Entanglement & Superposition Probabilities

[bulx]

TL;DR Summary

Is it possible to prepare an entangled pair of particles such that their states are known to be opposite/orthogonal but their superposition has a higher likelihood of one of those states?

I cannot find a clear answer on the following beginner’s question on some QM fundamentals:

Suppose we have two particles, A and B. Let’s say we generated these as (or otherwise entangled them as) an entangled pair with opposite/orthogonal states. Perhaps horizontally and vertically polarized photons.

I am reading that it is possible to “prepare” this entangled pair with a probability of being in a certain state. Said differently: that we can filter (or otherwise prepare) pairs of entangled particles based on the probability of the state of their superposition.

Is this correct? It is possible to prepare a pair of entangled particles (e.g. vertically and horizontally polarized photon pair) such that their superposition probability of being in a certain state (e.g. vertically polarized) is high (e.g. >50%)?

 

[PeterDonis]

Let’s say we generated these as (or otherwise entangled them as) an entangled pair with opposite/orthogonal states. Perhaps horizontally and vertically polarized photons.

No. If one photon is horizontally polarized and the other is vertically polarized, they aren't entangled.

The entangled state I think you mean here is the singlet state, where the polarizations are known to be opposite (so total spin is zero), but the polarizations of both photons individually are completely uncertain (so a measurement of either one has a 50-50 chance of either polarization).

I am reading that it is possible to “prepare” this entangled pair with a probability of being in a certain state.

You have already specified the state: the singlet state. See above. That is the only state that is consistent with your description.

Said differently: that we can filter (or otherwise prepare) pairs of entangled particles based on the probability of the state of their superposition.

I'm not sure what you mean by this. Do you have a reference for where you got this from?

 

[Nugatory]

I can write down the state $\frac{\sqrt{3}}{2}|HV\rangle+\frac{1}{2}|VH\rangle$ which is an entangled state with unequal probabilities, but I don't know of any reasonable way of preparing that state.

 

[Hill]

I can write down the state $\frac{\sqrt{3}}{2}|HV\rangle+\frac{1}{2}|VH\rangle$ which is an entangled state with unequal probabilities, but I don't know of any reasonable way of preparing that state.

If we can prepare a single qubit state $\frac{\sqrt{3}}{2}|0\rangle+\frac{1}{2}|1\rangle$ then using Controlled-NOT gate ($C_{NOT}$) we get:
$C_{NOT}((\frac{\sqrt{3}}{2}|0\rangle+\frac{1}{2}|1\rangle)\otimes |1\rangle)=C_{NOT}(\frac{\sqrt{3}}{2}|01\rangle+\frac{1}{2}|11\rangle)=\frac{\sqrt{3}}{2}|01\rangle+\frac{1}{2}|10\rangle$

 

[Nugatory]

If we can prepare a single qubit state $\frac{\sqrt{3}}{2}|0\rangle+\frac{1}{2}|1\rangle$ then….

Yes, but how would we prepare that state?

 

[Hill]

Yes, but how would we prepare that state?

In case of photon polarization, for example, it needs to be polarized at angle $\alpha$ to $H$ direction, where $cos(\alpha)=\frac{\sqrt{3}}{2}$.

 

[bulx]

Apologies @PeterDonis for my confusing (layman’s!) initial language. Thanks @Nugatory for setting me straight and @Hill for the insightful preparation response. I am indeed curious about all this in the context of gates / logic / computing.

This answers my (intended) question, that such an entangled state is mathematically conceivable.

Then, taking:

$\frac {\sqrt 3} 2 \left| HV \right> + \frac 1 2 \left| VH \right>$

Is there any nuance when we use the term “probability” in this context? Is it: if we managed to prepare a trillion of these pairs, randomly selected a particle from each pair and measured its polarization, we can expect the results of those measurements to be more $H$ than $V$?

 

[PeterDonis]

This answers my (intended) question

Ok, good.

Is there any nuance when we use the term “probability” in this context? Is it: if we managed to prepare a trillion of these pairs, randomly selected a particle from each pair and measured its polarization, we can expect the results of those measurements to be more $H$ than $V$?

