I Is Polarisation Entanglement Possible in Photon Detection?

[zonde]

So do we have a statistical mixture of the pure states |0> and |1>, or do we have a statistical mixture of the states |0*> and |1*>?

The answer to this questions is 'yes'

Both are entirely equivalent descriptions of the same mixed state.

Then let's agree that we disagree and leave it at that.

 

[zonde]

• Now, suppose you are uncertain whether the actual state is |\psi_1\rangle or |\psi_2\rangle. For example, suppose someone generated a random number, and with probability p_1, produced a system in state |\psi\rangle, and with probability p_2, produced a system in state |\psi_2\rangle. Then what is the expected value of observable O? This is classical ignorance, and we can use classical reasoning. Letting \rho_1 = |\psi_1\rangle \langle \psi_1| and \rho_2 = |\psi_2\rangle \langle \psi_2|, if there is a probability p_1 of being in state |\psi_1\rangle, in which case the expected value is tr(\rho_1 O), and a probability of p_2 of being in state |\psi_2\rangle, in which case, the expected value is tr(\rho_2 O), then the expected value, not knowing the state is a weighted average: \langle O \rangle = p_1 tr(\rho_1 O) + p_2 tr(\rho_2 O) = tr(\rho O), where \rho = p_1 \rho_1 + p_2 \rho_2. [Note: it's actually only this simple if |\psi_1\rangle and |\psi_2\rangle are orthogonal states. I'm not sure off the top of my head how to account for the possibility that they are overlapping.]
  
• So to incorporate classical probability, due to ignorance, you just create a weighted average of density matrices. It turns out that every density matrix can be written as a weighted average of pure state density matrices (although there can be more than one way to write it that way).

And, do you say that with this classical ignorance approach we can change basis of mixed state so that in this new mixed state we have pure states in this new basis? But what if this mixture is not 50/50 but say 75/25? In diagonal basis it would have to be 50/50 mixture but then measurement in H/V basis for that new mixture won't produce 0.75/0.25 probabilities.

 

[rubi]

Well, if we say that mixed state is statistical mixture of pure states then it makes sense as the information how to split mixed state into two subensembles of pure states is just a classical hidden variable. And Simon in his post #21 is calling mixed states by that name "statistical mixture".
So if you do not agree with that then of course my arguments do not apply.

Simon said that the system is described by a statistical mixture and he is right about that. It doesn't matter, whether the state arises from a classical lack of information or from a partial trace. Both scenarios are described by exactly the same mixed state.

Then let's agree that we disagree and leave it at that.

This isn't something one can legitimately disagree with. Simon is objectively right. It can be proven mathematically.

[Note: it's actually only this simple if |\psi_1\rangle and |\psi_2\rangle are orthogonal states. I'm not sure off the top of my head how to account for the possibility that they are overlapping.]

Just as a side note: It might look suspicious at first, but it also works in this case.

 

[Simon Phoenix]

In diagonal basis it would have to be 50/50 mixture

No - if we begin with a mixed state of the form a|0><0| +b|1><1| then we can say this is equivalent to a statistical mixture of the pure states |0> occurring with probability a and the pure states |1> occurring with probability b = 1-a

This density operator is already diagonal in this basis and the eigenvalues are a and b. If you transform to another basis it will not in general give you a diagonal matrix representation for the operator - but even if it did the eigenvalues (that is the probabilities) would still be a and b.

 

[zonde]

This isn't something one can legitimately disagree with. Simon is objectively right. It can be proven mathematically.

Let's take example I gave in post #45
Say we take two pulsed lasers so that their pulses do not overlap in time. We polarize one beam with H polarizer but the other with V polarizer and then mix them with beamsplitter. Obviously we can tell apart photons from combined beam by looking at detection time of photon.

I suppose you do not suggest that we can describe this setup as a statistical mixture of pure states in arbitrary basis (with states being pure in that basis) and that you can prove it mathematically.

 

[rubi]

I suppose you do not suggest that we can describe this setup as a statistical mixture of pure states in arbitrary basis (with states being pure in that basis) and that you can prove it mathematically.

Yes of course I can prove this. Simon has already done it in a previous post. $2\rho = \left|h\right>\left<h\right|+\left|v\right>\left<v\right|=2\mathrm{id}=\left|h^\prime\right>\left<h^\prime\right|+\left|v^\prime\right>\left<v^\prime\right|$, where the primed basis vectors are arbitrarily rotated.

 

[zonde]

$2\rho = \left|h\right>\left<h\right|+\left|v\right>\left<v\right|=2\mathrm{id}=\left|h^\prime\right>\left<h^\prime\right|+\left|v^\prime\right>\left<v^\prime\right|$, where the primed basis vectors are arbitrarily rotated.

Sorry but this is way too cryptic. I mean the part "2id"

 

[rubi]

Sorry but this is way too cryptic.

That's a standard calculation, there is nothing cryptic about it. But here you have a fully verbose version:
Let $\rho=\frac{1}{2}\left| h\right>\left< h\right|+\frac{1}{2}\left| v\right>\left< v\right|$ and $U$ be an arbitrary unitary operator ($U^\dagger U = U U^\dagger = \mathrm{id}$). Since $\left|h\right>$ and $\left|v\right>$ are orthogonal, the expression for $\rho$ is just the completeness relation of $\mathbb C^2$, multiplied by $\frac{1}{2}$. Hence $\rho = \frac{1}{2}\mathrm{id}$. But then $\rho=\frac{1}{2}U U^\dagger$. We can insert $\mathrm{id}$ between the $U$'s and pull the factor in between as well. Then we can insert $\rho$ again: $\rho=U\frac{1}{2}\mathrm{id}U^\dagger = U\rho U^\dagger = U\left(\frac{1}{2}\left| h\right>\left< h\right|+\frac{1}{2}\left| v\right>\left< v\right|\right)U^\dagger$. So we get $\rho=\frac{1}{2}U\left| h\right>\left< h\right|U^\dagger+\frac{1}{2}U\left| v\right>\left< v\right|U^\dagger$. With the definitions $\left|h^\prime\right>=U\left|h\right>$ and $\left|v^\prime\right>=U\left|v\right>$, we find $\rho=\frac{1}{2}\left| h^\prime\right>\left< h^\prime\right|+\frac{1}{2}\left| v^\prime\right>\left< v^\prime\right|$.

