Q: What is the “monogamy of entanglement”?

Posted on October 26, 2018 by The Physicist

Physicist: The Monogamy of Entanglement is the statement that “maximally entangled” particles only show up in pairs. Entanglement is a sliding scale, so things can be non-entangled or a little entangled, but when the quantum states of two things are completely tied up in each other, there’s just no room for a third. This is more about how entanglement is defined mathematically and less about a dramatic insight into the true nature of reality.

Monogamy of Entanglement: Two maximally entangled particles cannot be entangled with any others. But the exact meaning of entanglement, like matrimony, is a matter of definition.

One particle: Quantum stuff can be (and generally is) in a “superposition”, multiple states at the same time. For example, an electron belonging to Alice (or anyone else), can be both “spin up” and “spin down” in equal parts; a superposition that’s written $|\uparrow\rangle_A+|\downarrow\rangle_A$ . Of course, you never look at something and notice that it’s in more than one state (superpositions need to be inferred using tricks like the double slit experiment). Before measuring its spin, the electron can be in the superpostion $|\uparrow\rangle_A+|\downarrow\rangle_A$ , but after measuring, its state is definite, either $|\uparrow\rangle_A$ or $|\downarrow\rangle_A$ (depending on what the result was). Which of these results you see is fundamentally random.

Now here’s the thing. Quantum randomness is relative. One person’s superposition is another person’s definite state. For example, that superposition from earlier is also a definite state $|\uparrow\rangle_A+|\downarrow\rangle_A=|\rightarrow\rangle_A$ . The spin right/left states are literally a superposition of the spin up/down states (this was as surprising and weird when it was first figured it out as it should be for you right now). If you measure $|\rightarrow\rangle_A$ in the vertical direction it’s a superposition of spin up and down, but if you measure it sideways it’s a definite state.

Spin causes particles to act like tiny magnets, which gives us a handy way to measure it. Left: When you measure spin vertically, there’s no way to tell the difference between a string of spin right electrons (top) and a string of random spin up/down electrons (bottom). Right: But if you measure horizontally, the string of spin right electrons always get sorted correctly, while the up/down electrons continue to produce random results.

If some devious ne’er-do-well sent you a never-ending stream of $|\uparrow\rangle_A$ or $|\downarrow\rangle_A$ states, flipping coins to determine which every time, it would look the same as a stream of $|\rightarrow\rangle_A$ states so long as you measured both vertically. But if you measured sideways, then the stream of $|\rightarrow\rangle_A$ states would always yield the same predictable result ( $|\rightarrow\rangle_A$ ), and the stream of random up/down states would still yield random results. In exactly the same way that the left/right spin states are superpositions of the up/down states, the up/down states are superpositions of the left/right states (so they produce random results when measured horizontally).

Here’s the point. If you’re always looking at the same, definite state, then you should be able to find some measurement that always produces a definite result. But if you’re being fed a random set of states, there’s no way to do a measurement that produces consistent results. I suppose you could turn the detector off and walk away, but that’s dirty pool.

Quick aside: I’m dropping the coefficients in front of the states that would normally dictate their probability of being measured (the state $|\uparrow\rangle_A+|\downarrow\rangle_A$ where both results show up 50% of the time, should have been written $\frac{1}{\sqrt{2}}|\uparrow\rangle_A+\frac{1}{\sqrt{2}}|\downarrow\rangle_A$ ). I figure if you notice they’re missing, then you probably already know how to put them back, and if you don’t notice they’re missing, then you probably don’t want to be distracted.

Two Particles: When you need to worry about multiple systems (particles, quantum computers, needlessly diverse collections of scarves, or whatever else), you don’t worry about their individual superpositions of states, you worry about superpositions of their collective states. So if you have a second electron, under the custodianship of Bob, you wouldn’t first talk about the states of Alice’s electron ( $|\uparrow\rangle_A$ and $|\downarrow\rangle_A$ ) and then, having done that, talk about the states of Bob’s electron ( $|\uparrow\rangle_B$ and $|\downarrow\rangle_B$ ). Instead, you’d talk about superpositions of their collective states ( $|\uparrow\rangle_A|\uparrow\rangle_B$ , $|\downarrow\rangle_A|\uparrow\rangle_B$ , $|\uparrow\rangle_A|\downarrow\rangle_B$ , and $|\downarrow\rangle_A|\downarrow\rangle_B$ ).

