As per Wikipedia:

Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently.

Also if we consider a bipartite mixed state:

we say that a mixed state is separable if it can be written as: $$ \rho = \sum_i p_i \rho_i^A \otimes \rho_i^B $$

So if we consider a mixed state: $$ \frac{1}{2} | 00 \rangle \langle 00 | + \frac{1}{2} | 11 \rangle \langle 11 |$$

As per the definition of mixed entangled quantum state (stated above) we can say that given state is not entangled, but isn't the measurement (in standard basis) on 1st particle will determine the outcome of measurement (in standard basis) of 2nd particle? If yes, doesn't it mean that given mixed state is entangled?

I am new to quantum mechanics, so it might be a silly question but it will be great if someone can explain it to me.


2 Answers 2


The problem is that entanglement doesn't mean that a measurement on the first particle determines the outcome of the second. Although this is often perpetuated, it's not the gist of entanglement.

Entanglement is (mostly?) nonclassical correlations. Bipartite states are correlated, when the outcome of the measurement on one particle tells us something about the outcome of the measurement on the other particle. In your case, this is true and thus the states are correlated. The key difference between correlations and entanglement is the question, whether the states are nonclassically correlated.

In your case, you could imagine the following protocol: Charlie picks a bit (0 or 1) at random from a uniform distribution, duplicated it and gives one of the bits in a hidden fashion to Alice and the other to Bob. The resulting state of the sytem is the one of your question and, if both Alice and Bob know what Charlie has done, "measuring" their bit (i.e. looking what it is) reveals to them what Bob will measure. Since I didn't use quantum mechanics at all, this is what we would consider as classical correlations.

So what's the business with entanglement? To see this, let's have a look at a truly entangled state:

$$ \rho= 1/2(|00\rangle+|11\rangle)(\langle 00|+\langle 11|) =1/2(|00\rangle\langle 00|+|11\rangle\langle 00|+ |00\rangle\langle 11|+ |11\rangle\langle 11|)$$

If you now measure in the $|0\rangle,|1\rangle$-basis on Bob's side, you will - as before - know the configuration of the system on Alice's side immediately. The states are, once again, fully correlated.

But now let's take the basis $\frac{1}{\sqrt{2}}(|0\rangle \pm |1\rangle)$ on Bob's side. Annd this is where the difference comes in. If you measure your state (let's call it $\rho_{clas}$ in this basis and the outcome is that it is in state $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, you can easily calculate that at this moment, Alice's state is given by

$$ \frac{1}{2} |0\rangle\langle 0|+|1\rangle\langle 1| $$

so if Alice measures in the same basis, she will obtain any one of the the states $\frac{1}{\sqrt{2}}(|0\rangle \pm |1\rangle)$ with equal probability (as before) and you can't predict anything from Bob's measurement outcome.

For my state $\rho$, this is completely different. Here, the outcome on Alice's side after Bob obtained $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ is

$$ \frac{1}{2}(|0\rangle+|1\rangle)(\langle 0|+\langle 1|) $$

hence Alice will obtain the same state as Bob, when she measures. And this is it: You can prove that correlations like this cannot occur with separable states. It's just not possible to have this kind of perfect correlation when you measure in different bases (see Bell inequalities).

The above picture is a bit simplified. In particular, the complete link between nonclassical correlations and entanglement is not completely clear. For instance, not every entangled state violates a Bell-inequality, so the correlations might be a bit more intricate, but the above should tell you enough to answer your question. Also, it is not clear whether separable states might not also sometimes have nonclassical correlations and maybe there is a better measure of "nonclassical correlations" than entanglement (see quantum discord), but this is still debated upon.


$\newcommand{\HH}{\mathcal{H}}$As Martin says, entanglement is correlation rather than anything like "determining" the state of the other particle. We don't necessarily even need to talk about correlations, though, although they are one of the primary interesting features of entangled states.

More precisely, your quote

Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently.[emphasis mine]

needs to be interpreted carefully to get its correct meaning:

Given two systems with spaces of states $\HH_1,\HH_2$, the space of states of the combined system is the tensor product $\HH_1\otimes\HH_2$. If we are given two bases $a_i$ and $b_j$ of the subsystem spaces, it is spanned by all possible combinations $a_i\otimes b_j$ (its dimension is hence the product of the dimensions of its components).

Entanglement is, in its essence, now the simple observation that there are states in $\HH_1\otimes\HH_2$ which cannot be written as $\psi\otimes\phi$ for $\psi\in\HH_1$ and $\phi\in\HH_1$, since the space of such states would be only the Cartesian product $\HH_1\times\HH_2$, which is strictly smaller than the tensor product if the spaces are more than two-dimensional, since its dimension is only the sum of the dimensions of its components.

Those states which cannot be written as the tensor product of states of the subsystem are called entangled or inseparable, because they cannot be described by independent states of the subsystems.

This extends to the mixed state/density matrix formalism by observing that not every matrix on the tensor product is a sum of tensor products of matrices on the subsystem (again, the dimensions don't match), so a general mixed state is more than weighted sums of mixed states of the subsystems.

But, at the end, entanglement here doesn't ever talk about particles, or measurement. It is solely an abstract property of the state, and it is a form of combined state that is classically non-existent, since classically, the Cartesian product of configuration spaces gives the total configuration space. This difference means that entangled states often exhibit "non-classical behaviour" in some form or other, but they are not defined through that behaviour.

  • $\begingroup$ Does that mean that one could check if there is any spooky interaction by making 2 experiments : if one measures only one side "without" measuring the other subsystem the light should be unpolarized, but if one measures subsystem B then the light in A should be polarized, for example ? $\endgroup$ Commented Feb 8, 2022 at 7:08
  • $\begingroup$ Since the reduced density matrix of a singlet state is a mixed state it is unpolarized ? But what happen to the density matrix if one measures in both places ? $\endgroup$ Commented Feb 8, 2022 at 7:16

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