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


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:

enter image description here

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.

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!

  • 2
    $\begingroup$ The first two trits of each ket are going to be measured, so you can do different things based on what they are. Look at each case separately, e.g. $b |0\rangle + a |1\rangle + c|2\rangle$ for $01$. How would you restore this to the state $a |0\rangle + b |1\rangle + c|2\rangle$? $\endgroup$ Commented Sep 28, 2018 at 21:11
  • $\begingroup$ @CraigGidney, like I said, I see how measuring the second trit tells you which two amplitudes need to be transposed, and the first trit tells you how to "fix" the $w$'s after making the necessary transposition. The part I'm having trouble with (assuming my computations are correct) is that Aaronson says "Depending on the value of 0,1, or 2 Bob can apply a ? gate 0, 1, or 2 times." I only see how to do with with Bob choosing between 3 different transpositions and three ways of fixing $w$'s (one of which is doing nothing). Does that clarify my confusion? $\endgroup$
    – Joe
    Commented Sep 28, 2018 at 21:20
  • 1
    $\begingroup$ There are two distinct fixup operations, one for each ?. One of them should be 'add k mod 3' or 'subtract k mod 3' where k is the measurement result. $\endgroup$ Commented Sep 28, 2018 at 21:49
  • 1
    $\begingroup$ Ok, I see how to deal with the $w$'s. You can use the matrix $\text{diag}(1,w,w^2)$ $k$ times, where the first trit measures to $|k \rangle$. However, I still don't see how to achieve the transposition of the amplitudes. Especially since two amplitudes need to be transposed in every case, meaning "applying the operation 0 times (or adding 0 modulo 3)" will never achieve the desired output. $\endgroup$
    – Joe
    Commented Sep 28, 2018 at 22:06
  • 3
    $\begingroup$ The amplitudes should not be transposed, they should be cyclically shifted. Retry the calculation of the state. $\endgroup$ Commented Sep 28, 2018 at 22:19

1 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):

trit teleportation

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.

  • $\begingroup$ Craig this is fantastic. Not only did you answer my question, but now I know about this simulator too. Thank you so much! $\endgroup$
    – Joe
    Commented Oct 1, 2018 at 13:38

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