I am given the scenario that instead of the two parties (A & B) sharing the bell state $|\phi_+\rangle$ they share the mixture $\rho_\lambda = \lambda|\phi_+\rangle\langle\phi_+|+(1-\lambda)\frac{\mathbb{1}}{2}\otimes \frac{\mathbb{1}}{2}$ in the quantum teleportation protocol where A wants to TP $|\psi\rangle_{in}$. I'm asked to find B state at the end.

Now I know that we can take this on a case by case basis and then take the mixture of the two cases. with probability $\lambda$ they share the bell state and the teleportation is carried in the usual case and B state would be $|\psi\rangle_{in}$. However I'm lost as to what happens in the other case? With probability $1-\lambda$ the share $\frac{\mathbb{1}}{2}\otimes \frac{\mathbb{1}}{2}$ however what is B state at the end of the protocol? I'm confused since I don't know how to work through the protocol with states represented as density operators instead of vector states. The solution says B state is still $\frac{\mathbb{1}}{2}$ since A and B are independent, however I don't understand how this was derived. I do understand that at the beginning B state is $\frac{\mathbb{1}}{2}$ since we partially trace over the full shared state but then I don't know how to proceed from there.


2 Answers 2


So, basically you're asking how to compute the output of a teleportation operation if the maximally entangled state is replaced by the maximally mixed state.

There are a couple of ways that you might do this. The first is brute force calculation. Remember that for each unitary $U$ in the circuit, the density matrix transforms as $\rho\mapsto U\rho U^\dagger$. I don't propose to do that calculation here, but will suggest a couple of shortcuts.

A second option is to recognise that the whole point of the teleportation circuit is to measure the first two qubits in the bell basis $|\Psi_x\rangle$ for $x\in\{0,1\}^2$. Consider one of these. The state after measurement is $$ (\langle \Psi_x|\otimes I) |\psi\rangle\langle\psi|\otimes I\otimes I(|\Psi_x\rangle\otimes I). $$ What should be very clear here is the tensor product structure of (Alice's 2 qubits)$\otimes$(Bob's qubit). Practically, this means, without even doing the calculation, that Bob's state is going to be unaffected by everything else: he just gets $I/2$ as the outcome.

A final way to see the same thing is to decompose the maximally mixed state in terms of the Bell basis: $$ I\otimes I=\sum_x|\Psi_x\rangle\langle \Psi_x|. $$ So, I can think of using the maximally mixed state as choosing, with probability $1/4$, each of the Bell states. But each of those is just the "standard" Bell state that you'd use for teleportation with one of the 4 Pauli matrices $I/X/Y/Z$ applied on Bob's qubit. Thus, the output is $$ \frac14\left(|\psi\rangle\langle\psi|+X|\psi\rangle\langle\psi|X+Y|\psi\rangle\langle\psi|Y+Z|\psi\rangle\langle\psi|Z\right)=I/2. $$

  • $\begingroup$ for the last method say we think of $\psi_+$ which is $\phi_+$ with $X$ applied to B then in the last step of the protocol, B applies a unitary on his side to remake $\psi$ however B state before the last operation unitary is $X|\psi'\rangle$ so after unitary applied it is $UX|\psi'\rangle$ which is not equal to X times the input state since U and X dont necessarily commute. Where am i going wrong? $\endgroup$
    – Luca Ion
    Nov 16, 2022 at 9:47
  • $\begingroup$ You mean in order to adapt for the different measurement results that Alice gets? OK, I skimmed over that part :) Again, take each of the 4 different results separately. The corrections are $I/X/Y/Z$, meaning that all they do is permute the 4 different possible actions. Hence, the sum remains the same. $\endgroup$
    – DaftWullie
    Nov 16, 2022 at 9:55
  • $\begingroup$ right, but you could get corrections s.t one correction permutes say $X|\psi\rangle\langle\psi|X$ into $Y|\psi\rangle\langle\psi|Y$ but the other correction on $Y|\psi\rangle\langle\psi|Y$ leaves it unchanged i.e applies identity; then in this case the sum changes from $X|\psi\rangle\langle\psi|X+Y|\psi\rangle\langle\psi|Y$ to $Y|\psi\rangle\langle\psi|Y+Y|\psi\rangle\langle\psi|Y$ which clearly changes it. $\endgroup$
    – Luca Ion
    Nov 16, 2022 at 20:05
  • $\begingroup$ But if you take the total set of 16 possibilities (from 4 corrections x 4 Bell states), each appears 4 times $\endgroup$
    – DaftWullie
    Nov 17, 2022 at 7:27
  • $\begingroup$ Okai, and one more question, why is that sum equal to Identity/2? $\endgroup$
    – Luca Ion
    Nov 18, 2022 at 14:44

