenta of atom and photon need not change much in the collision. In order for the total momentum to be conserved, while still obeying the laws of quantum mechanics, the state describing the atom and photon together must be entangled.

 

[Edward Wij]

The quantum state of a pair of particles is said to be entangled if you can't express it as a mixture of product states (as opposed to a superposition)

If you have a single wavefunction for two particles that is non-factorizeable, that means you cannot express it as the product of two wavefunctions, one for each particle.

if a photon scatters off an atom, the photon and atom interact in the scattering process. The quantum state describing the two of them together becomes entangled. The way that this can create entanglement can be seen in terms of conservation laws.

The collision between atom and photon must conserve total momentum, but the uncertainties in the momenta of atom and photon need not change much in the collision. In order for the total momentum to be conserved, while still obeying the laws of quantum mechanics, the state describing the atom and photon together must be entangled.

In decoherence.. what else comprise the environment that can entangle with the system besides photons? maybe Microwave Background Radiation? what else? It appears that it is hard to create entangled pair like in EPR or para down conversion in photons, yet in decoherence the environment can easily get entangled with the system, why is that?

 

[jfizzix]

Any quantum object in "the environment" that is capable of interacting with a system at all is capable of being entangled with it.

So, cosmic rays, space dust, photons of all energies (from radio, to microwave, to IR, to visible, to UV, X-rays, gamma rays, and beyond), particles in the solar wind, all could interact with a quantum system, and entangle with it.

The difficulty is not so much in creating entanglement.
Rather, it is in isolating a quantum system from the environment, so that a pair of entangled particles stay entangled and don't decohere.
Photons in parametric downconversion are nice because they don't interact strongly with the environment. Other quantum systems are harder to isolate, and so are harder to work with.

 

[Edward Wij]

Any quantum object in "the environment" that is capable of interacting with a system at all is capable of being entangled with it.

So, cosmic rays, space dust, photons of all energies (from radio, to microwave, to IR, to visible, to UV, X-rays, gamma rays, and beyond), particles in the solar wind, all could interact with a quantum system, and entangle with it.

The difficulty is not so much in creating entanglement.
Rather, it is in isolating a quantum system from the environment, so that a pair of entangled particles stay entangled and don't decohere.
Photons in parametric downconversion are nice because they don't interact strongly with the environment. Other quantum systems are harder to isolate, and so are harder to work with.

I'd like to understand entanglement using purely the image of waves, in superposition the book says

"Now, suppose two waves are heading towards each other, with amplitudes in the range such that the superposition principle applies. When the waves meet, a complex wave is created according to the rule given above. If the waves then separate, the original amplitudes are retained (barring friction). That is, no distortion was introduced, and the wave is linear. Thus, a complex linear sstem can be decomposed without destroying the system. You can see that linear systems lend themselves to mathematical treatment because they can be reduced to a simple superposition of elements."

What is the wave equivalent in entanglement such that the original amplitudes are no longer retained?

 

[stevendaryl]

What is the wave equivalent in entanglement such that the original amplitudes are no longer retained?

For simplicity, let's just consider one spatial dimension, $x$. Then a general wave function for a two-particle state would be:

$\psi(x_1, x_2)$

The meaning of this is that $|\psi(x_1, x_2)|^2$ is the probability density of finding the first particle at position $x_1$ and the second particle at position $x_2$.

Such a state is "factorizable" if there are two functions, $A(x)$ and $B(x)$ such that:

$\psi(x_1, x_2) = A(x_1) B(x_2)$

It's "entangled" if that's not possible. So for example, $\psi(x_1, x_2) = \frac{1}{\sqrt{2}} (A_1(x_1) B_1(x_2) + A_2(x_1) B_2(x_2))$ is entangled, if $A_1$ and $A_2$ are orthogonal states, and $B_1$ and $B_2$ are orthogonal states.

 

[Nugatory]

What is the wave equivalent in entanglement such that the original amplitudes are no longer retained?

