# Does the type of entangled pair make any difference in quantum teleportation?

From what I understand, in order to teleport the state of a qubit from place A to place B I need an entangled pair whose entanglement state will be consumed to perform the teleportation. All the resources I found about teleportation explain the procedure with a Bell pair as the entangled pair. However, I understand that we could use any type of entangled pair, not just the Bell one.

Would it make any difference to not use a Bell pair as entangled pair for the underlying swapping procedure? I think I'd need a different set of local correction rules to be performed at B but I'm not sure about it.

I apologize if I wrote the concept in a imprecise way. I am just an engineer, it's my first time reading about quantum mechanics but I need to understand things in order to simulate them with a software.

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Commented Oct 28, 2022 at 2:22

Any maximally entangled state can be used to perform teleportation. This is true when the goal is teleporting a single qubit, but also more generally in arbitrary dimensions.

The general idea is as follows:

1. Let $$|\Psi\rangle$$ be a maximally entangled bipartite state. In some choice of local bases, you can always write this as $$|\Psi\rangle=\sum_i|i,i\rangle$$. This is the "resource state". It's the entangled state shared by Alice and Bob that allows performing teleportation.
2. Now think of Alice's pov. She wants to send over some state $$|\psi\rangle$$, and has available one part of $$|\Psi\rangle$$ to do as she pleases. She then performs a nonlocal measurement on $$|\psi\rangle$$ and her piece of $$|\Psi\rangle$$. More specifically, she needs to perform a measurement in a(ny) basis of maximally entangled states (cf this answer on qc.SE for more info on this). It just so happens that given an orthonormal (wrt the Hilbert-Schmidt inner product) basis of unitaries $$\{U_a\}_a$$, one can write such a basis of maximally entangled states as $$\{(I\otimes U_a)|\Psi\rangle\}_a$$. Note that, as discussed here, one can always find such an orthonormal basis. The post-measurement outcome on Bob's side conditional on Alice finding the outcome $$a$$ is therefore (I'm neglecting phases and normalisation here): $$\big[(\langle \Psi|(I\otimes U_a^\dagger))\otimes I\big](|\psi\rangle\otimes|\Psi\rangle) \equiv \sum_{ij} (\langle i|\otimes \langle i|\otimes I)(I\otimes U_a^\dagger\otimes I)(|\psi\rangle\otimes|j\rangle\otimes|j\rangle) \\ = \sum_{ij} \psi_i (U_a^\dagger)_{ij} |j\rangle \equiv \bar U_a|\psi\rangle.$$
3. We just showed that by performing a measurement on the basis of maximally entangled states given above, whenever Alice finds the outcome $$a$$, Bob ends up with the state $$\bar U_a |\psi\rangle$$. Therefore Bob can simply perform the correcting operation $$\bar U_a^\dagger$$ on his state, conditionally to Alice telling him she found the outcome $$a$$, and teleportation will be achieved.

This reduces to the protocol you might be more familiar with choosing the unitaries $$U_a$$ to be the Pauli operators (and the identity matrix).

Some related posts are:

This is probably outside the scope of this question, but it is interesting to note that the above reasoning can be generalised even further, when the shared state is not necessarily maximally entangled. In such cases one ends up performing some sort of "channel teleportation", meaning that instead of the state $$|\psi\rangle$$ being faithfully transferred to Bob, he receives the state $$\Phi(\mathbb{P}(\bar U_a |\psi\rangle))$$ conditionally to Alice measuring the outcome $$a$$, where $$\Phi$$ is a channel linked in a precise way to the initial shared state.

More precisely, let $$\Phi$$ be an arbitrary channel, and let $$J(\Phi)\equiv (I\otimes\Phi)\mathbb{P}_\Psi$$ be the corresponding Choi operator (again, I'm neglecting normalisation factors). Here I'm using the shorthand notation $$\mathbb{P}_\Psi\equiv \mathbb{P}(|\Psi\rangle)\equiv |\Psi\rangle\!\langle\Psi|$$ for projectors, and again denoting with $$|\Psi\rangle\equiv\sum_i|i,i\rangle$$ the maximally entangled state (of relevant dimension). Note that you could have also equivalently said that the shared state is some arbitrary bipartite $$\rho$$, and then defined the channel $$\Phi$$ as the one such that $$J(\Phi)=\rho$$.

We can now do the exact same calculation carried out above: measure in the basis of maximally entangled states built as $$(I\otimes U_a)|\Psi\rangle$$ and work out Bob's state conditional to each outcome $$a$$. In this case we're dealing with generally non-pure states, so we have to change the formalism accordingly, and the expression becomes: $$\operatorname{Tr}_{AB}\left\{ \left[\mathbb{P}((I_A\otimes U_a)|\Psi\rangle)\otimes I_C\right](\mathbb{P}_\psi\otimes J(\Phi) ) \right\} = \sum_{ij} \psi_j \bar\psi_i \operatorname{Tr}_B[ (U_a |i\rangle\!\langle j| U_a^\dagger \otimes I_C) J(\Phi)] \\ = \sum_{ij,k\ell} \psi_j\bar\psi_i \operatorname{Tr}[U_a |i\rangle\!\langle j| U_a^\dagger |k\rangle\!\langle \ell|] \,\,\Phi(|k\rangle\!\langle\ell|) = \Phi(\bar U_a \mathbb{P}_\psi \bar U_a^\dagger).$$ I labeled here with $$A,B,C$$ the three relevant spaces. So $$|\psi\rangle$$ leaves in $$A$$, and $$J(\Phi)$$ in $$BC$$ (so in particular note that when writing $$\operatorname{Tr}_B$$ above, the partial trace is performed on the first of the two spaces in the inner expression).

We get back the standard teleportation case when $$\Phi=\operatorname{Id}$$ is the identity channel, in which case $$J(\Phi)=\mathbb{P}_\Psi$$ and we recover the same results as before. Another notable case is when $$\Phi$$ is a unitary channel that commutes with the unitaries $$U_a$$, in which case it is again easy to correct the operations on Bob's side, and we perform what is often referred to as "quantum gate teleportation" (cf eg https://quantumcomputing.stackexchange.com/q/25999/55 and https://quantumcomputing.stackexchange.com/q/1806/55).

A closely related post about the relation between Choi and teleportation is What's the intuition behind the Choi-Jamiolkowski isomorphism?.