You can work out explicitly how teleportation works for a generic shared state $\rho$.

Standard teleportation protocol — In the standard description, you have a state $|\psi_A\rangle$ that you want to teleport, and a shared (maximally entangled) state $|\Psi_{AB}\rangle=d^{-1/2}\sum_{i=1}^d |i,i\rangle$ that is shared between the two parties, and is consumed to achieve teleportation. More precisely, Alice performs a joint measurement on $|\psi_A\rangle$ (which she initially holds) and her share of $|\Psi_{AB}\rangle$. The outcome of this measurement is then communicated to Bob, which uses it to decide the operation to apply on his share of $|\Psi_{AB}\rangle$ in order to obtain $|\psi\rangle$. Formally, this process can be described saying the overall initial state $|\psi_A\rangle\otimes|\Psi_{AB}\rangle$ is measured projectively with some $\{\Pi_a\}_a$, where $\Pi_a\equiv| u_a\rangle\!\langle u_a|$. And when the measurement outcome is $a$, the corresponding post-measurement (unnormalised) state is $$(\langle u_a\rvert\otimes I_B)(|\psi_A\rangle\otimes|\Psi_{AB}\rangle).$$ If Bob then applies a correction operation that is a unitary $U_a$ (thus conditional to the outcome $a$), the final state is $$(\langle u_a\rvert\otimes U_a)(|\psi_A\rangle\otimes|\Psi_{AB}\rangle).$$ Note that here $\langle u_a\rvert$ acts on the first two (Alice's) spaces, while $U_a$ only acts on Bob's space. We achieve perfect teleportation when the above is (a scalar multiple of) $|\psi\rangle$ itself. Or equivalently, when we have the condition $$(\langle u_a\rvert\otimes U_a)(|\psi_A\rangle\otimes|\Psi_{AB}\rangle) = p_a |\psi\rangle,$$ where $p_a$ is the probability of getting the outcome $a$ when measuring.

Reformulation with mixed states — Let's reformulate the above in the more general formalism of mixed states, channels, and POVMs. Note that I'm not actually generalising anything yet, I'm just writing what the formulas would look like in this different, but still equivalent for now, formalism. We can still consider, without loss of generality, the state to send over as pure, and denote it with $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$. The shared state will now be written as $\mathbb{P}_\Psi$, and let's write the correction unitaries $U_a$ as the corresponding channels $\Phi_a(X)\equiv U_a X U_a^\dagger$. The relation defining successful teleportation will now read $$\operatorname{tr}_A[(\Pi_a\otimes \Phi_a)(\mathbb{P}_\psi\otimes\mathbb{P}_\Psi)] = p_a \mathbb{P}_\Psi.$$

Generalised teleportation — The above immediately suggests a way to frame a "generalised teleportation protocol": given some shared bipartite state $\rho$ and projective measurements $\{\Pi_a\}_a$, when do there exist correction channels $\Phi_a$ such that $$\operatorname{tr}_A[(\Pi_a\otimes \Phi_a)(\mathbb{P}_\psi\otimes \rho)] = p_a \mathbb{P}_\psi.$$ We can immediately see that this cannot be possible in general. For example, if $\rho$ is a separable state, so $\rho=\sum_j q_j \rho_j\otimes\sigma_j$ for some set of probabilities $q_i$ and states $\rho_j,\sigma_j$, then the LHS would give $$ \operatorname{tr}_A[(\Pi_a\otimes \Phi_a)(\mathbb{P}_\psi\otimes\rho)] = \sum_j q_j \operatorname{tr}[\Pi_a(\mathbb{P}_\psi\otimes\rho_j)] \Phi_a(\sigma_j), $$ and it's clear that this cannot give back $\mathbb{P}_\psi$ for all $|\psi\rangle$.

