Constructing two-qubit density matrix given expectation values of all products of Pauli operators

Question

I think my question breaks down into two parts.

Let's say you have a two qubit system and you can perform projective measurements.

Each round of measurements will consist of results looking like below.

\begin{array}{|c|c|c|c|} \hline & \sigma_x \otimes \sigma_x& \sigma_x \otimes \sigma_y & \sigma_x \otimes \sigma_z& \sigma_y \otimes \sigma_x &\sigma_y \otimes \sigma_y&\sigma_y \otimes\sigma_z&\sigma_z \otimes \sigma_x&\sigma_z \otimes \sigma_y&\sigma_z \otimes \sigma_z \\ \hline round 1 &\uparrow \downarrow &\uparrow \uparrow &\downarrow \uparrow &\uparrow \uparrow&\downarrow \uparrow&\downarrow \downarrow&\uparrow \downarrow&\downarrow \uparrow& \uparrow\uparrow\\ \hline \end{array}

You repeat multiple rounds of this sort of measurements.

So you have experimentally acquired values of $\langle \sigma_i \otimes \sigma_j \rangle$ (expectation values of products of Pauli matrices) with i and j being x,y, and z, not including the identity.

Given what I found in this paper, I think you can construct the density matrix of the two-qubit system if you have expectation values of all possible products of Pauli matrices, including the identity matrix.

So you need to know $\langle \sigma_I \otimes \sigma_I \rangle$, $\langle \sigma_I \otimes \sigma_x \rangle$, $\langle \sigma_I \otimes \sigma_y \rangle$, $\langle \sigma_I \otimes \sigma_z \rangle$, $\langle \sigma_x \otimes \sigma_I \rangle$,$\langle \sigma_x \otimes \sigma_x \rangle$,$\langle \sigma_x \otimes \sigma_y \rangle$, $\langle \sigma_x \otimes \sigma_z \rangle$,$\langle \sigma_y \otimes \sigma_I \rangle$ ,$\langle \sigma_y \otimes \sigma_x \rangle$, $\langle \sigma_y \otimes \sigma_y \rangle$, $\langle \sigma_y \otimes \sigma_z \rangle$, $\langle \sigma_z \otimes \sigma_I \rangle$, $\langle \sigma_z \otimes \sigma_x \rangle$, $\langle \sigma_z \otimes \sigma_y \rangle$, $\langle \sigma_z \otimes \sigma_z \rangle$, for full tomograhy.

1. Can I just get away with measuring expectation values of the products of Pauli operators without the identity matrix to figure out expectation values of all 16 products of two Pauli operators?

2. Can I construct the full density matrix of a two-qubit system if I know the expectation values of all 16 products of two Pauli operators?

Answer 1

You need $N^2$ real parameters to characterise a generic $N\times N$ Hermitian operator. In the case of quantum states, due to the normalisation constraint, you need $N^2-1$.

In the specific case of a two-qubit system, you have $N=2^2=4$, and thus need $15$ expectation values to totally reconstruct the density matrix. Knowing these values, the density matrix reads $$\rho = \frac{I + \sum_{i,j=0}^3 \operatorname{Tr}[\rho (\sigma_i\otimes\sigma_j)](\sigma_i\otimes\sigma_j)}{4},$$ where the sum on the right does not include the $i=j=0$ term, and $\sigma_0\equiv I$.

More generally, given any complete orthogonal set of traceless Hermitian operators $\{\sigma_i\}$ such that $\operatorname{Tr}(\sigma_i\sigma_j)=\delta_{ij}N$, you can write the density matrix as $$\rho=\frac{I + \sum_i \operatorname{Tr}(\sigma_i \rho)\sigma_i}{N}.$$

Can I just get away with measuring expectation values of the products of Pauli operators without the identity matrix to figure out expectation values of all 16 products of two Pauli operators?

No. To see it consider for example the expectation value $\operatorname{Tr}\big((\sigma_x\otimes I)\rho\big)$. All products of two Pauli operators give zero contribution for this value, so you really need to know the value of $\langle \sigma_x\otimes I\rangle$ to have a complete description of $\rho$.

Answer 2

Just a partial answer to add to that of glS.

Remember that when you make a measurement of $\langle\sigma_i\rangle$ you actually have to make two projective measurements associated with the two eigenstates of $\sigma_i$. Let's call them $|\nu_{i1}\rangle$ and $|\nu_{i2}\rangle$. They are respectively associated with eigenvalues $1$ and $-1$. The measurement of $\sigma_i$ is then given by $$ \langle\sigma_i\rangle = \langle\nu_{i1}|\rho|\nu_{i1}\rangle - \langle\nu_{i2}|\rho|\nu_{i2}\rangle . $$ It turns out that one can use the same two projective measurements to get the measurement for the identity, because the two eigenstates are also eigenstates of the identity, both with the eigenvalues $1$. Hence, $$ \langle I\rangle = \langle\nu_{i1}|\rho|\nu_{i1}\rangle + \langle\nu_{i2}|\rho|\nu_{i2}\rangle \equiv 1. $$

In a practical experiment, one would measure these quantities in terms of coincidence counts, which are only proportional to these expectation values. Therefore, one can compute the actual expectation values in terms of these coincidence counts by dividing them by the sum of counts associated with the identity.