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I'm reading Qiskit's documentation. https://qiskit.org/textbook/ch-gates/multiple-qubits-entangled-states.html#3.2-Entangled-States- and they show qubits entangled as $|00\rangle$ (both qubits spin down) or $|11\rangle$ (both qubits spin up). Also written as

$$ \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) $$

From my understanding, entanglement requires one particle to be spin up and the other spin down. How is the above entangled state possible?

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  • $\begingroup$ "From my understanding, entanglement requires one particle to be spin up and the other spin down" - that's actually not the case! Entanglement can occur even if both paricles are spin up or both are spin down. It just requires that if you measure one of the particles, the state of the other particle changes. Another equivalent way of thinking about it is that two particles are entangled if they are not "seperable" i.e. that you cannot write down their state as a tensor product state $|a\rangle \otimes |b\rangle$. $\endgroup$ Jul 8 at 9:48
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Singlet and other entangled states

I think you’re getting confused by the singlet state. This is the state $\frac{1}{\sqrt{2}}(|10\rangle - |01\rangle)$. This is a special state because it has an overall spin quantum number $s=0$, from which other interesting properties arise. For QC, the most interesting properties about this state have to do with expectation values. For an example look at this answer.

However, the singlet state is not the only entangled state. You can define entangled states as those for which you can know everything about the composite system but nothing about the individual parts of it. In other (more mathematical) words, those states that cannot be expressed as product states.

Take a look at the Bell states, they are the maximally entangled states for two qubits. The first Bell state, $|\Phi^+\rangle$, is the one you mention in your question.

Another, very simple, way of think about entangled states is by thinking of states in which the result of measuring one qubit tells you something about the other qubit(s) (see below). This relation doesn’t necessarily need to be an inverse relationship like in the singlet state.

Classical and quantum correlation

As @Adam Zalcman pointed out in the comments, “the result of measuring one qubit tells you something about the other qubit(s)” can also apply to separable states like $\frac12|00\rangle\langle00|+ \frac12|11\rangle\langle11|$. This last state is a mixed state, not an entangled state. What it is saying is that the only information we have about the system is that it is in the $|00\rangle$ or $|11\rangle$ state, each with $1/2$ probability.

In this case, measuring a $0$ in one qubit will tell you the second qubit is also $0$, and the same for $1$. However, this doesn’t mean the state is entangled. The correlation arises from our ignorance about the system, and it is therefore a classical correlation, not a quantum correlation like with entangled states.

For some more on mixed and entangled states, check out this answer from Physics.SE.

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  • $\begingroup$ Nice answer overall. However, the remark that for entangled states "the result of measuring one qubit tells you something about the other qubit(s)" is misleading because it applies equally well to separable states such as $\frac12|00\rangle\langle 00| + \frac12|11\rangle\langle 11|$. IOW, the description fails to distinguish between classical and quantum correlations. $\endgroup$ Jul 2 at 5:35
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    $\begingroup$ @AdamZalcman thanks for pointing that out! I wasn’t very careful when saying it. What I tried to say was that for the Bell states, measuring the state of one qubit will tell you the state of the other. But yeah, should’ve been more careful with that remark. I’ll make edit that now. $\endgroup$
    – epelaaez
    Jul 2 at 5:41
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    $\begingroup$ FWIW, I think it's very difficult to explain entanglement without any math at all. Most natural language descriptions fall into the trap of lumping classical and quantum correlations together. In fact, earlier in your answer you already did an excellent job of pointing out the key difference: while the state of the composite system is definitive, the states of individual parts are not. Therefore, an entirely good way to fix the answer is to just remove the final paragraph :-) $\endgroup$ Jul 2 at 5:53
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    $\begingroup$ @AdamZalcman thanks for the feedback! I added a remark about what you mentioned earlier, I tried to keep it simple. Let me know if you think this is appropriate. If not, I’ll just stick with the answer given in the earlier paragraphs. $\endgroup$
    – epelaaez
    Jul 2 at 5:58
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    $\begingroup$ BTW, I don't think it is really necessary to mark edits explicitly in text. If anyone is interested in answer history then that is available by clicking the "edited" link. The explicit "edit" markers stand in the way of readability and exposition. Ultimately, the goal is to write comprehensible answers that explain the question :-) $\endgroup$ Jul 2 at 6:06
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As @epelaaez noted in their answer, by definition, a state is entangled if it cannot be expressed as the product of smaller qubit states. For example, $$ \frac{1}{2}\left(|00\rangle + |01\rangle + |10\rangle + |11\rangle\right) =\left[\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)\right]\left[\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)\right]$$ is not entangled because it factors into the product of $\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)$ with itself.

However, you cannot factor $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ since $$\begin{aligned}\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) &= \frac{1}{\sqrt{2}} (|a_0\rangle + |a_1\rangle)(|b_0\rangle + |b_1\rangle)\\ &= \frac{1}{\sqrt{2}} (|a_0b_0\rangle + |a_0b_1\rangle + |a_1b_0\rangle + |a_1b_1\rangle)\end{aligned}. $$ To get the $|00\rangle$ term, you need $a_0=0$ and either $b_0$ or $b_1$ to equal $0$ (with the other one not present): $$\frac{1}{\sqrt{2}} (|0\rangle + |a_1\rangle)(|0\rangle),$$ but then there is no way to get the $|11\rangle$ term from the factorization.

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