Optimising state tomography for fully entangled states

Question

As tomography methods are usually inefficient, it's interesting to find good approximation.

I was wondering the following:

Assume one wants to estimate a state $\rho$ on $n$-qubits. Given a basis of measurements, e.g. $\{X,Y,Z\}$, the number of experimental settings is $3^n$.

My question regards the case where $\rho$ is a fully entangled state, e.g. a generalised GHZ state.

Can we suppress the number of experimental settings to just $3$ and still getting a good estimation? Since we know that $\rho$ is fully entangled, could be enough to just consider 3 settings measuring $X^{\otimes n}$, $Y^{\otimes n}$ and $Z^{\otimes n}$?

Answer 1

This idea totally works, and is often used. The simplest example is if you have a two-qubit state, and measure $\langle XX\rangle = \langle ZZ\rangle = 1$. Then the only possible solution is $|\phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt2$, so you did full state tomography with only two settings.

Of course, this will never happen in reality, so a less trivial example is obtaining $\langle XX\rangle = \langle ZZ\rangle = v$. You can still say something nontrivial about the quantum state you produced experimentally $\rho$, namely that its fidelity with $\phi^+$ is lowerbounded by $v$, that is, $$F(\rho, \phi^+) = \langle \phi^+|\rho|\phi^+ \rangle \ge v.$$ See e.g. this paper for a more serious exploration of this technique.

Answer 2

You cannot fully characterise a general state without measuring a full basis of operators. To characterise an $m$ dimensional $\rho$ you need $m^2-1$ (independent) measurement outcomes. Measure anything less than that, and you won't know the projection of the state on some axis (in general).

If say you only measure $XX,YY,ZZ$ on a two-qubit state, you won't be able to know what's $\langle XY\rangle$ etc. This might change if there are underlying assumptions on the state.

It is true that measuring $XXX,YYY,ZZZ$ is sufficient to fully characterise your state if you assume that it is of "GHZ-class". By which I mean, if the state is $|\psi_{a,b}\rangle= a|000\rangle+b|111\rangle$, then $$\langle XXX\rangle = 2ab, \qquad \langle YYY\rangle = 0, \qquad \langle ZZZ\rangle = a^2-b^2,$$ which means you can retrieve $a,b$ from the measurement results (and in fact only $XXX$ and $ZZZ$ are needed). However, this is only because you made a (quite strong) assumption on the form of the input state.

If you consider different types of maximally entangled states, say, $W$ states, then this already stops being true. Note for example how all states of the form $$a|001\rangle + b|010\rangle + c|100\rangle$$ give $\langle XXX\rangle=\langle YYY\rangle=0$ and $\langle ZZZ\rangle=-1$, regardless of the values of the coefficients $a,b,c$.