Left: In classical physics a system’s state is the aggregation of each part taken individually. Right: In quantum physics the system as a whole has a state that’s shared between each part.

For example, if both electrons are in the same equal superposition of spin up and down from earlier, then their collective state is

If Alice measures her electron and finds that it’s spin up, then the state of the pair is now

$|\uparrow\rangle_A\left(|\uparrow\rangle_B+|\downarrow\rangle_B\right)$

and if she finds that her electron is spin down, then the overall state is

$|\downarrow\rangle_A\left(|\uparrow\rangle_B+|\downarrow\rangle_B\right)$

In other words, in this situation it doesn’t matter what Alice sees; Bob’s state is independent of it. This particular collective state can be described first by considering Alice’s electron and then considering Bob’s (or vice versa), because knowing the result of a measurement on one part tells you nothing about what you’ll see when you measure the other. A state like this is called “separable”, because it can be separated mathematically. Physicists are very clever namers of things, which is why they often refer to themselves as namerators.

Non-seperable states are entangled; a measurement on one part gives you information about what the result of a measurement on another part will be. For example,

$|\uparrow\rangle_A|\uparrow\rangle_B+|\downarrow\rangle_A|\downarrow\rangle_B$

is definitely not separable. If Alice were to measure her electron the state of the system as a whole would be either $|\uparrow\rangle_A|\uparrow\rangle_B$ or $|\downarrow\rangle_A|\downarrow\rangle_B$

In other words, by looking at her own electron, Alice knows what Bob will measure, whenever he gets around to it. This state is “maximally entangled” because knowing one measurement allows you to perfectly predict the other.

But once Alice has measured her electron, and the system as a whole is now either $|\uparrow\rangle_A|\uparrow\rangle_B$ or $|\downarrow\rangle_A|\downarrow\rangle_B$ , then as far as Bob is concerned, his state is randomly either $|\uparrow\rangle_B$ or $|\downarrow\rangle_B$ . Unlike the single-particle case with $|\rightarrow\rangle$ , there’s no “correct” way to measure these randomly selected states to make the result definite.

This randomness isn’t a symptom of Alice having done a measurement, it’s a feature of entangled states in general. After all, Bob and his pet electron could be anywhere. Despite the nearly universal, but categorically false, belief that entangled particles affect each other, nothing that Alice does or doesn’t do with her electron will ever have any impact on Bob’s electron.

If Alice measures her particle, Bob gets a random result. But since the particles can’t communicate and it doesn’t actually matter what Alice does, Bob gets a random result anyway.

Entangled particles follow the same rule that single particles follow: wrong measurement = random result, correct measurement = definite result. But in order to do the “correct measurement” you always need to have access to the entire state. If you have both entangled particles in front of you, then you can easily do a measurement to determine which maximally entangled state you have (for example $|\uparrow\rangle_A|\uparrow\rangle_B+|\downarrow\rangle_A|\downarrow\rangle_B$ and $|\uparrow\rangle_A|\downarrow\rangle_B+|\downarrow\rangle_A|\uparrow\rangle_B$ are both maximally entangled, but very different states). These “correct” joint measurements always involve a step where the two particles are forced to interact, which is something you can’t do if they’re on opposite sides of the universe.

If you only have access to part of an entangled state, you’ll get random results no matter how you measure it. But if you have access to all of an entangled state, there is a “correct” measurement with predictable results.

This leads to a rather profound fact: entangled states always appear random when you only measure part of them. That is, they never seem to be in a superposition of states, they always appear to be a randomly selected state, since there’s no measurement that makes them predictable (see the one particle case).

