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Considering just the single qubit case, the four possible operators you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - iY) \end{align} and also that \begin{align} X &= |+\rangle\langle +| - |-\rangle\langle -|\\ Y &= |+i\rangle\langle +i| - |-i\rangle\langle -i|\\ \end{align} so the two off-diagonal elements of the natural basis can be written entirely in terms of the states $|\pm\rangle$ and $|\pm i \rangle$.

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I'll explain a bit more about state tomography, because process tomography follows readily from it.

We wish to reconstruct an unknown $\rho$ from some set of measurements. We start from the Born rule that says the probability of getting outcome $i$ upon measurement

\begin{align} p_i = Tr(M_i \rho) \end{align}

Mathematically, $M_i$ and $\rho$ are both Hermitian, positive semidefinite matrices, and while we assume to know $M_i$, we do not know $\rho$. If we choose to measure in the $Z$ basis, then we have chosen the measurement set ("positive operator-valued measure")

\begin{align} \{ M_0 = |0\rangle\langle 0|, M_1 = |1\rangle\langle 1| \} \end{align}

and in retrospect when we obtain outcome 0/1, we say we measured with $M_{0/1}$. Consider what $p_0$ actually is symbolically

\begin{align} p_0 &= Tr(M_0 \rho) \\ &= Tr(|0\rangle\langle 0| \rho)\\ &= Tr(\langle 0 | \rho | 0 \rangle)\\ &= \langle 0 | \rho | 0 \rangle\\ &= \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} \rho_{00} & \rho_{01}\\\rho_{10} & \rho_{11}\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}\\ &= \rho_{00} \end{align}

So if I measure in the $Z$ basis many times, I can estimate $p_0$, which is actually element $(0,0)$ of $\rho$. Since 0 and 1 are mutually exclusive, whenever I don't get 0, I must get 1, and by similar logic can estimate $\rho_{11}$. The diagonals are easily obtained in the $Z$ basis, but we will need to measure in the $X$ and $Y$ bases to get the off-diagonals, as shown above. When measuring in the $X$ basis, the POVM then is

\begin{align} \{ M_+ = |+\rangle\langle +|, M_- = |-\rangle\langle -| \} \end{align}

I'll just tabulate what quantities you can estimate given these new outcomes/bases

\begin{align} p_+ = \frac{1}{2}(\rho_{00}+\rho_{01}+\rho_{10}+\rho_{11})\\ p_i = \frac{1}{2}(\rho_{00}-\rho_{01}-\rho_{10}+\rho_{11})\\ p_{+i} = \frac{1}{2}(\rho_{00}+i\rho_{01}-i\rho_{10}+\rho_{11})\\ p_{-i} = \frac{1}{2}(\rho_{00}-i\rho_{01}+i\rho_{10}+\rho_{11})\\ \end{align}

With (complex) linear combinations of the above quantities, we can fix the off-diagonals. Thus, measuring in the $X,Y,Z$ bases is enough to constrain $\rho$. Process tomography for some process $\mathcal{E}$ is the same thing, but the Born rule becomes

\begin{align} p_{ij} = Tr(M_i \mathcal{E}(\rho_j)) \end{align}

and we can play the same game. For example, preparing the state $|0\rangle\langle 0|$ and obtaining outcome $|1\rangle\langle 1|$ fixes a particular element of the process matrix. To keep this answer "short", check out section II.B of this paper (which uses a convenient vectorized notation) for more details.

Considering just the single qubit case, the four possible operators you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - iY) \end{align} and also that \begin{align} X &= |+\rangle\langle +| - |-\rangle\langle -|\\ Y &= |+i\rangle\langle +i| - |-i\rangle\langle -i|\\ \end{align} so the two off-diagonal elements of the natural basis can be written entirely in terms of the states $|\pm\rangle$ and $|\pm i \rangle$.

---

I'll explain a bit more about state tomography, because process tomography follows readily from it.