Yes, that's what the state you wrote down is telling you. [Edit: I misread the question as it was posted. See post #19 for a correction.] The respective probabilities are the squares of the coefficients, so the state you wrote has a 3/4 probability of $HV$ and a 1/4 probability of $VH$, so about 3/4 of the trillion pairs would be $HV$ and about 1/4 of them would be $VH$.

 

[Hill]

if we managed to prepare a trillion of these pairs

Each pair has a first particle and a second particle. If you measure the first particle in a pair, you get $H$ with 75% probability and $V$ with 25%. If you measure the second particle in a pair, you get $V$ with 75% probability and $H$ with 25%.

 

[f95toli]

Yes, but how would we prepare that state?

I might be missing something, but why would that be so difficult? it is just another point on the Bloch sphere.
Photons are (as always ) probably not ideal for this, but I don't see why it wouldn't be possible using a coupled qubit system (e.g. two coupled ions or superconducting qubits). It would just be a couple of single qubit gates followed by the abovementioned CNOT, or?

 

[vanhees71]

Apologies @PeterDonis for my confusing (layman’s!) initial language. Thanks @Nugatory for setting me straight and @Hill for the insightful preparation response. I am indeed curious about all this in the context of gates / logic / computing.

This answers my (intended) question, that such an entangled state is mathematically conceivable.

Then, taking:

$\frac {\sqrt 3} 2 \left| HV \right> + \frac 1 2 \left| VH \right>$

Is there any nuance when we use the term “probability” in this context? Is it: if we managed to prepare a trillion of these pairs, randomly selected a particle from each pair and measured its polarization, we can expect the results of those measurements to be more $H$ than $V$?

In QT it's very important to clearly state which observable you measure. Then the state clearly predicts probabilities for this measurements.

Also note that the notation here is a shortcut, i.e., a product vector like $|HV \rangle$ stands really for $\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2,V)|\Omega \rangle$, where $|\Omega \rangle$ is the vacuum of the (free) electromagnetic field and $\hat{a}^{\dagger}(\vec{k},H)$ is the creation operator for a photon with momentum $\vec{k}$ and polarization state H etc. It describes the situation that you precisely have 1 photon with momentum $\vec{k}_1$ which is H-polarized and one photon with momentum $\vec{k}_2$ that is V-polarized.

So your state more precisely reads
$$|\Psi \rangle=\left ( \frac{\sqrt{3}}{2} \hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2,V) + \frac{1}{2} \hat{a}^{\dagger}(\vec{k}_1,V) \hat{a}^{\dagger}(\vec{k}_2,H) \right )|\Omega \rangle.$$
Now you can ask precise questions.

The photons are distinguished by their momenta, and you can measure them by puting a detector far enough from the photon source detecting photons coming from the direction $\vec{k}_1/|\vec{k}_1|$ and a detector detecting coming from the direction $\vec{k}_2/|\vec{k}_2|$, and you can measure whether the photon is in H-polarization or in V-polarization by, e.g., putting a polarizing beam splitter in according orientation before each of the two detectors. In this way you can determining the polarization state of the photon with $\vec{k}_1$ and that of the photon with $\vec{k}_2$.

Now you can calculate all probabilities concerning the measurement of any polarization state of both photons. It's most simple if you ask for polarizations measuring the states you used to write down the state the photons are prepared in: The probabilities for the four possible outcomes are (now using again the more convenient shorthand notation):
$$\begin{split}
P(H,H) &=|\langle HH|\Psi \rangle=|\langle HH |\sqrt{3}/2 HV \rangle + \langle HH |1/2 VH \rangle|^2=0,\\
P(H,V)&=|\langle HV|\Psi \rangle=|\langle HV |\sqrt{3}/2 HV \rangle + \langle HV |1/2 VH \rangle|^2=3/4,\\
P(V,H)&=|\langle VH|\Psi \rangle=|\langle VH |\sqrt{3}/2 HV \rangle + \langle VH |1/2 VH \rangle|^2=1/4,\\
P(V,V) &= |\langle VV|\Psi \rangle=|\langle VV |\sqrt{3}/2 HV \rangle + \langle VV )|1/2 VH \rangle|^2=0.
\end{split}$$
This tells you that you either find that the photon with momentum $\vec{k}_1$ is H polarized and that with momentum $\vec{k}_2$ is V polarized, which happens with probability $3/4$ or that he photon with momentum $\vec{k}_1$ is V polarized and that with momentum $\vec{k}_2$ is H polarized, which happens with probability $1/4$. The other two possibilities do not occur, i.e., have probabilities 0.

