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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 $|\psi\rangle_{in}$. Im 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 im 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? Im confused since I dont 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 dont understand how this was derived. I do understand that at the beginning B state is $\frac{\mathbb{1}}{2}$ since we partial trace over the full shared state but then I dont know how to proceed from there.

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2 Answers 2

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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 $$ \frac14\left(|\psi\rangle\langle\psi|+X|\psi\rangle\langle\psi|X+Y|\psi\rangle\langle\psi|Y+Z|\psi\rangle\langle\psi|Z\right)=I/2. $$

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  • $\begingroup$ 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 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 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 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 at 7:27
  • $\begingroup$ Okai, and one more question, why is that sum equal to Identity/2? $\endgroup$
    – Luca Ion
    Nov 18 at 14:44
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You can also consider the general case of how teleportation works out for a generic shared state via Choi operators.

Suppose the shared state is some $\rho$, which corresponds to some channel $\Phi$ via $\rho=J(\Phi)$, with $J(\Phi)$ denoting the Choi state of $\Phi$. More precisely, let $\Phi$ be an arbitrary channel, and let $J(\Phi)\equiv (I\otimes\Phi)\mathbb{P}_\Psi$ be the corresponding Choi operator (again, I'm neglecting normalisation factors). Here I'm using the shorthand notation $\mathbb{P}_\Psi\equiv \mathbb{P}(|\Psi\rangle)\equiv |\Psi\rangle\!\langle\Psi|$ for projectors, and again denoting with $|\Psi\rangle\equiv\sum_i|i,i\rangle$ the maximally entangled state (of relevant dimension).

In such cases one ends up performing some sort of "channel teleportation", meaning that instead of the state $|\psi\rangle$ being faithfully transferred to Bob, he receives the state $\Phi(\mathbb{P}(\bar U_a |\psi\rangle))$ conditionally to Alice measuring the outcome $a$.

We now measure in the basis of maximally entangled states built as $(I\otimes U_a)|\Psi\rangle$, and compute Bob's state conditional to each outcome $a$. In this case we're dealing with generally non-pure states, so we have to change the formalism accordingly, and the expression becomes: $$ \operatorname{Tr}_{AB}\left\{ \left[\mathbb{P}((I_A\otimes U_a)|\Psi\rangle)\otimes I_C\right](\mathbb{P}_\psi\otimes J(\Phi) ) \right\} = \sum_{ij} \psi_j \bar\psi_i \operatorname{Tr}_B[ (U_a |i\rangle\!\langle j| U_a^\dagger \otimes I_C) J(\Phi)] \\ = \sum_{ij,k\ell} \psi_j\bar\psi_i \operatorname{Tr}[U_a |i\rangle\!\langle j| U_a^\dagger |k\rangle\!\langle \ell|] \,\,\Phi(|k\rangle\!\langle\ell|) = \Phi(\bar U_a \mathbb{P}_\psi \bar U_a^\dagger). $$ The idea of this expression is to take the input total state $(\mathbb{P}_\psi\otimes J(\Phi) )$, and project the first two spaces onto $(I_A\otimes U_a)|\Psi\rangle$, which in the density matrix formalism is done multiplying with the corresponding projector and partial tracing. I labeled here with $A,B,C$ the three relevant spaces. So $|\psi\rangle$ leaves in $A$, and $J(\Phi)$ in $BC$ (so in particular note that when writing $\operatorname{Tr}_B$ above, the partial trace is performed on the first of the two spaces in the inner expression).

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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