Information diagram for entangled particles?

[Heidi]

Hi Pf

I read that when two particles are maximally entangled , all the information is in the correlations between the particles. If we need 1 bit to describe one particles. Are two bits in the correlations?
three particles may be entangled and then we need to considere 3 bits. As entanglement may be not maximal, I wonder if we can use a diagram
https://en.wikipedia.org/wiki/Information_diagram
with 7 pieces to say the first piece contains this fraction of the 3 bits, the second this other fraction ans so on. at the end the sum would be equal to 3.
If it is not possible like that, is it possible with oriented graphs or with another picture?

 

[jambaugh]

IMNSHO, you are focusing on the wrong things. Your first sentence is correct. But it does not depend on how many "bits" each particle can encode. The only caveat is that two maximally entangled particles ( each maximally entangled over all possible other systems) is that they must have the same degrees of freedom. So, if you have an n-state particle with, therefore, $log_2(n)$ "bits" of information, then a second equivalent particle has $log_2(n)$ "bits" and the two together can encode $log_2(n^2)=log(n)+log(n)$ "bits" with their $n^2$ total degrees of freedom.

The only question then, is whether those "bits" of information are encoded, half and half, on the pairs separately (not entangled at all) or encoded in their relative "states" (maximally entangled) or something in-between, (partially entangled). Also remember that "entanglement" is not a state variable, it is a meta-variable, relative qualifier describing something about which variables one is considering.

 

[jambaugh]

Addendum: Here's a concrete example that may help clarify. Consider two spin-1/2 particles. Label them Particle A and Particle B. Each encodes one quantum bit or qubit (two "states" per). The two together encode 2-qubits (specifically $log_2(2)+log_2(2)=2log_2(2)$ because the two together have 4 states (specifically $log_2(2\times 2)=2log_2(2)$).

(Similarly with a pair of 5-state systems, each encoding $\log_2(5)\approx 2.322$ qubits each with a total of $log_2(5)+log_2(5) = log_2(5\times 5) \approx 4.644$ qubits.)

Now I can encode 2 bits in these qubits in terms of whether particle A has spin-z +1/2 or -1/2 and independently whether particle B has spin-z +1/2 or -1/2.

In this case zero information is encoded in the spin-x and spin-y components of the pair. Since each bit is encoded one in-toto per particle they are unentangled.

Alternatively, for example, I can encode the two bits in terms of different observables indicating whether the two spin-x components are correlated vs anti-correlated...
(the "state" vector resides in the 2-dim subspace spanned by $\left\lvert \uparrow \uparrow\right\rangle \pm \left\lvert \downarrow\downarrow\right\rangle$ where $\uparrow = \sigma_x = +1/2$)
and, independently whether the two spin-y components are correlated vs anti-correlated (selecting one 1-dim subspace). In this case there is no new information encoded in the relationship between or separate values of the spin-z components. (Any observation of these will be intrensically 50%-50% random.) Knowing both the correlation vs anti-corr. of spin x and spin y would be a case of maximal entanglement.

In a third alternative, I can know that total spin is either 1 or 0. If 0, there's no additional independent information and, again the particles are maximally entangled. Knowing this is the same as knowing each of the x, y, and z components of spin is anti-correlated.

But the alternative, total-spin = 1 still allows an additional variable with three possibilities. Is the total x-component of spin +1, or 0, or -1? Once that is decided then the relationships of y, and z spins tells us nothing new. Note that by knowing that the total spin is 1, one does not know, necessarily that the spin component in a given direction is correlated or not. But there is still a restriction on how much information can be simultaneously encoded in, either the separate spin components in a given direction, or on the correlation vs anti-correlation of the spins in multiple directions. We know this fact by virtue of what observables commute with which.

A final observation. Here I am treating the variable specifying whether we are referring to particle A vs particle B as if it were a classical variable. Imagine two boxes and that we have restricted the composite system so that the total number of particles in each box is exactly 1. With this extension of the system, and considering the total spin zero case (total anti-correlation), we can refer to "the A particle" and "the B particle" and observe total anti-correlation of, say, the spin-z components. Or we can refer to the "spin-z up particle" and the "spin-z down particle" and observe total anti-correlation of where the two particles are, A vs B. There is a relativity in whether we are thinking of our encoded bits as entangled via whether we encoding whether the spin is "locally" specified according to whether we define "locally" via our classically treated A vs B variable, or alternatively (but meaning the same thing) whether we are encoding the (quantum) location of our particle (A vs B) "locally" specified according to whether we define "local" via a classical treatment of the spin-z up vs spin-z down variable given we know we have exactly one of each.

