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I am a little confused whether the entangled state and Bell state are the same thing? If they have a bit of contrast, what is the difference between them?

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    $\begingroup$ The four Bell states are regarded as maximally entangled quantum states of two qubits. However, there are unaccountably infinitely many more entangled states. As analogy, $\sqrt{2}$ and $\pi$ are irrational numbers, but we know that they are just two out of the uncountable irrational numbers out there. In similar vein, the four bell states are just an example of four entangled states out of the uncountable infinite number of entangled states. However, they are special so they have their own name. Just like how $\pi$ and $e$ have their own name and symbol. $\endgroup$
    – KAJ226
    Mar 21 at 17:31
  • $\begingroup$ Also, wrt the 4 Bell states of 2 qubits , we can say that once we know the state of one of the qubit, we are sure of the state of the other qubits. For e.g. considering 1/sq rt 2 (|00> +|11>) , once we measure one entangled qubit, with measured value = 0, the other has to be 0. $\endgroup$
    – Kittu A
    yesterday
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A bipartite state $\rho_{AB}$ is called entangled if it cannot be written as $$ \rho_{AB} = \sum_i \lambda_i \sigma_A^i \otimes \sigma_B^i, $$ for some $\lambda_i\geq 0$, $\sum_i \lambda_i = 1$ and $\sigma_A^i$, $\sigma_B^i$ are states on systems $A$ and $B$ respectively, for all $i$. In other words, a state is entangled if it cannot be written as a convex combination of product states.

The four Bell-states are usually referring to the states $$ | \Phi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle) \qquad |\Psi^{\pm}\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle) $$ The Bell states are all entangled in the sense above with for example $\rho = |\Phi^{\pm}\rangle \langle \Phi^{\pm}|$ or similarly for the other states. However, not all entangled states are Bell-states. For example, the states $|\psi\rangle = \cos(\theta) |00\rangle + \sin(\theta) |11\rangle$ for $\theta \in (0, \pi/4)$ are all entangled but are not Bell-states.

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Put in simply words you could say that $$bell \implies entangled$$ but not the opposite. Namely the Bell state is a precise case of the entangled state.

We have that an entangled state is such that it cannot be expressed as a tensor product, i.e. $$\nexists |x⟩ |y⟩: |\phi⟩ = |x⟩\otimes |y⟩$$ in the two qbits case.

On the other hand, the Bell State is a precise state obtained by a precise sequence of transformations resulting in: $$|\psi⟩ = \dfrac{1}{\sqrt{2}}|00⟩+\dfrac{1}{\sqrt{2}}|11⟩$$ Which is indeed entangled (you can prove it) but not the only case of entanglement.

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A simpler answer with fewer mathematical symbols:

The Bell states are examples of entangled states, but there's other examples of entangled states that are not Bell states, for example the W state:

$$ \tag{1} \frac{1}{\sqrt{3}}\left(|001\rangle + |010\rangle + | 100\rangle\right), $$

the GHZ states for example:

$$ \tag{2} \frac{1}{\sqrt{2}}\left(|000\rangle + |111\rangle\right), $$

and many more.

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