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In an answer to a previous question, Generalization for n quantum teleportations, Craig Gidney states:

The more complicated way to generalize teleportation is figuring out how to make it work on qutrits and qudits instead of only qubits. Basically, instead of using a "basis" made up of tensor products of X and Z matrices, you need to switch to a basis based on clock and shift matrices.

How can quantum teleportation be generalized for qudits?

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Let's define the shift and clock matrices (the generalisation of the Pauli X and Z matrices) as $$ X=\sum_{i=0}^{d-1}|i+1\text{ mod }d\rangle\langle i|\qquad Z=\sum_{i=0}^{d-1}\omega^i|i\rangle\langle i| $$ where $\omega=e^{2\pi \sqrt{-1}/d}$. Now we can define a maximally entangled orthonormal basis (the equivalent of the Bell basis): $$ |\Psi_{ij}\rangle=X^iZ^j\otimes\mathbb{I}\frac{1}{\sqrt{d}}\sum_{k,l=0}^{d-1}|k\rangle|l\rangle. $$ (Reader exercise: verify that $\langle\Psi_{ij}|\Psi_{kl}\rangle=\delta_{ik}\delta_{jl}$.)

The teleportation setup is basically the same as for qubits. Alice holds an unknown qubit state $|\psi\rangle\in\mathbb{C}^d$, and Alice and Bob share the two-qudit state $|\Psi_{00}\rangle$.

Alice performs a measurement between her two qudits using the basis $|\Psi_{ij}\rangle$, and gets an answer $(ij)$. Let's assume that the answer is $(00)$. In this case, Bob receives the state $$ d^2\text{Tr}_A\left(|\Psi_{00}\rangle\langle\Psi_{00}|_A\otimes\mathbb{I}_B\cdot|\psi\rangle\langle\psi|\otimes|\Psi_{00}\rangle\langle\Psi_{00}|\right)=|\psi\rangle\langle\psi|, $$ i.e. the state is perfectly teleported. What happens for the other measurement results? We need to calculate $$ d^2\text{Tr}_A\left(\left(X^iZ^j\otimes\mathbb{I}|\Psi_{00}\rangle\langle\Psi_{00}|Z^{-j}X^{-i}\otimes\mathbb{I}\right)_A\otimes\mathbb{I}_B\cdot|\psi\rangle\langle\psi|\otimes|\Psi_{00}\rangle\langle\Psi_{00}|\right) $$ But this is the same as $$ d^2\text{Tr}_A\left(|\Psi_{00}\rangle\langle\Psi_{00}|_A\otimes\mathbb{I}_B\cdot\left(Z^{-j}X^{-i}|\psi\rangle\langle\psi|X^iZ^j\right)\otimes|\Psi_{00}\rangle\langle\Psi_{00}|\right) $$ so it's as if we're teleporting the state $Z^{-j}X^{-i}|\psi\rangle$ and getting measurement result $(00)$. So, we know that Bob receives $Z^{-j}X^{-i}|\psi\rangle$, so when Alice sends Bob the 2-dit message of her measurement result $(ij)$, he can apply the correction $X^iZ^j$, and he's perfectly received $|\psi\rangle$, no matter what Alice's measurement result was.

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  • $\begingroup$ Do you happen to have a reference where I might be able to learn more about the topics covered in this answer? $\endgroup$ – meowzz Dec 24 '18 at 11:06
  • $\begingroup$ Not really. Just open any text book that you like and really understand the qubit case. This really is a trivial generalisation. $\endgroup$ – DaftWullie Dec 25 '18 at 8:30

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