I've been taught that measuring the spin of an entangled particle causes it to become unentangled, but not how or why, so I'd like to know by what process this occurs.

Is the cause known? If so, please explain why observation causes the particles to become unentangled. Thanks!

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    $\begingroup$ I once stumbled on a post on the xkcd forums which helps me a lot with these questions. Basically, for observations to happen, there has to be an interaction between particles, or as the post put it less/more(?) eloquently, whenever a physicist says "observe", mentally replace it with "hit with sh*t" $\endgroup$
    – Sanchises
    Mar 27, 2015 at 9:14
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    $\begingroup$ Well, they don't really become unentangled - the relation between the two particles is still there (not necessarily always, of course, but usually). However, a lot of new entanglements develop, so the two particles are no longer entangled exclusively - the relation between the two original particles is now lost in the noise of all the others as well. $\endgroup$
    – Luaan
    Mar 27, 2015 at 9:48

6 Answers 6


"Entanglement" is a term describing economically the quantum mechanical state of a system of particles. It is a short hand way of saying : these particles are described by the solution of the Schrodinger equation, with a wave function that can predict the probability of finding the individual particles in a specific (x,y,z) each with specific quantum numbers". The conserved quantum numbers accompany the particles after the state interacts, and any measurement is an interaction.

A measurement is one instant in the accumulation of the probability function, of experimentally checking the theory. Many measurements will give the square of the wavefunction for this system of particles, which means we should prepare an ensemble of same condition particles.

When one measures the quantum mechanical condition of a particle , the original wavefunction is no longer valid, the particle is now in a new quantum mechanical state, no longer entangled in a coherent wavefunction with the others, as the boundary conditions are changed by the measurement of this one particle.

Quantum numbers that are conserved allow to know what are the possible collective quantum numbers of the remaining particles by the values carried by the particle measured.


The answer is simple: measurement causes the wave-functions to collapse.

It can be said that one of the fundamental properties that makes Quantum mechanics so strange is the idea of superposition, which is the property that if you have two physically valid descriptions of a state, then it is physically just as valid for a system to be in any linear combination of both states at the same time (think Schrödinger's cat).

Entanglement is just a particular example of a superposition of states. For instance you can describe the spin of two particles as both being up or both being down. Entanglement is a state where they are in a superposition of both spins being up and both spins down at the same time.

Now if you make a measurement of the spin of one or both of the particles, then this will cause your superposition to collapse into being either both up or both down thus destroying the special superposition state that is the entangled state.

  • $\begingroup$ Not all formulations of quantum mechanics need collapse. You should mention that this also happens in decoherence/einselection approaches, although they know of no collapse. $\endgroup$
    – ACuriousMind
    Mar 27, 2015 at 13:29
  • $\begingroup$ @ACuriousMind Not all formulations of QM require collapse as a postulate, but they all need collapse (or whatever equivalent name you wish to give to the measurement process). The decoherence process derived by people such as W. Zurek (whose work I greatly admire) provide a micro description of this process. This is analogous to how statistical mechanics provides a micro description of thermodynamics, and just as StatMech explains rather than invalidates thermodynamics, so also decoherence better explains rather than invalidates the phenomena of collapse. $\endgroup$ Mar 27, 2015 at 17:08

What "entangled" means

You probably have put too much in your head for the meaning of the word "entangled." Let me fix that for you:

Two systems are entangled in quantum mechanics if the results of separate experiments on those two systems display strange correlations when you bring them back together and compare them.

Notice that not all correlations can necessarily be realized by QM. You see this a lot when you're looking at Bell's inequality violations: often the "maximum violation" allowed by QM in some situation is not the biggest that you could imagine if you could 100% play God with the measurement outcomes.

Notice also that there are not-strange correlations: for example, I have one ball, I drop it down a tube, it hits a divider and goes into one or another sealed opaque box without my knowing which one it's in. Now experiments like "do I hear a rattle when I shake the box?" are correlated: if I hear a rattle in this box, I don't hear it in that box. We could talk about entanglement here, but it is a "classical" entanglement.

