# Quantum teleportation of a mixed state through a pure state?

Let's assume we have a register of qubits present in a mixed state $$\rho = \sum_i^n p_i|\psi_i\rangle \langle \psi_i|$$

and we want to teleport $$\rho$$ through a random pure state $$|\phi\rangle$$. What would be the result of this teleportation?

Additionally, what would happen if $$|\phi\rangle$$ is one of the pure states $$|\psi_i\rangle$$, $$\rho$$ is composed of?

I do struggle a little with the density matrix formalism of QC, therefore, I could use some help here.

• What do you mean by "teleport through a random pure state"? Do you know the state? Is it the Bell state used for teleportation? Could you provide a quantum circuit showing what you mean? May 4 at 20:56

The teleportation should behave just the same with a mixed state as it does with a pure state. I'm going to assume a bit of familiarity with how teleportation works for pure states, as you can find many resources addressing that problem.

We can describe the task of teleportation as follows: systems $$M$$, $$A$$, and $$B$$, our goal is to prepare the state $$\rho_M$$ (subscript denotes that its currently stored in $$M$$) in system $$B$$, using only local operations over $$MA$$ plus some preexisting entanglement between $$AB$$. Its simplest to begin with a Bell state prepared over $$AB$$: $$\tag{1} |\Phi_0\rangle_{AB} = \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB})$$

and so the initial state defined over $$MAB$$ is given by: $$\tag{2} \rho_M \otimes |\Phi_0\rangle\langle \Phi_0|_{AB}= \left(\sum_{i=1}^n p_n |\psi_i\rangle\langle \psi_i|_M \right)\otimes |\Phi_0\rangle\langle \Phi_0|_{AB}$$

When one works out teleportation for a pure state $$|\psi_i\rangle$$ its straightforward (but tedious) to show that $$\tag{3} |\psi_i\rangle_M\otimes |\Phi_0\rangle_{AB} = \frac{1}{2}\sum_{\ell=0}^3 |\Phi_\ell\rangle_{MA} \otimes \sigma_\ell|\psi_i\rangle_B$$

where $$\{\sigma_\ell\}$$ are the pauli operators $$\{I,X,Y,Z\}$$ and $$\{|\Phi_\ell\rangle \}$$ are the Bell basis states$$^1$$ for $$\ell=0,1,2,3$$. But since the tensor product is linear we can just substitue (3) into (2) to rewrite the initial state before teleportation as: \begin{align}\tag{4} \rho_M \otimes |\Phi_0\rangle\langle \Phi_0|_{AB} &= \sum_{i=1}^n p_n \sum_{\ell=0}^3 |\Phi_\ell\rangle\langle \Phi_\ell|_{MA} \otimes \sigma_\ell|\psi_i\rangle\langle \psi_i|_B \sigma_\ell \\ &= \sum_{\ell=0}^3 |\Phi_\ell\rangle\langle \Phi_\ell|_{MA} \otimes\left(\sum_{i=1}^n p_n\sigma_\ell|\psi_i\rangle\langle \psi_i|_B \sigma_\ell\right) \end{align}

So now the teleportation protocol can proceed just the same as with pure states: Perform the measurement $$\{|\Phi_\ell\rangle\langle \Phi_\ell|\}$$ on $$MA$$, classically transmit the measured bit $$\ell$$ to $$B$$, and then have $$B$$ perform a recovery operation $$\sigma_\ell$$ to get back the correct state.

It might help to view how this works using an ensemble description of $$\rho$$ where we can describe $$\rho$$ as a classical distribution over states $$|\psi_i\rangle$$ that are drawn with probabilities $$p_i$$: $$\tag{5} \rho = \{(p_1, |\psi_1\rangle), \dots, (p_i, |\psi_i\rangle), \dots (p_n, |\psi_n\rangle)\}$$ From this perspective, every time you run the teleportation protocol you can imagine that you are actually randomly drawing and teleporting a pure state $$|\psi_i\rangle$$ with probability $$p_i$$. Then the the distribution of teleported states is still just $$\{(|\psi_i\rangle, p_i)\}$$. This makes it clear that the mixed state teleportation will work regardless of what entangled state over $$AB$$ you begin with.

Additionally, what would happen if $$|\phi\rangle$$ is one of the pure states $$|\psi_i\rangle$$ that $$\rho$$ is composed of?

I don't think this make sense. To teleport an $$s$$-dimensional state you will generally require entanglement over a $$2s$$-dimensional system $$AB$$. For example, it would not make sense for us to include $$|\Phi_0\rangle \langle \Phi_0|$$ in the ensemble $$\rho$$ because the dimensionalities of systems $$M$$ and $$AB$$ are off by a factor of two

$$^1$$ There are many choices for Bell basis states; for concreteness you can consider the basis $$\{|\Phi_0\rangle, (\sigma_1\otimes I)|\Phi_0\rangle, i(\sigma_1\otimes \sigma_3)|\Phi_0\rangle, (I\otimes \sigma_3)|\Phi_0\rangle\}$$

What do you mean by teleportation "through" some state?

In essence, teleportation just swaps the states of stationary qubits:

$$|\psi\rangle_A | \Phi_{start} \rangle_{AB} ~~\longrightarrow~~ |\Phi_{end}\rangle_{AA} |\psi\rangle_B$$ Here $$| \Phi_{start} \rangle$$ and $$|\Phi_{end}\rangle$$ are some Bell states.
Two of 3 particles belong to Alice and one to Bob.

If you put some mixed state into the algorithm then the result will be the same: $$\rho \otimes |\Phi_{start} \rangle \langle \Phi_{start} | ~~\longrightarrow~~ |\Phi_{end}\rangle \langle \Phi_{end} | \otimes \rho.$$

To understand this intuitively you can think of a mixed state $$\rho$$ as a classical probability distribution over states $$|\psi_i\rangle$$ (it doesn't matter that the choice of $$\psi_i$$ is not unique in general). So, the initial state of teleportation will be a mixture of $$|\psi_i\rangle_A | \Phi_{start} \rangle_{AB}$$ with the same weights as in $$\rho$$. What we'll get after Alice's measurement? Measuring a mixed state is equivalent to measuring each component with the assumption that results will coincide. So, after Alice obtained one of 4 possible results we can think that the measurement of each component $$i$$ gave the same result. The post-measurement state will be the mixture of post-measurement states for each component, e.g. a mixture of $$|\Phi_{end}\rangle_{AA} |r_i\rangle_B$$, again with weights from $$\rho$$. Since the results are the same for each component then Bob's final correction will also be the same. That is, the final state will be a mixture of $$|\Phi_{end}\rangle_{AA} |\psi_i\rangle_B$$ – with the same weights as in $$\rho$$.