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From my understanding, a qubit is entangled when the state of one qubit depends on the other, and vice versa. Can the following bell states have probability amplitudes other than 1/2 and still be entangled?:

$ |\Phi^\pm\rangle = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B \pm |1\rangle_A \otimes |1\rangle_B) $

${\displaystyle |\Psi ^{\pm }\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |1\rangle _{B}\pm |1\rangle _{A}\otimes |0\rangle _{B})} $

For example, is it possible to have bell pairs with probability amplitudes that are not $ \dfrac{1}{\sqrt{2}}$, but rather something like this: $|\Psi\rangle = \frac{\sqrt{3}}{2} (|00 \rangle + \frac{1}{2} |11\rangle)$

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Absolutely.

Given an arbitrary two-qubit state $|\psi \rangle $, it is NOT entangled if we can write $|\psi \rangle $ as: $$ |\psi \rangle = |a\rangle \otimes |b\rangle \hspace{.75 cm} \textrm{where} \ \ |a\rangle, |b \rangle \in \mathbb{C}^2 $$

Thus, there are many entangled states! Essentially, if you pick a random two-qubit pure state, it is most likely to be an entangled state. Now, not all entangled state are equal. Some are more entangled than other. For example, Bell states are maximal entangled state, but a state like $|\psi \rangle = \dfrac{\sqrt{3}}{2}|00\rangle + \dfrac{1}{2}|11\rangle $ is less entangled than the Bell states. In fact, we can quantify this using the concept of "Concurrence" which is directly related to the concept of "Entanglement of Formation" (Here is the original paper on these concepts).

For the Bell state $|\psi_{Bell} \rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}} $, the Concurrence measurement is $1$, which is the same for entanglement of formation.

For another state, say the one you interested in, $|\phi \rangle = \dfrac{\sqrt{3}}{2}|00\rangle + \dfrac{1}{2}|11\rangle $, you can work out the Concurrence measurement value for this state to be $\dfrac{3}{4}$ which is less than $1$. Thus, one would say that this state is less entangled than the Bell state $|\psi_{Bell} \rangle$.

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  • $\begingroup$ Why are people so obsessed with concurrence? Not to mention that for pure bipartite states, there is no point using is. Just use the entanglement entropy. Note that saying "the concurrence is smaller" is not the real reason why to say it is less entangled. $\endgroup$ – Norbert Schuch Jan 27 at 22:17
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They won't be Bell-pairs though(by definition). Because Bell-pairs are maximally entangled. Meaning, if you take a partial trace over one of the subsystems, the resulting state must have maximum entropy. Which, with unequal amplitudes, would not be possible.

But they can be entangled. Assuming that you wanted a state like: $|\Psi\rangle = \frac{\sqrt{3}}{2} |00\rangle + \frac{1}{2} |11\rangle$, it is indeed entangled. You can look up Peres-Horodecki criterion to detect the presence of entanglement in smaller dimensions. The amplitudes do not have to be equal, the only thing that matters is that there is no way to write it as a product of two states.

The state that you have mentioned at the end, $|\Psi\rangle$, is not a valid quantum state. The squared sum of amplitudes must be 1.

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  • $\begingroup$ There is no point in using the Perez-Horodecki criterion for pure states. $\endgroup$ – Norbert Schuch Jan 27 at 22:17

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