A good example to keep in mind

To give an example of a "strange" correlation I often like to give an example of a 3-player game I call Betrayal. The idea is that there are 3 players on a team which endures a million "rounds" of experiments. In each round, we separate the 3 into completely isolated rooms, flash a command on a screen, and they have a few seconds to hit either a button labeled 1 or a button labeled 0. If they do "the right thing" as a team in all of the million rounds, they all get a billion dollars: so they're all working together as a team. The trick is that we sometimes try to force one person, who we'll call a "traitor", to work at cross-purposes to the other two. So there are two sorts of rounds. In control rounds we simply give them all the command, "make the sum of your team's chosen numbers even", and they all win the round if, when we take the three 1's or 0's that they press and sum them together, the result of this is even. In test rounds we choose a traitor at random and give them the same command as before -- make the sum even -- but we give the other two people the opposite command: "this is a traitor round, you make the sum of your team's chosen numbers odd." The team passes the round only if the sum of the 3 numbers they hit is odd.

You can prove with classical random variables that there is no 100% solution to this puzzle. (Basically you get $X_o + Y_o + Z_e \equiv X_o + Y_e + Z_o \equiv X_e + Y_o + Z_o \equiv 1 ~~\text{(mod 2)}$, sum those three equations together to get $$X_e + Y_e + Z_e + 2 (X_o + Y_o + Z_o) \equiv X_e + Y_e + Z_e \equiv 3 \equiv 1 ~~\text{(mod 2)},$$which contradicts the requirement that $X_e + Y_e + Z_e \equiv 0~~\text{(mod 2)}.$ So it does not matter what joint probability distribution you choose among these six random variables or how they correlate: classical probability cannot simultaneously give you all of these equivalences. Essentially the problem is that in classical probability we could ask a player $X$ for both answers $X_o, X_e$ and they could provide us with both values; it "doesn't matter" for a joint probability distribution that we didn't ask for the other one.

You can also easily prove with quantum mechanics that there is a 100% solution with three qubits in a GHZ state. Let the Hadamard basis be $|+\rangle = |0\rangle + |1\rangle$ and $|-\rangle = |0\rangle - |1\rangle$, then we have $$|+++\rangle = |000\rangle + |001\rangle + |010\rangle + |011\rangle + |100\rangle + |101\rangle + |110\rangle + |111\rangle$$ $$|---\rangle = |000\rangle - |001\rangle - |010\rangle + |011\rangle - |100\rangle + |101\rangle + |110\rangle - |111\rangle$$so that conveniently in the computational basis,$$|+++\rangle + |---\rangle = |000\rangle + |011\rangle + |101\rangle + |110\rangle$$ $$|+++\rangle - |---\rangle = |001\rangle + |010\rangle + |100\rangle + |111\rangle$$In other words, you all start off with the Hadamard-GHZ state $|+++\rangle + |---\rangle$ and anyone who is asked to make the sum odd simply performs a controlled-phase rotation in the Hadamard-basis: the unitary transform such that $|+\rangle\rightarrow|+\rangle, ~~ |-\rangle \rightarrow i |-\rangle$. If two people do this to their qubits in isolation, we flip from a state where all of the measurements have an even number of 1's to the state $|+++\rangle + i^2 |---\rangle = |+++\rangle - |---\rangle$, where all of the measurements have an odd number of 1's.

Why measurement destroys entanglement

There is a type-error in the question: a measurement doesn't destroy entanglement, it reveals it. You don't get to see entanglement unless you measure a bunch of things and bring the measurements together afterwards.

But measuring the wrong thing can destroy entanglement, sure. And that's for a familiar reason: in QM, once you measure X, often you cannot measure Y in the same way afterwards. You measure a spin up-down, you find it is up; then you measure it left-right, you find it is left; now you measure it up-down again: shockingly, sometimes it will be down now.

Similarly, if one of our hopeful candidates in the 3-member team accidentally measures in the Hadamard basis instead of the computational one, and they find their qubit is in the state $|+\rangle$, they have all lost the ability to do the measurements which display entanglement: the resulting system will just look like an unentangled $|+++\rangle$ system to all further experiments. One way to phrase this is that the entanglement was destroyed, but really what happened was that you performed a measurement which made it impossible for further measurements to reveal the entanglements in the system.

The Quantum Eraser Experiment

To give a great example of how this stuff can play out, let us consider a slightly different experiment: the quantum eraser. This is usually described in terms of a double-slit experiment where the states $|0\rangle$ and $|1\rangle$ are evolved into bell curves which overlap on our detectors, but the state $|+\rangle$ is evolved into an "interference pattern". The full treatment of what the pattern on the detectors looks like requires "state matrices" and "tracing out qubits" to really understand.