You have the two waves $|A+B-\rangle$ and $|A-B+\rangle$ propagating happily in the superimposed state $\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)$, just as in your "waves heading towards each other" situation you have the superposition $\psi=\frac{\sqrt{2}}{2}(\psi_L+\psi_R)$ where $\psi_L$ and $\psi_R$ are the left-moving and right-moving waves.

The superposition lasts until you perform a measurement.

In the $\psi=\frac{\sqrt{2}}{2}(\psi_L+\psi_R)$ case, a measurement of the position yielding $X_0$ causes the wavefunction to collapse into $\psi=\delta(x-X_0)$, which is not a superposition in the position basis (warning: there are a number of mathematical subtleties about the continuous spectrum of the position operator that I have glossed over). In the $\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)$ case, a measurement of either $A$ or $B$ (for definiteness, we can assume that these are measurements of the spin of two particles and $|A+B-\rangle$ is the state in which particle A is spin-up and its entangled partner B is spin-down) will cause the wavefunction to collapse into either $|A+B-\rangle$ or $|A-B+\rangle$, neither of which is a superposition in this basis.

(These post-measurement states are still superpositions in other bases. "Going North" is not a superposition if I've using North and East as my basis whereas "going Northeast" is (it's superposition of North and East). However, if I use Northeast and Northwest as my basis, it's the other away around).

 

[bhobba]

I'd like to understand entanglement using purely the image of waves, in superposition the book says

If anyone can explain it that way - I will tip my hat to them.

The very concept of superposition and entanglement is mathematical. It is this - if |a> and |b> are states then c1*|a> + c2*|b> where c1 and c2 are complex numbers is also a state.

Entanglement is an extension of superposition to different systems. Suppose two systems can be in state |a> and |b>. If system 1 is in state |a> and system 2 is in state |b> that is written as |a>|b>. If system 1 is in state |b> and system 2 is in state |a> that is written as |b>|a>. But we now apply the principle of superposition so that c1*|a>|b> + c2*|b>|a> is a possible state, The systems are entangled - neither system 1 or system 2 are in a definite state - its in a peculiar non-classical state the combined systems are in.

It impossible, utterly impossible to explain it in English. You must do the math.

I have previously given you a link that with a minimum of math explains all this stuff:
http://quantum.phys.cmu.edu/CQT/index.html

Griffith's has been kind enough to make his book freely available. It is written not only for mathematicians and physicists, but philosophers as well. It avoids advanced math as much as possible - but since it can't be explained any other way some math is required.

If you want to understand this you must put the effort into come to grips with even a minimal amount of math. There is no other way. English and images is simply not up to the task.

Thanks
Bill

 

[Edward Wij]

If anyone can explain it that way - I will tip my hat to them.

The very concept of superposition and entanglement is mathematical. It is this - if |a> and |b> are states then c1*|a> + c2*|b> where c1 and c2 are complex numbers is also a state.

Entanglement is an extension of superposition to different systems. Suppose two systems can be in state |a> and |b>. If system 1 is in state |a> and system 2 is in state |b> that is written as |a>|b>. If system 1 is in state |b> and system 2 is in state |a> that is written as |b>|a>. But we now apply the principle of superposition so that c1*|a>|b> + c2*|b>|a> is a possible state, The systems are entangled - neither system 1 or system 2 are in a definite state - its in a peculiar non-classical state the combined systems are in.

Yes. I wholeheartedly agree with you that one needs to understand the math. That's what I'm doing now. But it will take 3 month to read Griffith's book. That is why I just need to know something now which is why I made this thread. What is the meaning of "|b>"? , why is there a vertical line and an arrow to the right with b at middle? Is it differential algebra or differential geometry or what branch of math does that fall under? Also in the following I know that the "+" or plus in "c1*|a>|b> + c2*|b>|a>" is that there is interference. But what is c1*|a>|b>? I know c1 is some probability but what is |a>|b>? Does it mean you multiply |a> with |b>? Why. Can you put actual values into them so I know what exacty the equation is doing? That is all I need to know for know so in your reply pls. just answer the above without adding others so I won't be confuse. Thank you.