 

[zonde]

That's a standard calculation, there is nothing cryptic about it. But here you have a fully verbose version:
Let $\rho=\frac{1}{2}\left| h\right>\left< h\right|+\frac{1}{2}\left| v\right>\left< v\right|$ and $U$ be an arbitrary unitary operator ($U^\dagger U = U U^\dagger = \mathrm{id}$). Since $\left|h\right>$ and $\left|v\right>$ are orthogonal, the expression for $\rho$ is just the completeness relation of $\mathbb C^2$, multiplied by $\frac{1}{2}$. Hence $\rho = \frac{1}{2}\mathrm{id}$. But then $\rho=\frac{1}{2}U U^\dagger$. We can insert $\mathrm{id}$ between the $U$'s and pull the factor in between as well. Then we can insert $\rho$ again: $\rho=U\frac{1}{2}\mathrm{id}U^\dagger = U\rho U^\dagger = U\left(\frac{1}{2}\left| h\right>\left< h\right|+\frac{1}{2}\left| v\right>\left< v\right|\right)U^\dagger$. So we get $\rho=\frac{1}{2}U\left| h\right>\left< h\right|U^\dagger+\frac{1}{2}U\left| v\right>\left< v\right|U^\dagger$. With the definitions $\left|h^\prime\right>=U\left|h\right>$ and $\left|v^\prime\right>=U\left|v\right>$, we find $\rho=\frac{1}{2}\left| h^\prime\right>\left< h^\prime\right|+\frac{1}{2}\left| v^\prime\right>\left< v^\prime\right|$.

It seems like you are saying $\mathrm{id}=U\mathrm{id}U^\dagger$. Did I get it right?

 

[rubi]

It seems like you are saying $\mathrm{id}=U\mathrm{id}U^\dagger$. Did I get it right?

Yes. $U U^\dagger=\mathrm{id}$ because of unitarity. But applying $U^\dagger$ and then $U$ is the same thing as applying $U^\dagger$, then doing nothing, and then applying $U$. So $\mathrm{id}=U U^\dagger = U \mathrm{id} U^\dagger$.

 

[Simon Phoenix]

Say we take two pulsed lasers so that their pulses do not overlap in time. We polarize one beam with H polarizer but the other with V polarizer and then mix them with beamsplitter

You're getting into much deeper waters here - with pulsed beams you're really going to need a continuum mode description for the fields here. We could do a single mode treatment which is a bit of a simplified model, but which nevertheless retains many salient features. If you do that, then we could model the laser beams as single mode coherent states (pure states). Putting coherent states in the input arms of a beamsplitter gives us coherent states in the output arms. So I don't think this model is going to give you what you think it does.

 

[Zafa Pi]

Definitely not. If you have the situation in which you don't know what the photon's polarization is, then you don't use superpositions, you use mixed states, which are represented by density matrices, not pure states.

The distinction is the difference between column matrices and square matrices. If you use \left( \begin{array}\\ 1 \\ 0\end{array} \right) to represent a horizontally polarized photon, and \left( \begin{array}\\ 0 \\ 1\end{array} \right) to represent a vertically polarized photon, then a superposition of horizontal and vertical photons would be represented by the general column matrix: \left( \begin{array}\\ \alpha \\ \beta \end{array} \right) where |\alpha|^2 + |\beta|^2 = 1. An observable would be represented in this simple model by a 2x2 matrix O. If the photon is in the state U (a column matrix), then the expected value of a measurement of O would be given by: \langle O \rangle = U^\dagger O U.

In contrast, a mixed state consisting of a probability p_1 of being horizontally polarized and p_2 of being vertically polarized would be represented by the 2x2 matrix: \left( \begin{array}\\ p_1 &amp; 0 \\ 0 &amp; p_2 \end{array} \right). If the photon is in the mixed state D (a 2x2 matrix), then the expectation value for a measurement of O would be: tr(D O), where D O means matrix multiplication, and tr means the trace operator (add up the diagonal values of the resulting 2x2 matrix).

You chose "in contrast" to represent the mixed state with the density matrix, but the pure state |U> can also be represented by the density matrix |U><U| = D' . And once again the expected value of measuring with O would be tr(D'O). I admit the calculation in doing it this way could be more cumbersome, but it is a more uniform approach.

 

[jfizzix]

Thanks very much jfizzix. I'm still a bit confused about mixed states and how they apply to single photons. Suppose a measurement is made and the photon is found to have a horizontal polarisation would all later measurements on that same photon still have an evens chance of being either horizontally or vertically polarised?

The short answer is no.
Once a photon is found to have a horizontal polarization, all subsequent measurements of it in the horizontal/vertical basis will give the horizontal result 100 percent of the time.

That being said, it is understandable to think that assigning a mixed state to a single photon makes no sense.
Certainly if every single photon has some definite quantum state, then mixed states could only describe mixtures of photons.
However mixed states can apply to single photons as well.
In particular, if you consider a polarization-entangled pair of photons, then either single photon is described with a mixed state.
This agrees with experiment as well, as polarization-entangled light looks unpolarized when you look at just one half of each entangled pair.

 

[stevendaryl]

You chose "in contrast" to represent the mixed state with the density matrix, but the pure state |U> can also be represented by the density matrix |U><U| = D' . And once again the expected value of measuring with O would be tr(D'O). I admit the calculation in doing it this way could be more cumbersome, but it is a more uniform approach.

Right, you can consider a pure state as a special case of a density matrix.

 

[stevendaryl]

You chose "in contrast" to represent the mixed state with the density matrix, but the pure state |U> can also be represented by the density matrix |U><U| = D' . And once again the expected value of measuring with O would be tr(D'O). I admit the calculation in doing it this way could be more cumbersome, but it is a more uniform approach.

The point of the "in contrast" is that I was responding to a comment where someone said that if you don't know the polarization of a photon, then maybe you should consider it to be in a superposition of all possible polarizations. I responded saying that you should consider it to be in a mixed state, rather than a superposition.

 

[zonde]

Yes. $U U^\dagger=\mathrm{id}$ because of unitarity. But applying $U^\dagger$ and then $U$ is the same thing as applying $U^\dagger$, then doing nothing, and then applying $U$. So $\mathrm{id}=U U^\dagger = U \mathrm{id} U^\dagger$.

