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Consider the standard quantum teleportation scheme. Let the first register hold the state $\newcommand{\ket}[1]{|#1\rangle}\ket\psi$ to be teleported, and the second register be the one shared between the two parties. We can then summarise the protocol as follows:

  1. Start with the state $\ket{\psi}_1\ket{0,0}_{23}$
  2. Apply the unitary $U\equiv \operatorname{CNOT}(H\otimes I)$ between second and third register, thus evolving $\ket{00}$ to $\ket{\Psi^+}\simeq \ket{00}+\ket{11}$.
  3. Apply $U^{-1}$ between first and second registers.
  4. Measure the first two registers in the computational basis, and observe that for every outcome there is a local operation that can be applied to the third register that gives back $\ket\psi$.

My question is about the choice of this specific unitary $U$: Consider the following generalisation of the teleportation scheme:

enter image description here

Here, we again start with $\ket\psi\otimes\ket{0,0}$, but now apply some unitary $B$ between second and third register and then $B^{-1}$ between first and second. We then measure the first two registers in the computational basis.

What are the possible unitaries $B$ such that this circuit works as a teleportation protocol? More specifically, for what choices of $B$ can we always find a unitary to apply to the third register to get back $\ket\psi$ regardless of the measurement results $a,b$?

Intuitively, $B$ must be some gate creating maximal entanglement between the qubits, but requiring that $B$ sends $|00\rangle$ to a maximally entangled state is also not enough. As a counterexample, we can consider the following $B$: $$B\equiv \begin{pmatrix} 0&0&1&0 \\ 1/\sqrt2 & 1/\sqrt2 & 0 & 0 \\ 1/\sqrt2 & -1/\sqrt2 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}.$$ It is not hard to check that with this choice teleportation is not possible, despite the two parties sharing a maximally entangled state. For example, the outcome $a=b=1$ projects the third qubit to $\ket0$, which contains no information about $\ket\psi$.

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    $\begingroup$ Do you require applying $B$ and then $B^{-1}$, or could the second unitary have some other relation to the first unitary? $\endgroup$
    – Mark S
    Mar 2, 2020 at 21:43
  • $\begingroup$ @MarkS I think both cases are interesting. I used $B^{-1}$ in analogy to what is used in the standard scheme, but if the answer is given more naturally lifting this constraint that is also fine. I suspect understanding the answer to one problem also easily leads to the answer of the other though. $\endgroup$
    – glS
    Mar 2, 2020 at 23:59
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    $\begingroup$ What a neat question. I'm inspired to think that $\vert\psi\rangle_1$ could be a transmon qubit; Further $\vert0,0\rangle_{23}$ could be trapped ion qubits, and $B$ could be a gate acting homogeneously among trapped ion qubits.. As I understand it nothing precludes the existence of a gate $B'$ acting heterogeneously therebetween on qubits $_{12}$. You could teleport a transmon qubit into a trapped ion qubit. $\endgroup$
    – Mark S
    Mar 3, 2020 at 0:34

2 Answers 2

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A necessary and sufficient condition on the unitary $B$ is that its columns all correspond to maximally entangled states.

There also does not need to be any relationship between the two unitaries labeled $B$ and $B^{-1}$ in your figure: as long as you start with a maximally entangled state of systems 2 and 3, and then measure systems 1 and 2 with respect to an orthonormal basis of maximally entangled states, there will be an operation (which depends on the measurement outcome) on system 3 that recovers the original state.

Moreover, this is all true not only for qubits, but for any single dimension $d$ shared by systems 1, 2, and 3.

These facts were proved by Werner:

R. Werner. All teleportation and dense coding schemes. Journal of Physics A 34: 7081-7094, 2001 (link to paper).

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  • $\begingroup$ Thanks for interesting answer. Could you please provide an example of another teleportation protocol? Is there any practical reason for using a protocol with different unitaries than that with preparing Bell state? $\endgroup$ Mar 3, 2020 at 5:44
  • $\begingroup$ As the paper cited in my answer shows, you get an example of a teleportation protocol from any orthonormal basis of maximally entangled states. If you have one such basis, say the Bell basis as in the usual qubit teleportation protocol, you can always generate a new example by multiplying each basis vector by $V\otimes W$ for any choice of unitaries $V$ and $W$. Up to global phases, this already gives you every orthonormal basis of maximally entangled states in the qubit case. $\endgroup$ Mar 3, 2020 at 21:06
  • $\begingroup$ For larger local dimensions, you can get an othonormal basis of maximally entangled states through the so-called discrete Weyl operators, and there are other constructions, including ones based on Latin squares. A search for the key words "unitary error bases" will reveal several examples. $\endgroup$ Mar 3, 2020 at 21:07
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    $\begingroup$ Regarding practicality, I suppose one may consider (as an example) that it is more efficient to teleport a qutrit using a qutrit-qutrit maximally entangled state and a measurement with 9 outcomes than it would be to embed the qutrit state into the state of two qubits, which require two maximally entangled qubit-qubit states and a measurement with 16 outcomes to teleport -- but this sort of advantage is small and comes at the cost of a more complicated protocol. In terms of asymptotic cost, the standard teleportation protocol is already maximally efficient in some sense. $\endgroup$ Mar 3, 2020 at 21:08
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Looking back at this question, and with the hindsight of the other given answer, I realised the original framing is somewhat ill-posed. Thinking of teleportation with as in the figure with $B$ and $B^{-1}$ is a bit misleading: what matters is only that the two parties share a maximally entangled state, and there needs be no relation between the unitary used to generate said entanglement and the one used to perform the measurement with $|\psi\rangle$, the state to send over.

For a more thorough description of the protocol hinted at in this other answer, see also this other answer of mine.

The gist is: projecting on Alice's side onto a maximally entangled state of the form $|\Phi\rangle=(U\otimes I)|\Psi\rangle$ (where $|\Psi\rangle\equiv\sum_i |i,i\rangle$ is some reference maximally entangled state), results in Bob's holding the state $U^\dagger |\psi\rangle$, which is then easily corrected by applying the (local) unitary operation $U$.

Thus, if we can find a basis of maximally entangled state on which to measure on Alice's side, we're through. Luckily, this can be done taking the states $|\Psi_i\rangle\equiv(U_i\otimes I)|\Psi\rangle$ for any basis of unitary operations, thus satisfying $\operatorname{Tr}(U_i^\dagger U_j)=d\delta_{ij}$ with $d$ dimension of the space, and with $i=1,...,d^2$. Thus, measuring on this basis, gives some outcome $i$, corresponding to which Bob's state is $U_i^\dagger |\psi\rangle$. Alice sends Bob the value of $i$, and he uses it to apply the correct operation and recover $|\psi\rangle$.

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