Is Polarisation Entanglement Possible in Photon Detection?

[stevendaryl]

Let's deal with this first. Where do you see I said there is no unique way how to recover full state? I said there is absolutely no way (unique or not) how you can match prediction worked out from full state to any possible prediction worked out from statistical mixture of separate states in particular case considered.

I think everybody agrees, you cannot (except by guessing) construct the composite states from the mixed states of the parts. The terminology "there is no unique way" just means that there is more than one composite state that corresponds to the pair of mixed states. The composite state is not unique, given the component states. It's just another way of saying that the function from composite states to component states is many-to-one, so the inverse is one-to-many.

 

[zonde]

The terminology "there is no unique way" just means that there is more than one composite state that corresponds to the pair of mixed states. The composite state is not unique, given the component states. It's just another way of saying that the function from composite states to component states is many-to-one, so the inverse is one-to-many.

There certainly is no communication problem with the things you say.

 

[rubi]

If you have a state $\Psi\in\mathcal H_1\otimes\mathcal H_2$, then there exist mixed states $\rho_i$ on $\mathcal H_i$ such that for all operators $A_i:\mathcal H_i\rightarrow\mathcal H_i$, it is true that $\left<\Psi,A_1 \otimes \mathrm{id}_{\mathcal H_2} \Psi\right> = \mathrm{Tr}_{\mathcal H_1}\left(\rho_1 A_1\right)$ and $\left<\Psi,\mathrm{id}_{\mathcal H_1} \otimes A_2 \Psi\right> = \mathrm{Tr}_{\mathcal H_2}\left(\rho_2 A_2\right)$. Then obviously, $\Psi$ is an example of a state on $\mathcal H_1\otimes\mathcal H_2$ such that all predictions worked out from the $\rho_i$ match the predictions of $\Psi$. The correspondence is just given by $A_1\rightarrow A_1\otimes\mathrm{id}_{\mathcal H_2}$ and $A_2\rightarrow\mathrm{id}_{\mathcal H_1} \otimes A_2$.

 

[Dadface]

If a photon is in a superposition, then there is one way to measure the photon (i.e., a measurement basis) where you will get the same outcome 100 percent of the time.

For example, if you have a diagonally polarized photon, it can be expressed as a superposition of horizontal and vertical polarization. If you measure in the horizontal/vertical basis, you will have a 50/50 chance of getting either result.
On the other hand, if you measure in the diagonal/antidiagonal basis, you will measure the diagonal result 100 percent of the time.

If a photon is in a mixed state, then there is no such basis where you can get the same outcome all the time. This for example corresponds to the statistics of unpolarized light, where you get 50/50 chances of either measurement result in all measurement bases.

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?

 

[zonde]

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?

If a photon passes one polarizer then it will pass any other polarizer with the same orientation (if you don't change it's polarization using wave plate or something else).

 

[zonde]

@Simon Phoenix and @stevendaryl I understand what you are saying about combining back mixed states into composite state and many to one relationship. So there is no problem with that. But I didn't try to combine mixed states back into composite state. Instead I took projections for separate components of mixed states and only then tried to combine them back. It seems you somehow missed that step.
So let me try different approach to the problem I see with mixed states.
Tracing out is done in particular polarization basis. But we can change the basis for composite state to different polarization basis and then get corresponding mixed states by tracing out the other part. That way we can get different mixed states that represent one side of composite state.
Are those different mixed states supposed to represent the same physical situation? My answer is that they don't and it implies that composite state to mixed states of components is one to many mapping. But it seems at odds with what Simon said: "what you have is described by a statistical mixture".

 

[rubi]

But I didn't try to combine mixed states back into composite state. Instead I took projections for separate components of mixed states and only then tried to combine them back. It seems you somehow missed that step.

So you're saying that you take completely different states and combine them into a composite state? That's of course not a legal operation. If you change the state in the middle of a calculation, you shouldn't be surprised if you get inconsistent results. Moreover, "taking a component of a mixed state" doesn't make much sense, since there usually is no unique decomposition of a mixed state into projectors.

So let me try different approach to the problem I see with mixed states.
Tracing out is done in particular polarization basis. But we can change the basis for composite state to different polarization basis and then get corresponding mixed states by tracing out the other part. That way we can get different mixed states that represent one side of composite state.

You will get exactly the same mixed state, since the partial trace is a basis independent operation. For a given state $\Psi\in \mathcal H_1\otimes \mathcal H_2$, there is a unique $\rho$ such that for all $A$, you have $\left<\Psi,A\otimes\mathrm{id}_{\mathcal H_2}\Psi\right> = \mathrm{Tr}_{\mathcal H_2}\left(\rho A\right)$. This $\rho$ doesn't depend on any choice of basis.

 

[zonde]

Moreover, "taking a component of a mixed state" doesn't make much sense, since there usually is no unique decomposition of a mixed state into projectors.

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.

 

[Dadface]

Thanks zonde I know that (post 40). What confuses me is how the definition of mixed state applies to a single photon.

