How to measure the sign of quantum amplitudes

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

Lets say the qubits are in an initial state $|psirangle = sum_{0}^{2^n-1}alpha_i|irangle$. 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 $|psirangle = sum_{0}^{2^n-1}alpha_i|irangle$ to $D|psirangle = sum_{0}^{2^n-1}(2mu-alpha_i)|irangle$ 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 $|2mu-alpha_i| forall i$

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

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 $mu = frac{|alpha_i| pm|2mu-alpha_i|}{2}$ So at max there can only be $2$ values of $mu$. Hopefully we have narrowed down to a single value of $mune0$. If we have then we can use it to easily calculated $alpha_i$ from $|alpha_i|$ and $|2mu-alpha_i|$ 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 $|jrangle$ for some $j$ which has an amplitude $alpha_jne0$.
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.

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 $(|2nranglepm|2n+1rangle)/sqrt{2}$. Since $alpha_{2n}^2approxalpha_{2n+1}$, 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 $(|2nranglepm|2n-1rangle)/sqrt{2}$ 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.