We can always expand our perspective to include the classical index variable as another binary quantum variable, consider the composite system as a whole, and then pick a new index variable in such a way that the composite system as subdivided by that new index var., "is entangled" or alternatively "is not entangled". Entanglement is not a property of the physical system but rather a quality of how we decompose that physical system into parts. (But of course we have a strong bias, given the local causality assumption of system dynamics to decompose using position.)

Those are my bourbon induce thoughts on the matter in any event.

 

[Heidi]

I can use the tensor product $$H_1 \otimes H_2$$ for the pairs of particles. I take the orthonormal basis
$$\downarrow \downarrow \ \downarrow \uparrow \ \uparrow \downarrow \ \uparrow \uparrow$$
when i describe the pair by a choice of the four complex components a,b,c,d:
$$a\downarrow \downarrow +b \downarrow \uparrow +c \uparrow \downarrow +d \uparrow \uparrow$$
there is nothing to add, all is in these four numbers.
I can compute the corresponding 4*4 density matrix, trace it out
for Bon and Alice and so on. i can do that.
It is the same for three particles
My probleme is graphical.
what is the best picture to show the results?

 

[Heidi]

For two particles, it is possibe to use a oriented circle. A and B are opposite points on the circle. say at -1 and +1 on the x axis. We have the oriented upper half circle going from A to B and the lower from B to A.
we can label these four elements with four real numbers summming to 2.
Can we do the same thing with 3 particles (a unital vector in the tensor product of 3 particles is described by 8 complex numbers (and 1 relation between them)?

 

[vanhees71]

One should keep in mind that entropy as a measure of missing information given a state description is relative to what's defined as "complete knowledge". In QT "complete knowledge" means that a system is prepared in some pure state. This can be achieved, e.g., by preparing the particle in a state where a complete set of compatible observables takes determined values (eigenvalues of the corresponding self-adjoint operatos).

That's why the entropy of any pure state is 0, because if we know that a system is prepared in a known pure state, we have complete knowledge about the system.

Now, if you consider the two spins as a whole and prepare them in the pure state
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle-|-1/2,1/2 \rangle),$$
the entropy of the two-spin system is 0.

If you now, however ask, what's the state of one of the spins, you have to use the reduced statistical operator for the state description, and for this you get
$$\hat{\rho}_{1}=\mathrm{Tr}_{2} |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1},$$
and for this you get indeed
$$S=-\mathrm{Tr} \hat{\rho} \ln \hat{\rho}=\ln 2,$$
and this is the state of maximum entropy, i.e., of least possible knowledge.

That's a specific feature of entanglement between the observables of two subsystems prepared in an entangled pure state. It's what Einstein dubbed "inseparability", i.e., though you have maximal complete knowledge about the system as a whole (being prepared in a pure state) you may have maximal entropy, i.e., minimal possible knowledge about one of its subsystems, and this was the main quibble Einstein had about QT.

 

[Heidi]

Yes but what about the graphical question ? look at my first post.
i gave a link to a Venn diagram. when we have a Bell pair i can label
the intersection with 2 and by 0 on the two other pieces.
I would like to extend this to 3 or more particles.

 

[vanhees71]

I'm not sure, what you want to achieve with this diagram. Given the state, you have to define what you want to measure on the system. For any physically possible experiment you then get the probabilities in accordance with standard probability theory.

