# Generalizing quantum teleportation for qudits

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?

## 2 Answers

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): $$|\Psi_{ij}\rangle=(X^iZ^j\otimes\mathbb{I})\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1}|k\rangle|k\rangle.$$ (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.

• Do you happen to have a reference where I might be able to learn more about the topics covered in this answer? – user820789 Dec 24 '18 at 11:06
• Not really. Just open any text book that you like and really understand the qubit case. This really is a trivial generalisation. – DaftWullie Dec 25 '18 at 8:30
• @DaftWullie is the sum in the second equation correct? Shouldn't there be a single index there? Otherwise that's the same as $(\sum_k \lvert k\rangle)\otimes(\sum_j \lvert j\rangle)$, which isn't entangled – glS Jun 3 '19 at 11:43
• @glS quite probably! I’ll sort later... – DaftWullie Jun 3 '19 at 16:24

Quantum teleportation actually works with any unitary basis.

Let $$H$$ be Hilbert space of dimension $$d$$.
Suppose the set of unitaries $$\{ W_{i}, i=1,2,..,d^2 \}$$ is a unitary basis in $$\mathcal{L}(H)$$, that is $$\text{Tr}(W_{i}^\dagger W_j) = d \cdot \delta_{ij}.$$ Then $$|\Psi_i \rangle = \frac{1}{\sqrt{d}} \sum_{j=0}^{d-1} W_i |j\rangle \otimes |j\rangle$$

is an orthonormal basis in $$H \otimes H$$.

Suppose Alice wants to teleport a state $$|\phi \rangle_C \in H_C$$, while with Bob they share an entangled state $$|\Psi_{00}\rangle_{AB} = \frac{1}{\sqrt{d}} \sum_{j=0}^{d-1} |j\rangle_A \otimes |j\rangle_B$$ in the space $$H_A \otimes H_B$$, so the total state is $$|\phi\rangle_C \otimes |\Psi_{00}\rangle_{AB}.$$ It's not hard to show that this total state equals to $$\frac{1}{d} \sum_{i=1}^{d^2} | \Psi_{i} \rangle_{CA} \otimes W_{i}^{-1} | \phi \rangle_B .$$ So, Alice just needs to perform a measurement in the basis $$\{ |\Psi_{i} \rangle_{CA} \}$$ and transmit the result $$i$$ to Bob. Then Bob applies $$W_{i}$$ to his state to obtain $$| \phi \rangle_B$$.