There are still some clever things you can do with entangled particles that take advantage of their quantum correlations, like quantum teleportation, but two facts remain immutable: 1) when you measure one particle alone the result is random and 2) the two particles never ever influence each other in any way.

How entanglement is defined: When you don’t have access to both entangled particles and are forced to measure them one at a time, you get random results. You can describe how random those results are in terms of how much information it takes to describe them: two possible results (up or down) takes 1 bit of information (0 or 1). When you do have access to both entangled particles, you can do a measurement that will always produce the same, predictable result. One possible result takes zero bits of information. After all, if you had flipped a two-headed coin, would you really need to write down the results at all? Zero bits.

This difference between the randomness of the particles when they’re apart minus the randomness when they’re together is how entanglement is defined. A maximally entangled pair of particles has “1 ebit” of entanglement and a separable pair of particles has 0 ebits.

And yes, there are states in between. For example, $|\uparrow\rangle_A|\uparrow\rangle_B+|\uparrow\rangle_A|\downarrow\rangle_B+|\downarrow\rangle_A|\downarrow\rangle_B$ is partially entangled, because if Alice looks at her electron, she’ll learn a little about Bob’s electron (and vice versa), but not everything. This state has a modest 0.55 ebits of entanglement.

Three Particles: If Alice’s electron is in a superposition of spin up and down, its state is $|\uparrow\rangle_A+|\downarrow\rangle_A$ . If Alice and Bob each have electrons that are entangled such that they’re spin up and down together, then their state is $|\uparrow\rangle_A|\uparrow\rangle_B+|\downarrow\rangle_A|\downarrow\rangle_B$ .

So what’s stopping Carol and her electron from also being both spin up and down in tandem with Alice and Bob’s? Not a damn thing.

Today we can create states like this with not just three, but dozens of particles. No big deal. Any reasonable person (with at least a passing familiarity with quantum notation) would say that state looks pretty entangled.

With the third-act-introduction of Carol, suddenly they’re working with one of two randomly selected states ( $|\uparrow\rangle_A|\uparrow\rangle_B$ or $|\downarrow\rangle_A|\downarrow\rangle_B$ instead of a nice, pure, entangled state $|\uparrow\rangle_A|\uparrow\rangle_B+|\downarrow\rangle_A|\downarrow\rangle_B$ ). These states are random when Alice and Bob measure them on their own and when they bring their electrons together, they’re still looking at one of two random states, so even together the result of any measurement is going to end up random.

However! Entanglement is narrowly defined is in terms of how much more random your particles are when they’re apart vs. when they are together. Same amount of randomness means zero entanglement. So because there’s a third (or fourth or fifth…) party involved the state, no two parts can be maximally entangled.

Entangled states are random apart and predictable together. That difference is how entanglement is defined. If a state is divided among more than two places, no pair can have the whole state, so the results are always random.

So it’s not that you can’t do fancy quantum states with many particles at once, and it’s not that something terribly profound happens when you move from using two particles to using three, but going by the stringent definition of entanglement: if two parties share a maximally entangled state with each other, they don’t share it with anyone else. One might even go so far as to say that entanglement is monogamous.

Like many seemingly arbitrary mathematical definitions, it turns out that the monogamy of entanglement is a powerful, useful statement. It is the key behind why quantum security is so secure (and quantum). By sharing a string of maximally entangled states with each other, then measuring some and sharing the results, Alice and Bob can check to see if there’s a third party corrupting their state. If there isn’t, they can dive into their private conversation. If there is, then they’ve caught themselves an eavesdropper. Quantum cryptography is Carol-proof!

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12 Responses to Q: What is the “monogamy of entanglement”?