 

[Heidi]

In the way i see this information diagram there is not a lot of things to say about the measurements:
The clever experimentalist (CE) says to Alice and Bob (they are far from her and she is in the middle) I will send you two level particles (say qbits) . You are free to measure what you want in any directions, at the end you will calculate the averages, the entropy , the fringe visibility and so on. and we will verify if I have well predicted your calculations.
She chooses a vector in
$$H_1 \otimes H_2$$
As this space has four dimensions she needs two bits to encode this choice . She have the entire information: 2 bits.
she calculates that they will receive density matrices with the same entropy r.
she draws a diagram. Alice and Bob have the same quantity of information: 1 - r bit. 2r bits have still to be placed
she can draw an arrow from A to B and another from B to A. it is in the case where there would be an assymetry and calculat how to label the arrows (i think that here it is with r and r.

this is for 2 particles.
CE could also be at the center of an equilareral triangle with Alice Bob and Charles at the vertices and do the same thing.
we have to label the vertices , there are 6 arrows but is it enough?
Is it more complicated?
Look at the Venn diagram on the link in post 1. there are 7 pieces. 3 for A B and C, 3 for their 2 by 2 intersection but also a piece in the middle.
What about the arrows i began to draw?

 

[vanhees71]

The important point is that you cannot discuss properly such questions if not very carefully specifying who is doing which measurement on which state (preparation) of the three-qbit system.

 

[Heidi]

Do you agree with what i wrote in the 2 particles case?
CE draws the diagram and calculate the labels on her computer before Bob and Alice begin their measurements. It is about entropy and quantity of information, not on results of peculiar mesurements.
it is the same with A B and C
she chooses 8 complex numbers to determine a unit vector in a 8 dimention tensor space. we are not interested in the details , if she uses SPDCs mirrors or things like that. maybe the energy has to be precised.

 

[vanhees71]

I don't understand what you wrote. You don't specify which state is prepared nor who measures what. So I can't say what you are talking about. All quantum theory tells you are the probabilities for measurement results on a system in a given state.

 

[Heidi]

What can i say to be more precise. look at post 4
the state is chosen in the tensor space
i considered an orthonormal basis . the state has components a b c and d.
i thought it was enough. i can add thar up is relative to a vercical direction and down also. that the first is to Alice and the second to Bob.
CE compute his pure density matrix , trace it out for Bon and Alice and uses a device to send them states corresponding to these reduced matrices.

 

[Heidi]

We can always expand our perspective to include the classical index variable as another binary quantum variable, consider the composite system as a whole, and then pick a new index variable in such a way that the composite system as subdivided by that new index var., "is entangled" or alternatively "is not entangled". Entanglement is not a property of the physical system but rather a quality of how we decompose that physical system into parts. (But of course we have a strong bias, given the local causality assumption of system dynamics to decompose using position.)

@jambaugh
Excuse me for this late answer.
I wonder if you are right when you say that entanglement is not
a property of the whole composite system.
there are cases when physics do not let you consider other point of view that the A particle and the B particle. consider that a particle decays in two identical particles neat a black hole. one falls
inside and the other escapes. You have no access to one of them.
it is not a point of view. You receive particles in a density matix state. Am i wrong?

 

[vanhees71]

Of course, entanglement is a property of the entire composite system. That's because it describes the "inseparability" of the parts of the composite system prepared in such an entangled state and it's causing the "stronger-than-classical correlations" between observations on these constituents (in the sense of Bell's theorem).

 

[Heidi]

Hi Vanhees71
i asked you i you agree to post 5
Has this diagram a sense?
i label the two points by 1 - r (r is the entropy that alice or bob can measure) and each half oriented circle by r.

 

[vanhees71]

No, for me the diagram doesn't make sense without a definition in which state the system is prepared and which measurement is done on an ensemble of such prepared systems. Only then can you analyze the possible outcomes and their probabilities.

 

[Heidi]

Bob and Alice measure the density matrix (in the up down basis relative to vertical direction)
the experimentalist sends them this state with a device
such that these desity matrix is partial trace of the pure vector
with a, b , c, d component in the dd, du, ud, uu basis same signification for her as for alice and bob.
is the device the problem for you.
can you give me examples of thinfs i can add.

 

[Heidi]

If she sends bell pairs i put 0 on the points and 1 on each half circle
if she sends independent pure statd i put 1 on the points and 0 on the half circles. and so on for the in between

 

[vanhees71]

Bob and Alice measure the density matrix (in the up down basis relative to vertical direction)
the experimentalist sends them this state with a device
such that these desity matrix is partial trace of the pure vector
with a, b , c, d component in the dd, du, ud, uu basis same signification for her as for alice and bob.
is the device the problem for you.
can you give me examples of thinfs i can add.

