Undergrad Is Polarisation Entanglement Possible in Photon Detection?

[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.

 

[zonde]

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.

I agree that there something wrong with this discussion.
Phrases "state $\rho$" and "mixed state is a density matrix" make a category error. A state is a vector. Density matrix is not a vector or anything similar.

 

[Simon Phoenix]

I don't understand what's going on here, at all.

I'm having the same difficulty. I think what's happening is that there's this feeling that because a component part has come from an entangled pair then somehow this entanglement must be 'reflected' in the 'state' of the component part - so it has to be more than just a classical-like mixture. I think this is the sticking point.

Or not, I'm not totally sure lol.

 

[zonde]

I think what's happening is that there's this feeling that because a component part has come from an entangled pair then somehow this entanglement must be 'reflected' in the 'state' of the component part - so it has to be more than just a classical-like mixture. I think this is the sticking point.

I think one problem is with "component part" of entangled state. As much us I understand what a product space is there can be no independent descriptions for each side ("component parts") within that space. So the closest thing to "component part" that we can have is a trace that gives us density matrix. But such density matrix is not a component of entangled state (a vector).

 

[vanhees71]

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.

That's of course true, but for this you need to entire (in this case the two-photon state). It's not described by the reduced states of either photon.

 

[rubi]

I agree that there something wrong with this discussion.
Phrases "state $\rho$" and "mixed state is a density matrix" make a category error. A state is a vector. Density matrix is not a vector or anything similar.

No that's not correct. If anything deserves the name "state" in quantum mechanics, it's the density matrix. Vectors are ambiguous due to the arbitrary phase factor. If we didn't acknowledge this, we couldn't even explain spin or the Galilei invariance of the hydrogen atom, so it has practical importance to use the term correctly, i.e. not for vectors. In practice, we are often lazy and refer to vectors as states, but it is strictly speaking not correct. Moreover, we could also represent density matrices as vectors if we cared to purify them.

I think one problem is with "component part" of entangled state. As much us I understand what a product space is there can be no independent descriptions for each side ("component parts") within that space. So the closest thing to "component part" that we can have is a trace that gives us density matrix. But such density matrix is not a component of entangled state (a vector).

Well, you understand incorrectly. Everything that can be known about a component of a system is contained in the mixed state computed by the partial trace operation. The situation is completely analogous to the composite system, which is also just a component of an even larger system. Again, I urge you to confront a linear algebra textbook to clear up this misunderstanding. Your reasoning is mathematically erroneous.

 

[vanhees71]

The point is that taking the partial trace over a part of a composite system means that you forget about all of the traced out part. What you get is a statistical operator that describes the statistics for the outcome of measurements on this partial system. By definition you don't care about how it's entangled with the rest of the world, and that's the key issue to resolve all the apparent "problems" with entangled states brought up by EPR and much better a bit later by Einstein himself.

Take the standard example of the polarization entangled two-photon states. Only looking at the polarizations and keeping in mind that it refers to photons measured far distant places A (Alice) and B (Bob), you have the pure entangled state vector
$$|\Psi \rangle=\frac{1}{\sqrt{2}}(|HV \rangle + |VH \rangle),$$
and the state is described by the statistical operator
$$\hat{\rho}_{AB}=|\Psi \rangle \langle \Psi|.$$
The state of the single photons is given by tracing out the other photon, and as shown way up in the thread this leads to
$$\hat{\rho}_{A}=\hat{\rho}_B=\frac{1}{2} \hat{1}.$$
Both photons are perfectly unpolarized. In practice this means A and B measure on an ensemble of such prepaared photons with 50% probability H and 50% probability V polarized photons, and no matter to which direction of the polarization filter this measurement refers. There's this and only this information you can get measuring only the single-photon polarizations, and it doesn't matter, how in detail you prepare this state. They could as well come from a thermal source with a filter concerning the energy (frequency) of the photons.

