Before measuring a quantum particle(photon) it exists in a superposition state, once we observe(measure) it, it settles in one of the possible states(destroying superposition). For entangled particles, does measuring destroy entanglement as well?


5 Answers 5


Yes: If you measure an entangled property, you destroy the entanglement, always.

I think @OndřejČernotík has already already nicely answered this one, but I think your question included the assumption that the measurement and the entanglement were talking about the same property, e.g. spin polarization. And if so, the answer is just a simple yes.

Superposition and entanglement slight variants of the same phenomenon, that being the ability of a quantum system to contain more than one possible state at the same time. In superposition the multiple states mostly share a single relatively small region in xyz space, such as an atom, whereas in entanglement the states may be quite large.

Here's as more specific example of that commonality, first given by Einstein: If you have a very large wave function and then find the particle at one distant part of that wavefunction, how is it that that probability of finding that same particle at some other large region of the wave function a light second "instantly" drops to zero? How did that distant part of the wave function "know" at an apparently superluminal velocity that the particle had already been found?

Einstein's example was just another form of entanglement, not of conserved angular momentum but of mass-energy. The universe insists on absolute conservation of both, so in both cases finding the property (angular momentum or mass-energy) at one well-defined location requires that that property be balanced out or removed from the rest of the universe, no matter how large the wave function has become.

In terms of superposition, such examples show that any quantum wave function contains entanglement, even if it is "just" a smooth, simple wave packet for the particle location. It's just that the entanglement issue does not show up clearly until you make the wave packet so large that its curious conflicts with the speed of light become apparent.


Entanglement is a matter of perspective. It depends on how you split a system into subsystems. Say you have a system $S$ with an associated Hilbert space $\mathcal{H}$, which can be written as a tensor product of two factor spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ with their associated subsystems $S_1$ and $S_2$: $$\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$$ Then you can classify the states in $\mathcal{H}$ as either entangled or separable with respect to these subsystems. Specifically, if there is a pure state representation for either subsystem, then the total state is separable, otherwise it is not and the subsystems are entangled. Whether a measurement destroys this entanglement depends on what you measure. Any measurement that happens entirely in $S_2$ or $S_2$ will result in a pure state in the measured subsystem therefore destroy the entanglement.

However you can also perform a measurement on both systems at the same time which does not necessarily lead to a pure state in the subsystems, but only a pure state in the total system. In that case the entanglement of the two subsystems can be preserved.

If the two subsystems are spatially separated like in a two remote particle setup, then you can naturally only measure locally on one of the subsystems and destroy the entanglement between the two particles.


That depends on what kind of measurement you have in mind. Don't forget that often what is a measurement outcome from one point of view is a superposition from another. So some kind of superposition can still be preserved.

Take, for example, a Bell state measurement, which projects a pair of particles (not necessarily entangled before the measurement!) onto one of the maximally entangled Bell states. But when you take just one of the particles, say, polarization entangled photons, and measure its polarization, the entanglement is destroyed by the measurement.


I will make a more concrete example. For two electrons it is possible to have a state where two particles are entangled, such that their spins are antiparallel.

In this context, the Hilbert space of a composite system $H$ can be viewed as the direct sum of the subspaces $H_S$ and $H_A$ consisting respectively of the symmetric and antisymmetric vectors: $$H = H_S \oplus H_A = (H1 \otimes H2)_S \oplus (H1 \otimes H2)_A.$$ For example a generic state state $\chi$ in the composite Hilber space $H$can be $\chi = a(|+\rangle \otimes |+\rangle ) + b(|+\rangle \otimes |-\rangle )$, with the condition that $|a|^2+|b|^2=1.$

Using the notation $|s \ m_s\rangle$, like the one used to describe the Helium atom, two entangled states, which will always result in antiparallel spin are the following ones: $$ |1 0\rangle = \frac{1}{\sqrt{2}} \bigl( |+\rangle \otimes |-\rangle + |-\rangle \otimes |+\rangle \bigl) = \frac{1}{\sqrt{2}} \bigl( |+-\rangle + |-+\rangle \bigl) $$ $$ |1 0\rangle = \frac{1}{\sqrt{2}} \bigl( |+\rangle \otimes |-\rangle - |-\rangle \otimes |+\rangle \bigl) = \frac{1}{\sqrt{2}} \bigl( |+-\rangle - |-+\rangle \bigl) $$

Let's note that $|10\rangle$ is a symmetric vector while $|0 0\rangle$ is an antiysmmetric one, so they are eigenvectors of the permutation operator $\Pi$.

If at a given time the state of the system consisting of the two electrons is represented by a vector in $H_S$ , since all the observables (related to the composite system) commute with $\Pi$, the state of the system will always remain in $H_S$, as a consequence of either time evolution or of the perturbation due to any measurement processes.

For the same reason, if initially the state is represented by a vector in $H_A$, it will always remain in $H_A$, no matter what will happen to the system.

Entanglement states for spin, are exactly these states belonging to $H_S$ and $H_A$, and thus are not destroyed by measurements on the composite systems (measurement of a nondegenerate observable).


A measurement by some detector is when one of your entangled qubits interacts with your detector in some way such that it triggers a reaction in the detector that you can measure, for photons you use Avalanche photo diode's (APD).

When you're photon hits the APD, it triggers an electron to move which in turn should cause more electrons to do, and this 'avalanche's' until you have a measurable current.

What happens to the photon however is that it interacts with the atoms in the APD in many, probabilistic ways. This is how that, after the APD hit, measurements on the photons entangled pair will become probabilistic, because there is no way to know what is going on with the first photon.

Perhaps they are theoretically still entangled and have corrolations between all the tiny measurements made on the first photon by the APD, but it would be impossible to tell. It's easier to think that the entanglement is just gone, immeasurably so.


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