The density matrix with respect to which basis? That's the information that's lacking here! You cannot draw diagrams depicting the probabilities without telling what you measure, i.e., wrt. which basis you want to get the density matrix for. Maybe it's just a semantic misunderstanding. So here is what I mean.

The state is described by a statistical operator. In your case it's a pure state and thus written as
$$\hat{\rho}=|\Psi \rangle \langle \Psi |.$$
If I understand you right, you consider three spins/qbits, and thus any state is of the form
$$|\Psi \rangle = \sum_{\sigma_1,\sigma_2,\sigma_3} c_{\sigma_1 \sigma_2 \sigma_3} |\sigma_1 \rangle \otimes |\sigma_2 \rangle \otimes |\sigma_3 \rangle, \quad \sigma_j \in \{1/2,-1/2 \}.$$

The density matrix refers to a measurement, where you meausure some set of possible observables, defining a unique orthonormal basis $|u_j \rangle$ and the density matrix is given by the matrix elements of the statistical operator wrt. this basis,
$$\rho_{ij}=\langle u_i|\hat{\rho} u_j \rangle.$$

 

[Heidi]

I thank you to spend so much time with my questions. we use several words differently.
You say that you have measured a density matrix if having a basis you are able to write the components.
For me something is measured if you can descibe its components in any basis.
And we can easily agree when what is measured is basis independant.
if i am interested in the module of a complex number i can use x,y or r, theta it will be the same.
Here l use the density matrix to find its entropy. It is like its trace. No need to speak of a basis

 

[vanhees71]

No, I'm not saying, I'm measuring a density matrix. I'm measuring observables on an ensemble of independently and equally prepared systems to get the statistics described by the density matrix.

The entropy (to be precise, the Shannon-Jaynes-von-Neumann entropy) is indeed basis independent and defined as
$$S=-\mathrm{Tr} (\hat{\rho} \ln \hat{\rho}),$$
but what has this to do with your diagram, whose meaning I still don't understand.

 

[Heidi]

My idea is to find quantum equalities (or inéqualities) which explain why severall no go sentences are true.
the first example is: one cannot get perfect path information and perfect fringe visibility. another one: one get no fringes with Bell pairs (even if no path detector is used)

I give you a case where it works.
consider a pure two level state
$$cos(\alpha)|d> + sin(\alpha) |u>$$

its module is equal to one and i suppose that $$0 < \alpha < pi/4.$$

I take its density matrix in the u,d basis
$$\rho = \begin{pmatrix} cos^2(\alpha) & sin(\alpha)cos(\alpha)\\ sin(\alpha)cos(\alpha) & sin^2(\alpha) \end{pmatrix}$$look at
https://www.google.fr/url?sa=t&rct=...f/1905.00917&usg=AOvVaw13xWMPflAeK2lDLLsk7qCo
Eq (5) reads: : $$V = 2 | \rho_{12}|$$
here sin(2 alpha)

we have $$cos^2(\alpha) - sin^2(\alpha) = cos(2 \alpha)$$
and $$2sin(\alpha)cos(\alpha)= (sin(2 \alpha)$$
$$V = 2 | \rho_{12}|$$ here sin(2 alpha)
So $$cos^2(2 \alpha) + sin^2(2 \alpha) = 1$$
it is the Englert equality for pure state $$P^2 + V^2 = 1$$
It is obvious that V comes from the off diagonao elements of the density matrix,
I write a matrix (a diagram) so that the sum of the elements is 1.
the off diagonal a function of $$\rho$$ and the diagonal elements proportional to the diagonal elements $$\rho_{12}$$
$$\begin{bmatrix} cos^2(\alpha)(1- V^2) & \frac{V^2}{2} \\ \frac{V^2}{2} & sin^2(\alpha)(1- V^2) \end{bmatrix}$$
on the graph with a oriented circle, the opposite points would correspond to the diagonal elements and the curved arrows to the off diagonal elements

I had in mind that the entropy should be in it but it does not seem so.

this is for a pure state.
it illustrates a no go sentence for a 2*2 pure matrix
I would like to do the same for the 4*4 pure density matrix for a pair of particles in $$H_1 \otimes H_2$$

look at the link above , they try things like that.

