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.

  • 4
    $\begingroup$ 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? $\endgroup$ May 4, 2021 at 20:56
  • $\begingroup$ @CraigGidney Teleportation "through" a state feels like fairly common terminology to me: Instead of using a maximally entangled state for the teleportation (you teleport "through" the state -- it is what establishes the link in teleportation), you use some other (e.g. non-maximally entangled) state. -- For instance, this allows for a very elegant perspective on the Choi-Jamiolkowski isomorphism (see e.g. physics.stackexchange.com/questions/270032/274739#274739), and this can also be seen as underlying protocols such as gate teleportation (incl. implem. of T gates using magic states). $\endgroup$ Feb 26 at 13:18

2 Answers 2


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 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,m=0}^3 |\Phi_\ell\rangle\langle \Phi_m|_{MA} \otimes \sigma_\ell|\psi_i\rangle\langle \psi_i|_B \sigma_m\\ &= \sum_{\ell,m=0}^3 |\Phi_\ell\rangle\langle \Phi_m|_{MA} \otimes\left(\sum_{i=1}^n p_n\sigma_\ell|\psi_i\rangle\langle \psi_i|_B \sigma_m\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 $MA$ with outcome $\ell$ leaving system $B$ in the state $\sigma_\ell \rho \sigma_\ell$, 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 $$ \{|\Phi_0\rangle, (\sigma_1\otimes I)|\Phi_0\rangle, i(\sigma_1\otimes \sigma_3)|\Phi_0\rangle, (I\otimes \sigma_3)|\Phi_0\rangle\} $$

  • 1
    $\begingroup$ I've added a full derivation of (3) here: quantumcomputing.stackexchange.com/questions/21130/… $\endgroup$
    – forky40
    Sep 10, 2021 at 16:34
  • $\begingroup$ when n = 1, (4) has to be a pure state, but it failed to match. if you substitue (3) into (2) correctly, you will get a term like: $|\Phi_\ell\rangle\langle \Phi_m|_{MA}$ $\endgroup$
    – Maxwell
    Feb 26 at 8:47
  • $\begingroup$ yes you're right, it should be fixed now $\endgroup$
    – forky40
    Feb 26 at 15:18

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 $\rho$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.