I can understand that it's possible to get a mixture of photons where, for example, half are vertically polarised and the other half horizontally polarised. I can understand that if you take one photon from that mixture there's a 50% chance it will be vertically polarised and a 50% chance it will be horizontally polarised. In other words I can understand how the term mixed state can be applied to mixture of photons but not to a single photon. Expressing it differently...when the single photon leaves the mixture completely can we still describe it as being in a mixed state?

 

[zonde]

Thanks zonde I know that. What confuses me is how the definition of mixed state applies to a single photon.

I can understand that its possible to get a mixture of photons where, for example, half are vertically polarised and the other half horizontally polarised. I can understand that if you take one photon from that mixture there's a 50% chance it will be vertically polarised and a 50% chance it will be horizontally polarised. In other words I can understand how the term mixed state can be applied to mixture of photons but not to a single photon.

I can understand it as classical hidden variable of single photon (whether it belongs to one or the other subensemble). Say we take two pulsed lasers so that their pulses do not overlap in time. We polarize on 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. This is certainly statistical mixture.
Ok, but what if we mix two initially coherent beams that are then polarized with orthogonal polarizers and mixed afterwards. Will it be different from my first example? I can't answer this but as I understand rubi's position he says it will be different.

 

[stevendaryl]

@Simon Phoenix and @stevendaryl I understand what you are saying about combining back mixed states into composite state and many to one relationship. So there is no problem with that. But I didn't try to combine mixed states back into composite state. Instead I took projections for separate components of mixed states and only then tried to combine them back. It seems you somehow missed that step.

I don't understand the distinction between what you're saying you didn't try, and what you did try.

So let me try different approach to the problem I see with mixed states.
Tracing out is done in particular polarization basis.

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

 

[stevendaryl]

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.

Just to make it clear what is meant by calling mixed states a "statistical mixture":

1. A pure state $|\psi\rangle$ corresponds to the density operator $\rho = |\psi\rangle \langle \psi|$.
2. The significance of a density matrix is that it gives expected values for observables. Mathematically, if a system is in a state described by a density matrix $\rho$, then the expected value for the observable corresponding to operator $O$ is: $\langle O \rangle = tr(\rho O)$, where $tr$ is the trace operator.
   
3. 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)[/itex], 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.]
   
4. 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).

A mysterious aspect of quantum mechanics is that there is mathematically no difference between a proper mixed density matrix you obtain by taking into account classical uncertainty, and an improper mixed density matrix that you obtain by tracing out one component of a composite density matrix. So an EPR experiment in which Alice's and Bob's particles are correlated, the density matrix that Bob uses for his particle is the same, whether or not you believe that Alice has "collapsed the wave function". If she hasn't collapsed the wave function, then the density matrix is an improper mixture, which you get from tracing out Alice's particle. If she has collapsed the wave function (and Bob hasn't been told what the result was), then the exact same density matrix is interpreted as a proper mixture, which you get from classical uncertainty about which way the wave function collapsed.

The fact that Bob's density matrix is the only thing relevant for Bob conducting local experiments, plus the fact that that matrix doesn't change when Alice "collapses the wave function" is what allows us to say that Alice's measurement has no effect on Bob's measurement.

 

[Simon Phoenix]

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

Yes a mixed state is equivalent to a statistical mixture of pure states. However, you've touched on a subtlety here.

Let's suppose we have a spin-1/2 particle in the mixed state described by |0><0| + |1><1| (the omitted normalization constant is 1/2, but I really must learn LaTex one of these days) where the |0> and |1> are eigenstates of the spin-z operator. Now transform basis to the eigenstates of the spin-x operator which we label as |0*> and |1*>. In this new basis the density operator is |0*><0*| + |1*><1*|. 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. In fact for this particular mixed state it can always be written as |down><down| + |up><up| in any spin direction you choose.

As has been noted above, the trace operation, and expectation values, are completely independent of the basis in which we choose to represent our mixed state, and independent of the basis in which we choose to perform our trace. Normally we try to choose bases that make our calculations simpler, but we could be perverse and choose bases that turn everything into an algebraic goulash

 

[stevendaryl]

Thanks zonde I know that (post 40). What confuses me is how the definition of mixed state applies to a single photon.

I can understand that it's possible to get a mixture of photons where, for example, half are vertically polarised and the other half horizontally polarised. I can understand that if you take one photon from that mixture there's a 50% chance it will be vertically polarised and a 50% chance it will be horizontally polarised. In other words I can understand how the term mixed state can be applied to mixture of photons but not to a single photon. Expressing it differently...when the single photon leaves the mixture completely can we still describe it as being in a mixed state?

Yes. If a photon has a probability $p_1$ of being horizontally polarized, and a probability $p_2$ of being vertically polarized, then that situation is described by the mixed state $p_1 |H\rangle \langle H| + p_2 |V\rangle\langle V|$.

That's one of the strange things about quantum mechanics, that the same mixed state describes two different situations:

1. You have a pure state of a composite system, and you trace out one of the components to get a mixed state for the other.
2. You have a single system whose state is unknown, and you use the probabilities for the various possibilities to construct a mixed state.

 

[vanhees71]

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

 

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