Quantum State Tomography from IQ plane data

Question

Background:

I am given to understand that the steps of Quantum State Tomography (QST) are as follows for a single qubit:

1. The qubit is in the state $\psi=a_0|0\rangle+a_1|1\rangle$ with density matrix $\rho = |\psi\rangle\langle\psi|$.
2. Decompose $\rho$ in terms of the Pauli basis: $$ \rho = \frac{1}{2}\left( I_{2\times 2} + \sum_{A}\langle \sigma_A \rangle \sigma_A \right) $$ where $A=x,y,z$ and $\sigma_A = (\sigma_x,\sigma_y,\sigma_z)$ the 3d vector of 2x2 Pauli matrices (generators of SU(2)).
3. Then there are the "measurement equations": $$ \begin{pmatrix}p_0\\p_1 \end{pmatrix} = \begin{pmatrix}\beta_I^{|0\rangle} & \beta_A^{|0\rangle}\\ \beta_I^{|1\rangle} & \beta_A^{|1\rangle} \end{pmatrix}\begin{pmatrix} \langle I_{2\times 2} \rangle\ \\ \langle \sigma_A \rangle \end{pmatrix} $$ where the parameters $\beta$ tell us about the bias of our system and allow us to convert probabilities of the system being in the $|0\rangle$ or $|1\rangle$ state to (unphysical) measurement expectations.

I took this from this source (click).

Question:

I understand steps 1 and 2.

• I am very unclear as to what step 3 entails. That is, how one obtains any of the info in the equation of step 3? I understand that we want to somehow use the matrix of coefficients $\beta$ together with the probabilities $p_0,p_1$ in order to estimate the rightmost vector with the expectations?

• In order to do so, if this is the case, one needs the $p_0,p_1$ values. They are read from the IQ plane (if transmon or xmon are used). But what about the $\beta$ matrix? Where do its values come in order to obtain what we really want, the expectation values of the measurement operators?

• Finally, having these expectation values, how do we proceed to estimate the state? I.e., to do tomography?

Answer 1

We know what a measurement of a particular operator should yield: measuring $\sigma_A$ should yield the value $\langle\sigma_A\rangle$, and measuring $I$ should yield 1. Now, I assume the quoted protocol actually measures overlaps with the various eigenstates of the operators, and that $p_0$ and $p_1$ are the probabilities of being in the $+$ or $-$ eigenstates of the various operators...

Measuring in the $z$ basis, with eigenstates $|0\rangle$ and $|1\rangle$, we expect the probabilities $p_0=(1+\langle\sigma_z\rangle)/2$ and $p_1=(1-\langle\sigma_z\rangle)/2$. Similar results will hold in the $x$ or $y$ basis, if we consider overlaps with the eigenstates such as $|\pm x\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$: $$p_{0}^{basis \,A}=\frac{1+\langle\sigma_A\rangle}{2}\quad \mathrm{and}\quad p_{1}^{basis \,A}=\frac{1-\langle\sigma_A\rangle}{2}.$$ This is what we expect from tomography: do a few different measurements, get some probabilities that depend on the state in question. One we know all of these probabilities, we can immediately infer the values of $\langle\sigma_A\rangle$, which uniquely determine the values of $a_0$ and $a_1$ up to a global phase.

Now, in a perfect world, the above description implies that, for all $A$, $$\begin{pmatrix}\beta_{I}^{|0\rangle} & \beta_{A}^{|0\rangle} \\ \beta_{I}^{|1\rangle} & \beta_{A}^{|1\rangle}\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$ However, there may be biases in the measurement device, for example always adding a little bit to $p_0$ and subtracting a little bit from $p_1$. This implies that the $\beta$ matrix must change in order to properly relate the measured values to the actual parameters of the state. By measuring the probabilities $p_i$ in the various bases, one can guess what the parameters of the state must be, and by finding biases in the measurements, one can define the true relation between the measurements and the state parameters.