The photons are entangled concerning the polarization, which implies that if the photon with momentum $\vec{k}_1$ is found to be H polarized, then necessarily the photon with momentum $\vec{k}$ must be V polarized and vice versa.

The single photons are not in a determined polarization state. For the photon with momentum $\vec{k}_1$ the probabilities to find it in either the H or the V polarization state are
$$P_1(H)=P(H,H)+P(H,V)=3/4, \quad P_1(V)=P(V,H)+P(V,V)=1/4,$$
and for the photon with momentum $\vec{k}_2$
$$P_2(H)=P(H,H)+P(V,H)=1/4, \quad P_3(V)=P(H,V)+P(V,V)=3/4.$$
More precisely the polarization state of the photon with momentum $\vec{k}_1$ is given by a mixed state, which is determined by "tracing" the original two-photon state over the other photon's polarization states, i.e., the socalled reduced state.

The original state is given by the statistical operator
$$\hat{\rho}_{12}=|\Psi \rangle \langle |\Psi \rangle = \frac{3}{4} |HV \rangle \langle HV| + \frac{\sqrt{3}}{4} (|HV \rangle \langle VH|+|VH \rangle \langle HV |) + \frac{1}{4} |VH \rangle \langle VH|.$$
Then the polarization state of photon 1 (i.e., the one with momentum $\vec{k}_1$) is
$$\hat{\rho}_1=\mathrm{Tr}_2 \hat{\rho}_{12} = \sum_{p_1,p_1', p_2 =H,V} |p_1 \rangle \langle p_1,p_2 |\hat{\rho}_{12}|p_1',p_2 \rangle \langle p_1'|=\frac{1}{4} |H \rangle \langle H| + \frac{3}{4} |V \rangle \langle V|.$$
In the same way you get the polarization state for photon 2:
$$\hat{\rho}_2=\frac{3}{4} |H \rangle \langle H| + \frac{1}{4} |V \rangle \langle V|.$$
You can of course also ask for any other probabilities concerning polarization measurements on the two photons or on each of the single photons.

E.g. you can ask, whether photon 1 is left-circular or right-circular polarized for this you have to put some quarter-wave plate in the right orientation before the polarizing beam splitter. The corresponding polarization states are
$$|L \rangle=\frac{1}{\sqrt{2}} (|H \rangle + \mathrm{i} |V \rangle), \quad |R \rangle=\frac{1}{\sqrt{2}} (|H \rangle -\mathrm{i} |V \rangle).$$
Now you can use directly the reduced statistical operator for photon 1 to get the corresponding probabilities for this measurement
$$P_1(L)=\langle L|\hat{\rho}_1|L \rangle=\frac{1}{2} (\langle H|-\mathrm{i} \langle V|) \hat{\rho}_1 (|H \rangle + \mathrm{i} |V \rangle) = \frac{1}{2} \left (\frac{3}{4}+\frac{1}{4} \right)=\frac{1}{2}.$$
Then of course $P_1(R)=1/2$ too.

 

[f95toli]

Also note that the notation here is a shortcut, i.e., a product vector like |HV⟩ stands really for a^†(k→1,H)a^†(k→2,V)|Ω⟩, where |Ω⟩ is the vacuum of the (free) electromagnetic field and a^†(k→,H) is the creation operator for a photon with momentum k→ and polarization state H etc. It describes the situation that you precisely have 1 photon with momentum k→1 which is H-polarized and one photon with momentum k→2 that is V-polarized.