Quick overview: we start in the state $|00+\rangle$, we measure that third qubit in the Hadamard basis, we discover $+$, which is an "interference pattern", and we are happy. Actual details: the operator associated with this measurement is $$1 \otimes 1 \otimes |+\rangle\langle+| = |00+\rangle\langle00+| + |01+\rangle\langle01+| + |10+\rangle\langle10+| + |11+\rangle\langle11+|,$$and when we measure that (with appropriate factors of $\sqrt{1/2}$ inserted) we find an expected value of 1, which means that it's in the state $|+\rangle$.

Now we measure "which path does it go?" by performing a CNOT (quantum controlled-not gate) from qubit 3 to 2. This takes us to the state $|000\rangle + |011\rangle$. It turns out that the operator above expects only a value of $1/2$ in this state. This ultimately means "no interference pattern". We measured "which way" the photon went through the slits 0 and 1, and so we lose the glorious interference pattern due to it going through both with a quantum mixture $|+\rangle$ of 0 and 1.

But now suppose that the first two qubits were not starting in the state $|00\rangle$ but the state $|00\rangle + |11\rangle$. Then the first experiment is unaffected due to lack of entanglement, but the CNOT now puts us in the state: $$|000\rangle + |011\rangle + |110\rangle + |101\rangle = |+++\rangle + |---\rangle$$. Now we can again calculate the expectation value of the above operator and we see $1/2$: no interference pattern.

But: there is now a measurement we can make on a remote qubit which "erases" the which-way information: measure the first qubit in the Hadamard basis to find an explicit $|+\rangle$ or $|-\rangle$, and then you must see an interference pattern when you look at the screen. Interesting, no?


A situation like this already happens in classical physics.

Suppose that I write either "0" or "1" on two pieces of paper, the same thing on both pieces, and I put them into two separate envelopes. Then, if you don't know which thing I wrote, the contents of the two envelopes are "correlated": opening one tells you about the state of the other.

But once you've opened up one of the envelopes you then know what's inside both of them, so there's no remaining correlation between the envelopes. Making the observation has destroyed the correlation.

Now, quantum entanglement is different from classical correlation (Bell's Theorem shows that there are patterns of correlation that quantum mechanics can create but classical physics can't). However, the way in which quantum entanglement is destroyed upon making an observation is pretty much the same as the way that the classical correlation is destroyed when you open one of the envelopes.

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    $\begingroup$ I think this still too deeply misses the point of quantum entanglement. It would be like there is no number in the envelope until you open it - and in fact, depending on who opens it, you could force it to be a "0" or "1", or an "A" or "B" $\endgroup$
    – Señor O
    Mar 27, 2015 at 17:38
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    $\begingroup$ Sure, there are some ways in which entanglement is very different from correlation. It's just that the way it's destroyed by observations isn't one of them. $\endgroup$ Mar 27, 2015 at 17:41
  • $\begingroup$ I think I see now. Entangled particles share a rhythm, but the rhythm is unknown until you listen to it through some electromagnetic means. But because electromagnetism introduces more beats to the rhythm, as soon as you can possibly know the rhythm of one, you've altered that rhythm, and the particles no longer share the same mad beats. $\endgroup$ Jul 14, 2015 at 13:12
  • $\begingroup$ For you not knowing what's in the envelopes is the correlation between them? I'd say the correlation is there upon creation of correlation and stays forever. $\endgroup$
    – psycho brm
    Jan 3, 2017 at 17:41

Suppose that system $S_1$ is entangled with system $S_2$. An observation is just coupling the measured system to some other system that allows you to easily make records of experimental results. Everything that happens after that can explained by unitary evolution of the joint system of the measuring instrument, the measured system and the environment, see:


Now, about this particular issue. If you measure $S_2$ with measurement apparatus $M$ which is in contact with environment $E$, then $S_1$ is now entangled with the joint system $S_2ME$. The system $S_2$ is relatively easy to manipulate and measure in systems set up to test for entanglement. You can do tests for entanglement in photon polarisations by rotating the polarisation of one or both of of the photons or measuring them with a specific polarisation filter. But if the system with which $S_1$ is entangled include some measuring instrument and the environment, you won't be able to manipulate them easily, nor will you know which observable to measure to test for the entanglement. See



First I'll describe what happens when you measure the spin of just one particle. Say you setup a some magnets to make a region of inhomogeneous field so that when a spin up particle heads forward into the region it ends up deflect right. And a spin down heads forward it ends up deflected left. Then if you have a superposition of up and down such as $\alpha|+\rangle = \beta |-\rangle$ the wavefunction evolves into a capital Y where the fraction of squared norm heading right or left is $|\alpha|^2/(|\alpha|^2+|\beta|^2)$ and $|\beta|^2/(|\alpha|^2+|\beta|^2)$ respectively.

