# 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?

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, 2018 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, 2018 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, 2019 at 11:43
• @glS quite probably! I’ll sort later... Jun 3, 2019 at 16:24
• @DaftWullie how do you get the first expression for the state Bob received after alice measured 00. How to find these density matix? Dec 23, 2021 at 11:28

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.

The question has already been answered. Still, I figured I won't hurt if I provide another description of how you get to the general protocol. This will essentially be a rephrasing of this other excellent answer. I'm posting it separately from my other answer here because this is from a completely different angle, and I think I would just make it more confusing if both things are discussed in the same post.

# Approach #1

## Setup

We have a shared maximally entangled state, call it $$|\Psi\rangle=\sum_i |i,i\rangle$$ (I'll ignore normalisation factors throughout this, as they're not pivotal to the discussion).

Let the state to send over be some $$|\psi\rangle$$, held by Alice. The initial state can thus be written as $$|\psi\rangle\otimes|\Psi\rangle \equiv |\psi\rangle_A\otimes|\Psi\rangle_{A'B}.$$ Here, $$A'$$ is the second system held by Alice. Idea being that $$|\psi\rangle$$ is initially held by Alice, while $$|\Psi\rangle$$ is initially held by both Alice and Bob. We want to somehow "send" $$|\psi\rangle$$ from Alice to Bob by only using classical communication (exploiting the state $$|\Psi\rangle$$ which Alice and Bob share).

## Effect of projecting on maximally entangled state

Let $$|\Phi\rangle$$ be some other maximally entangled state, which can thus again be written as $$|\Phi\rangle=\sum_j |u_j,v_j\rangle$$ for some orthonormal bases $$\{|u_j\rangle\},\{|v_j\rangle\}$$. What happens if we project the first two degrees of freedom of $$|\psi\rangle\otimes|\Psi\rangle$$ onto $$|\Phi\rangle$$? Well, we get $$(\langle\Phi|_{AA'}\otimes I_{B})(|\psi\rangle_A\otimes|\Psi\rangle_{A'B}) = \sum_{ij} \langle u_j|\psi\rangle \langle v_j|i\rangle \,|i\rangle_B = \underbrace{\left(\sum_{ij} |i\rangle\!\langle v_j|i\rangle\!\langle u_j|\right)}_{\equiv W} |\psi\rangle,$$ where we can easily notice that $$W$$ is a unitary operation, as $$W=\bar V U^\dagger$$, with $$U,V$$ the unitaries whose columns are $$|u_j\rangle$$ and $$|v_j\rangle$$, respectively. It's also worth noting that these unitaries are tightly related to $$|\Phi\rangle$$, as $$|\Phi\rangle=(UV^T\otimes I)|\Psi\rangle.$$ So, rephrasing the above in a slightly different way, we:

1. Observe that any maximally entangled state $$|\Phi\rangle$$ can be written as $$|\Phi\rangle=(U\otimes I)|\Psi\rangle$$ for some unitary $$U$$.
2. Observe that projecting $$AA'$$ onto $$|\Phi\rangle$$ gives $$(\langle\Phi|_{AA'}\otimes I_{B})(|\psi\rangle_A\otimes|\Psi\rangle_{A'B}) = U^\dagger |\psi\rangle.$$ In conclusion, projecting onto any maximally entangled state, gives an outcome that equals $$|\psi\rangle$$, up to a unitary operation.

## Use a basis of maximally entangled states

The only problem with the above is that one does not simply "project" a state onto something. Rather, one chooses a measurement basis, and sees what comes out. In other words, the projection we computed above tells us what happens when specific states are found as a result of a measurement (in a suitable measurement basis).

So, to use the above, we clearly want to find a measurement basis whose components are all maximally entangled states. If we can do that, we're essentially set. But again, we already mentioned that all maximally entangled states are writable as $$|\Psi_U\rangle\equiv (U\otimes I)|\Psi\rangle$$ for some $$U$$. We then further notice that if $$\operatorname{Tr}(U^\dagger V)=0$$, then $$\langle \Psi_U|\Psi_V\rangle=0$$, which follows immediately from the following general property: $$\langle\Psi|(A\otimes I)|\Psi\rangle = \operatorname{Tr}(A).$$ Thus, if we can find a basis (wrt the $$L_2$$ inner product, for the space of linear operators) of $$d^2$$ unitary operators, i.e. a collection of unitaries such that $$\operatorname{Tr}(U_i^\dagger U_j)=d \delta_{ij}$$, we correspondingly get a basis of maximally entangled states. Denote this basis with $$\{|\Psi_i\rangle\}_i$$.

In conclusion, measuring $$AA'$$ on this basis, when the $$i$$-th outcome is found, the corresponding post-measurement state is $$U_i^\dagger |\psi\rangle$$. It is then enough for Alice to tell Bob (via classical communication or whatever) the value of $$i$$, so that Bob can then apply the unitary $$U_i$$ on his state, and thus retrieving $$|\psi\rangle$$.

# Approach #2

Here's another approach, again heavily inspired by this other answer, which I rather like.

Notice that we kinda want to go from $$|\psi\rangle_A|\Psi\rangle_{A'B}$$ to $$|\Psi'\rangle_{AA'}|\psi\rangle_B$$ for some final $$|\Psi'\rangle$$. One generally thinks of this $$|\Psi'\rangle_{AA'}$$ as consumed in the measurement process, but there's not loss in generality in thinking it to be the post-measurement state or such. Point of this observation being, that the teleportation protocol sort of amounts to a swap operation between $$A$$ and $$B$$.

Here's an interesting observation: we can write the swap operation $$\operatorname{SWAP}_{13}$$ between first and third space, in a generic tripartite system, as $$\operatorname{SWAP}_{13}= \frac{1}{d}\sum_i (U_i\otimes I \otimes U_i^\dagger),$$ for any basis of unitary operations for the space of linear operators over $$\mathbb{C}^d$$, meaing these unitaries satisfy $$\operatorname{Tr}(U_i^\dagger U_j)=d\delta_{ij}$$.

Using this, we can write $$|\psi\rangle|\Psi\rangle = \operatorname{SWAP}_{13}|\Psi\rangle|\psi\rangle = \frac1{d}\sum_{i=1}^{d^2} (U_i\otimes I)|\Psi\rangle\otimes (U_i^\dagger|\psi\rangle) \simeq \sum_{i=1}^{d^2} |\Psi_i\rangle\otimes (U_i^\dagger |\psi\rangle),$$ which makes it clear that projecting $$AA'$$ on $$|\Psi_i$$ and applying $$U_i$$ on $$B$$ achieves teleportation.