Applying $U$ to some matrix rotates the matrix by some angle $\theta$ and then applying $U^\dagger$ will rotate the matrix back by the same angle $\theta$. Hmm, then this is tautology. Tautology of course proves nothing.

 

[Zafa Pi]

\

Why do you say this?

You're not correct, because that's exactly what we have - each of the individual particles in an entangled state is described by a mixed state that is formally identical to a statistical mixture of pure states.

Look at vanhees' post number 8 above - which bit of that post don't you get?

I believe that zonde in post #16 is absolutely correct. I will give a simple model that illustrates the nature of various the types of states mentioned in this thread.

We let the 2D vector [1,0] be the state |0⟩ = |0º⟩, [0,1] = |1⟩ = |90º⟩ (q-information notation).
Both of those are pure states, if we measure |0⟩ with the Pauli Z operator we will get 1 for sure, and if we measure |1⟩ with Z we -1 for sure.
The superposition √½|0⟩ + √½|1⟩ = |45º⟩ when measured with Z yields ±1 with probability ½ each.
However if we measure |45º⟩ with Pauli X we get 1 for sure. |45º⟩ is thus a pure state.
A pure state |ψ⟩ has density matrix |ψ⟩⟨ψ|.
If we have have a pair of states |0⟩ and |1⟩, and each can occur with probability ½, then we have a mixed state.
This state has density matrix ρ = ½|0⟩⟨0| + ½|1⟩⟨1|.
If we measure this state with Z we get ±1 (pr ½ each), if we measure with X we get the same. Whatever operator we measure ρ with we get ±1.
ρ is a mixed state (or rarely called a proper mixed state). An arbitrary mixed state can have an ensemble of many pure states with associated probabilities.
Lastly, if we have a pair (left and right) with joint state |J⟩ = √½(|00⟩ + |11⟩) and we measure the left of the pair with Z we get ±1, we get ±1 with any operator.
It seems as though we are measuring the mixed state ρ, or perhaps the mixed state of |45º⟩ and |-45º⟩ (pr ½ each) which gets the same results.
The same is true for the right of the pair. However neither the left or right is a mixed state.
The reason for this is subtle. If both left and right were mixed states (even with different mixtures) we could have the states separated and prove a Bell inequality.
Yet we know with the proper measurements we can violate that inequality.
Left and right are called improper mixed states. What does this mean? Nothing more than to say they are not mixed stares or pure states.
I think these magical entities should be called stateless or perhaps Karana.

 

[Simon Phoenix]

The same is true for the right of the pair. However neither the left or right is a mixed state.

That's simply wrong - I refer you back to Vanhees' post above

A state is a state, and it's described by a statistical operator. There's no physical distinction between "proper" or "improper" states.

I'm guessing you would (incorrectly) disagree with this statement?

The reason for this is subtle.

And wrong

 

[Zafa Pi]

That's simply wrong - I refer you back to Vanhees' post above
I'm guessing you would (incorrectly) disagree with this statement?
And wrong

Look at post #19 or improper mixed states on line. Do I need to prove a Bell inequality for you when Alice and Bob have mixed states, they are classical you realize.
Indeed vanhees71is mistaken. The physical distinction lies in the entangled correlations that cannot be replicated with "proper" mixed states.

 

[rubi]

Applying $U$ to some matrix rotates the matrix by some angle $\theta$ and then applying $U^\dagger$ will rotate the matrix back by the same angle $\theta$. Hmm, then this is tautology. Tautology of course proves nothing.

No, this is not a tautology. $U U^\dagger = \mathrm{id}$ is a non-trivial condition on a matrix. (Moreover, even if it were a tautology, this would not be a problem. Tautologies are used regularly in proofs. Check out the http://www.people.vcu.edu/~rhammack/BookOfProof/Other.pdf, which uses the tautology $A\vee\neg A =\top$, where $A=x \text{ is irrational}$.) And the fact that you can insert an identity matrix in between any matrix multiplication is something people usually learn within the first five hours of their university studies, so if you don't understand this step, you should first learn basic linear algebra before tackling quantum theory. Linear algebra is essential for quantum theory.

 

[Simon Phoenix]

Do I need to prove a Bell inequality for you when Alice and Bob have mixed states, they are classical you realize.

You're missing the point. The tracing operation is saying "I'm only going to look at the properties of this particle and completely ignore anything to do with the other one"

Whether this particle has been generated as a partner of an entangled state, or simply prepared in a (proper) statistical mixture is of monumental insignificance for the state of the particle considered on its own.

Of course, when we look at correlations between things we can now distinguish which of these 2 situations we have - but we're no longer considering properties of the particles independently of one another - we're looking at joint properties. Absolutely the same physical state for the individual particles in both cases - different global state.

 

[Simon Phoenix]

Do I need to prove a Bell inequality for you when Alice and Bob have mixed states, they are classical you realize.

Consider the following situation. Alice prepares spin-1/2 particles in states chosen uniformly at random from one of the six eigenstates of the spin operators at 0, 60 and 120 degrees. She sends these particles to Bob.

Bob measures spin in one of the directions 0, 60 and 120, chosen at uniformly at random, for each incoming particle.

There is a violation of a Bell inequality between Alice's state preparation data and Bob's measurement data.

There are no correlated particles here, or entanglement, and there's no question here that Alice has been preparing 'proper' mixed states

 

[zonde]

You're missing the point. The tracing operation is saying "I'm only going to look at the properties of this particle and completely ignore anything to do with the other one"

You are missing the point too. Idea behind improper mixed state is that the reasoning about properties of this particle alone is meaningless. So when you talk about properties of this particle alone you actually assume that any entanglement of this particle with the rest of global state is FAPP random and only then you get appearance of mixed state. But if that assumption does not hold (entanglement is not random FAPP) you can't refer to that particle as being in mixed state. And that is the case with one side of entangled pair.

 

[Simon Phoenix]

any entanglement of this particle with the rest of global state is FAPP random

I'm really sorry zonde, but I have absolutely no idea what this statement even means. Is there another way you could express this?

Idea behind improper mixed state is that the reasoning about properties of this particle alone is meaningless.

Well that's certainly not the conventional idea behind the distinction 'proper' and 'improper'.

Whether I have a 'proper' or 'improper' mixture I can certainly perform experiments on one particle alone can't I? So why is it meaningless to reason about the results of such experiments?

