What I understand...

A tomographically complete set of operators is a set of operators such that measuring the probability distributions of an unknown quantum state over the spectrum of each of these operators allows you to write down the pre-measurement quantum state (of course, one needs infinitely many copies of the identically prepared unknown quantum state to carry this out). For example, for a two-dimensional Hilbert space of spin half particles, the spin operators in $x$, $y$, and $z$ directions form such a set which I can verify by explicitly writing a unique state consistent with a given set of probability distributions over the spectrum of these three operators.

What I'm looking for...

I'm not sure if I understand the mathematical conditions that I can write down for a set of operators to tell me if it's a tomographically complete set of operators or not. Intuitively, I expect it to be something like "a largest set of non-commuting operators" because such a set would give me all the information about the phases which would be hidden if I perform measurements over a commuting set of operators. But what is the precise mathematical definition/criterion for such a largest set of non-commuting operators?

Wikipedia says that a tomographically complete set of operators forms an "operator basis on the Hilbert space". I don't think I understand this statement, for example, the three spin operators constitute a tomographically complete set of operators but I can't write down $S^2$ as a linear combination of $S_x, S_y, S_z$ which is something I should be able to do if $S_x, S_y, S_z$ formed a basis for all operators in the Hilbert space, or so do I think. Or, is this not what an operator basis on the Hilbert space supposed to mean?

  • $\begingroup$ I believe there are different definitions around. Are you looking specifically for the condition that the expectations of the operators identify the state, or more generally that the full probability distributions over the outcomes of each operator identify the state? $\endgroup$
    – pglpm
    Jun 5, 2020 at 15:50
  • $\begingroup$ @pglpm Good point, I had the full probability distributions over the outcomes of each operator in mind. I've edited the question. $\endgroup$
    – user87745
    Jun 5, 2020 at 15:56
  • $\begingroup$ My impression – but I'm not fully sure at the moment – is that you can consider the mixture of the operators in question, so as to form a positive-operator-valued measure. Then the rank of the latter tells you if each state leads to a distinct probability distribution of the outcomes of this POVM $\endgroup$
    – pglpm
    Jun 5, 2020 at 16:02
  • $\begingroup$ Why do you want the full probability distributions? What does that even mean? Eigenspaces? That's not what is meant by tomography - usually people think of expectation values. $\endgroup$ Jun 5, 2020 at 16:02
  • 1
    $\begingroup$ @DvijD.C. I guess it depends how you like your quantum mechanics ;) You can do both in tomography, but expectation values are enough (cf. my answer). Of course, the $p_i$ are just expectation values of eigenspace projectors, so all good. -- But this tells you that in case you have the $p_i$, you only need that the eigenspace projectors form a basis. $\endgroup$ Jun 5, 2020 at 16:14

1 Answer 1


The space of operators is a complex Hilbert space with scalar product $\langle X,Y\rangle = \mathrm{tr}[X^\dagger Y]$. Even further, the space of hermitian matrices is a real Hilbert space with the same scalar product.

In order to completely reconstruct a vector $\rho$ in a Hilbert space, you thus want the scalar product with a set of hermitian operators $\{B_i\}$ which span the full space of hermitian matrices (as a real vector space). Then, you can write $$ \rho = \sum C_i \, \langle B_i,\rho\rangle = \sum C_i\,\mathrm{tr}[B_i^\dagger\rho]\ , $$ where the $\{C_i\}$ and the $\{B_i\}$ form a bi-orthogonal set, $\langle B_i,C_j\rangle = \delta_{ij}$.

You might wonder whether there are constraints from positivity which simplify the problem, but you can see that this will not be the case by considering $\rho+\lambda I$ in case of a non-positive $\rho$, which is positive for sufficiently large $\lambda$ and requires the same amount of information to be reconstructed (since $\langle B_i,I\rangle=\mathrm{tr}\,B_i$ contains no information).

On the other hand, we also know that $\mathrm{tr}\,\rho=1$, and this does provide a constraint: Specifically, if you choose your basis that it contains the identity $I$, then you know $\langle I,\rho\rangle = 1$.

Now to your example above: What you forget (in case you talk about spin 1/2) is the identity operator. Only with the identity operator, the Paulis form a basis for the space of operators, and allow you to express e.g. $S^2=3 I/4$.

  • $\begingroup$ Thus i take it that for the usual $(x, p)$ moving-particle-in-space scenario the tomographically complete set will be uncountably many operators, right? I.e. it is all operators $|x\rangle\langle x|$ for every position value $x$? $\endgroup$ Jun 5, 2020 at 16:16
  • $\begingroup$ How is that, though? Can't you just measure $x$ repeatedly and build up a histogram? Or what do you mean, "exact position"? Are you thinking of imperfect measurements? Or measurements of limited resolution? $\endgroup$ Jun 5, 2020 at 16:18
  • $\begingroup$ I'm not sure the whole tomography thing makes sense in non-separable spaces. I wouldn't even bet much on what I say above for separable spaces, though things probably go through. $\endgroup$ Jun 5, 2020 at 16:19
  • $\begingroup$ @The_Sympathizer I suppose the point is that you cannot capture the information of the complete histogram (the probability distribution over various possible outcomes of the measurement of position operator) in the expectation value of the position operator. So if you want to express the state using some expectation values, you'd need to consider the expectation values of the projection operators $\vert x\rangle \langle x\vert$ which contain the information of the full probability distribution over the eigenvalues of the position operator. $\endgroup$
    – user87745
    Jun 5, 2020 at 17:31
  • $\begingroup$ This is in general a sufficient but not a necessary condition if you know something about your density matrix. If you know, for example, that it is made up of at most r linearly independent pure states (that is you have a rank <= r constraint) then this is far too strong. The above would require n^2 observables when in fact 2 n r - r^2 well chosen observables would suffice. $\endgroup$ Aug 10, 2022 at 17:05

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