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I have read before that classical correlations between qubits does not guarantee entanglement, but is the opposite also true?

Consider a 3 qubit system, prepared as follows:

$1-$ A Hadamard gate is applied to qubit 0
$2-$ A CNOT gate is applied where the target is qubit 1 and the control is qubit 0
$3-$ A Hadamard gate is applied to qubit 2
$4-$ A CNOT gate is applied where the target is qubit 1 and the control is qubit 2

Is this state entangled? $$\frac12 (|000\rangle + |011\rangle + |101\rangle + |110\rangle)$$ If it is, how would I know, given that I do not see how the measurement of any one qubit gives me information about the states of the other qubits?

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    $\begingroup$ How about when you measure two qubits. Do you know anything about the third? What does that say about the entanglement? $\endgroup$
    – Mark Spinelli
    Commented Oct 3, 2023 at 23:54
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    $\begingroup$ sure there is correlation in that system. It's the same correlation you'd have for 3 bits that can be in one of the states 000, 011, 101, 110 with equal probability. You have maximal correlation between the first bit and the other two (taken together). Or more generally between any bit and the other two. It's true that there is no correlation between any pair of individual bits, but that doesn't mean lack of correlation altogether $\endgroup$
    – glS ♦
    Commented Oct 4, 2023 at 2:28
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    $\begingroup$ I think this may come down to how you define classical correlation. Do you insist on always measuring both qubits in the Z basis? If so, a state like $|0+\rangle+|1-\rangle$ might be a better example? It is clearly maximally entangled (apply Hadamard on second qubit to give Bell state), but a $Z$ measurement on either qubit always has 50:50 outcomes on the other qubit, no matter the measurement result. $\endgroup$
    – DaftWullie
    Commented Oct 4, 2023 at 8:31
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    $\begingroup$ The state you showed is indeed entangled. A necessary and sufficient mathematical condition for a state to be entangled is that it cannot be written as a product state, e.g. in your example, you cannot write your state as $(a|0\rangle+b|1\rangle) \otimes (c|0\rangle+d|1\rangle) \otimes (e|0\rangle+f|1\rangle)$ $\endgroup$
    – sheesymcdeezy
    Commented Oct 4, 2023 at 13:00
  • $\begingroup$ Okay, I know get that we can notice the correlation when measuring two qubits, but is that always the case? For example, I read here that entangled qubits can be quantum correlated without being classically correlated. Is there a way to think about this intuitively or is the main way is to try and see whether the states can be decomposed into product states? $\endgroup$
    – yousef elbrolosy
    Commented Oct 4, 2023 at 14:35

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Is this state entangled? $$\frac12 (|000\rangle + |011\rangle + |101\rangle + |110\rangle)$$

Yes.

If it is, how would I know,

You know by comparing to the most general three-qubit unentangled state: $$ \left(a_0|0\rangle+a_1|1\rangle\right)\times \left(b_0|0\rangle+b_1|1\rangle\right)\times \left(c_0|0\rangle+c_1|1\rangle\right) $$ $$ =a_0 b_0 c_0 |000\rangle + a_0 b_0 c_1 |001\rangle + a_0 b_1 c_0 |010\rangle + a_0 b_1 c_1 |011\rangle + a_1 b_0 c_0 |100\rangle + a_1 b_0 c_1 |101\rangle + a_1 b_1 c_0 |110\rangle + a_1 b_1 c_1 |111\rangle $$ $$ \stackrel{?}{=} \frac{1}{2}|000\rangle + \frac{1}{2}|011\rangle + \frac{1}{2}|101\rangle + \frac{1}{2}|110\rangle $$

In other words, in matrix form, you are comparing: $$ \left( \begin{matrix} a_0b_0c_0\\ a_0b_0c_1\\ a_0b_1c_0\\ a_0b_1c_1\\ a_1b_0c_0\\ a_1b_0c_1\\ a_1b_1c_0\\ a_1b_1c_1 \end{matrix} \right) \stackrel{?}{=} \left( \begin{matrix} 1/2\\ 0\\ 0\\ 1/2\\ 0\\ 1/2\\ 1/2\\ 0 \end{matrix} \right) $$

If your state was unentangled, it would require that $a_0\neq 0$ and $b_0\neq 0$ and $c_0 \neq 0$ (since the $000$ coefficient is 1/2).

Since the $001$ coefficient is zero, if your state was unentangled, this implies that $c_1=0$.

Since the $010$ coefficient is zero, if your state was unentangled, this implies that $b_1=0$.

Since the $011$ coefficient is $\frac{1}{2}$, if your state was unentangled, this implies that $a_0 b_1 c_1 = \frac{1}{2}$, but this is a contradiction since we have already established that $b_1=0$ and $c_1=0$ (so $a_0 b_1 c_1 = 0$).

Since we have arrived at a false statement (namely that $0=\frac{1}{2}$) we know that our premise (that the state is unentangled) must be false.

Therefore the state is entangled.

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