Remember that when I've done the trace it means that I'm ignoring anything about the other system, about joint properties, and so on. Or we could say that I simply don't care about anything else other than the bit I'm focusing on. So, of course, it can't matter whether I have an improper or proper mixture - because I'm ignoring everything else - including where I've got my particle from. That's what the trace means. I simply don't care about the 'rest of the world' in this perspective.

Now, of course, if I got curious and asked whether my object of interest had been derived from an entangled state or not - then I would have to examine joint properties to try to answer this - I would then have to care about the rest of the world - and things can no longer be worked out from the individual components.

To try to bring in a classical analogy here - suppose I have a joint distribution for 2 random variables so $P(a,b)$. Now clearly I can 'trace out' one of the variables here and find the marginal distributions $P(a)$ and $P(b)$. What does this mean? Well $P(a)$ can be measured by just looking at the variable $a$ alone, completely ignoring anything about $b$.

Are you saying that this procedure is only legitimate if I don't have correlations? After all, I can construct $P(a)$ which would be the analogue of a 'proper' mixed state. Or I could consider $P(a)$ to have been derived from some $P(a,b)$ so analogous to an 'improper' mixed state.

If you have issues with the quantum version why don't you have issues with the classical version? Following the same logic of your arguments about density operators we would conclude that there are 2 different kinds of $P(a)$ that are not equivalent depending on whether they've been derived from some correlated distribution or not.

I'm sorry but I'm really struggling here to understand what your difficulty is and why you think there's a difference between proper and improper mixtures. Indeed, just as it is sometimes useful to think about an 'improper' mixture as a 'proper' mixture - it goes the other way too; sometimes it is useful to consider a mixed state to be an improper mixture (this is the process of 'purification' where we expand the Hilbert space so instead of working with a mixed state we work with a pure state of a larger space).

 

[Dadface]

Thank you very much jfizzix, stevendaryl and others. It looks like there's an interesting discussion going on around here but my particular inquiry has been answered.

 

[stevendaryl]

Lastly, if we have a pair (left and right) with joint state |J⟩ = √½(|00⟩ + |11⟩) and we measure the left of the pair with Z we get ±1, we get ±1 with any operator.
It seems as though we are measuring the mixed state ρ, or perhaps the mixed state of |45º⟩ and |-45º⟩ (pr ½ each) which gets the same results.
The same is true for the right of the pair. However neither the left or right is a mixed state.
The reason for this is subtle. If both left and right were mixed states (even with different mixtures) we could have the states separated and prove a Bell inequality.

I think this is just a matter of semantics. When people say that Alice's photon "is" a mixed state, they mean that the predictions for the results of Alice's measurements on her photon are given by the single-photon mixed state:

\frac{1}{2} |0\rangle \langle 0| + \frac{1}{2} |1\rangle \langle 1|

This density matrix is a mathematical object that captures exactly your summary:

If we measure this state with Z we get ±1 (pr ½ each), if we measure with X we get the same. Whatever operator we measure ρ with we get ±1.

The same is true for Bob's photon. So mathematically, we can capture this situation by describing Alice's photon as being in a mixed state described by a particular density matrix, and Bob's photon as being in a mixed state described by another density matrix (actually, they're both the same).

Now, beyond asking what Alice's statistics are and what Bob's statistics are, we can also ask how Alice's results are correlated with Bob's result. The answer is that you can't determine that from their separate density matrices. Information about correlations is lost when you compute density matrices for components from the pure state for the composite system. We all agree with that.

Your conclusion, that those correlations prove that the photons are not really in mixed states, seems to be a matter of you reading more into the phrase "mixed state" than is actually meant. It means nothing other than a way to summarize the statistics for measurements on that particle.

 

[zonde]

I'm really sorry zonde, but I have absolutely no idea what this statement even means. Is there another way you could express this?

Let's say we have entanglement between two ensembles A and B. Now while member particles of one ensemble A are part of a single beam that over time is exposed to the same local environment member particles of the other ensemble are spreading out and over time are exposed to vastly different local environments. This is how I understand random entanglement.

Well that's certainly not the conventional idea behind the distinction 'proper' and 'improper'.

Whether I have a 'proper' or 'improper' mixture I can certainly perform experiments on one particle alone can't I? So why is it meaningless to reason about the results of such experiments?

Of course it is not meaningless to reason about the results of such experiments. But you are not talking about measurements only. You are putting forward particular model that is behind these measurements. The problem is with this model.

Remember that when I've done the trace it means that I'm ignoring anything about the other system, about joint properties, and so on. Or we could say that I simply don't care about anything else other than the bit I'm focusing on. So, of course, it can't matter whether I have an improper or proper mixture - because I'm ignoring everything else - including where I've got my particle from. That's what the trace means. I simply don't care about the 'rest of the world' in this perspective.

Yes, I understand that. What I say is that there is "influence" there that you are ignoring. And when you say that the model you propose is adequate (and you can ignore that "influence") I say that it is adequate only when that "influence" follows certain pattern. And this is not the case with entangled particles.
Of course the idea about this "influence" comes from the very case we are discussing so the arguments have to be independent from existence/non existence of such "influence". And therefore arguments against your model come from different side. In particular the argument is that it is inconsistent. And for that argument to work we have to look at the situation from two different sides.

To try to bring in a classical analogy here - suppose I have a joint distribution for 2 random variables so $P(a,b)$. Now clearly I can 'trace out' one of the variables here and find the marginal distributions $P(a)$ and $P(b)$. What does this mean? Well $P(a)$ can be measured by just looking at the variable $a$ alone, completely ignoring anything about $b$.

Are you saying that this procedure is only legitimate if I don't have correlations? After all, I can construct $P(a)$ which would be the analogue of a 'proper' mixed state. Or I could consider $P(a)$ to have been derived from some $P(a,b)$ so analogous to an 'improper' mixed state.

If you have issues with the quantum version why don't you have issues with the classical version? Following the same logic of your arguments about density operators we would conclude that there are 2 different kinds of $P(a)$ that are not equivalent depending on whether they've been derived from some correlated distribution or not.

But this analogue is not adequate. You read $P(a)$ and $P(b)$ directly. You do not propose model of "a" and "b" inferred from number of different readings (using different external parameters) on identically prepared "a"s and "b"s.