Arthur Snyder says:

October 26, 2018 at 7:31 pm

Does the same argument apply to spacial entanglement or does it have something to do with there only being two states (up/down) to consider in the spin case?
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October 26, 2018 at 11:08 pm

@Arthur Snyder
Any kind of entanglement. Although with spacial entanglement, defining a maximally entangled state is about as difficult as getting a particle to be any any particular place exactly. The nice thing about the examples here is they involve discrete states (which are generally a lot easier to work with and think about).
UpDownUpDownLeftRightStart says:

October 30, 2018 at 12:11 am

are the two states just before “are both maximally entangled, but very different states” the same? maybe the second could be |u>A|d>B+|d>A|u>B
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October 30, 2018 at 9:04 am

@UpDownUpDownLeftRightStart
Absolutely!
The article was beginning to run the risk of being too complicated, so I skirted around talking about the other maximally entangled states. There are four “Bell states”. The one in the post, the one you have there, and then two more that involve a phase shift: $|\uparrow\rangle_A|\uparrow\rangle_B-|\downarrow\rangle_A|\downarrow\rangle_B$ and $|\uparrow\rangle_A|\downarrow\rangle_B-|\downarrow\rangle_A|\uparrow\rangle_B$ .
Joe Parkhill says:

November 26, 2018 at 11:58 pm

Hi, this is more a question of the very last part of your article regarding how Alice and Bob can check if a third party is eavesdropping. The question is a departure from the main content of the article into something more specific, but you seem to actively be answering questions, and I can’t find the answer I’m looking for.

There’s going to be some slight set up to make sure my question is clear, but feel free to skip to the bottom if there’s no need.

Setup:
————————————–

In the E91 protocol, which uses quantum entanglement, there is a device which creates entangled pairs of photons and sends one member of this pair to Alice, and the other to Bob.

Alice then randomly picks a base with which to measure her photon, and Bob also randomly picks a base with which to measure his photon. This proceeds for some time, and each time Alice and Bob note their measurement base and their result.

After some time Alice and Bob announce the bases they used to measure their photons. All of the measurements where Alice and Bob used different bases for their photon are added to the Rejected Key, and all of the measurements where Alice and Bob used the same base are kept hidden and made the Secret Key.

The values of the Rejected Key are announced by both Alice and Bob.

From the Rejected Key Alice and Bob determine whether their entanglement maximally violates Bell’s inequality, if this is true, then there was no tampering by Eve (because interference from a third party would decrease entanglement), else if Bell’s inequality is satisfied they abort communication and try again.

My question:
————————————————-
Doesn’t Alice and Bob’s measuring of their entangled photons corrupt the entanglement just as much as if some third party was doing it? If so then how they can use their rejected key to see if Bell’s inequality is satisfied or not?

When you said “By sharing a string of maximally entangled states with each other, then measuring some and sharing the results, Alice and Bob can check to see if there’s a third party corrupting their state.” What would this corruption be? And why isn’t this corruption naturally introduced by Alice and Bob?

Thanks!
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November 27, 2018 at 9:57 am

@Joe Parkhill
You’re exactly right; any direct measurement, including those by Alice and Bob, breaks the entanglement between the photons. However, the entanglement isn’t broken a moment before the measurement, it’s broken by the measurement itself.
One of the key properties of entanglement, and the one that makes it useful for quantum crypto, is that measurements of the pair are the same when the same measurement is done on both. The probability of getting the same result is $\cos^2(\theta)$ , where $\theta$ is the difference in angles between the two measurement devices (e.g., if they measure at right angles to each other, the probability of them getting the same result is $\cos^2(90)=0$ ).
Notice that the polarization of the photons themselves is not considered.
If a photon has a known polarization, then the chance that your detector will positively identify that polarization is $\cos^2(\phi)$ , where $\phi$ is the difference in the angle between the photon’s polarization and the alignment of the detector.
For non-entangled but correlated photons, you do have to worry about the polarizations of the photons. If you send a string of photons to Alice and Bob that are always either both vertical or both horizontal, then the results will be perfectly correlated when the detectors aligned vertically or horizontally. But if the detectors are aligned diagonally (45 degrees to both the horizontal and vertical), they’ll each produce random results: every photon coming down the pipes will have a $\cos^2(45)=\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}$ chance of producing either result, and so only a 50% chance of producing the same result as the other photon. Correlated in the correct alignment, but random in the wrong alignment. Entangled photons are correlated in every alignment of the detectors. When an Eavesdropper breaks the entanglement when measuring, the new photons they send along to Alice will not be entangled with the ones that get to Bob.
Alice and Bob also “corrupt” the entanglement in exactly the same way, but at the same time that they do, the entanglement has done its job for the two of them.
Joe Parkhill says:

November 28, 2018 at 11:52 am

Ahh, this all makes a lot of sense now, thank you, but could you expand a little more on “Entangled photons are correlated in every alignment of the detectors”?