It impossible, utterly impossible to explain it in English. You must do the math.

I have previously given you a link that with a minimum of math explains all this stuff:
http://quantum.phys.cmu.edu/CQT/index.html

Griffith's has been kind enough to make his book freely available. It is written not only for mathematicians and physicists, but philosophers as well. It avoids advanced math as much as possible - but since it can't be explained any other way some math is required.

If you want to understand this you must put the effort into come to grips with even a minimal amount of math. There is no other way. English and images is simply not up to the task.

Thanks
Bill

 

[bhobba]

Yes. I wholeheartedly agree with you that one needs to understand the math. That's what I'm doing now. But it will take 3 month to read Griffith's book. That is why I just need to know something now which is why I made this thread. What is the meaning of "|b>"? , why is there a vertical line and an arrow to the right with b at middle?

Its called the Dirac Bra-Ket notation. Griffith explains it all:
http://quantum.phys.cmu.edu/CQT/chaps/cqt03.pdf

You are over thinking this - all its saying is the states form a vector space - that's it, that's all.

After you anderstand that you can then come to grips with what it means physically.

That such is the case ie the states form a vector space is one of the rock bottom essences of QM. It can be explained by deeper principles, but that is just by the by, to start with just accept its how QM is.

Diracs explanation may help you:
http://www.informationphilosopher.com/solutions/scientists/dirac/chapter_1.html

Thanks
Bill

 

[Edward Wij]

For simplicity, let's just consider one spatial dimension, $x$. Then a general wave function for a two-particle state would be:

$\psi(x_1, x_2)$

The meaning of this is that $|\psi(x_1, x_2)|^2$ is the probability density of finding the first particle at position $x_1$ and the second particle at position $x_2$.

Such a state is "factorizable" if there are two functions, $A(x)$ and $B(x)$ such that:

$\psi(x_1, x_2) = A(x_1) B(x_2)$

It's "entangled" if that's not possible. So for example, $\psi(x_1, x_2) = \frac{1}{\sqrt{2}} (A_1(x_1) B_1(x_2) + A_2(x_1) B_2(x_2))$ is entangled, if $A_1$ and $A_2$ are orthogonal states, and $B_1$ and $B_2$ are orthogonal states.

If c1*|a>|b> is dirac bra-ket notation. what is this $\psi(x_1, x_2) = A(x_1) B(x_2)$, what notation or math is that? linear algebra or differential algebra? just tell me why do you have to multiply $A(x_1)$ by $B(x_2)$, is it just to signify supeposition? I know the Schroedinger Equation and what $\psi$ means and squaring psi to get probability.

 

[bhobba]

If c1*|a>|b> is dirac bra-ket notation. what is this $\psi(x_1, x_2) = A(x_1) B(x_2)$, what notation or math is that?

Again Griffiths explains it - please make the time and effort to go through it.

Technically the wave-function is the expansion of the state in terms of position eigenvectors. This leads to |a>|b> in the dirac bra-ket notation becoming $\psi(x_1, x_2)$.

Me telling you that however will not make sense without the background to understand it. There is no short-cut - you must start from the start and progress.

Please read Griffiths from the start and post with any questions - do not try to skip.

Thanks
Bill

 

[Edward Wij]

Again Griffiths explains it - please make the time and effort to go through it.

Technically the wave-function is the expansion of the state in terms of position eigenvectors. This leads to |a>|b> in the dirac bra-ket notation becoming $\psi(x_1, x_2)$.

Me telling you that however will not make sense without the background to understand it. There is no short-cut - you must start from the start and progress.

Please read Griffiths from the start and post with any questions - do not try to skip.