I'm sorry but I'm really struggling here to understand what your difficulty is

I'm not sure I believe you. It's in your post #48 and my answer in post #51.
You are claiming that it is fine to look at two mutually exclusive (as I see) descriptions as a single consistent description. Obviously to view some model as inconsistent I'm relying on the (hopefully) shared meaning of concept "inconsistent". And I'm not fine with giving up the concept of "inconsistent".

 

[stevendaryl]

You are missing the point too. Idea behind improper mixed state is that the reasoning about properties of this particle alone is meaningless.

Why do you say that? The improper mixed state for Alice's particle that you get from tracing over Bob's particle is a precise mathematical summary of the statistics for measurements performed by Alice. It's perfectly meaningful. It does not tell you anything about correlations between Alice's results and Bob's results, but it's not intended to do that.

So when you talk about properties of this particle alone you actually assume that any entanglement of this particle with the rest of global state is FAPP random and only then you get appearance of mixed state. But if that assumption does not hold (entanglement is not random FAPP) you can't refer to that particle as being in mixed state. And that is the case with one side of entangled pair.

What do you mean "you can't refer to that particle as being in a mixed state"? It's just a definition. The mixed state for Bob's particle is simply a mathematical object that summarizes the statistics for Bob's measurements on that particle.

 

[zonde]

The mixed state for Bob's particle is simply a mathematical object that summarizes the statistics for Bob's measurements on that particle.

Do you agree that two mathematical objects $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}\frac{1}{\sqrt2}&-\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}&\frac{1}{\sqrt2}\end{pmatrix}$ as written in the same basis are two different mathematical objects?

 

[stevendaryl]

Of course it is not meaningless to reason about the results of such experiments. But you are not talking about measurements only. You are putting forward particular model that is behind these measurements. The problem is with this model.

What model are you talking about? The mathematics of density matrices, traces, and so forth is simply mathematical tools for calculating probabilities for measurement results. It's not a "model" in the sense of an interpretation of the mathematics, it's just the mathematics itself. (Well, I guess it has an operational interpretation, in the sense that "If a system is described by density matrix \rho, and you perform a measurement of an observable corresponding to the operator O, then the expectation value of your result will be tr(\rho O))

What I say is that there is "influence" there that you are ignoring. And when you say that the model you propose is adequate (and you can ignore that "influence") I say that it is adequate only when that "influence" follows certain pattern. And this is not the case with entangled particles.

As far as measurements performed by Bob on his particle, the density matrix corresponding to the improper mixed state for his particle predicts exactly the statistics for the results. Whatever is going on with Alice's particle has no impact on those statistics. So what is your notion of "adequate" under which the improper density matrix is inadequate?

 

[stevendaryl]

Do you agree that two mathematical objects $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}\frac{1}{\sqrt2}&-\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}&\frac{1}{\sqrt2}\end{pmatrix}$ as written in the same basis are two different mathematical objects?

What are you talking about? Yes, those are certainly two different matrices. What does that have to do with the claim that for Bob's measurements of his particle, the density matrix \frac{1}{2} |0\rangle \langle 0| + \frac{1}{2} |1\rangle \langle 1| predicts the statistics for his results?

 

[zonde]

What are you talking about? Yes, those are certainly two different matrices. What does that have to do with the claim that for Bob's measurements of his particle, the density matrix \frac{1}{2} |0\rangle \langle 0| + \frac{1}{2} |1\rangle \langle 1| predicts the statistics for his results?

Oh, I just remembered what you said in post #46 so I thought I would ask you as you are around.

Ah! I love it when an issue has a simple, definitive resolution! The trace operation is independent of basis.

As I understand trace operation can produce mixed states corresponding to both these matrices depending on the basis in which we write entangled state.

 

[stevendaryl]

Oh, I just remembered what you said in post #46 so I thought I would ask you as you are around.

As I understand trace operation can produce mixed states corresponding to both these matrices depending on the basis in which we write entangled state.

No, you understand incorrectly. The tracing operation is independent of basis.

 

[zonde]

No, you understand incorrectly. The tracing operation is independent of basis.

So the tracing operation can produce mixed state corresponding to only one of these two martices, right?

 

[stevendaryl]

So the tracing operation can produce mixed state corresponding to only one of these two martices, right?

Yes. It would be worth going through this yourself, but here's my calculation:

The pure spin state for an anti-correlated pair of spin-1/2 particles is: \frac{1}{\sqrt{2}} (|U_z\rangle |D_z\rangle - |D_z\rangle |U_z\rangle). The corresponding density matrix is

\sum_{a i b j} \rho_{a i b j} |a\rangle |i \rangle \langle b| \langle j|

where a, b refers to the first particle, and i, j refers to the second particle, and all the indices range over the possibilities U_z, D_z, and where the nonzero coefficients \rho_{a i b j} are given by:

• \rho_{U_z D_z U_z D_z} = \rho_{D_z U_z D_z U_z} =\frac{1}{2}
• \rho_{U_z D_z D_z U_z} = \rho_{D_z U_z U_z D_z} =-\frac{1}{2}

The reduced matrix for particle 2 is:

\sum_{i j} \rho_{i j} |i\rangle \langle j|

where \rho_{ij} \equiv \sum_{a} \rho_{a i a j}. The nonzero values for \rho_{ij} are:

• \rho_{U_z U_z} = \frac{1}{2}
• \rho_{D_z D_z} = \frac{1}{2}

So \rho = \frac{1}{2} |U_z\rangle \langle U_z| + \frac{1}{2} |D_z\rangle \langle D_z|

Using the usual two-component spinors: |U_z\rangle = \left( \begin{array} \\ 1 \\ 0 \end{array} \right) and |D_z\rangle = \left( \begin{array} \\ 0 \\ 1 \end{array} \right), this means:

\rho = \ \left( \begin{array} \\ \frac{1}{2} &amp; 0 \\ 0 &amp; \frac{1}{2} \end{array} \right).

Now, go through the whole thing again using the basis

|U_x\rangle = \frac{1}{\sqrt{2}} (|U_z\rangle + |D_z\rangle)
|D_x\rangle = \frac{1}{\sqrt{2}} (|U_z\rangle - |D_z\rangle)

In this basis, the pure state is \frac{1}{\sqrt{2}}(|U_x\rangle |D_x\rangle - |D_x\rangle |U_x\rangle). (You can work this out.) This is exactly the same as for the z-basis, except the names of the states have changed.

