Let $\rho$ be a single-qubit state. A standard way to characterise $\rho$ is to measure the expectation values of the Pauli matrices, that is, to perform projective measurements in the three mutually unbiased bases corresponding to the Pauli matrices, and collect the corresponding statistics.

Each such measurement can be described by a projective measurement. Generally speaking, measuring in an orthonormal basis $\{|u_1\rangle, |u_2\rangle\}$ corresponds to the POVM $$\{|u_1\rangle\!\langle u_1|, |u_2\rangle\!\langle u_2|\}.$$

Is there a way to describe the entire process of measuring the state in the various possible bases using a single POVM? In other words, what does a POVM which is informationally complete for a single qubit look like? Is there one that can be considered to directly correspond to the standard state tomography scheme?

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    $\begingroup$ You could always just take the POVM $\{\tfrac13 |0\rangle \langle 0|, \tfrac13 |1 \rangle \langle 1 |, \tfrac13 | + \rangle \langle + |, \tfrac13 |-\rangle \langle - |, \tfrac13 |i\rangle \langle i |, \tfrac13 |-i\rangle\langle -i|\}$. $\endgroup$
    – Rammus
    Commented Mar 17, 2021 at 21:02
  • $\begingroup$ @Rammus ah, that makes sense, thanks. I wonder though, that's clearly not an "optimal" POVM, in the sense that it uses more elements than strictly required. Equivalently, its components are not linearly independent. Does that mean the tomography scheme is, to some degree, inefficient? I'm not sure how to make this idea more precise; I'm thinking of inefficiency along the lines of there possibly being POVMs that can reconstruct the state within a given accuracy with a smaller number of samples. $\endgroup$
    – glS
    Commented Mar 17, 2021 at 23:59
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    $\begingroup$ Have you heard of SIC-POVMs? They are optimal (in your sense). Standard (1-qubit) QST is a bit inefficient because you're gathering statistics for all $6$ Pauli eigenstates - but it's just much easier to actually measure in the $3$ Pauli bases than it is to implement a (S)IC-POVM $\endgroup$
    – JSdJ
    Commented Mar 18, 2021 at 10:26
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    $\begingroup$ For QST on general quantum states I am not aware of any choices of POVM for which your statistical error vanishes more rapidly compared to others - if you add some assumptions or expectations on your quantum states there are better and worse choices though $\endgroup$
    – JSdJ
    Commented Mar 18, 2021 at 10:30
  • $\begingroup$ Obviously, to get the full information about a vector in a $d$-dimensional vector space you need the scalar product with $d$ linearly independent vectors. Since those are scalar product between hermitian operators (which are real), we are talking about real vector spaces, so for qu-d-its a POVM with $d^2$ elements is both necessary and sufficient. That linear independence can be achieved with rank-1 projectors should be clear. $\endgroup$ Commented Mar 19, 2021 at 23:12

2 Answers 2


Quantum state tomography owes its power and flexibility to the fact that it supports a wide class of measurements. Any informationally complete POVM, i.e. one whose elements span the space $L_H(\mathcal{H})$ of Hermitian operators on the target system's Hilbert space $\mathcal{H}$ qualifies for use in QST.

One way to highlight the generality of QST with respect to the choice of POVM is to cast it as a vector reconstruction problem. Every orthonormal basis $\{v_k\}$ of a vector space $V$ has the reconstruction property that $u = \sum_k \langle u, v_k\rangle v_k$ for any vector $u$. It turns out that certain sets of vectors other than bases, namely frames, also have a variant of the property. More precisely, for any vector $u$ and a frame $\{v_k\}$ we have

$$u = \sum_k \langle u, v_k\rangle\tilde{v_k}\tag1$$

where $\{\tilde{v_k}\}$ is a frame dual to $\{v_k\}$. The dual frame can be computed as $\tilde{v_k} = S^{-1}v_k$ where $S$ is the frame operator defined as $S: u \mapsto \sum_k\langle u, v_k\rangle v_k$.

Quantum state tomography can be analyzed as an application of frame theory to the task of reconstructing an element of $L_H(\mathcal{H})$. Suppose that positive operators $E_k$ sum to identity and span $L_H(\mathcal{H})$, i.e. $\{E_k\}$ is both a POVM and a frame. Let $\{F_k\}$ denote the frame dual to $\{E_k\}$. Then by $(1)$ for any operator $\rho$ we have

$$ \rho = \sum_k \mathrm{tr}(\rho E_k)F_k.\tag2 $$

This enables complete characterization of a quantum state, because the coefficients $\mathrm{tr}(\rho E_k)$ are accessible experimentally while operators $F_k$ can be derived using frame theory summarized above.

