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?


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.

  • $\begingroup$ Do you happen to have a reference where I might be able to learn more about the topics covered in this answer? $\endgroup$ – user820789 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
  • $\begingroup$ @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 $\endgroup$ – glS Jun 3 '19 at 11:43
  • $\begingroup$ @glS quite probably! I’ll sort later... $\endgroup$ – 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 $| \phi \rangle_B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.