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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 $Z^{\otimes n}$?

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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.

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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 $a,b,c$.

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  • $\begingroup$ Great! Perhaps for the $W$ states we just need a different basis to optimise the QPT..? $\endgroup$
    – Daniele Cuomo
    May 25, 2022 at 9:54
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    $\begingroup$ @DanieleCuomo sure. You can always find a suitable measurement with 4 outcomes that characterises all states of the form $a|\psi\rangle+b|\phi\rangle$ for, some fixed choice of $|\psi\rangle,|\phi\rangle$, etc. The choice of measurement effectively allows to characterse some subspace of the space of states (in the Bloch representation), so any set of states in that subspace will be characterised, and everything not in there won't. This shouldn't be surprising: the more you already know about the state, the less information you need to acquire $\endgroup$
    – glS ♦
    May 25, 2022 at 10:20
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    $\begingroup$ just imagine the situation on single qubits in the Bloch sphere. Measuring $X$ you find the $x$ coordinate of the state. If you have a priori knowledge that tells you that the state is for sure sitting on the $x$ axis, then you don't need to measure anything else $\endgroup$
    – glS ♦
    May 25, 2022 at 10:29
  • $\begingroup$ Hence, for the W state, we may just need to measure $\langle ZXX \rangle$, $\langle XZX \rangle$, $\langle XXZ \rangle$. $\endgroup$
    – Daniele Cuomo
    May 25, 2022 at 10:57
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    $\begingroup$ @DanieleCuomo as I said, probably, yes. If not those, there will be other measurements that will work to that effect. But the point is that this has nothing to do with entanglement. The same argument can be made about any other class of states $\endgroup$
    – glS ♦
    May 25, 2022 at 10:58

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