How to measure the sign of quantum amplitudes

Question

I have a quantum state on $ n $ qubits ($ 2^n $ amplitudes) for which I know the amplitudes are real numbers. I want to take the state out as a vector. I can estimate the magnitude of the amplitudes by doing some measurements and taking the square root of the probabilities, but I loose the sign information.

What kind of measurements do I need to make to recover the sign information? I read a little about state tomography, but it looks really unpractical for $n>2$ (my scale is $n > 10$). Is there an easier way?

Answer 1

An empirical solution could be to use the Grover's Diffusion Operator $D$.

Lets say the qubits are in an initial state $|\psi\rangle = \sum_{0}^{2^n-1}\alpha_i|i\rangle$. Since global phase/sign is irrelevant. We can assume that phase/sign of $\alpha_0$ is + for the sake of convenience (If $\alpha_0=0$ choose the lowest index with non-zero amplitude).
We can find the constants $|\alpha_i|\forall i$ by taking square roots of probabilities and hence we can assume their knowledge.

Grover's Diffusion Operator maps $|\psi\rangle = \sum_{0}^{2^n-1}\alpha_i|i\rangle$ to $D|\psi\rangle = \sum_{0}^{2^n-1}(2\mu-\alpha_i)|i\rangle$ where $\mu = \sum_{0}^{2^n-1}\alpha_i$. We can find the probability distribution of this state and let us say we now also have knowledge of $|2\mu-\alpha_i| \forall i$

Using values of $|\alpha_i|$ and $|2\mu-\alpha_i|$ we get 4 possible values of $\mu = \frac{\pm|\alpha_i| \pm|2\mu-\alpha_i|}{2}$.

Remember we have $2^n$ values of $i$ each which can give us a group of $4$ possible values of $\mu$. We find the common value of $\mu$ across all these $2^n$ groups of $4$.

Since we assumed $\alpha_0>0$ we only get 2 possible values of $\mu = \frac{|\alpha_i| \pm|2\mu-\alpha_i|}{2}$ So at max there can only be $2$ values of $\mu$. Hopefully we have narrowed down to a single value of $\mu\ne0$. If we have then we can use it to easily calculated $\alpha_i$ from $|\alpha_i|$ and $|2\mu-\alpha_i|$ thus giving us the sign information for all $i$.

If $\mu=0$ or there are $2$ possible $\mu$ then we must modify the original state. A possible solution is too selectively flip the sign (using $Controlled$ $Z$ gates) if and only if the state is $|j\rangle$ for some $j$ which has an amplitude $\alpha_j\ne0$.
This will result in a new $\mu'$ which cannot be zero if $\mu=0$. Applying same procedure on this state will yield 1/2 value(s) which can be used to deduce original $\mu$. Since $Z$ gate only changes sign but not magnitude of amplitude the probability distributions will remain same.

I know this isn't a complete formal solution but hopefully it helps.

Answer 2

Being restricted to real amplitudes means that you don't need to go for full on tomography. If you were looking at a single qubit, for example, to do full tomography with projective measurements, you'd need to make $X$, $Y$ and $Z$ measurements, while for the real-only version, you'd only need to make $X$ and $Z$ measurements.

The question, then, is what's a good tactic? It's not something I've thought/read about previously. Here's a couple of options depending on how complex you want to make your experiment:

• Hadamard every qubit and repeat your amplitude determination step. The results should be sufficient to reverse engineer the signs, it's "just" a classical computation (I make no promises that it's an easy computation).

• Assume the weights are $\alpha_i^2$, and that these are ordered. Apply a measurement with projectors onto states $(|2n\rangle\pm|2n+1\rangle)/\sqrt{2}$. Since $\alpha_{2n}^2\approx\alpha_{2n+1}$, it shouldn't take many measurements (relatively!) to determine the relative signs of the amplitudes $\alpha_{2n},\alpha_{2n+1}$. Repeat using projectors onto states $(|2n\rangle\pm|2n-1\rangle)/\sqrt{2}$ and that's enough to globally reconstruct the phases.

• I wonder if there's a smarter method, similar to the previous one, but incorporating a "divide and conquer" strategy where you group the amplitudes into two sets with total weights as close to 1/2 as possible. But I don't immediately see it...

• the answer of user1294287 looks plausible (aside from some normalisation issues), although I wonder what accuracy one has to achieve.