# Calculating bipartite state from joint probability distribution

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?

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.

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.

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).

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.

• I think the question was how to perform this the other way around - rather than calculating the expectation values for these operators, I interpreted the question as 'if I have these probabilities/expectation values, how do I calculate/reconstruct $\rho$'?
– JSdJ
Jul 31 '20 at 10:39
• @JSdJ probably you are right. Jul 31 '20 at 11:42
• Thank you @DavitKhachatryan Jul 31 '20 at 16:49
• @Omkar you are welcome. I think I misunderstood the question and this is not a relevant answer. Somebody even downvoted the answer...So I guess I should delete this post. Aug 1 '20 at 9:16
• I found it helpful @DavitKhachatryan. So i don't think you should delete this answer. Aug 1 '20 at 13:45