Only if you make measurements on the two-photon system you can figure out the entanglement. For that you must measure both photons at A and B and make sure to relate only photon pairs which were prepared in the entangled pair. The latter is usually done by doing a coincidence measurement, i.e., A and B keep precise enough time stamps of their registration events and then can figure out the correlations described by the entanglement.

This is important, because with all the wild speculations about differences between states that are described by the same statistical operator QT wouldn't be a consistent description, and you'd have to find a better definition of states. To my knowledge there's not the slightest hint that this is necessary.

 

[kith]

If two systems are entangled, we cannot assign a state vector to the subsystems individually. So we have two choices:
1) Simply accept that the individual subsystem doesn't have a quantum state on its own.
2) Trace out the other subsystem to obtain the reduced density matrix and broaden the concept of what a quantum state is to include it.

@zonde and @Zafa Pi, you don't have to agree with option 2) but you have to acknowledge that this is the standard terminology used in quantum physics. You have both been here for a while and there have been quite a few fruitless discussions which have been at least complicated by terminology issues. Why don't you learn and use the standard terminology in order to discuss the physical content you have in mind?

 

[kith]

Here is also an argument, why the distinction between proper mixed states (obtained by mixing ensembles) and improper mixed states (obtained by tracing out a part of an entangled system) isn't as clear cut as one might think.

If we want to mix two ensembles physically, we need to interact with them somehow to bring them together. How can we be sure that we have obtained a proper mixed state? We can't because it would involve checking correlations with the environment for which it is almost always impossible to measure all the relevant degrees of freedom. And because interaction typically leads to entanglement, we have every reason to expect that we have obtained an improper mixed state.

So if one wants to make this distinction, he needs to outline how to obtain a proper mixed state in the first place.

 

[forcefield]

I think the root cause of confusion here is the simple fact that getting first the measurement result "vertical" and then the measurement result "horizontal" is not, physically, the same as getting first "horizontal" and then "vertical". However, QM (of course) makes the same probabilistic prediction for both sequences. And discussing about (order of) individual measurement results goes beyond QM.

 

[rubi]

Trace out the other subsystem to obtain the reduced density matrix and broaden the concept of what a quantum state is to include it.

I wouldn't say that it is a broadening of the concept of state. If zonde isn't happy with a matrix $\rho=\frac{1}{2}\left(\begin{array}{cc}1 & 0 \\ 0 & 1 \\\end{array}\right)$ on $\mathcal H=\mathbb C^2$, I will just give him $\mathcal H = \mathbb C^4$ and $\left|\Psi\right>=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ 0 \\ 0 \\1 \end{array}\right)$. Instead of calculating expectation values with the trace formula $\mathrm{Tr}(\rho A)$, he can calculate them with the standard formula $\left<\Psi\right|\tilde A\left|\Psi\right>$, where $\tilde A = \left(\begin{array}{cc}A & 0 \\ 0 & A \\\end{array}\right)$ is just the block matrix with $A$ on the diagonal. Of course all of this is completely unnecessary, but if zonde wants a state vector, he gets a state vector.

I think the root cause of confusion here is the simple fact that getting first the measurement result "vertical" and then the measurement result "horizontal" is not, physically, the same as getting first "horizontal" and then "vertical".

We're not talking about sequential measurements here, so that's not an issue.

 

[forcefield]

We're not talking about sequential measurements here, so that's not an issue.

Well, I meant the usual QM measurement, which requires an ensemble.

 

[stevendaryl]

I think one problem is with "component part" of entangled state. As much us I understand what a product space is there can be no independent descriptions for each side ("component parts") within that space. So the closest thing to "component part" that we can have is a trace that gives us density matrix. But such density matrix is not a component of entangled state (a vector).

Well, in quantum mechanics, both vectors and density matrices are referred to as "states". But since a vector is a special case of a density matrix (or corresponds to a special case of a density matrix), we may as well view density matrices as the general notion of a quantum-mechanical state, and pure states (ones that arise from vectors) are a special case. A "state" for quantum mechanics is just a mathematical object that summarizes the statistics for possible measurement results on the system. That broad definition covers density matrices and vectors.

