How to measure an unknown state produced by a source of qubits?

What kind of experiment can allow me to measure an unknown state produced by a source of qubits? For example: the state of photon polarization. But it can be another one.

I have no information about the qubits. The state is unknown. Is this a problem, to apply Quantum State Tomography?

My idea:

Make measurements in many different basis, after applying some transformations on the qubits, like applying a Hadamard Gate, and then reconstructing the state with the data from these measurements in the many different basis.

Is this the best approach or is the problem more basic? The way the question was given to me is just, "how to measure an unknown state produced by a source of qubits?". I don't know if the question was poorly stated, because naively I can say "just take a measurement instrument and measure". But, if I want information about the state, then many measurements in different bases are able to give me the information I need to write down its density matrix.

Is this right?

2 Answers

I thought I'd elaborate a little on @Chris E's answer with some details about what kind of precision we can expect in our approximation as we increase the number of measurements we make. I'll be elaborating on the single-qubit state tomography example from Wikipedia, since I think this example is really enlightening but the terse way it's described on Wikipedia can be hard to follow.

Say we have a source which produces unlimited copies of a qubit in some mixed state described by a (to us, unknown) density matrix $$\rho$$. One way of representing this matrix generally would be as a statistical combination of two orthogonal states $$\lvert\psi\rangle = \alpha\lvert 0\rangle + \beta\lvert 1\rangle$$ and $$\lvert\psi_\perp\rangle = -\bar\beta\lvert 0\rangle + \bar\alpha\lvert 1\rangle$$ with respective probabilities $$r$$ and $$(1-r)$$, in which case the density matrix would be given by $$\rho = r\lvert \psi\rangle\langle\psi\rvert + (1-r)\lvert\psi_\perp\rangle\langle\psi_\perp\rvert = \begin{bmatrix}r|\alpha|^2 + (1-r)|\beta|^2 & (2r-1)\alpha\bar\beta \\ (2r-1)\bar\alpha\beta & (1-r)|\alpha|^2 + r|\beta|^2\end{bmatrix}$$ On the other hand, the Wiki explanation prefers to use the following general representation of a density matrix: $$\rho = \frac{I+\vec r\cdot \vec\sigma}{2} = \frac{1}{2}\begin{bmatrix}1+r_z & r_x-ir_y \\ r_x+ir_y & 1 - r_z\end{bmatrix}$$ Imagine that we perform a large number $$N$$ of measurements on this mystery state in the standard basis $$\lvert 0\rangle, \lvert 1\rangle$$. The respective probabilities of the results being $$\lvert 0\rangle$$ and $$\lvert 1\rangle$$ are: $$r|\langle 0\lvert\psi\rangle|^2 + (1-r)|\langle 0\lvert\psi_\perp\rangle|^2 = r|\alpha|^2 + (1-r)|\beta|^2$$ $$r|\langle 1\lvert\psi\rangle|^2 + (1-r)|\langle 1\lvert\psi_\perp\rangle|^2 = (1-r)|\alpha|^2 + r|\beta|^2$$ Hence, performing $$N$$ measurements in this basis should result in approximately $$\tfrac{1+r_z}{2}N$$ measurements resulting in $$\lvert 0\rangle$$ and approximately $$\tfrac{1-r_z}{2}N$$ measurements resulting in $$\lvert 1\rangle$$. This "approximately" can be quantified using the Central Limit Theorem - we can expect that the deviation from the expected proportion will decay proportionally to $$\mathcal O(N^{-1/2})$$. We can therefore approximate $$r_z$$ by the following quantity: $$\tilde{r_z} = \frac{(\text{count of }\lvert 0\rangle)-\text{count of }\lvert 1\rangle)}{N} = r_z + \mathcal O\Big(\frac{1}{\sqrt{N}}\Big)$$ and should and even more specific description of convergence be desired, we can calculate standard deviations to obtain bounds on the hidden constant involved in the $$\mathcal O$$ error term.

We can similarly calculate the results of measuring the mystery qubit in the Hadamard basis $$\lvert + \rangle, \lvert -\rangle$$ by rewriting $$\lvert\psi\rangle,\lvert\psi_\perp\rangle$$ as follows: \begin{align}\lvert\psi\rangle &= \tfrac{\alpha+\beta}{2}\lvert +\rangle + \tfrac{\alpha-\beta}{2}\lvert -\rangle \\ \lvert\psi_\perp\rangle &= \tfrac{\bar\alpha -\bar\beta}{2}\lvert +\rangle - \tfrac{\bar \alpha + \bar\beta}{2}\lvert -\rangle \end{align} from which we may calculate (using some complex arithmetic manipulations) the following probabilities of measuring $$\lvert +\rangle$$ and $$\lvert -\rangle$$ respectively: \begin{align}r|\langle +\lvert\psi\rangle| + (1-r)|\langle +\lvert\psi_\perp\rangle| &= \frac{|\alpha|^2+|\beta|^2}{2} + (2r-1)\frac{\alpha\bar\beta + \bar\alpha\beta}{2} \\ &= \frac{|\alpha|^2+|\beta|^2}{2} + (2r-1)\text{Re}(\bar\alpha\beta)\end{align} \begin{align}r|\langle -\lvert\psi\rangle| + (1-r)|\langle -\lvert\psi_\perp\rangle| &= \frac{|\alpha|^2+|\beta|^2}{2} - (2r-1)\frac{\alpha\bar\beta + \bar\alpha\beta}{2} \\ &= \frac{|\alpha|^2+|\beta|^2}{2} - (2r-1)\text{Re}(\bar\alpha\beta)\end{align} Note that this quantity $$(2r-1)\text{Re}(\bar\alpha\beta)$$ is precisely $$r_x$$, according to our two equivalent matrix representations of $$\rho$$. Hence, we can approximate $$r_x$$ as the following proportion: $$\tilde{r_x} = \frac{(\text{count of }\lvert +\rangle) - (\text{count of }\lvert -\rangle)}{N} = r_x + \mathcal O\Big(\frac{1}{\sqrt{N}}\Big)$$ again, using the Central Limit Theorem to justify the $$\mathcal O(N^{-1/2})$$ convergence. We can do something very similar to approximate $$r_y$$, using the $$Y$$-basis according to the Wiki article - but I'll leave the algebra for that one to you to verify.

In any case, $$\mathcal O(N^{-1/2})$$ convergence is not very fast convergence - to give you an idea of how slow it is, if we wanted to get just a single additional decimal point of precision in our approximation to $$\rho$$, we would need to do $$100\times$$ the number of measurements. But this does show how your intuition that measuring in a couple different bases (in fact, just three!) allows us to obtain asymptotically convergent approximations to the density matrix of the unknown qubit. (As a side note, if there happens to be any tomography algorithm that does better than $$\mathcal O(N^{-1/2})$$, or a proof that no such algorithm exists, I would be really interested to hear about it in the comments!)

• That was a very interesting and useful answer. I really appreciate it. Thanks for your time and effort. Mar 13, 2023 at 19:08
• @Dimitri Sure thing, I'm glad it was helpful! :-) Mar 13, 2023 at 20:34

That's quite right. If you had many copies of the photon you wanted to characterize, quantum state tomography would be the process to do it.

However, it's worth noting that doing all the measurements you mention requires many copies, so you're really characterizing your photon source. If you only have a single photon, characterizing its full state is impossible since you can only make a single measurement of it before you change it's state*.

*: Technically there are some weak measurements protocols that might let you infer a bit more information probabilistically, but you still can't get the full information about the state with only a single copy of it.

• hey, thanks for the answer! Mar 13, 2023 at 19:08