The composite density matrix is analogous to the case for the z-basis. Then when you perform the trace, you end up with the reduced matrix:

\rho = \frac{1}{2} |U_x\rangle \langle U_x| + \frac{1}{2} |D_x\rangle \langle D_x|

Using the spinor representation: |U_x\rangle = \left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) and |D_x\rangle = \left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{array} \right), this means:

|U_x\rangle \langle U_x| = \left( \begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) \left( \begin{array} \\ \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} &amp; \frac{1}{2} \\ \frac{1}{2} &amp; \frac{1}{2} \end{array} \right)|D_x\rangle \langle D_x| = \left( \begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{array} \right) \left( \begin{array} \\ \frac{1}{\sqrt{2}} &amp; \frac{-1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} &amp; \frac{-1}{2} \\ \frac{-1}{2} &amp; \frac{1}{2} \end{array} \right)

\rho = \frac{1}{2} |U_x\rangle \langle U_x| + \frac{1}{2} |D_x\rangle \langle D_x| = \left( \begin{array} \\ \frac{1}{2} &amp; 0 \\ 0 &amp; \frac{1}{2} \end{array} \right)

You end up with the same reduced density matrix, regardless of which basis you work with.

 

[zonde]

Yes. It would be worth going through this yourself, but here's my calculation:

The pure spin state for an anti-correlated pair of spin-1/2 particles is: \frac{1}{\sqrt{2}} (|U_z\rangle |D_z\rangle - |D_z\rangle |U_z\rangle). The corresponding density matrix is

\sum_{a i b j} \rho_{a i b j} |a\rangle |i \rangle \langle b| \langle j|

where a, b refers to the first particle, and i, j refers to the second particle, and all the indices range over the possibilities U_z, D_z, and where the nonzero coefficients \rho_{a i b j} are given by:

• \rho_{U_z D_z U_z D_z} = \rho_{D_z U_z D_z U_z} =\frac{1}{2}
• \rho_{U_z D_z D_z U_z} = \rho_{D_z U_z U_z D_z} =-\frac{1}{2}

The reduced matrix for particle 2 is:

\sum_{i j} \rho_{i j} |i\rangle \langle j|

where \rho_{ij} \equiv \sum_{a} \rho_{a i a j}. The nonzero values for \rho_{ij} are:

• \rho_{U_z U_z} = \frac{1}{2}
• \rho_{D_z D_z} = \frac{1}{2}

So \rho = \frac{1}{2} |U_z\rangle \langle U_z| + \frac{1}{2} |D_z\rangle \langle D_z|

Using the usual two-component spinors: |U_z\rangle = \left( \begin{array} \\ 1 \\ 0 \end{array} \right) and |D_z\rangle = \left( \begin{array} \\ 0 \\ 1 \end{array} \right), this means:

\rho = \ \left( \begin{array} \\ \frac{1}{2} &amp; 0 \\ 0 &amp; \frac{1}{2} \end{array} \right).

Now, go through the whole thing again using the basis

|U_x\rangle = \frac{1}{\sqrt{2}} (|U_z\rangle + |D_z\rangle)
|D_x\rangle = \frac{1}{\sqrt{2}} (|U_z\rangle - |D_z\rangle)

In this basis, the pure state is \frac{1}{\sqrt{2}}(|U_x\rangle |D_x\rangle - |D_x\rangle |U_x\rangle). (You can work this out.) This is exactly the same as for the z-basis, except the names of the states have changed.

The composite density matrix is analogous to the case for the z-basis. Then when you perform the trace, you end up with the reduced matrix:

\rho = \frac{1}{2} |U_x\rangle \langle U_x| + \frac{1}{2} |D_x\rangle \langle D_x|

Using the spinor representation: |U_x\rangle = \left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) and |D_x\rangle = \left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{array} \right), this means:

|U_x\rangle \langle U_x| = \left( \begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) \left( \begin{array} \\ \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} &amp; \frac{1}{2} \\ \frac{1}{2} &amp; \frac{1}{2} \end{array} \right)|D_x\rangle \langle D_x| = \left( \begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{array} \right) \left( \begin{array} \\ \frac{1}{\sqrt{2}} &amp; \frac{-1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} &amp; \frac{-1}{2} \\ \frac{-1}{2} &amp; \frac{1}{2} \end{array} \right)

\rho = \frac{1}{2} |U_x\rangle \langle U_x| + \frac{1}{2} |D_x\rangle \langle D_x| = \left( \begin{array} \\ \frac{1}{2} &amp; 0 \\ 0 &amp; \frac{1}{2} \end{array} \right)

You end up with the same reduced density matrix, regardless of which basis you work with.

But your choice to write
$|U_z\rangle$ as $\left( \begin{array} \\ 1 \\ 0 \end{array} \right)$ and $|D_z\rangle$ as $\left( \begin{array} \\ 0 \\ 1 \end{array} \right)$
and
$|U_x\rangle$ as $\left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right)$ and $|D_x\rangle$ as $\left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{array} \right)$
is arbitrary. I could swap the labels "z" and "x" and I would have different mathematical objects describing the same thing. So what determines particular representation? It is your choice of eigenbasis, right? And what kind of transformation do you use to get from one representation to the other? It's rotation, right?

 

[stevendaryl]

But your choice to write
$|U_z\rangle$ as $\left( \begin{array} \\ 1 \\ 0 \end{array} \right)$ and $|D_z\rangle$ as $\left( \begin{array} \\ 0 \\ 1 \end{array} \right)$
and
$|U_x\rangle$ as $\left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right)$ and $|D_x\rangle$ as $\left(\begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{array} \right)$
is arbitrary. I could swap the labels "z" and "x" and I would have different mathematical objects describing the same thing. So what determines particular representation? It is your choice of eigenbasis, right? And what kind of transformation do you use to get from one representation to the other? It's rotation, right?