Only if the system under consideration is photonic, or?
The OP was a general question about OP, it has nothing to do with photons. While photons is -for historical reasons- a popular system to teach QM, they are also really, really complicated objects. For the OP it is probably better to consider a pair of coupled two-level system (=qubits)

 

[vanhees71]

Ah sorry, I thought it's about polarization of photons, because the OP used the H, V notation for the qbit states (which usually stands for horizontally and vertically photon-polarization states). The very same treatment of course works for identical spin-1/2 particles (which however are fermions of course, but this doesn't play so much a role for polarization/spin measurements in this context).

 

[PeterDonis]

it is just another point on the Bloch sphere.

No, it isn't, because the system is two qubits, not one. The Bloch sphere is for one qubit.

 

[PeterDonis]

Each pair has a first particle and a second particle. If you measure the first particle in a pair, you get $H$ with 75% probability and $V$ with 25%. If you measure the second particle in a pair, you get $V$ with 75% probability and $H$ with 25%.

This isn't quite sufficient as a description, though, because the particles are correlated; if you measure the first particle of a pair to be $H$, the second must be $V$, and if the first is $V$, the second must be $H$. That's why I stated the probabilities in terms of the pairs.

 

[vanhees71]

I'm still confused, whether you discuss about photons or other particles. The important point is that the notation $|HV \rangle$ is a very condensed notation which may confuse beginners. Of course photons (or identical particles for that matter) are more complicated because of their indistinguishability, which is why I used the creation-operator notation, where the symmetrization (for bosons) or antisymmetrization (for fermions) is built in automatically. This, of course, is for the here discussed issues of entanglement not very important.

So let's work with distinguishable spin 1/2 particles, e.g., an electron and a positron. Then it's clear that the short-hand notation $|\sigma_1 \sigma_2 \rangle$ with $\sigma_1,\sigma_2 \in \{1/2,-1/2 \}$ means the product state of the spin-$z$-component eigenstates of the electron and the positron, i.e.,
$$|\sigma 1 ,\sigma_2 \rangle=|\text{e}^-,\sigma_1 \rangle \otimes |\text{e}^+,\sigma_2 \rangle$$,
i.e., the state in question is in this more concise notation
$$|\psi \rangle=\frac{\sqrt{3}}{2} |\text{e}^-,+1/2 \rangle \otimes |\text{e}^+,-1/2 \rangle + \frac{1}{2} |\text{e}^-,-1/2 \rangle \otimes |\text{e}^+,+1/2 \rangle.$$
Then the concise measurements are to measure the spin $z$ component of the electron and that of the positron, which gives a precise definition of what's measured on which particle.

Of course after this has become clear you can use the more convenient short-hand notation $|\sigma_1,\sigma_2 \rangle$ for these product states again.

The examples of my previous posting are of course exactly the same here: Instead of "photon with momentum $\vec{k}_1$" simply read "electron" and instead of "photon with momentum $\vec{k}_2$" simply read "positron".

 

[Hill]

This isn't quite sufficient as a description, though, because the particles are correlated; if you measure the first particle of a pair to be $H$, the second must be $V$, and if the first is $V$, the second must be $H$. That's why I stated the probabilities in terms of the pairs.

This is certainly so. I just wanted to add an answer regarding a single measurement because the OP asked,

selected a particle from each pair and measured its polarization

 

[PeterDonis]

the OP asked...

Ah, yes, I see; selecting a single particle only from each pair raises the question of which particle. It would be better to measure both particles from each pair, to see the correlations.

 

[PeterDonis]

if we managed to prepare a trillion of these pairs, randomly selected a particle from each pair and measured its polarization, we can expect the results of those measurements to be more $H$ than $V$?

Yes, that's what the state you wrote down is telling you.

Actually, @Hill makes a good point, and I didn't read your description carefully enough.

If you only select one particle randomly from each pair, then the answer to your question as you ask it above is no, because random selection will give you the first particle half the time and the second particle the other half of the time, so your final set of results will be an average of "more $H$ than $V$" from the first particle and "more $V$ than $H$" from the second particle. Since the proportions in both cases are the same (just with $H$ and $V$ swapped), the final results will be an equal mixture of $H$ results and $V$ results.

