In the thesis "Efficient Simulation of Random Quantum States and Operators" on page 25 there is a portion of text explaining a method for quantum process tomography. It claims that states from an orthonormal basis for density matrices is easy to generate experimentally. This seems odd... since some of the matrices it includes in the basis are not hermitian...

They define the basis as follows:

$B_{\sigma}=\{\sigma^{(i, j)}=|i\rangle\langle j|: 1 \leq i, j \leq d \}$

but they probably actually meant

$B_{\sigma}=\{\sigma^{(i, j)}=|i\rangle\langle j|: 0 \leq i, j \leq d - 1 \}$

regardless, it generates matrices that aren't hermitian, and thus shouldn't be viable quantum states.

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    $\begingroup$ edited title. i think an orthonormal basis implies not all matrices are Hermitian. i'm not sure what the proof would be though. $\endgroup$ Mar 12, 2021 at 18:38

2 Answers 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 a concise summary of state, measurement, and process tomography.

  • $\begingroup$ "so the two off-diagonal elements of the natural basis can be written entirely in terms of the states..." i think you forgot to include $|+i\rangle$ and $|-i\rangle$. you need them to represent $Y$ $\endgroup$ Mar 12, 2021 at 17:22
  • $\begingroup$ yep sorry, just added $\endgroup$
    – chrysaor4
    Mar 12, 2021 at 17:24
  • $\begingroup$ "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⟩" can you elaborate more on this? i'm not familiar with some of the terminology you are using here. Do each of those projectors always result in 1 or -1 outcomes after measuring some qubit? What do those output values correspond to? $\endgroup$ Mar 12, 2021 at 17:28
  • $\begingroup$ why does choosing Z result in including the projector onto |1⟩? $\endgroup$ Mar 12, 2021 at 17:28
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    $\begingroup$ the answer to this question: quantumcomputing.stackexchange.com/questions/16408/… may have much of what you are trying to explain. $\endgroup$ Mar 12, 2021 at 17:33

Yes, $|i\rangle\langle j|$ is not a density matrix in general. But those matrices indeed form a basis of the space of all matrices. And if we know those $d^2$ values $\mathcal E(|i\rangle\langle j|)$ then we can fully reconstruct the action of $\mathcal E$. So, this is a sort of "mathematical" tomography.

Note, however, that we can pick $d^2$ density matrices that will form a basis of all matrices (it won't be orthogonal). For example, we can take traceless Hermitian Gell-Mann_matrices, add a multiple of $I$ to each of them and scale them, so they will become density matrices.

Another example of such basis is a SIC-POVM, but it's not yet proved that it exists in every dimension $d$.

A simpler basis of density matrices is described here.


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