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I have seen that a pure state, $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$$, can be teleported from Alice to Bob using Bell measurements and an operation, provided Alice and Bob share the EPR state, $$\frac{1}{\sqrt{2}}\left(|0\rangle_{A}|0\rangle_{B}+|1\rangle_{A}|1\rangle_{B}\right)$$ where the subscripts $A$ and $B$ indicate that the qubit is in Alice or Bobs Node respectively. Suppose instead Alice wishes to transmit one of the qubits in an entangled state in her node, $$|\Psi\rangle=a|00\rangle+b|11\rangle$$ Since we can no longer describe the state of the either qubit as a pure state, is it still possible to perform the teleportation on one of the qubits, such that the entangled state $|\Psi\rangle$ is shared between Alice and Bobs nodes. If so, how can I show this?

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Yes, standard teleportation preserves entanglement in the way that you are looking for. You're basically doing entanglement swapping, wherein Alice holds $|\Psi\rangle_{A_1A_2}$ and half of the Bell state $|\Phi_0\rangle_{A_3 B}$, a Bell measurement is applied to $A_2 A_3$, and then the result is your original state $|\Psi\rangle_{A_1 B}$ split between the two parties.

You can prove that the teleportation works using the same derivation for regular teleportation. The main trick is to expand the initial joint state on $|\Psi\rangle_{A_1 A_2} |\Phi_0\rangle_{A_3 B}$ in terms of a Bell basis on $A_2 A_3$ (see this answer for more details). That looks like: \begin{align} |\Psi\rangle_{A_1 A_2} |\Phi_0\rangle_{A_3 B} &= \frac{1}{\sqrt{2}}(a|00\rangle + b|11\rangle)(|00\rangle + |11\rangle)\\ &= \frac{1}{\sqrt{2}} (a|0\color{red}{00}0\rangle + a|0\color{red}{01}1\rangle + b|1\color{red}{10}0\rangle + b|1\color{red}{11}1\rangle)\\ &= \frac{1}{2} \left(a|0\rangle\color{red}{\left(|\Phi_0\rangle + |\Phi_3\rangle\right)} |0\rangle + a|0\rangle\color{red}{\left(|\Phi_1\rangle +i|\Phi_2\rangle\right)}|1\rangle \right. \\&\qquad\qquad\qquad+ \left. b|1\rangle\color{red}{\left(|\Phi_1\rangle -i |\Phi_2\rangle\right)}|0\rangle + b|1\rangle\color{red}{\left(|\Phi_0\rangle - |\Phi_3\rangle\right)}|1\rangle \right),\\ \end{align}

where $\{|\Phi_\ell\rangle\}$ is a particular choice of maximally entangled basis. Then rearrange the order of the kets to write this as \begin{align} |\Psi\rangle_{A_1 A_2} |\Phi_0\rangle_{A_3 B} &= \frac{1}{2}\left[ (a|00\rangle_{A_1 B} + b|11\rangle_{A_1 B} ) \otimes |\Phi_0\rangle_{A_2 A_3}\right. \\&\qquad+\left. (a|01\rangle_{A_1 B} + b|10\rangle_{A_1 B} ) \otimes |\Phi_1\rangle_{A_2 A_3} \right. \\&\qquad +\left. i(a|01\rangle_{A_1 B} - b|10\rangle_{A_1 B} ) \otimes |\Phi_2\rangle_{A_2 A_3} \right. \\&\qquad +\left. (a|00\rangle_{A_1 B} - b|11\rangle_{A_1 B} ) \otimes |\Phi_3\rangle_{A_2 A_3} \right] \\&= \sum_{\ell=0}^3 (I \otimes \sigma_\ell)|\Psi\rangle_{A_1 B} \otimes |\Phi_\ell\rangle_{A_2 A_3} \end{align}

where $\{\sigma_\ell\}_{\ell=0}^3$ are the Pauli operators. In this form it is clear that a Bell-basis measurement on $A_2 A_3$ (Alice's half of the ebit and half of her entangled state) will 'teleport' that half of $|\Psi\rangle$ to Bob, resulting in $|\Psi\rangle_{A_1 B}$ up to some single-bit recovery operation $\sigma_{\ell}$ to be applied by Bob once he receives $\ell$ (two classical bits).

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