We wish to reconstruct an unknown $\rho$ from some set of measurements. We start from the Born rule that says the probability of getting outcome $i$ upon measurement

\begin{align} p_i = Tr(M_i \rho) \end{align}

Mathematically, $M_i$ and $\rho$ are both Hermitian, positive semidefinite matrices, and while we assume to know $M_i$, we do not know $\rho$. If we choose to measure in the $Z$ basis, then we have chosen the measurement set ("positive operator-valued measure")

\begin{align} \{ M_0 = |0\rangle\langle 0|, M_1 = |1\rangle\langle 1| \} \end{align}

and in retrospect when we obtain outcome 0/1, we say we measured with $M_{0/1}$. Consider what $p_0$ actually is symbolically

\begin{align} p_0 &= Tr(M_0 \rho) \\ &= Tr(|0\rangle\langle 0| \rho)\\ &= Tr(\langle 0 | \rho | 0 \rangle)\\ &= \langle 0 | \rho | 0 \rangle\\ &= \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} \rho_{00} & \rho_{01}\\\rho_{10} & \rho_{11}\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}\\ &= \rho_{00} \end{align}

So if I measure in the $Z$ basis many times, I can estimate $p_0$, which is actually element $(0,0)$ of $\rho$. Since 0 and 1 are mutually exclusive, whenever I don't get 0, I must get 1, and by similar logic can estimate $\rho_{11}$. The diagonals are easily obtained in the $Z$ basis, but we will need to measure in the $X$ and $Y$ bases to get the off-diagonals, as shown above. When measuring in the $X$ basis, the POVM then is

\begin{align} \{ M_+ = |+\rangle\langle +|, M_- = |-\rangle\langle -| \} \end{align}

I'll just tabulate what quantities you can estimate given these new outcomes/bases

\begin{align} p_+ = \frac{1}{2}(\rho_{00}+\rho_{01}+\rho_{10}+\rho_{11})\\ p_i = \frac{1}{2}(\rho_{00}-\rho_{01}-\rho_{10}+\rho_{11})\\ p_{+i} = \frac{1}{2}(\rho_{00}+i\rho_{01}-i\rho_{10}+\rho_{11})\\ p_{-i} = \frac{1}{2}(\rho_{00}-i\rho_{01}+i\rho_{10}+\rho_{11})\\ \end{align}

With (complex) linear combinations of the above quantities, we can fix the off-diagonals. Thus, measuring in the $X,Y,Z$ bases is enough to constrain $\rho$. Process tomography for some process $\mathcal{E}$ is the same thing, but the Born rule becomes

\begin{align} p_{ij} = Tr(M_i \mathcal{E}(\rho_j)) \end{align}

and we can play the same game. For example, preparing the state $|0\rangle\langle 0|$ and obtaining outcome $|1\rangle\langle 1|$ fixes a particular element of the process matrix. To keep this answer "short", check out section II.B of this paper (which uses a convenient vectorized notation) for a concise summary of state, measurement, and process tomography.

Considering just the single qubit case, the four possible projectors you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - iY) \end{align} and also that \begin{align} X &= |+\rangle\langle +| - |-\rangle\langle -|\\ Y &= |+i\rangle\langle +i| - |-i\rangle\langle -i|\\ \end{align} so the two off-diagonal elements of the natural basis can be written entirely in terms of the states $|\pm\rangle$ and $|\pm i \rangle$. For single-qubit state tomography, it is sufficient to measure with just three projectors: $|0\rangle$, $|+\rangle$, $|+i\rangle$. For the requirement that the trace be 1, we also need to count the -1 outcomes in one of those three bases, of which Z is usually chosen, so we also should include the projector onto $|1\rangle$. Process tomography is just an extension of state tomography, where such a complete basis is produced from both the prepared states and chosen measurement operators, to reconstruct the process matrix $\chi$, which has dimension $d^2 \times d^2$.

Considering just the single qubit case, the four possible operators you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - iY) \end{align} and also that \begin{align} X &= |+\rangle\langle +| - |-\rangle\langle -|\\ Y &= |+i\rangle\langle +i| - |-i\rangle\langle -i|\\ \end{align} so the two off-diagonal elements of the natural basis can be written entirely in terms of the states $|\pm\rangle$ and $|\pm i \rangle$.

---

I'll explain a bit more about state tomography, because process tomography follows readily from it.