The key point about this construction is that any POVM that is also a frame qualifies for use in quantum state tomography, so we need to make additional choices to exhibit a specific POVM.

Example 1: Standard QST

Any orthonormal basis is a frame which is its own dual. A well-known example of an orthonormal basis in $L_H(\mathbb{C}^2)$ is the set $\{I/\sqrt{2}, X/\sqrt{2}, Y/\sqrt{2}, Z/\sqrt{2}\}$. In this case equation $(2)$ takes the form

$$ \rho = \frac12\left(\mathrm{tr}(\rho)I + \mathrm{tr}(\rho X)X+ \mathrm{tr}(\rho Y)Y + \mathrm{tr}(\rho Z)Z\right). $$

However, the Pauli operators are not positive and hence not a POVM. We can obtain a POVM by replacing each operator in the set with its two eigenprojectors and renormalizing

$$ \left\{\frac{|0\rangle\langle 0|}{3}, \frac{|1\rangle\langle 1|}{3}, \frac{|+\rangle\langle +|}{3}, \frac{|-\rangle\langle -|}{3}, \frac{|{+i}\rangle\langle {+i}|}{3}, \frac{|{-i}\rangle\langle {-i}|}{3}, \right\}. $$

This set spans $L_H(\mathbb{C}^2)$, because the original orthonormal basis does and is therefore an informationally complete POVM. The advantage of this POVM is that it is often relatively simple to realize experimentally.

Example 2: Minimal POVM

A frame spans its vector space, so the smallest informationally complete POVM has at least $\dim L_H(\mathbb{C}^2) = 4$ elements. There are many examples that attain this minimum. For instance, define

$$ \begin{align} |\psi_0\rangle &= |0\rangle \\ |\psi_1\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}|1\rangle \\ |\psi_2\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{2\pi i}{3}}|1\rangle \\ |\psi_3\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{4\pi i}{3}}|1\rangle \end{align} $$

and set $E_k = \frac12|\psi_k\rangle\langle\psi_k|$ (see also this picture). Following standard frame theory we calculate

$$ \begin{align} F_0 &= \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix} \\ F_1 &= \begin{pmatrix} 0 & \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix} \\ F_2 &= \begin{pmatrix} 0 & -\frac{1+i\sqrt{3}}{\sqrt{2}} \\ -\frac{1-i\sqrt{3}}{\sqrt{2}} & 1 \end{pmatrix} \\ F_3 &= F_2^T \end{align} $$

which together with $E_k$ and equation $(2)$ provides a complete description for single-qubit QST using a minimal POVM.

  • $\begingroup$ interesting; I'd never heard of frames, thanks. How exactly are you using "frame theory" in the second example? Isn't "frame theory" about linear decompositions in terms of linearly dependent vectors? Here you have four linearly independent ones. Even though they are not orthogonal, decomposition in terms of these only requires to use the standard dual basis construction. $\endgroup$
    – glS
    Commented Mar 22, 2021 at 23:05
  • $\begingroup$ although I'm a bit puzzled by this frame theory thing I think. Let $\{u_k\}$ be some (not necessarily linearly independent) set of vectors whose span equals the underlying space. Say we want the coefficients $\alpha_k$ in the decomposition $x=\sum_ku_k \alpha_k$ for some $x$. This amounts to solving the linear problem $x=U\alpha$, with $U$ the matrix whose columns are $u_k$. The solution (space) is $\alpha\in U^+ x+\ker(U)$. Do we need something more than this in this context? Unless the framework is mostly used for infinite-dimensional spaces, where I guess this might not hold $\endgroup$
    – glS
    Commented Mar 22, 2021 at 23:15
  • $\begingroup$ Re "using frame theory": I meant that I computed $F_k$ using the formula $F_k = S^{-1}(E_k)$ where $S$ is the frame operator introduced earlier. Re linear dependence: Like a basis, frame must span the whole space, but unlike a basis, it may or may not be linearly independent. If it has $\dim V$ elements (as in the second example) then they are linearly independent. If it has more elements (as in the first example) then they are not. $\endgroup$ Commented Mar 22, 2021 at 23:20
  • $\begingroup$ Dual basis construction is subsumed in the dual frame construction, e.g. in the second example we have $\mathrm{tr}(E_i^\dagger F_j) = \delta_{ij}$ so $F_k$ is the dual basis to $E_k$ (it's readily visible for $i=0$ due to the zeros in top left corners of $F_j$ for $j=1,2,3$). $\endgroup$ Commented Mar 22, 2021 at 23:50
  • $\begingroup$ Re pseudo-inverse: these two formulations of vector reconstruction are equivalent. On one hand, pseudo-inverse is in this case $U^+ = U^\dagger (UU^\dagger)^{-1}$ so setting $\alpha = U^+x$ (i.e. picking the zero vector in $\ker U$) we get $U\alpha = UU^+x = UU^\dagger (UU^\dagger)^{-1}x=Ix=x$. On the other hand, the frame operator is $S=UU^\dagger$, so $\tilde{u}_k = S^{-1}u_k = (UU^\dagger)^{-1}u_k$ and defining $\tilde{U}=[\tilde{u}_1, \dots, \tilde{u}_m]=S^{-1}U$ the equation $(1)$ above takes the form $\tilde{U}U^\dagger x = S^{-1}UU^\dagger x = (UU^\dagger)^{-1} UU^\dagger x = Ix = x$. $\endgroup$ Commented Mar 23, 2021 at 0:47