Thanks
Bill

Thanks. Last question before I finish reading it which would take a year in March 2016 (because I have to finish other books). It's great to know the relationship between |a>|b> and $\psi(x_1, x_2)$ being "Technically the wave-function is the expansion of the state in terms of position eigenvectors. This leads to |a>|b> in the dirac bra-ket notation becoming $\psi(x_1, x_2)$." I know the meaning of eigenvectors being familiar with hilbert space. Now in Nugatory $\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)$, what kind of math is that? why is there A + B -, how does it relate to both |a>|b> and $\psi(x_1, x_2)$? It's ok if Nugatory or Stevendaryl (thanks to both too) will answer because Bhobba will just tell me to read the whole book and know the answer next year when just want a *rough* idea of it now.

 

[jfizzix]

I'd like to understand entanglement using purely the image of waves,
What is the wave equivalent in entanglement such that the original amplitudes are no longer retained?

If you just had two particles, each described by their own wavefunction, and there was no interaction between the two, you would see the two waves interfere, and separate out again.

In this non-interacting case, the two particles would be described independently by their own wave equations.

In order for there to be entanglement, there must be an interaction between the two particles.

To accommodate this interaction, you can no longer describe the behavior of the two particles by two different wave equations. You would have to have one bigger wave equation that can't separate into two because of extra terms describing the interaction between the two particles.

 

[rkastner]

I have written a popular book on this, that explains it with hardly any math, in conceptual terms. See my website rekastner.wordpress.com for info.

 

[bhobba]

I know the meaning of eigenvectors being familiar with hilbert space. Now in Nugatory $\psi=\frac{\sqrt{2}}{2}(|A+B-\rangle+|A-B+\rangle)$, what kind of math is that? why is there A + B -, how does it relate to both |a>|b> and $\psi(x_1, x_2)$?

Its vector space math.

You might think you know what a Hilbert Space is, but if that is unclear you don't - its simple vector space math and Hilbert spaces are vector spaces:
http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/Hilbert-Spaces.pdf

Thanks
Bill

 

[naima]

As the subject is "the math of entanglement", I am surprised to see that the word Hamiltonian does not appear!
A two level system to be measured is prepared as a |s0> + b|s1>.
A measurement device for this obsevable is in its rest state $|S_0>$ so we start with $(a |s_0> + b|s_1) > \otimes |S_0>$
We have to associate an hamiltonian H(t) so that this vector evolves to
$a |s_0>\otimes |S_0> + b|s_1 > \otimes |S_1>$ Here S_1 is a state of the apparatus coupled to the second level of the system.
This describes the premeasurement or entanglement.
I have here a question. If the two levels are the eigenvectors of an hermitian operator, they are orthogonal. What about $|S_0>$ and $|S_1>$?
Decoherence will occur further. What has to be added so that it will choose this preferred basis (what about an orthonormal basis?)

 

[stevendaryl]

If c1*|a>|b> is dirac bra-ket notation. what is this $\psi(x_1, x_2) = A(x_1) B(x_2)$, what notation or math is that?

It's a two particle wave function. A wave-function is a function on configuration space. Configuration space has one coordinate for each dimension of space and for each particle; so 2 particles in 3D space produces a 6-dimensional configuration space: $x_1, y_1, z_1, x_2, y_2, z_2$

linear algebra or differential algebra? just tell me why do you have to multiply $A(x_1)$ by $B(x_2)$, is it just to signify supeposition? I know the Schroedinger Equation and what $\psi$ means and squaring psi to get probability.

No, it has nothing to do with superpositions. You've never worked with the Schrodinger equation for more than one particle? As I said, in general, the Schrodinger describes a wave function, which is a function on configuration space.

Suppose you have two particles with a harmonic oscillator interaction. Then the Schrodinger equation for such a system would be given by:

$H \psi(x_1, x_2) = (-\frac{\hbar^2}{2m_1} \frac{d^2}{dx_1^2} -\frac{\hbar^2}{2m_2} \frac{d^2}{dx_2^2} + \frac{k}{2} (x_2 - x_1)^2) \psi(x_1, x_2)$

 

[naima]

Maybe the intrication between apparatus and environment does not only lead to projective measurements but to POVMs?

