Calculating bipartite state from joint probability distribution

Question

We can calculate single qubit state by measuring it in pauli observables {$\sigma_{x},\sigma_{y},\sigma_{z}$} and then looking at its probability distribution. How to do this when we are having joint probability distribution for multiqubit state? Specifically, I want to know the procedure for calculating 2-qubit state when it's measured in $\sigma_{x}\otimes\sigma_{x}$, $\sigma_{y}\otimes\sigma_{y}$ and $\sigma_{z}\otimes\sigma_{z}$ observables?

Answer 1

This would not be enough information to reconstruct the bi-partite state.

Single-qubit case

For the one-qubit case, reconstruction of the state (which we describe as $\rho$) works, because the single-qubit Pauli observables $\{\sigma_{x},\sigma_{y},\sigma_{z}\}$ together with the $\sigma_{I}$-operator creates a basis for the space of single-qubit density matrices. If our probabilities are $\{p_{I}=1,p_{x},p_{y}.p_{z}\}$, we reconstruct as: $$ \rho = \sum_{i \in \{I,x,y,z\}} p_{i}\sigma_{i}. $$

Bi-partite case

In principle we can do this also for a bi-partite state, but the operators over which we sum still need to form a basis for the space of (now bi-partite) density matrices. A simple count of the dimensions involved tells us that there should be $16$ elements in this basis, and not the four in $\{I, \sigma_{x}\otimes \sigma_{x},\sigma_{y}\otimes \sigma_{y},\sigma_{z}\otimes \sigma_{z}\}$.

The most obvious (and used) choice is to also include the cross products of all the Paulis. We then get a set of $4^{2} = 16$ elements, which is called the two-qubit Pauli group $\mathcal{P}^{2}$:

$$ \mathcal{P}^{2} = \{\sigma_{I},\sigma_{x},\sigma_{y},\sigma_{z}\} \otimes \{\sigma_{I},\sigma_{x},\sigma_{y},\sigma_{z}\}. $$ If we would have all $16$ $p_{i}$'s, the reconstruction is as straightforward for the one-qubit case: $$ \rho_{2} = \sum_{i \in \{I,x,y,z\}\times \{I,x,y,z\}} p_{i}\sigma_{i}. $$

You said you already have $p_{x,x}, p_{y,y}$ and $p_{z,z}$. $p_{I,I}$ is a freebie because it needs to be $1$; so you still need the $12$ other probabilities.

Some intuition

Basically, the above analysis tells us that to completely characterize our system of two qubits, knowing only what they do in this symmetric-coupling sense is not enough information. Basically, we are missing two sets of information:

• We need to know how these two qubits act under aymmetric coupling. That is to say, we need the probabilities for measurements of e.g. $\sigma_{x} \otimes \sigma_{y}$, or $\sigma_{z} \otimes \sigma_{x}$. Of course, there are $|\{x,y,z\}\times \{x,y,z\}|=9$ different elements here, but we already counted the three symmetric ones.
• We also need to know what they do individually: if we measure 'nothing' on the first qubit but we measure the second qubit in any of the Pauli bases, we still learn something about the second qubit. These are the operators $\sigma_{I}\otimes \{\sigma_{x},\sigma_{y},\sigma_{z}\}$ and vice-versa: there are $6$ of them.

This gives a total of $1$ (for $\sigma_{I}\otimes \sigma_{I}$) + $3$ (for our original $3$ operators) + $9-3 = 6$ (for the asymmetric coupling operators) + $6$ (for the individual operators). This sums up to $16$, so we now have accounted for all of the operators.

Then how to actually get these other probabilities?

This question deals with the same problem, and there I also explain how to obtain these $12$ other probabilities from experimental outcomes. Note that these $3$ measurements really are not enough, and that you will need at least $9$ different measurement results: you need all symmetric and asymmetric operators.

Final note + further reading

As a last remark, the techniques of reconstructing density matrices from probability distributions (or a finite number of measurement outcomes) are collectively known as quantum state tomography or QST (It even has a Wikipedia page, hurray!). There are many more advanced techniques, but I won't go into them here - if you ever want to learn more googling the term QST is a good start, but of course you should also feel free to ask any questions on the stack exchange.

Answer 2

1. Calculating $\langle \sigma_z \otimes \sigma_z \rangle$

$$\langle \sigma_z \otimes \sigma_z \rangle = Tr\big(\sigma_z \otimes \sigma_z \rho\big) = \rho_{11} - \rho_{22} - \rho_{33} + \rho_{44}$$

As can be seen from this answer $\rho_{11}$, $\rho_{22}$, $\rho_{33}$ and $\rho_{44}$ are the probabilities of measuring $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$ correspondingly. This can be seen for example by calculating $Tr(P_{01} \rho) = \rho_{22}$, where $P_{01} = |01\rangle \langle 01|$ is the projector for the $|01\rangle$. Note, that $\rho_{11}$, $\rho_{22}$, $\rho_{33}$ and $\rho_{44}$ can be calculated from repeated experiments by applying $\sigma_z$ basis measurements (also it is described in this answer).

2. Calculating $\langle \sigma_x \otimes \sigma_x \rangle$

$$\langle \sigma_x \otimes \sigma_x \rangle = Tr\big(\sigma_x \otimes \sigma_x \rho\big) = Tr\big((H \otimes H) \; \sigma_z \otimes \sigma_z \; (H \otimes H) \; \rho\big) = \\ = Tr\big( \sigma_z \otimes \sigma_z \; (H \otimes H) \; \rho \; (H \otimes H) \big) = Tr\big( \sigma_z \otimes \sigma_z \; \rho'\big) = \rho'_{11} - \rho'_{22} - \rho'_{33} + \rho'_{44}$$

Because $H\sigma_z H = \sigma_x$ and the cyclic property of the trace. Here $\rho' = H \otimes H \; \rho \; H \otimes H$. So, after applying $H\otimes H$ the the initial $\rho$ we just need to calculate $Tr\big( \sigma_z \otimes \sigma_z \; \rho'\big)$ that we already know how to do.

3. Calculating $\langle \sigma_y \otimes \sigma_y \rangle$

The same works, but instead of $H$, we take $H S^{\dagger}$:

$$ \langle \sigma_y \otimes \sigma_y \rangle = Tr\big( \sigma_z \otimes \sigma_z \; \rho''\big) = \rho''_{11} - \rho''_{22} - \rho''_{33} + \rho''_{44} $$

where $\rho'' = \big(HS^{\dagger} \otimes HS^{\dagger} \big) \; \rho \; \big(SH \otimes SH \big)$, because $SH\sigma_z HS^\dagger = \sigma_y$ as can be seen from this answer.

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In a slightly different way, the same logic works for the other Pauli terms (we just need to apply such gates, after which we will have either $\sigma_z$ or $I$ in the trace). But of course, if we can measure also in $\sigma_x$ and $\sigma_y$ basis directly the gates before the measurements will be unnecessary. As was pointed out in this answer we should calculate all $16$ Pauli terms in order to estimate the density matrix (the mentioned three are not enough).

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It is possible to combine measurements for $\langle \sigma_y \otimes \sigma_y \rangle$ and $\langle \sigma_x \otimes \sigma_x \rangle$, by measuring in Bell basis as was discussed in this question.