# Why can't we do pure state tomography using interference with a known reference state?

I am new to quantum tomography, and right now I am trying to understand tomography with pure states. It's my understanding if we have the amplitude magnitudes as well as relative phases, we can completely reconstruct a pure state.

We can obtain magnitudes by simply measuring all the qubits in the state and generating a bitstring distribution.

As per this answer, we can find relative phases via interference patterns with a known state.

My question is, why isn't pure state tomography this simple? If we choose our known state carefully, we don't need to construct different measurement bases and we can just measure all qubits at once. To my knowledge, assuming the reference state is chosen carefully, why doesn't this scale O(1) in terms of measurement bases?

• you can absolutely perform full state tomography measuring in a single measurement basis. You just need the POVM describing the measurement to have a sufficient number of outcomes. Which in terms of qubits amounts to saying you need the initial state to interact with a larger number of "ancillary" qubits which are then measured. As a simple example, if you evolve an input qubit state through a random 4x4 unitary evolution (technically, a 4x2 isometry) then measuring the output qubits in their computational basis gives you enough information to fully reconstruct the input state\
– glS
Dec 3, 2023 at 15:42
• there's many relevant references but a good recent one using SIC-POVMs is journals.aps.org/prxquantum/abstract/10.1103/…
– glS
Dec 3, 2023 at 15:55

## 1 Answer

There are two caveats that make this task more difficult. First, measurement via an interference pattern is actually already a measurement with a continuous number of reference states, so people looking for efficient tomography schemes need to find more efficiencies. Second, there are many relative phases to be measured for a general quantum state, and a direct measurement of each of those requires a different reference state.

Let's begin. If you have a state $$|\psi\rangle=\sqrt{p_0}|0\rangle+\sqrt{p_1} e^{i\theta_1}|1\rangle$$ for real parameters $$p_i$$ and $$\theta_i$$ and where we've set the global phase to 0, a measurement of the two probabilities $$p_0$$ and $$p_1$$ gives you almost all the information, while interfering the state with a reference $$|\phi_{01}\rangle=\frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt{2}}$$ tells you $$|\langle\phi_{01}|\psi\rangle|^2=\frac{p_0+p_1+2\sqrt{p_0 p_1}\cos(\phi-\theta_1)}{2}.$$ If you do that measurement for a continuous set of parameters $$\phi$$ that range from 0 to $$2\pi$$ and plot the distribution, you can read off the maximum $$(\sqrt{p_0}+\sqrt{p_1})^2/2$$, the minimum $$(\sqrt{p_0}-\sqrt{p_1})^2/2$$, and the phase $$\theta_1$$ as the value of $$\phi$$ where the distribution achieves its maximum. That certainly requires multiple measurements using multiple reference states with different values of $$\phi$$, while efficient tomography schemes might tell you exactly three values of $$\phi$$ to use that will give you all of the requisite information.

Then, notice that I didn't use $$p_0+p_1=1$$, because I wanted to list more general states like $$|\psi\rangle=\sqrt{p_0}|0\rangle+\sqrt{p_1}e^{i\theta_2}|1\rangle+\sqrt{p_2}e^{i\theta_2}|2\rangle.$$ We can learn about $$p_0$$ and $$p_1$$ using the above interference with $$|\phi_{01}\rangle$$. So now I'll define a general $$|\phi_{jk}\rangle=\frac{|j\rangle+e^{i\phi}|k\rangle}{\sqrt{2}}$$ and say you can interfere your state with that to find the relative phase between the $$|j\rangle$$ and $$|k\rangle$$ components of your state. So now using the reference state $$|\phi_{02}\rangle$$ will let you learn the value of $$\theta_2$$ by looking at the maximum of the interference pattern, where again you actually need multiple values of $$\phi$$ for the reference state in order to find that. (Note: these states could be shorthand for anything: it might be that $$|0\rangle=|000\rangle$$, $$|1\rangle=|100\rangle$$, $$|2\rangle=|010\rangle$$, etc.)

Lastly, if you have mixed states, there are even more free parameters. So for pure states you have to remember that there are multiple relative phases and that each relative phase requires a continuous number of reference states in order to be measured via interferometry, so quantum-state tomography looks for clever ways of avoiding redundant measurements in order to most efficiently find the independent parameters of a state.