In order to get "more $H$ than $V$" you would need to only measure the first particle of each pair, not randomly select one.

My previous response assumed that what you really meant to ask was, if you measure both particles of each pair, will you get more $HV$ than $VH$. The answer "yes" that I gave is really an answer to that question.

 

[PeterDonis]

The OP was a general question about OP, it has nothing to do with photons.

I disagree; the OP says:

Perhaps horizontally and vertically polarized photons.

There is a "perhaps" there, but the notation the OP chooses is the one standardly used for photon polarizations, so I think the OP effectively adopts that case for discussion.

 

[f95toli]

No, it isn't, because the system is two qubits, not one. The Bloch sphere is for one qubit.

I was referring to the post by Hill which required single qubit gate operations followed by a CNOT. At least at first sight I don't see why the operations suggested there couldn't be done since with qubits (=any controllable 2-level system) you can in principle prepare whatever physical 2-qubit state you want using a suitable combination of single qubit gates, tunable couplers and 2-qubit gates.

There is a "perhaps" there, but the notation the OP chooses is the one standardly used for photon polarizations, so I think the OP effectively adopts that case for discussion.

Indeed, but my point (which I explained badly) was that photons are a bad system when thinking about problems like this. They are -as I pointed out above- complicated beasts and manipulating them is in practice really tricky (not that it can't be done, but the schemes used are really complicated).
I think it is very unfortunate that they are still used as "model systems" in many QM textbooks.

 

[PeterDonis]

I was referring to the post by Hill which required single qubit gate operations

Ah, sorry, I missed that.

 

[PeterDonis]

photons are a bad system when thinking about problems like this.

Yes, "qubits realized by photon polarizations" would be a better term for what I think the OP actually meant and what the thread discussion has actually focused on.

 

[vanhees71]

Ah, yes, I see; selecting a single particle only from each pair raises the question of which particle. It would be better to measure both particles from each pair, to see the correlations.

Exactly that's what I tried to explain using math to be clear: You need an additional labels to specify which particle has been measured. For the case of the indistinguishable photons this additional labels are the momenta of the photons, selected by the positions of the equipment used to measure these single photons. For the case of distinguishable particles any kind of "particle ID" is sufficient. The latter is the more simple case, but in the here discussed case it doesn't make much of a difference.

 

[vanhees71]

Indeed, but my point (which I explained badly) was that photons are a bad system when thinking about problems like this. They are -as I pointed out above- complicated beasts and manipulating them is in practice really tricky (not that it can't be done, but the schemes used are really complicated).
I think it is very unfortunate that they are still used as "model systems" in many QM textbooks.

No, why? Here it doesn't make much difference whether you use photons or some other realization of qbits. It's the beauty of qbits that almost everything fundamental as discussed here is independent of the specific physical realization of them.

The only subtlety is to discuss entanglement for indistinguishable particles vs. for distinguishable ones. This quibble however can be easily resolved by simply using annihilation and creation operators implementing the Bose or Fermi symmetrization or antisymmetrization automatically.

 

[bulx]

Thanks, all.

The explanation by @vanhees71 in post #11 was extremely helpful and cleared me up on the "Particle ID" concept. That was necessary for my follow-up question, below.

Let's adopt @PeterDonis 's phrasing, though it doesn't have to be photons,

"qubits realized by photon polarizations" would be a better term for what I think the OP actually meant and what the thread discussion has actually focused on.

and for the sake of argument take @Hill 's qubit preparation method (post #4 and post #6) or some similar series of actions.

It actually turns out I was stuck on the statistics, not the physics. This got me sorted (after I wrote out the math for myself):

If you only select one particle randomly from each pair, then the answer to your question as you ask it above is no, because random selection will give you the first particle half the time and the second particle the other half of the time, so your final set of results will be an average of "more $H$ than $V$" from the first particle and "more $V$ than $H$" from the second particle. Since the proportions in both cases are the same (just with $H$ and $V$ swapped), the final results will be an equal mixture of $H$ results and $V$ results.

