Problem about qutrit teleportation protocol

Question

I'm working through Scott Aaronson's Quantum Information Science problem sets, and I'm having trouble with a specific problem in ps5 (PDF). Specifically the following problem:

A “qutrit” has the form $a|0\rangle+b|1\rangle+c|2\rangle$, where $|a|^2+|b|^2+|c|^2=1$. Suppose Alice and Bob share the entangled state $(|00\rangle+|11\rangle+|22\rangle)/\sqrt 3$. Then consider the following protocol for teleporting a qutrit $|\psi〉=a|0\rangle+b|1\rangle+c|2\rangle$ from Alice to Bob: first Alice applies a CSUM gate from $|\psi〉$ onto her half of the entangled pair, where

$$\operatorname{CSUM}(|x\rangle\otimes|y\rangle) =|x\rangle\otimes|y+x \bmod 3\rangle$$

Next, Alice applies the unitary matrix $F$ to the $|\psi\rangle$ register, where

$$F=\frac{1}{\sqrt{3}}\begin{bmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega^3\end{bmatrix}$$

and $\omega=e^{2i\pi/3}$ so that $\omega^3= 1$. She then measures both of her qutrits in the $\{|0\rangle,|1\rangle,|2\rangle\}$ basis, and sends the results to Bob over a classical channel. Show that Bob can recover a local copy of $|\psi\rangle$ given these measurement results.

This quantum circuit summarizes the protocol:

Here the double-lines represent classical ‘trits’ being sent from Alice to Bob. Depending on the value of 0, 1 or 2 Bob can apply a ? gate 0, 1, or 2 times. Prove that $|\psi〉=|\psi_\text{out}\rangle$ for appropriately chosen ? gates for all possible measurement results. Hint: You could explicitly work out all 9 possible cases, but you could also save time by noticing a general pattern that lets you handle all the cases in a unified way.

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Here's what I've done: after applying the CSUM gate to the mixed state of three qutrits, Alice and Bob share the state: \begin{align*} & \frac{a}{\sqrt{3}}(|000 \rangle + |011 \rangle + |022 \rangle) \\ + & \frac{b}{\sqrt{3}}(|110 \rangle + |121 \rangle + |102 \rangle) \\ + & \frac{c}{\sqrt{3}}(|220 \rangle + |201 \rangle + |212 \rangle) \end{align*} After Alice applies $F$ to the first qutrit in the shared state, they're left with: \begin{align*} & \frac{|00 \rangle}{3}(a|0 \rangle + c|1 \rangle + b|2 \rangle)\\ + & \frac{|01 \rangle}{3}(b|0 \rangle + a|1 \rangle + c|2 \rangle)\\ + & \frac{|02 \rangle}{3}(c|0 \rangle + b|1 \rangle + a|2 \rangle) \\ + & \frac{|10 \rangle}{3}(a|0 \rangle + w^2c|1 \rangle + wb|2 \rangle)\\ + & \frac{|11 \rangle}{3}(wb|0 \rangle + a|1 \rangle + w^2c|2 \rangle)\\ + & \frac{|12 \rangle}{3}(w^2c|0 \rangle + wb|1 \rangle + a|2 \rangle)\\ + & \frac{|20 \rangle}{3}(a|0 \rangle + wc|1 \rangle + w^2b|2 \rangle) \\ + & \frac{|21 \rangle}{3}(w^2b|0 \rangle + a|1 \rangle + wc|2 \rangle) \\ + & \frac{|22 \rangle}{3}(wc|0 \rangle + w^2b|1 \rangle + a|2 \rangle) \\ \end{align*}

So after Alice measures her qutrits, whatever is inside the parenthesis is the state that Bob holds. However, I don't wee what operations can be used multiple times to "fix" the output.

It seems to me that Alice could communicate to Bob with one trit which two coefficients need to be transposed, and use one trit to tell Bob how to fix the remaining $w$'s. That doesn't seem to fit the desired protocol though, making me doubt the computations that I have performed above. If anyone could help me out (or point out a better approach), it would be appreciated. Thanks!

Answer 1

I tested this out in Quirk by embedding each qutrit into two qubits, and I get a simular result to you, where in addition to the cyclic shift fixup and the phasing operation you need to transpose states 1 and 2. Presumably there's some simple change to the circuit that fixes this, such as picking a different F, but I did't check too hard to see if it was possible.

Here's the circuit (and in the simulator itself):

I've put displays throughout the circuit to have a better idea that it's working. In the bottom left area the blue rectangle with the three circles along its diagonal is showing that, yes, we're in the 00+11+22 state.

In the top middle I'm preparing some state to teleport. It doesn't really matter what it is, so I used gates that keep changing what they're doing. The important bit is that the blue rectangle here looks the same as the one in the bottom right, indicating the state moved.

In the bottom right you can see the fixups being applied. The cyclic shift (-A mod 3), the mystery transpose (implemented by the swap), and then the phasing operations.

I didn't include the measurements, but if you add some before the controls in the top right and just before the Input A of the -A mod R operation the circuit will behave identically.