Suppose we have a pair of entangled photons, call them p1, and p2, and we fire them each through separate incorrectly aligned sensors. Given that we measure p1 to have polarization A, what can we say about the polarization of p2 and with what probability?

If the photons were not entangled, the answer should be “nothing”, since there is no correlation, but the photons are entangled, and there is correlation, so certainly we should be able to tell more?
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November 28, 2018 at 5:20 pm

@Joe Parkhill
When you measure polarization, the measurement takes the form “is it this polarization or 90 degrees off?”, you only get one of those two results. Quantum measurements are all like this (quantized), so it’s different from weighing something or finding the angle between things.
For entangled photons, the probability that both detectors will register the same thing (both “this polarization” or both “90 degrees off”) is equal to $\cos(\theta)$ where $\theta$ is the angle between the detectors’ orientation (their “this polarization”).
Classical things can be correlated, for example, two coins (covered and unknown) that are always the same side up. Photons can also be correlated (both vertical or both horizontal, but you don’t know which). Entanglement is kinda “a level up” from regular correlation. Entanglement includes superposition of states, while correlation is about merely not knowing which state is present.
It’s a subtle difference.
Yousef Elbrolosy says:

May 21, 2020 at 3:59 pm

I have a question, now that only two electrons can be maximally entangled, i dont quite get how general n-qubit states are created. which from what i learned make use of entangled states to increase computational space to 2^n which gives quantum computers the power over classical ones. so for example take this three qubit general state

a1|000⟩+a2|001⟩+a3|010⟩+a4|100⟩+a5|110⟩+a6|101⟩+a7|111⟩+a8|011⟩

If only qubits out of these 3 are entangled how can we represent it in such a form?

same question for any n-qubit general state

∣ψ⟩ n-qubit= a1∣000…000⟩+ a2∣000…001⟩+a3 ∣000…010⟩+⋯.
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May 21, 2020 at 7:17 pm

@Yousef Elbrolosy
Monogamy of entanglement basically boils down to the technical, mathematical definition of how “entanglement” is defined: in terms of a relationship between exactly two systems. So qubit 1 could be maximally entangled to the system of qubits 2 and 3 together.
What you’re talking about there is a general superposition of states (which are practically always entangled). In this case, a more useful (but less quantitative) way to describe entanglement is by asking the question “is this state separable?”. A separable state an be described in terms of just A and then just B. Nothing about the A part of the state tells you anything about the B part. Separability corresponds to zero entanglement, so being non-separable means “entangled, at least a little”.
Yousef Elbrolosy says:

May 22, 2020 at 11:33 am

okayy so it is safe to say that in this case all three qubits are entangled since they are inseperable to independent superpositions. And qubit 1 can be maximally entangled to qubit 2 and 3 together am i correct?

a1|000⟩+a2|001⟩+a3|010⟩+a4|100⟩+a5|110⟩+a6|101⟩+a7|111⟩+a8|011⟩

in the general n-qubit case is there is also maximally entangled qubits? (since there may be possibly many qubits) and what does maximally entangled differ from “not maximally entangled” or entangled a little bit?
Thank you in advance
Yousef Elbrolosy says:

May 22, 2020 at 11:43 am

Phrasing the question another way, it seems like not being maximally entangled is just adding for example other qubits that their coefficients are inseperable so become entangled. while the maximally entangled involves that the state of one qubit when measured has a direct instantanous influence on the other maximally entangled qubit, is that correct? is there any difference then between maximally entangled and non maximally entangled ones?