 

[Heidi]

Here i do not talk anymore about entropy.
So I agree completely with what you wrote before.
the thing which is measured is the fringe visibility.

 

[jambaugh]

@jambaugh
Excuse me for this late answer.
I wonder if you are right when you say that entanglement is not
a property of the whole composite system.
there are cases when physics do not let you consider other point of view that the A particle and the B particle. consider that a particle decays in two identical particles neat a black hole. one falls
inside and the other escapes. You have no access to one of them.
it is not a point of view. You receive particles in a density matix state. Am i wrong?

Excuse me also for the late reply. To be clear, by "property of the system" I specifically mean a physical observable. This is not to say entanglement isn't physically meaningful. By analogy time (for non-relativistic systems) and space-time (for relativistic ones) are not observables but rather parameters we use in system descriptions. More aptly entropy is not defined for a physical system. It rather is a property of the statistical description of a class of systems. This property of the description has physical meaning when the aforementioned class of systems corresponds to a given system constrained to be in that class (with a few other caveats.)

Let's take the specific example of a "singleton" state of a pair of spin-1/2 particles with positive total kinetic energy and total momentum zero. To further simplify let's restrain them to move in only one dimension, the x-direction.

In order to describe them as entangled we must choose a labeling variable, an observable we will treat as a classical index in order to differentiate the particles.

We could, for example utilize the sign of their x-momentum and refer to "the left particle" and "the right particle". In so doing we have chosen a frame and in that frame, "yes" they will be entangled i.e. the sharp state of the composite system cannot be factored into sharp spin-state of left particle $\otimes$ sharp spin-state of right particle. Similarly we could index the two particles by, say, their spin-z component and we will again find the description entangled when considering the momentum-states of "the spin up" vs "the spin down" particle.

The question then is whether all such factorizations lead to entanglement the answer is, I assert, "No". My loose argument is this. Consider the density matrix for the composite system. It resides in a 4-dim Hilbert space. Given it is in a "sharp state" that density matrix can be diagonalized to the form:
$$\rho_C=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)$$
This can be factored into a product:
$$\rho_A\otimes\rho_B = \left(\begin{array}{cc} 1& 0 \\ 0 & 0\end{array}\right)\otimes\left(\begin{array}{cc} 1& 0 \\ 0 & 0\end{array}\right)$$

More specifically there is one (in fact many) ways to re-express the composite system as a direct product of two factor systems where both are sharply defined. This is not a complete argument. A full exposition is worth doing on how we utilize a separate observable to allow us to factor the system in a way that the index variable can be treated classically, i.e. the factor systems are not "in a superposition" with respect to it's eigen-basis.

 

[Heidi]

My first aim was to find a visual diagram showing the informatin between the particles. A matrix may be seen as a diagram. Now i come to the idea that the diagram i was looking for is the density matrix (pure or mixed) of the global system.
The problem would be to contruct it from results of measurements.
I know how to do that for the density matrix of a qubit. i measure the probabilities of up in the 3 directions and it is $(Id + p1 \sigma_1 + p2 \sigma_2+ p3 \sigma_3)/2$ with the Pauli matrices.
Can we do like that when we have more than one qubit?

 

[Heidi]

My first aim was to find a visual diagram showing the informatin between the particles. A matrix may be seen as a diagram. Now i come to the idea that the diagram i was looking for is the density matrix (pure or mixed) of the global system.
The problem would be to contruct it from results of measurements.
I know how to do that for the density matrix of a qubit. i measure the probabilities of up in the 3 directions and it is $(Id + p1 \sigma_1 + p2 \sigma_2+ p3 \sigma_3)/2$ with the Pauli matrices.
Can we do like that when we have more than one qubit?

To be more correct the density matrix reads:
$$Id/2 + \langle S_1 \rangle \sigma_1 + \langle S_2 \rangle \sigma_2+ \langle S_3 \rangle \sigma_3$$
We measure the mean values of the spin projection in the 3 directions.
mfff