DrChinese said:
Are you sure? I admit that I open to enlightenment on the point, but don't believe I have seen an example previously.
Yes, I'm sure.
Given a teleportation operator ##T(\left| \psi \right\rangle \left\langle \psi \right| \otimes R_\text{epr}) = R_\text{junk} \otimes \left| \psi \right\rangle \left\langle \psi \right|## and a mixed state ##M = \sum_k \lambda_k \left| \psi_k \right\rangle \left\langle \psi_k \right|## we find:
##T(M \otimes R_\text{epr})##
##= T((\sum_k \lambda_k \left| \psi_k \right\rangle \left\langle \psi_k \right|) \otimes R_\text{epr})##
##= \sum_k \lambda_k T(\left| \psi_k \right\rangle \left\langle \psi_k \right| \otimes R_\text{epr})##
##= \sum_k \lambda_k R_\text{junk} \otimes \left| \psi_k \right\rangle \left\langle \psi_k \right|##
##= R_\text{junk} \otimes \sum_k \lambda_k \left| \psi_k \right\rangle \left\langle \psi_k \right|##
##= R_\text{junk} \otimes M##
We can also confirm by doing a full calculation of the teleportation process given an unknown mixed state.
Suppose Alice has a qubit in the state represented by the density matrix ##M = \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix}##. It might be mixed. She also shares an EPR pair ##P = \frac{1}{\sqrt{2}} \left| 00 \right\rangle + \frac{1}{\sqrt{2}} \left| 11 \right\rangle## with Bob. The density matrix for state of the system as a whole is:
##\psi_1 = M \otimes PP^\dagger
= \frac{1}{2} \begin{bmatrix}
M & 0 & 0 & M \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
M & 0 & 0 & M
\end{bmatrix}
= \frac{1}{2} \begin{bmatrix}
a & b & 0 & 0 & 0 & 0 & a & b \\
\overline{b} & c & 0 & 0 & 0 & 0 & \overline{b} & c \\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
a & b & 0 & 0 & 0 & 0 & a & b \\
\overline{b} & c & 0 & 0 & 0 & 0 & \overline{b} & c
\end{bmatrix}##
Now we apply the teleportation operations.
First, a controlled-not of Alice's qubit containing ##M## onto the first qubit of ##P##. The odd-index rows and columns (at indices 1, 3, 5, 7; you may be more used to calling them 2nd, 4th, 6th, and 8th) correspond to states where ##M## is ON. Toggling the first qubit of ##P## corresponds to pairing columns and rows whose index in binary have matching bits everywhere except in the second position (0-2, 1-3, 4-6, and 5-7) and swapping them. So a CNOT of ##M## onto the first qubit of ##P## swaps the 2nd and 4th columns, the 6th and 8th columns, the 2nd and 4th rows, and finally the 6th and 8th rows:
##\psi_{2}
= \frac{1}{2} \begin{bmatrix}
a & 0 & 0 & b & 0 & b & a & 0 \\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
\overline{b} & 0 & 0 & c & 0 & c & \overline{b} & 0 \\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\
\overline{b} & 0 & 0 & c & 0 & c & \overline{b} & 0 \\
a & 0 & 0 & b & 0 & b & a & 0 \\
0 & 0 & 0 & 0 & 0 & 0& 0 & 0
\end{bmatrix}##
Second, a Hadamard operation is applied to the qubit that started off storing ##M##. Group the rows by everything except ##M## (0 with 1, 2 with 3, 4 with 5, 6 with 7), then give the even-index row of each pair the pair's sum while the odd index gets the difference. Repeat for the columns. Gain a factor of 1/2.
##\psi_{3}
= \frac{1}{4} \begin{bmatrix}
a & a & b & -b & b & -b & a & a \\
a & a & b & -b & b & -b & a & a \\
\overline{b} & \overline{b} & c & -c & c & -c & \overline{b} & \overline{b} \\
-\overline{b} & -\overline{b} & -c & c & -c & c & -\overline{b} & -\overline{b}\\
\overline{b} & \overline{b} & c & -c & c & -c & \overline{b} & \overline{b} \\
-\overline{b} & -\overline{b} & -c & c & -c & c & -\overline{b} & -\overline{b}\\
a & a & b & -b & b & -b & a & a \\
a & a & b & -b & b & -b & a & a
\end{bmatrix}##
Third, a CNOT of ##P##'s first qubit onto its second qubit is performed. Swap the 3rd and 7th columns. And rows. Also 4th and 8th. Technically a measurement is supposed to happen beforehand, since ##P##'s first qubit was originally with Alice and now we're conditioning on its value in Bob-land. But
you can defer measurement when performing calculations; you'll get the same result at the end as long as the measured qubits are only used as controls in the interim.
##\psi_{4}
= \frac{1}{4} \begin{bmatrix}
a & a & a & a & b & -b& b & -b \\
a & a & a & a & b & -b& b & -b \\
a & a & a & a & b & -b& b & -b \\
a & a & a & a & b & -b& b & -b \\
\overline{b} & \overline{b}& \overline{b} & \overline{b} & c & -c & c & -c \\
-\overline{b} & -\overline{b}& -\overline{b} & -\overline{b} & -c & c & -c & c\\
\overline{b} & \overline{b}& \overline{b} & \overline{b} & c & -c & c & -c \\
-\overline{b} & -\overline{b}& -\overline{b} & -\overline{b} & -c & c & -c & c
\end{bmatrix}##
The final non-measurement operation is a controlled-Z of the qubit that started off storing ##M## onto ##P##'s second qubit. This negates the columns and rows whose indices are of the form 1X1. So negate the 6th and 8th columns, and rows:
##\psi_{4}
= \frac{1}{4} \begin{bmatrix}
a & a & a & a & b & b& b & b \\
a & a & a & a & b & b& b & b \\
a & a & a & a & b & b& b & b \\
a & a & a & a & b & b& b & b \\
\overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c \\
\overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c\\
\overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c \\
\overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c
\end{bmatrix}##
Which factors:
##\psi_{4} = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \otimes \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix}##
Measuring the first two qubits will give a uniformly random result (since they are each in the state ##\frac{1}{\sqrt{2}} \left| 0 \right\rangle + \frac{1}{\sqrt{2}} \left| 1 \right\rangle##), and drop them into the maximal mixed state. The actual protocol performs measurement earlier, but we used the deferred measurement principle to get the same final state despite delaying the measurement calculations until now:
##\psi_{5} = \frac{1}{4} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix} = \frac{1}{4} I_2 \otimes I_2 \otimes M##
As you can see, the third qubit (which corresponds to the second qubit of ##P##; i.e. Bob's qubit) has ended up in the state ##M##.
Also you can see that it's much easier to use the intuition that "if it works on pure states then it must work on mixed states since mixed states are like not knowing which pure state you're in" than it is to run out the full calculation.