As I've digested all of this and corrected my probabilities error, I am still curious:

Is there some other quantum phenomenon that exhibits "measurement order dependence"? Is it Werner states?

I had (mistakenly) taken the "first" particle and "second" particle to mean "the first that is measured" and "the second that is measured", which is different than a "Particle ID" (in which one particle is known and distinguished from the other, not just because of which we measure first).

Is this a valid Werner state ket notation:

$| \psi \rangle = \frac { \sqrt 3 } 2 | HH \rangle - \frac 1 2 | VV \rangle - \frac { \sqrt 2 } 2 | HV \rangle - \frac { \sqrt 2 } 2 | VH \rangle$

and does it mean:

"$75\%$ chance of $HH$ ($25\%$ $VV$) if the "first particle (ID)" is measured first but $50/50$ for $HV/VH$ if the "second particle (ID)" is measured first?"

---

Bell's Inequalities and Density Matrices: Revealing "Hidden" Nonlocality (Sandu Popescu, 1995)
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.74.2619

 

[PeterDonis]

Is there some other quantum phenomenon that exhibits "measurement order dependence"?

I'm not sure what you mean by "measurement order dependence".

Is it Werner states?

Do you mean these?

I had (mistakenly) taken the "first" particle and "second" particle to mean "the first that is measured" and "the second that is measured", which is different than a "Particle ID" (in which one particle is known and distinguished from the other, not just because of which we measure first).

Yes. The ordering of kets is based on what you call "Particle ID", not on measurement order.

Is this a valid Werner state notation?

$| \psi \rangle = \frac { \sqrt 3 } 2 | HH \rangle - \frac 1 2 | VV \rangle - \frac { \sqrt 2 } 2 | HV \rangle - \frac { \sqrt 2 } 2 | VH \rangle$

I can't say whether it's a valid Werner state, but it's not properly normalized. The squares of the coefficients don't add up to 1.

and does it mean:

"$75\%$ chance of $HH$ ($25\%$ $VV$) if the "first particle (ID)" is measured first but $50/50$ for $HV/VH$ if the "second particle (ID)" is measured first?"

No, it means nothing of the kind. As above, the ordering of the kets has nothing to do with the order of measurements.

If your state were properly normalized, it would be telling you that there is a 3/8 probability of $HH$, a 1/8 probability of $VV$, and a 1/4 probability for each of $HV$ and $VH$. Again, the ordering here is based on "Particle ID" and has nothing to do with which one is measured first.

 

[vanhees71]

I think, it's good to think about, what "state" and "measurement" means. Although this touches again some "interpretational issues", I think it's safe to say that the state, in general given by a statistical operator (self-adjoint positive semidefinite operator with trace 1), operationally describes a preparation procedure.

In your case that means you have somehow created a photon pair such that it's state is described by the pure state $\hat{\rho}=|\Psi \rangle \langle \Psi|$ with
$$|\Psi \rangle=\frac{1}{\sqrt{2}} \left [\frac { \sqrt 3 } 2 | HH \rangle - \frac 1 2 | VV \rangle - \frac { \sqrt 2 } 2 | HV \rangle - \frac { \sqrt 2 } 2 | VH \rangle \right].$$
I've renormalized your ket such that it's normalized to 1, as it should be for a state ket.

Now you can ask for probabilities when measuring polarization states of these photons. Of course, it depends on how you precisely set up these measurements. What's usually discussed in Bell tests is to independently measure the polarization states of the two photons. To make sure that one measurement cannot influence the other you can set up your measurement such that the outcome of the measurements are registered at space-like separated events.

To see whether that's a Werner state, check, whether you can write the statistical operator by checking whether it can be written in the way as described in the Wikipedia article, quoted by @PeterDonis in the previous posting, i.e., as a mixture of the projectors to the symmetric and antisymmetric state:

https://en.wikipedia.org/wiki/Werner_state

Think about, what happens when you exchange the states of the two photons in each of the kets you superimposed to define $|\Psi \rangle$!