Yes, the choice to represent |U_z\rangle as \left( \begin{array} \\ 1 \\ 0 \end{array} \right) was arbitrary. Different representations are related by unitary matrices. For any unitary 2x2 matrix U, you can let:

|U_z&#039;\rangle \equiv U \left( \begin{array} \\ 1 \\ 0 \end{array} \right)
|D_z&#039;\rangle \equiv U \left( \begin{array} \\ 0 \\ 1 \end{array} \right)
|U_x&#039;\rangle \equiv U \left( \begin{array} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array} \right)
|D_x&#039;\rangle \equiv U \left( \begin{array} \\ \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{array} \right)

For example, if we choose

U = \left( \begin{array} \\ \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} \end{array} \right), then in this representation:

|U_z&#039;\rangle = \left( \begin{array} \\ \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{array} \right)
|D_z&#039;\rangle = \left( \begin{array} \\ \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array} \right)
|U_x&#039;\rangle = \left( \begin{array} \\ 1\\ 0\end{array} \right)
|D_x&#039;\rangle = \left( \begin{array} \\ 0\\ 1\end{array} \right)

Then your density matrix would be changed from \rho to U \rho U^\dagger. But in our case, \rho is \frac{1}{2} I where I is the 2x2 identity matrix, and so U \rho U^\dagger = \frac{1}{2} U U^\dagger = \frac{1}{2} I.

The representation doesn't change.

 

[Simon Phoenix]

You are putting forward particular model that is behind these measurements. The problem is with this model.

First off, I'm only presenting the standard view of density operators you'll find in any good modern QM textbook. It's not a 'model' I'm putting forward, but the standard treatment of component parts of a larger system in QM, when we're only interested in the properties of the component parts.

Let me attempt to state the position as clearly as I can. If I have a quantum system $A$ described by a mixed state then I can either view this is a proper or improper mixture. In other words the descriptions
(1) mixed state = statistical mixture of pure states (proper)
(2) mixed state = result of tracing operation over some larger system (improper)
are entirely equivalent as far as the properties of $A$ are concerned.

The descriptions (1) and (2) have exactly the same mathematical form - they yield precisely the same mathematical expression for the state of $A$

There are no experiments I can do on $A$ alone that will allow me to distinguish between description (1) and description (2)

You say that description (1) as a means to understand the properties of $A$ alone is 'inconsistent' when we actually have an improper mixture. How so? What experimental consequences are there? How do we demonstrate this alleged inconsistency? How does it manifest itself in the mathematical description?

 

[zonde]

Then your density matrix would be changed from \rho to U \rho U^\dagger. But in our case, \rho is \frac{1}{2} I where I is the 2x2 identity matrix, and so U \rho U^\dagger = \frac{1}{2} U U^\dagger = \frac{1}{2} I.

The representation doesn't change.

I am lost. With mixed state we understand classical combination of (let's say) orthogonal pure states. Pure states are rays (sets of vectors with different complex phase). So a mixed state is combination of two sets of vectors. Vectors certainly can be rotated. How it comes that density matrix does not change under rotation?
Hmm, maybe it would make more clear if I would look at unbalanced classical combination of pure states.

 

[stevendaryl]

I am lost. With mixed state we understand classical combination of (let's say) orthogonal pure states. Pure states are rays (sets of vectors with different complex phase). So a mixed state is combination of two sets of vectors. Vectors certainly can be rotated. How it comes that density matrix does not change under rotation?
Hmm, maybe it would make more clear if I would look at unbalanced classical combination of pure states.

Yes, it's an oddity of density matrices that an equal mixture of "spin-up in the z-direction" and "spin-down in the z-direction" gives the same spin matrix as "spin-up in the x-direction" and "spin-down in the x-direction".

Let me work out a different density matrix. Suppose that there is a chance p_1 of being spin-up and a chance p_2 of being spin-down (in the z-direction). In the basis where spin in the z-direction is diagonal, this corresponds to the density matrix:
\left( \begin{array} \\p_1 &amp; 0 \\ 0 &amp; p_2 \end{array} \right)

Now, if you switch to the basis where spin in the x-direction is diagonal, this would correspond to the density matrix: \left( \begin{array} \\ \frac{1}{2} (p_1 + p_2) &amp; \frac{1}{2} (p_1 - p_2) \\ \frac{1}{2} (p_2 - p_1) &amp; \frac{1}{2} (p_1 + p_2) \end{array} \right)

So you get the same density matrix you started with in the special case p_1 = p_2.

 

[rubi]

You can also decompose density matrices of more complicated mixtures in several ways. Here is an example:
Let $\left|\psi_1\right> = \left|h\right>$ and $\left|\psi_2\right> = \frac{1}{\sqrt{2}}\left(\left|h\right>+\left|v\right>\right)$. Then $\rho_1=\frac{1}{2}\left|\psi_1\right>\left<\psi_1\right| + \frac{1}{2}\left|\psi_2\right>\left<\psi_2\right| =\frac{1}{4}\left(\begin{array}{cc}3 & 1 \\ 1 & 3 \\\end{array}\right)$.
Now consider $\left|\phi_1\right> = \frac{1+\sqrt{2}}{\sqrt{2 \left(2+\sqrt{2}\right)}}\left|h\right>+\frac{1}{\sqrt{2 \left(2+\sqrt{2}\right)}}\left|v\right>$ and $\left|\phi_2\right>=\frac{1-\sqrt{2}}{\sqrt{4-2 \sqrt{2}}}\left|h\right>+\frac{1}{\sqrt{4-2 \sqrt{2}}}\left|v\right>$. If we define $\rho_2=\frac{1}{4} \left(2+\sqrt{2}\right)\left|\phi_1\right>\left<\phi_1\right| + \frac{1}{4} \left(2-\sqrt{2}\right)\left|\phi_2\right>\left<\phi_2\right|$, we also find $\rho_2 = \frac{1}{4}\left(\begin{array}{cc}3 & 1 \\ 1 & 3 \\\end{array}\right)$. So we have $\rho_1 = \rho_2$, but both arise from completely different mixtures. (Homework: Confirm the calculation!)

 

[zonde]

Yes, it's an oddity of density matrices that an equal mixture of "spin-up in the z-direction" and "spin-down in the z-direction" gives the same spin matrix as "spin-up in the x-direction" and "spin-down in the x-direction".

Well, there might be one reasonable explanation for that oddity. If a density matrix does not represent mixed state but rather some kind of generalized measurement of mixed state then everything falls into place. Obviously measurements of equal mixture of "spin-up_z" and "spin-down_z" gives the same probabilities for any measurement as equal mixture of "spin-up_x" and "spin-down_x".
And measurements don't change with the change of basis.