So the beam comes in and forks into two beams. If you track the so called probability current, it actually forks just like the letter Y. We describe this algebraically by writing the spatial ($F$, $R$, $L$) and spin ($+, -$) degrees of freedom like $|+\rangle\otimes+F\rangle\mapsto|+\rangle\otimes |R\rangle$ and $|-\rangle\otimes|F\rangle\mapsto|-\rangle\otimes|L\rangle$ therefore $(\alpha|+\rangle+\beta|-\rangle)\otimes|F\rangle\mapsto\alpha|+\rangle\otimes |R\rangle+\beta|-\rangle\otimes |L\rangle.$

Now we can write the same thing for the entangled pair. It starts out with a spin like $|+\rangle_1|-\rangle_2-|-\rangle_1|+\rangle_2$ and each particle really has spatial degrees of freedom too so let's assume that particle 1 is moving in the forward direction, then we have $$|+\rangle_1|F\rangle_1|-\rangle_2-|-\rangle_1|F\rangle_1|+\rangle_2$$ which (if we do the measurement) evolves to $$|+\rangle_1|R\rangle_1|-\rangle_2-|-\rangle_1|L\rangle_1|+\rangle_2$$ and by placing detectors along the path of the left deflected beam and the right deflected beam we can make it so that $|+\rangle_1|R\rangle_1|-\rangle_2$ and $|-\rangle_1|L\rangle_1|+\rangle_2$ each act like they are the only thing in the world, each acts as if the other doesn't exist. You can do this by having the left and right deflected beams interact (directly or indirectly) with so many particles that it is just too hard to get the evolution of the beams to interact (which requires that they affect the so called probability current of the other one, which they won't do if they do not overlap ever again).

So you now either act like you are $|+\rangle_1|R\rangle_1|-\rangle_2$ or you act like you are $|-\rangle_1|L\rangle_1|+\rangle_2$ and each of those is a non entangled state. And in particular you can now precess the spins of one of them and thus change it's correlation with the other one. If you tried to do that before the measurement there would be no change in the correlation.

In a sense particle one evolves to have its spin become entangled with its spatial degree of freedom instead of being entangled with the other particle's spin, and then that spatial degree of freedom can become entangled with a wider environment, moving up the von Neumann chain.

To see the von Neumann chain imagine that in addition to the original $|F\rangle_1$ forward moving spatial degree of freedom there was a left detector in the $|S\rangle_L$ for "set" and a right detector in the state $|S\rangle_R$ for "set" and that they can switch to the states $|A\rangle_L$ and $|A\rangle_R$ for "activated" so that


evolves to $$|S\rangle_L|S\rangle_R|+\rangle_1|R\rangle_1|-\rangle_2-|S\rangle_L|S\rangle_R|-\rangle_1|L\rangle_1|+\rangle_2$$ which itself evolves to $$|S\rangle_L|A\rangle_R|+\rangle_1|R\rangle_1|-\rangle_2-|A\rangle_L|S\rangle_R|-\rangle_1|L\rangle_1|+\rangle_2$$ and as this cascades up to invovle many more particles, the ability to get them to overlap again and effective the probaiblity current of each other is ignored.

This is when it is OK to act like there is no more entanglement, it is also when you decide to pretend like a measurement outcome has occurred since it is consistent to treat each outcome as it's own world.

I wanted it to seem clear, but there is more to it than the clean notation suggests. For instance, that original singlet state could have been written in many bases, not just the $\pm$ basis, so it really was an equal (and entangled) mixture of every possible spin, not just an equal mixture of the $\pm$ basis that we happen to choose later to measure it with.

That's really where everything interesting happens. When they were entangled you could have measured them with a device that deflected $|+\rangle+i|-\rangle$ up (U) and deflected $|+\rangle-i|-\rangle$ down (D),$ so the entangled state has the ability to split into two branches for all kinds of different detectors and measurement devices.

And so back before you measured it you couldn't really change the spin of one in a way that changed the ratio of those later branches or the correlations of the outcome in each of those later two branches. And that's what the entanglement meant. Now that the branches act on their own, it is possible to change the polarity of particle one and of particle two on that same branch. That's what it really means that they aren't entangled anymore.


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