Undergrad Is Polarisation Entanglement Possible in Photon Detection?

[zonde]

Interesting - have fun trying to prove it is incorrect

Thanks
I will use counter example approach. It should be valid way to disprove general statement by showing that in some particular case this general statement does not hold.

So we will take entangled state:
$$|\Phi \rangle=\frac{1}{\sqrt{2}} (|H_1H_2 \rangle-|V_1V_2 \rangle),$$
for which each part separately is described by mixed states:
$$\hat{\rho_1}=\frac{1}{2} (|H_1 \rangle \langle H_1|+|V_1 \rangle \langle V_1|).$$
$$\hat{\rho_2}=\frac{1}{2} (|H_2 \rangle \langle H_2|+|V_2 \rangle \langle V_2|).$$
as these are statistical mixtures we can take each polarization state as separate subensemble and do the calculations on it alone.
So let us take subensembles $|H_1 \rangle \langle H_1|$ and $|H_2 \rangle \langle H_2|$. Now I would like to recover part of information concerning full state, in particular that for these two subensembles we can establish one to one correspondence between individual photon detections and later calculate correlations.
Next we look at expectation value for polarization measurement of $|H_1 \rangle \langle H_1|$ at an angle of 30^o. I am not so good with operators and matrices so I will simply use Malu's law and it gives us $p_1=\cos^2 \frac{\pi}{6}=0.75$.
Then we calculate expectation value for polarization measurement of $|H_2 \rangle \langle H_2|$ at an angle of -30^o. Again using Malu's law it gives us $p_2=\cos^2 -\frac{\pi}{6}=0.75$.
So without having full information we can't say how big is expectation value for coincidences but we can confidently say that it can't be less than 0.5 from measurements of two subensembles (c=0.75-(1-0.75)). We get the same result by considering subensembles $|V_1 \rangle \langle V_1|$ and $|V_2 \rangle \langle V_2|$. So average expectation value of coincidence rate of both pairs of statistical subensembles too can't be less than 0.5.
However prediction worked out using full entangled state gives us expectation value for coincidences at 60° (30° - -30°) $p=\cos^2 \frac{\pi}{3}=0.25$ that is less than 0.5. So there is no possible way how we could construct full entangled state from statistical subensembles of separate photons for particular case we considered. QED

 

[Simon Phoenix]

So there is no possible way how we could construct full entangled state from statistical subensembles of separate photons for particular case we considered

First off, I'm somewhat baffled by your logic. You appear to be saying that because there is no way to uniquely recover (or construct) the full density operator given only the reduced density operators (which is true, there is no unique way so that the reduced density operators correspond to more than one possible full density operator) then the reduced density operators cannot be a correct description for the individual components in an entangled pair. I do apologize but your logic here completely escapes me.

Secondly, if we have a photon in the state |H><H| + |V><V|, then measurement of linear polarization in any direction will give the result 'horizontal' or 'vertical' (in that direction) with equal probability. There's no way to infer anything about the coincidence counts from the reduced density operators - for that you have to examine correlations in the results, but then you are accessing some of the information about the global state. When you look at the two component parts individually - mathematically represented by the tracing operation - you are 'losing' or neglecting an amount of information equal to the mutual information of the global state. For an entangled system you can recover up to half of this information, but not all, by making measurements on each component part and comparing results (this is a general result for bipartite entangled systems).

 

[zonde]

First off, I'm somewhat baffled by your logic. You appear to be saying that because there is no way to uniquely recover (or construct) the full density operator given only the reduced density operators (which is true, there is no unique way) then the reduced density operators cannot be a correct description for the individual components in an entangled pair. I do apologize but your logic here completely escapes me.

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.

 

[Simon Phoenix]

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

Maybe there's a problem with semantics here, but if I have the density operators
|0><0| + |1><1| for particle 1 and a similar one for particle 2 then I can most certainly construct several possibilities for the global density operator.
|00><00| + |11><11| is one possibility
|0><0| + |1><1| ⊗ |0><0| + |1><1| is another
and so on
What I can't do is, just by looking at experimental results on each component system independently, distinguish between the possibilities. By independently here I mean by not considering any correlation - if we look at correlation we're no longer simply considering just the reduced density operators!

Of course you can't predict the statistics of measurements of the joint properties from consideration of the component parts independently! That's obvious - but it proves nothing more than the whole is more than the sum of its parts. You can't do this with classical correlations either. Tell me, can you construct a unique joint probability distribution P(a,b) from consideration of the marginal distributions P(a) and P(b) alone?

By your argument the impossibility of reconstructing P(a,b) would lead you to say that, therefore, the marginal distributions don't give us the correct statistics for the component systems independently - and are therefore incorrect. This, I hope you will agree, would be an absurd argument.

The reduced density operators are a kind of quantum analogue of the marginal distributions.

 

[stevendaryl]

But this is exactly what I said, no?

I was reacting to the sentence

Ok, but it should mean that we can combine A and B back together

It doesn't mean that. We can't combine A and B back together to get the state of the composite system.

Also, you said:

So if I show that it can't be done then your statement is false, right?

That's also not correct. It's true that it can't be done (the recombination), but that doesn't contradict anything Simon said.

 

[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 $\mathrm{id}=U U^\dagger = U \mathrm{id} U^\dagger$.