We wish to reconstruct an unknown $\rho$ from some set of measurements. We start from the Born rule that says the probability of getting outcome $i$ upon measurement

\begin{align} p_i = Tr(M_i \rho) \end{align}

Mathematically, $M_i$ and $\rho$ are both Hermitian, positive semidefinite matrices, and while we assume to know $M_i$, we do not know $\rho$. If we choose to measure in the $Z$ basis, then we have chosen the measurement set ("positive operator-valued measure")

\begin{align} \{ M_0 = |0\rangle\langle 0|, M_1 = |1\rangle\langle 1| \} \end{align}

and in retrospect when we obtain outcome 0/1, we say we measured with $M_{0/1}$. Consider what $p_0$ actually is symbolically

\begin{align} p_0 &= Tr(M_0 \rho) \\ &= Tr(|0\rangle\langle 0| \rho)\\ &= Tr(\langle 0 | \rho | 0 \rangle)\\ &= \langle 0 | \rho | 0 \rangle\\ &= \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} \rho_{00} & \rho_{01}\\\rho_{10} & \rho_{11}\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}\\ &= \rho_{00} \end{align}

So if I measure in the $Z$ basis many times, I can estimate $p_0$, which is actually element $(0,0)$ of $\rho$. Since 0 and 1 are mutually exclusive, whenever I don't get 0, I must get 1, and by similar logic can estimate $\rho_{11}$. The diagonals are easily obtained in the $Z$ basis, but we will need to measure in the $X$ and $Y$ bases to get the off-diagonals, as shown above. When measuring in the $X$ basis, the POVM then is

\begin{align} \{ M_+ = |+\rangle\langle +|, M_- = |-\rangle\langle -| \} \end{align}

I'll just tabulate what quantities you can estimate given these new outcomes/bases

\begin{align} p_+ = \frac{1}{2}(\rho_{00}+\rho_{01}+\rho_{10}+\rho_{11})\\ p_i = \frac{1}{2}(\rho_{00}-\rho_{01}-\rho_{10}+\rho_{11})\\ p_{+i} = \frac{1}{2}(\rho_{00}+i\rho_{01}-i\rho_{10}+\rho_{11})\\ p_{-i} = \frac{1}{2}(\rho_{00}-i\rho_{01}+i\rho_{10}+\rho_{11})\\ \end{align}

With (complex) linear combinations of the above quantities, we can fix the off-diagonals. Thus, measuring in the $X,Y,Z$ bases is enough to constrain $\rho$. Process tomography for some process $\mathcal{E}$ is the same thing, but the Born rule becomes

\begin{align} p_{ij} = Tr(M_i \mathcal{E}(\rho_j)) \end{align}

and we can play the same game. For example, preparing the state $|0\rangle\langle 0|$ and obtaining outcome $|1\rangle\langle 1|$ fixes a particular element of the process matrix. To keep this answer "short", check out section II.B of this paper (which uses a convenient vectorized notation) for more details.

Considering just the single qubit case, the four possible projectors you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - iY) \end{align} and also that \begin{align} X &= |+\rangle\langle +| - |-\rangle\langle -|\\ Y &= |+i\rangle\langle +i| - |-i\rangle\langle -i|\\ \end{align} so the two off-diagonal elements of the natural basis can be written entirely in terms of the states $|+\rangle$ and $|-\rangle$. For single-qubit state tomography, it is sufficient to measure with just three projectors: $|0\rangle$, $|+\rangle$, $|+i\rangle$. For the requirement that the trace be 1, we also need to count the -1 outcomes in one of those three bases, of which Z is usually chosen, so we also should include the projector onto $|1\rangle$. Process tomography is just an extension of state tomography, where such a complete basis is produced from both the prepared states and chosen measurement operators, to reconstruct the process matrix $\chi$, which has dimension $d^2 \times d^2$.

Considering just the single qubit case, the four possible projectors you list are \begin{align} |0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1| \end{align} and like you say, only the first and last are physical. Note, however, that \begin{align} |0\rangle\langle 1| = \frac{1}{2}(X + iY)\\ |1\rangle\langle 0| = \frac{1}{2}(X - iY) \end{align} and also that \begin{align} X &= |+\rangle\langle +| - |-\rangle\langle -|\\ Y &= |+i\rangle\langle +i| - |-i\rangle\langle -i|\\ \end{align} so the two off-diagonal elements of the natural basis can be written entirely in terms of the states $|\pm\rangle$ and $|\pm i \rangle$. For single-qubit state tomography, it is sufficient to measure with just three projectors: $|0\rangle$, $|+\rangle$, $|+i\rangle$. For the requirement that the trace be 1, we also need to count the -1 outcomes in one of those three bases, of which Z is usually chosen, so we also should include the projector onto $|1\rangle$. Process tomography is just an extension of state tomography, where such a complete basis is produced from both the prepared states and chosen measurement operators, to reconstruct the process matrix $\chi$, which has dimension $d^2 \times d^2$.