# How is quantum teleportation generalized to 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?

## 3 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? 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. 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... 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 $$| \phi \rangle_B$$.

Here's a way to formalise a general teleportation protocol without making any assumptions on dimensions or types of measurements and dynamics.

Alice and Bob share a state $$\rho\in\mathrm D(\mathcal H_A\otimes\mathcal H_B)$$ (here, $$\mathrm D(\mathcal H)$$ denotes the set of density matrices acting on a Hilbert space $$\mathcal H$$). Alice also has a state $$\sigma\in\mathrm D(\mathcal X)$$ that she wants to send Bob (assume $$\mathcal H_B\simeq\mathcal X$$, otherwise it becomes harder to see what "transmitting the state" means). Both she and Bob are allowed to do whatever they like to their states, but can only send each other classical information.

Suppose Alice performs some generic measurement, encoded in a POVM $$\mu\equiv\{\mu(a)\}_a\subset\mathrm{Pos}(\mathcal X\otimes\mathcal H_A).$$ Here, $$a$$ labels the set of possible outcomes. Bob's state conditioned to Alice finding the outcome $$a$$ (and assuming he knows which outcome was obtained) thus reads $$\newcommand{\tr}{\operatorname{Tr}}\rho(B|A=a) = N \tr_{\cal X\otimes H_A}[(\mu(a)\otimes I_{\cal H_B})(\sigma\otimes\rho)] %}{\operatorname{Tr}[(\mu(a)\otimes I_{\cal H_B})(\sigma\otimes\rho)]}. .$$ Note that this expression also assumes that Bob knows the initial state $$\rho$$, as well as the measurement $$\mu$$ Alice is going to perform (you can say they pre-arranged all of this before the transmission took place). Note also that for Bob to have the above state, he needs to know the outcome $$a$$. This means that Alice has to send this outcome. However, this is fine, because Alice is allowed to send classical information over to Bob.

The question we pose when discussing quantum teleportation is then: what operation (if any exists) should Bob perform on its state to end up with $$\sigma$$? More precisely, we are asking whether there is a collection of channels $$\Phi_a$$ such that $$\Phi_a(\rho(B|A=a)) = \sigma,\quad\text{for all } a\text{ and }\sigma.$$

Specialising to pure states, unitary evolutions, and projective measurements, the above becomes $$|\phi(B|A=a)\rangle = N (\langle a|\otimes I_{\mathcal H_B})(|\psi\rangle\otimes|\Psi\rangle),$$ and the final condition for teleportation is the existence of unitaries $$U_a$$ such that $$U_a |\phi(B|A=a)\rangle= |\psi\rangle.$$

Of course, this only defines the general formalism. Finding the precise conditions under which this can work, or finding gate decompositions to implement the corresponding unitaries once you found a viable set of unitaries and measurements, is a wholly different problem.