Algorithmic Assertions - Craig Gidney's Computer Science Blog

Puzzle: Deflipping a Quantum Coin

13 Apr 2014

Today I read the paper An Invitation to Quantum Game Theory. It's not very long, but it has a nice example of a situation where "going quantum" gives an unexpected advantage.

Quantum Q

Imagine a game between Captain Picard and Q (not sure where Alice and Bob are, but I'll stick with the characters used by the paper). The game is very simple: they will take turns either flipping or not flipping a coin without being able to see it. If the coin ends up heads, then Picard wins. Tails, Q wins.

The coin starts off tails, and the game will end after three turns. Q will get a chance to flip first, then Picard will get a chance, then Q will get to go one last time, then the coin will be revealed and the winner decided.

Classically speaking, Picard can guarantee a 50% chance of winning by randomizing his choice to flip or not flip the coin. But Q, being an omnipotent trickster, has snuck a quantum coin into the game. So Q isn't limited to just flipping the coin, he can apply any single-qubit operation he wants during his turns. Picard remains unaware and restricted to either doing nothing, or flipping the coin (flipping the coin applies an X gate to its state).

The question is: can Q use his quantum advantage to guarantee a win?

Thinking Space

Here's some space between the puzzle and my explanation of the solution.

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You can try ideas in this or this quantum circuit simulator. Picard is only allowed to use the X gate. Q wins if the output stays in state 0.

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Solution

Q can win 100% of the time by applying the Hadamard operation to the coin on each of his turns. With this strategy, a coin that started off tails will end up as tails whether or not Picard flips it.

Here's an animation showing a circuit of what's going on. Picard choosing to flip the coin is represented by the presence or absence of an X gate (the quantum version of a NOT gate). But instead of flipping the coin like he expects, he ends up doing an operation that spins the phase of the "head" component of the state:

HH and HXH circuits don't flip the output

I'm guessing that, for a lot of readers, the above animation didn't convey why the solution works. Maybe something more geometric?

Geometric Explanation

A nice way to visualize the state of a qubit is the Block Sphere.

Here's a diagram, modified from the one on Wikipedia, showing the state our system starts in:

Block sphere with current state being tails / up

The green blob shows the starting state: at the very top, 100% tails. Other points on the surface of the sphere correspond to other quantum states the coin can be in.

Flipping the coin corresponds to rotating by 180 degrees around the X axis. From the starting state, flipping the coin would move us to the very bottom at 100% heads. Like this:

Flipping the coin

But what Q has done is first rotate the state by 180 degrees around the diagonal axis X+Z. Starting from the tails state, this rotation moves the coin's state to the front of the sphere directly along the X axis:

Hadamarding the coin

Now if Picard goes to flip the coin by rotating around the X axis, nothing will happen:

Pointless flipping

And Q can restore the tails state by rotating around the X+Z axis by another 180 degrees:

Unhadamarding the coin

And so Q wins.

Summary

When you flip a quantum coin, you might not be doing what you think.

Part of what made this puzzle interesting to me is that I already knew $H \cdot X \cdot H = Z$, but it never occurred to me to use it as a trick. Quantum mechanics has a knack for producing surprising situations from trivial mathematical facts.

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