What's achievable with generalised teleportation — So, if we can't achieve teleportation for a generic shared state $\rho$, what can we achieve? One answer is obtained by simply carrying out the calculation assuming the measurement to be a projective measurement on the Bell basis (or more generally, a projective measurement in a basis of maximally entangled states). This means to use as measurement $$\Pi_a \equiv \mathbb{P}(|u_a\rangle)\equiv |u_a\rangle\!\langle u_a|, \qquad |u_a\rangle\equiv (U_a\otimes I)|\Psi\rangle,$$ where again $|\Psi\rangle$ is the "canonical" maximally entangled state, and $\{U_a\}$ is a set of unitaries that are orthonormal, meaning $\operatorname{tr}(U_a^\dagger U_b)=\delta_{ab}$. Note that any orthonormal set of maximally entangled states can be given this form, and vice versa any set of states thus written is a set of orthonormal maximally entangled states. So let's see what we get with this choice of measurement: $$\operatorname{tr}_A[(\Pi_a\otimes \Phi_a)(\mathbb{P}_\psi\otimes\rho)] = \operatorname{tr}_A[ (U_a\otimes I)\mathbb{P}_\Psi(U_a^\dagger\otimes I)\otimes \Phi_a) (\mathbb{P}_\psi\otimes\rho) ] \\ = \frac1d\langle\psi|U_a|i\rangle\!\langle j| U_a^\dagger |\psi\rangle \, (\langle j|\otimes I) (I\otimes\Phi_a)(\rho) (|i\rangle\otimes I)\\ = \frac1d \Phi_a\left( \langle j| U_a^\dagger \mathbb{P}_\psi U_a |i\rangle \rho_{jm,in} |m\rangle\!\langle n| \right) = \Phi_a(\Phi(U_a^\dagger \mathbb{P}_\psi U_a)), $$ where in the last identity we thought of $\rho$ as the Choi of $\Phi$, via the relation $$\rho = (I\otimes \Phi)\mathbb{P}_\Psi, \iff d \rho_{jm,in} = \langle m|\Phi(|j\rangle\!\langle i|)|n\rangle.$$

Conclusion — We thus concluded that for a generic shared state $\rho$, assuming projective measurements on a(ny) basis of maximally entangled states, though we do not always achieve ideal teleportation, we can precisely characterise when we do, because what we get on the other side after measurement and correction $\Phi_a$ is the (unnormalised) state $$\frac{1}{d^2}\Phi_a(\Phi(U_a^\dagger \mathbb{P}_\psi U_a)).$$ You can think of this as a sort of "channel teleportation", meaning that instead of the state $|\psi\rangle$ being faithfully transferred to Bob, he receives the state $\Phi(\mathbb{P}(U_a^\dagger |\psi\rangle))$ conditionally to Alice measuring the outcome $a$. Whether there are channels $\Phi_a$ that recover $|\psi\rangle$ from this for any $a$, depends on $\Phi$, and thus on $\rho$ itself.

Examples — We get back the standard teleportation case when $\Phi=\operatorname{Id}$ is the identity channel, in which case $J(\Phi)=\mathbb{P}_\Psi$ and we recover the same results as before. Another notable case is when $\Phi$ is a unitary channel that commutes with the unitaries $U_a$, in which case it is again easy to correct the operations on Bob's side, and we perform what is often referred to as "quantum gate teleportation" (cf eg How does quantum gate teleportation differ from state teleportation? and What is quantum gate teleportation?).

The case where the shared state is maximally mixed corresponds to using a fully depolarising channel $\Phi$, and we thus immediately see that no information is sent over.

A closely related post about the relation between Choi and teleportation is https://physics.stackexchange.com/q/270032/58382. This answer was in part copy pasted from this other answer of mine on physics.SE.


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