Is Polarisation Entanglement Possible in Photon Detection?

[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:

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

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}} & \frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \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}} & \frac{-1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} & \frac{-1}{2} \\ \frac{-1}{2} & \frac{1}{2} \end{array} \right)$

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

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:

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

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}} & \frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \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}} & \frac{-1}{\sqrt{2}} \end{array} \right) = \left( \begin{array} \\ \frac{1}{2} & \frac{-1}{2} \\ \frac{-1}{2} & \frac{1}{2} \end{array} \right)$

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

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:

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

For example, if we choose

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

$|U_z'\rangle = \left( \begin{array} \\ \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{array} \right)$
$|D_z'\rangle = \left( \begin{array} \\ \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array} \right)$
$|U_x'\rangle = \left( \begin{array} \\ 1\\ 0\end{array} \right)$
$|D_x'\rangle = \left( \begin{array} \\ 0\\ 1\end{array} \right)$

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:
$\left( \begin{array} \\p_1 & 0 \\ 0 & p_2 \end{array} \right)$

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

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.

 

[stevendaryl]

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.

Well, it sure seems to be that you're interpreting it in a way that is contrary to the way anyone else, so I was being generous. I guess I should have said: You're wrong.

I may have to leave the box and grab an X to find out more.

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.

Yes, it does. "State" in the sense of density matrices doesn't mean anything other than a statistical summary of possible measurement results.

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.

Why not?

You're aware of all this, so why am I writing it? Because I'm not sure why you wrote your post.

Okay, I'll try not to make the mistake of responding to you again.

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.

You say it's not a matter of semantics, but you're using terminology without defining what you mean. What does it mean to be in one state or another?

You are obfuscating the situation, the photons are not in a mixed state in spite of Alice in her box being unable to tell.

You argued that it's not a matter of semantics, but you seem to be using different meanings for words than anyone else is. It's a mixed state, because there is a mathematical definition of what it means to be in a mixed state, and it satisfies that definition.

You say that I'm obfuscating the situation, but I literally have no idea what you are talking about.

 

[stevendaryl]

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.

What does it mean to be "in a mixed state" to you? You seem to be using phrases with your own private definitions.

 

[stevendaryl]

Something has gone seriously wrong with this discussion. That tracing out one component of a two-component system produces a density matrix that looks like a mixed state is just simply a fact. Is it really a mixed state? Yes--a mixed state is a density matrix that cannot be written in the form $|\psi\rangle \langle \psi|$

There is no disputing these facts. They are simply facts. You can argue about what this implies about interpretations of quantum mechanics, but you can't dispute what's plainly true. I don't understand what's going on here, at all.

 

[rubi]

Facts are not very popular these days...

 

[Simon Phoenix]

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.

Well I won't derive the inequality for you, but I will give you an argument that justifies my comments.

Consider the normal Bell experiment set up. Something like $A \leftarrow S \rightarrow B$ where some source fires off particles to Alice and Bob measuring at 0, 60 and 120. You agree we'll see a violation of the Bell inequality if the source is generating entangled states. We'll assume a source of perfectly entangled particles.

Now although it's usual to assume Alice and Bob are (roughly) equidistant from the source and measure (roughly) simultaneously, it's not necessary to do so to see the violation. Things are set up in the 'usual' fashion so that we can draw some conclusions about local hidden variables. Alice and Bob will still see a violation if Bob stores the particles, makes several cups of tea, and then does his measurements. As long as we can associate Alice's measurement and Bob's measurement on the partner particles we'll still see the violation.

The distance of the source to the 2 parties is also irrelevant (for inequality violation) so we'll still get the inequality violation with the following set up
$$A \leftarrow S \rightarrow \rightarrow \rightarrow \rightarrow \rightarrow B$$

So we now imagine the source to be In Alice's lab. Now what happens when Alice makes a measurement? If she makes a measurement then she'll know, by virtue of the entanglement, what state is on its merry way to Bob. So all Alice has to do is to make a measurement, stop the particle going off to Bob, and now prepare a new single particle (unentangled with anything) in the state that would have gone on to Bob - which she knows, because she's made a measurement.

You should now be able to see that we can drop the entangled source and measurement part altogether. Alice simply prepares particles uniformly at random in one the six possible measurement eigenstates and sends them off to Bob. There'll be a Bell inequality violation between the preparation data of Alice and the measurement data of Bob.

Of course, we can't draw any of the important conclusions about local variables from doing things this way, but it does show us that we don't actually need entanglement to see a violation of the mathematical inequality - and also, if you think about it, does show us the equivalence of the proper/improper mixed states.