 

[Tyrosine]

Any quantum object in "the environment" that is capable of interacting with a system at all is capable of being entangled with it.

So, cosmic rays, space dust, photons of all energies (from radio, to microwave, to IR, to visible, to UV, X-rays, gamma rays, and beyond), particles in the solar wind, all could interact with a quantum system, and entangle with it.

The difficulty is not so much in creating entanglement.
Rather, it is in isolating a quantum system from the environment, so that a pair of entangled particles stay entangled and don't decohere.
Photons in parametric downconversion are nice because they don't interact strongly with the environment. Other quantum systems are harder to isolate, and so are harder to work with.

some questions:

Is entanglement transitive? If system A is entangled with system B, and which are coherent w.r.t. the environment, and then system C fully interacts with A, entangling with A, then does system B remain entangled to A, and if it does, then is B entangled to C?

Does lack of entanglement imply superposition?

Are the atoms in a molecule all entangled together?

TIA

 

[Edward Wij]

If you just had two particles, each described by their own wavefunction, and there was no interaction between the two, you would see the two waves interfere, and separate out again.

In this non-interacting case, the two particles would be described independently by their own wave equations.

In order for there to be entanglement, there must be an interaction between the two particles.

To accommodate this interaction, you can no longer describe the behavior of the two particles by two different wave equations. You would have to have one bigger wave equation that can't separate into two because of extra terms describing the interaction between the two particles.

It is said that measurement breaks entanglement, for example when you measure particle A in the EPR pair, the pair is broken and particle A and B are separated. How about in nucleus and electrons, these are entangled.. when you measure the electron.. does the electron become separated from the nucleus.. or are nucleus and electron still entangled?

 

[Tyrosine]

It is said that measurement breaks entanglement, for example when you measure particle A in the EPR pair, the pair is broken and particle A and B are separated. How about in nucleus and electrons, these are entangled.. when you measure the electron.. does the electron become separated from the nucleus.. or are nucleus and electron still entangled?

If it breaks the entanglement, then does that mean the other particle will remain coherent? If so, how is it that we can predict its spin from the measurement taken from the first particle?

 

[bhobba]

How about in nucleus and electrons, these are entangled..

I have defined entanglement - from that definition what do you think? And if you can't answer it maybe you need to understand the concepts better?

when you measure the electron.. does the electron become separated from the nucleus.. or are nucleus and electron still entangled?

So you are thinking they are entangled. But regardless of that how do you measure an electron in an atom?

Thanks
Bill

 

[bhobba]

Is entanglement transitive?

Look back to where I defined entanglement. Systems are no longer separate entities with a separate state. Thus its a meaningless question to ask if its transitive.

Does lack of entanglement imply superposition?

Go back to the definitions I gave. If a system is not entangled then it has a separate state. Any state is the superposition of other states in a myriad of different ways.

Are the atoms in a molecule all entangled together?

From the definition I gave what do you think?

Thanks
Bill

 

[Edward Wij]

I have defined entanglement - from that definition what do you think? And if you can't answer it maybe you need to understand the concepts better?
So you are thinking they are entangled. But regardless of that how do you measure an electron in an atom?

Thanks
Bill

You believe that measurements can't differentiate between mixed state and improper mixed state.. meaning you can't know if they are from entanglement or from the more ordinary superposition (there before observation). I was trying to find out if there are experimental setups that can distinguish it. In the nucleus - electron entanglement.. if you measure the electron and can know that it belongs to an atom.. then you can differentiate mixed state and improper mixed state. I think you measure it by sending photons to hit the electrons so if it's higher orbital you can distinguish it belongs to atoms and can say it's from entanglement (hence improper mixed state and not just mixed state). I'll leave it to Stevendaryl or Jfizzix or others to state their opinions to prove you assertions or disprove it (since I can't counter your opinions due to insufficient knowledge of all experimental setups).

 

[bhobba]

If it breaks the entanglement, then does that mean the other particle will remain coherent?

