# Why do we use the Bell state $|00\rangle+|11\rangle$ for quantum teleportation?

In the quantum teleportation protocol we use the Bell state given by $$\frac{1}{\sqrt{2}} \left( |00\rangle + |11\rangle\right).$$ My intuition tells me this works is because we can transform this Bell state to any other Bell states through a unitary transformation. And thus we can use the state $$|\phi \rangle = a|0 \rangle + b |1 \rangle$$ we want to teleport to perform controlled operations on it. I'm wondering is that the reason why we use a Bell state to perform quantum teleportation? And if there exists other states that we can use, what are the constraints?

• This paper might be helpful. Commented Feb 9, 2022 at 7:39

The constraint is essentially that the state you use has to be maximally entangled.

From a practical perspective (sufficiency), you can understand this in that, via the Schmidt decomposition, you can think of any maximally entangled state as the Bell state up to local unitaries (in, fact, a single unitary on, say, the output, is sufficient). You can easily compensate for whatever those unitaries are within your protocol. Just as, as you say, the transformation between the Bell basis propagates through and you can compensate with a correction at the end.

From a necessity perspective, one of the things you need is to not learn anything about your input state when you make the measurements. This means the outcome probabilities must be independent of the state, and therefore all $$1/4$$. This essentially imposes that the reduced density matrix of your entangled state must be maximally mixed which, if it's a pure state, means that it's maximally entangled.