POVM for standard QST in the Pauli bases

In standard single-qubit QST one measures in the Pauli bases, each with equal probability $\frac{1}{3}$. As @Rammus has pointed out, this corresponds to the POVM $$ \{E_{m}\} = \Big\{\tfrac{1}{3}|0\rangle\langle0|,\tfrac{1}{3}|1\rangle\langle1|,\tfrac{1}{3}|+\rangle\langle+|,\tfrac{1}{3}|-\rangle\langle-|,\tfrac{1}{3}|+i\rangle\langle+i|,\tfrac{1}{3}|-i\rangle\langle-i|\Big\} $$ for the usual Pauli eigenstates.

This is an informationally complete (IC-)POVM, as $\mathrm{span}\{E_{m}\} = \mathcal{L}(\mathcal{H})$ i.e. the POVM elements span the density operator space (of a single qubit) - and thus any valid quantum state can be reconstructed using the statistics of the outcomes of these POVMs.


However, there is some degeneracy here, as there are $6$ POVM elements, but a basis for $\mathcal{L}(\mathcal{H})$ needs only $4$ elements. There are plenty of minimal IC-POVMs, which just have four elements. They are often relatively impractical to actually implement, but can be interesting from a theoretical point of view.

One could say that any IC-POVM corresponds to a (state) tomography experiment, and any non IC-POVM cannot correspond to a proper QST experiment, as only the IC-POVM's span the density operator space.

Thus, any set of PSD operators $\{E_{m}\}$ for which:

  • $\sum_{m}{E_{m}} = I$

  • $\mathrm{span}(\{E_{m}\}) = \mathcal{L}(\mathcal{M})$

is a POVM that corresponds to a valid QST. Moreover, if $|\{E_{m}\}| = d^{2}$, the POVM is minimal.

Bonus: SIC-POVMs

Finally, as a sidenode due to the comments, a symmetric informationally complete POVM or SIC-POVM is a minimal IC POVM $\{G_{m}\}$ for which the elements are completely symmetric under the HS-inner product:

$$ \langle G_{m}, G_{m'}\rangle_{HS} = a \not = a(m,m') \,\,\,(\forall m,m'| m \not=m'). $$ In one-qubit systems, the canonical example is (the projectors of): $$ \{|0\rangle, \frac{1}{\sqrt{3}}(|0\rangle + \sqrt{2}|1\rangle, \frac{1}{\sqrt{3}}(|0\rangle + \sqrt{2}e^{\frac{2\pi i}{3}}|1\rangle, \frac{1}{\sqrt{3}}(|0\rangle + \sqrt{2}e^{\frac{4\pi i}{3}}|1\rangle\} $$

In higher dimensions, it's slightly more complicated - but that's ' a story for another time' :)

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    $\begingroup$ "In higher dimensions, it is more complicated"? - To the extent that not even their existence is known ... $\endgroup$ Commented Mar 19, 2021 at 23:11
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    $\begingroup$ @NorbertSchuch slightly was indeed an understatement meant as a light joke - although for some dimensions they are known, as you're probably well aware. $\endgroup$
    – JSdJ
    Commented Mar 20, 2021 at 12:24

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