 

[Zafa Pi]

I think this is just a matter of semantics. When people say that Alice's photon "is" a mixed state, they mean that the predictions for the results of Alice's measurements on her photon are given by the single-photon mixed state:

12|0⟩⟨0|+12|1⟩⟨1|12|0⟩⟨0|+12|1⟩⟨1|\frac{1}{2} |0\rangle \langle 0| + \frac{1}{2} |1\rangle \langle 1|

It is just a matter of semantics is a cop out. Whether a photon is in a mixed state or not (improper mixed state) isn't up for multiple interpretations.
In my post (#67) I said:
"Lastly, if we have a pair (left and right) with joint state |J⟩ = √½(|00⟩ + |11⟩) and we measure the left of the pair with Z we get ±1, we get ±1 with any operator.
It seems as though we are measuring the mixed state ρ, or perhaps the mixed state of |45º⟩ and |-45º⟩ (pr ½ each) which gets the same results."
So you must be aware that I am aware that by Alice merely measuring her own photon can't tell the difference of whether she has one from |J⟩ or ρ. This emphatically does not imply her photon has state ρ or or any other mixed state as Simon and vanhees claim. And you seem to as well.

So mathematically, we can capture this situation by describing Alice's photon as being in a mixed state described by a particular density matrix, and Bob's photon as being in a mixed state described by another density matrix (actually, they're both the same).

If I am in a "box" [Simon} and all I have to measure with is Z and find that measuring a sequence of photons in the same state and get results as if they had state ρ does not mean they are in that state. I may have to leave the box and grab an X to find out more. If Alice is in a room with an iron blob and no other ferromagnetic material and she says it's a magnet. Others in the room say it isn't because it can't pick anything up, for our purposes it's not a magnet because it doesn't behave like one. Alice says let me touch it to Bob's blob. They say, that doesn't matter right here and now it's not a magnet. Well if she were able to have it interact with Bob's blob and it stuck the others would be wrong.

You're aware of all this, so why am I writing it? Because I'm not sure why you wrote your post. From your post #19 it clear you know the distinction between a photon from |J⟩ or ρ. why couldn't you say so? You know that if others insist there is no distinction between mixed states and improper mixed they are wrong. There is no way that mixed states will be able to replicate the the combined correlations from |J⟩.

When you say:
"Your conclusion, that those correlations prove that the photons are not really in mixed states, seems to be a matter of you reading more into the phrase "mixed state" than is actually meant. It means nothing other than a way to summarize the statistics for measurements on that particle."
You are obfuscating the situation, the photons are not in a mixed state in spite of Alice in her box being unable to tell.

 

[Zafa Pi]

Consider the following situation. Alice prepares spin-1/2 particles in states chosen uniformly at random from one of the six eigenstates of the spin operators at 0, 60 and 120 degrees. She sends these particles to Bob.

Bob measures spin in one of the directions 0, 60 and 120, chosen at uniformly at random, for each incoming particle.

There is a violation of a Bell inequality between Alice's state preparation data and Bob's measurement data.

There are no correlated particles here, or entanglement, and there's no question here that Alice has been preparing 'proper' mixed states

I am familiar with a common Bell inequality where Alice and Bob each have 3 options. However that scenario requires that Alice and Bob's results will agree when they both select the same option (in your case the same observable). Unfortunately your model doesn't satisfy that condition.
I feel reasonably certain that you will not agree, and I won't comment further unless you derive the the particular Bell inequality you are referring to.

 

[rubi]

By that logic, it wouldn't make sense to say that the composite system is in the EPRB state either, since of course, the EPRB state is part of a much larger system called the "universe". If you were consistent, you would have to reject any mention of the phrase "the system is in the state $\rho$" unless the state of the whole universe is considered. And even then, it is in principle possible that parts of the universe might be entangled with another inaccessible parallel universe. Luckily, that's not how it works. It is exactly as valid to say that Alice's particle is in a mixed state as to say that the composite Alice/Bob system is in the EPRB state. Whenever the statistics of a particular system in consideration is consistent with a state $\rho$, we say that the particular system is in the state $\rho$.

 

[Zafa Pi]

Whenever the statistics of a particular system in consideration is consistent with a state ρρ\rho, we say that the particular system is in the state ρρ\rho

So if I measure a sequence of photons in the same state with Z and get statistics consistent the state ρ (from my post #67) the photons must be in that state rather than in the state √½(|0⟩ + |1⟩)?

 

[rubi]

So if I measure a sequence of photons in the same state with Z and get statistics consistent the state ρ (from my post #67) the photons must be in that state rather than in the state √½(|0⟩ + |1⟩)?

If you only measure one observable, you can't capture enough information to reconstruct the state completely. In that case, several states can be consistent with your observation. If you want a more accurate representation of the state, you would have to perform a quantum tomography. If you do that, you will find the system to be in the mixed state computed by the partial trace operation.

 

[Zafa Pi]

If you only measure one observable, you can't capture enough information to reconstruct the state completely. In that case, several states can be consistent with your observation. If you want a more accurate representation of the state, you would have to perform a quantum tomography. If you do that, you will find the system to be in the mixed state computed by the partial trace operation.

That was my point. When you say:

Whenever the statistics of a particular system in consideration is consistent with a state ρρ\rho, we say that the particular system is in the state ρρ\rho

I only have a problem with the word "we".

 

[rubi]

That was my point.

So your point was that Alice's particle is in the mixed state computed by the partial trace operation? Because that's what you get when you measure the state properly using quantum tomography. It sounded to me like you were rejecting this idea.

I only have a problem with the word "we".

By "we", I was referring to quantum physicists. If the statistics of a (sub-)system (of the multiverse) is consistent with some state $\rho$, quantum physicists say that it is in the state $\rho$. $\rho$ might be a pure state or a mixed state and is uniquely determined if you measure a tomographically complete set of observables for the particular (sub-)system (of the multiverse).

 

[Zafa Pi]

The bold part:

I think this is the central issue of debate in this thread. Simon, nubi, vanhees, and perhaps stevendayrl insist that if Alice in her box finds her measurements are consistent with a mixture then the state of what she is measuring is a mixture. They thus conclude that the state of entangled photons are mixed states, which are at the same time classical. That's a lot of intelligent weight against our position. I glad we don't live in the time of Giordano Bruno.
Of the things you've said that I've understood I agree, with the exception of the spelling of Malus. Keep up the fight I'm bowing out.