I don't know what you mean by coherent in this context.

If so, how is it that we can predict its spin from the measurement taken from the first particle?

Go back to the definition I gave. You are talking about EPR which concerns Bell states (look them up).

Consider the following Bell state 1/root 2*|a>|b> + 1/root2 |b>|a>. Now we measure system 1 and determine it is in state |a>. But since they are entangled it doesn't really have a separate state so what you have really done is measured the entangled system and got the answer |a>|b>. It is no longer entangled and system 2 is automatically in state |b>. That's an overview - you can consult the literature where what's going on in terms of observables is detailed.

That said I don't know what me saying this serves unless you understand the principles involved.

Thanks
Bill

 

[bhobba]

You believe that measurements can't differentiate between mixed state and improper mixed state.. meaning you can't know if they are from entanglement or from the more ordinary superposition (there before observation)

You are confused about the concepts of superposition and a mixed state. Being there prior to observation has nothing to do with superposition.

I will explain. Superposition is a concept applicable to pure states which are usually mapped to a Hilbert space. However states really are not vectors - they are in fact positive operators of unit trace. Pure states are of the form |u><u|. Mixed states are of the form Σ pi |ui><ui| where pi is the probability it will be i state |ui><ui| if you observe it. This follows from the Born rule and is why you can consider the state there before observing - the pi are probabilities - not superposition's. This is entirely different to superposition which only applies to pure states - not mixed states.

Please, please study the concepts before posting. I don't know why you keep posting about this. You will never understand it until you get to grips with he actual concepts which will require study. The above, while correct, is likely gibberish because you don't know the basics.

Thanks
Bill

 

[Edward Wij]

You are confused about the concepts of superposition and a mixed state. Being there prior to observation has nothing to do with superposition.

I will explain. Superposition is a concept applicable to pure states which are usually mapped to a Hilbert space. However states really are not vectors - they are in fact positive operators of unit trace. Pure states are of the form |u><u|. Mixed states are of the form Σ pi |ui><ui| where pi is the probability it will be i state |u><u| if you observe it. This follows from the Born rule and is why you can consider the state there bore observing - the pi are probabilities - not superposition's. This is entirely different to superposition which only applies to pure states - not mixed states.

Please, please study the concepts before posting. I don't know why you keep posting about this. You will never understand it until you get to grips with he actual concepts which will require study. The above, while correct, is likely gibberish because you don't know the basics.

Thanks
Bill

Yes I know the distinction, but just bad choice of words. I jus mentioned superposition in mixed state because the measuring device is entangling with either slit A or slit B and we don't know which one that is why we need the statistical operator in mixed state. So I shouldn't have mentioned superposition. Thanks for the details above though.

It still stands or the question still stands whether in measuring electrons in atoms, one can differentiate between mixed and improper mixed states.. or know the particle you are measuring comes from entanglement.. but since we know the nucleus and electron is entangled and you are measuring the electron then you can differentiate improper mixed state from mixed state. Although I think what you were saying is that mathematically and observationally they can't be distinguished.. but observationally if you detect it at orbitals.. you know it belongs to atoms so can differentiate. Hope others can comment on this.

 

[bhobba]

It still stands or the question still stands whether in measuring electrons in atoms

I have zero idea what you mean by measuring electrons in atoms.

However you seem to have correctly deduced it must be entangled with the nucleus.

Thanks
Bill

 

[Nugatory]

It's time to close this thread.

If you're looking for a layman-friendly math-free intuitive model of superposition and entanglement, you could do worse than reading The age of entanglement and Where does the weirdness go?.

However, if you want something that you can build a deeper understanding on, as opposed to taking our word for what the math says about a few toy examples, then there is no substitute for spending a few tens of hours with a decent textbook. That's clear from this thread - so far we've done nothing beyond demonstrating that it is impossible to develop an intuitive picture of superposition and entanglement in multi-particle systems without using the formalism of linear algebra and vector spaces.