The weird thing about quantum mechanics is that the same density matrix can arise from two very different processes: (1) tracing out the degrees of freedom of another component of a composite system, and (2) doing a classical "mixture" of two pure states to reflect ignorance about which pure state is applicable.

 

[zonde]

If two systems are entangled, we cannot assign a state vector to the subsystems individually. So we have two choices:
1) Simply accept that the individual subsystem doesn't have a quantum state on its own.
2) Trace out the other subsystem to obtain the reduced density matrix and broaden the concept of what a quantum state is to include it.

@zonde and @Zafa Pi, you don't have to agree with option 2) but you have to acknowledge that this is the standard terminology used in quantum physics. You have both been here for a while and there have been quite a few fruitless discussions which have been at least complicated by terminology issues. Why don't you learn and use the standard terminology in order to discuss the physical content you have in mind?

If you describe state using density matrix you are not broadening concept of state, you are changing it. And it can lead to quite a mess if you do not announce that clearly enough.

I tried to look up Ballentine and run across this:
"It allows us to discriminate between the two principal classes of interpretations.
A. A pure state $|\Psi\rangle$ provides a complete and exhaustive description of an individual system. A dynamical variable represented by the operator Q has a value (q, say) if and only if $Q|\Psi\rangle = q|\Psi\rangle$.
B. A pure state describes the statistical properties of an ensemble of similarly prepared systems."

Pay attention that in case A "state" is a state vector but in case B it's a state operator (at the start of the book Ballentine defines "state" as possible statistics of preparation procedure and describes it with a state operator).
So it seems that distinction between "state" as a vector or as an operator comes bundled with different interpretations.

 

[rubi]

If you describe state using density matrix you are not broadening concept of state, you are changing it. And it can lead to quite a mess if you do not announce that clearly enough.

As I have explained in post #115, density matrices and vectors are exactly equivalent. Nobody is changing the concept of state except yourself.

I tried to look up Ballentine and run across this:
"It allows us to discriminate between the two principal classes of interpretations.
A. A pure state $|\Psi\rangle$ provides a complete and exhaustive description of an individual system. A dynamical variable represented by the operator Q has a value (q, say) if and only if $Q|\Psi\rangle = q|\Psi\rangle$.
B. A pure state describes the statistical properties of an ensemble of similarly prepared systems."

Note, that he is talking about pure states in both cases. That's because he is not at all trying to saying what you are trying to put in his mouth. He is concerned with the difference between the Copenhagen interpretation and the ensemble interpretation, which he advocates in his book. Both interpretations work with pure states and mixed states.

Pay attention that in case A "state" is a state vector but in case B it's a state operator

No it's not. It's a "pure state" in both cases and you are free to write it as a matrix or a vector, since both notions are completely equivalent.

(at the start of the book Ballentine defines "state" as possible statistics of preparation procedure and describes it with a state operator).

Ballentine correctly defines a state as a density matrix, just like everybody else, because vectors contain an ambiguous phase factor. That has nothing to do with the quote you cited. According to Ballentine, a pure state is a density matrix (or a ray if you want) in both cases.

So it seems that distinction between "state" as a vector or as an operator comes bundled with different interpretations.

Maybe it seems like that to you, but quantum physicists universally don't agree with your personal opinion. And it's quite despicable that you're trying to put words in the mouth of a respected physicists just in order to defend your lost position.

 

[zonde]

rubi, you are contradicting yourself. In one sentence you write that state vectors and state operators are equivalent but in next sentence you write that vectors contain "ambiguous phase factor" and therefore state operator is correct expression for state and vector is not. So please make up your mind.

It's a "pure state" in both cases and you are free to write it as a matrix or a vector, since both notions are completely equivalent.

Ballentine correctly defines a state as a density matrix, just like everybody else, because vectors contain an ambiguous phase factor.