For the purposes of this question, suppose that the operator $R$ representing the observable $\mathsf{R}$ has nondegenerate eigenspaces.

In discussing state determination (for some given situation, how do we determine the corresponding state operator?) Ballentine writes

If we repeat the preparation–measurement sequence many times, and determine the relative frequency of the result $R = r_n$, we will in effect be measuring the probability $Prob(R = r_n|\rho)$, where $\rho$ denotes the unknown state operator. But, according to (2.26), we have $Prob(R = r_n|\rho) = \langle r_n|\rho|r_n\rangle$ for the case of a nondegenerate eigenvalue. Thus the measurement of the probability distribution of the dynamical variable $R$ determines the diagonal matrix elements of the state operator in this representation.

Perfect, so we can get the diagonal elements. What about the off-diagonal elements?

To this end, Ballentine writes

To obtain information about the nondiagonal matrix elements $\langle r_m|\rho|r_n\rangle $, it is necessary to measure some other dynamical variable whose operator does not commute with $R$.

It is this statement I don't follow. Why is it impossible for an operator which commutes with $R$ to give us the requisite information?

I know that, for any such commuting $A$, since $R$ is nondegenerate we have $A|r_n\rangle = \lambda |r_n\rangle$ for some $\lambda$. But how do I turn this into a "proof" that I can't obtain all the requisite matrix elements $\langle r_m|\rho|r_n\rangle $ of this operator?

  • 1
    $\begingroup$ Have a look on The Empirical Determination of Quantum States. William Band and James L. Park a paper form 1970. In this paper, they show, IIRC, that you need indeed 3 non-commuting observables (+identity operator) for a case of a two-dimensional Hilbert space to fully determine the density matrix. And they have follow up papers generalizing this result (again, IIRC). $\endgroup$ Commented Aug 4, 2023 at 20:47
  • $\begingroup$ To be sure, Ballentine gives a specific construction which lets me determine the state in general (Chapter 8.2 in his text). I am just wondering why there is no other choice involving a commuting operator. @TobiasFünke $\endgroup$
    – EE18
    Commented Aug 4, 2023 at 20:53
  • $\begingroup$ As I said, IIRC, the paper gives a proof for why, at least for a two-dimensional system, this has to be the case. I might be wrong here, you have to check. $\endgroup$ Commented Aug 4, 2023 at 21:01
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    $\begingroup$ @TobiasFünke see this discussion of MUBs which is quite relevant and complements Band&Park $\endgroup$ Commented Aug 5, 2023 at 1:30
  • $\begingroup$ Hi @ZeroTheHero, thanks for the link! $\endgroup$ Commented Aug 5, 2023 at 6:48

1 Answer 1


Commuting operators can be simultaneously diagonalized. This means that, for any commuting operator $A$ you design, it can be written in the eigenbasis of $R$ as a sum with no off-diagonal terms $$R=\sum_n R_n |r_n\rangle\langle r_n| \quad\mathrm{and}\quad A=\sum_n A_n |r_n\rangle\langle r_n|.$$ Measuring the expectation value of any such operator for the state $\rho=\sum_{mn} (\langle r_m |\rho |r_n\rangle) |r_m\rangle\langle r_n|$ simply gives $$\mathrm{Tr}(R\rho)=\sum_n R_n (\langle r_n |\rho |r_n\rangle)\quad \mathrm{and}\quad \mathrm{Tr}(A\rho)=\sum_n A_n (\langle r_n |\rho |r_n\rangle).$$ This can only ever be a function of $(\langle r_n |\rho |r_n\rangle)$ and not of any coefficient $(\langle r_n |\rho |r_m\rangle)$ with $m\neq n$ because any operator $A$ that commutes with $R$ will also not have any off-diagonal elements in $R$'s eigenbasis.

If you want an answer in the spirit of $A|r_n\rangle=\lambda_n|r_n\rangle$ as stated in the question, we proceed by again expressing $\rho$ in the eigenbasis of $R=\sum_n R_n |r_n\rangle\langle r_n|$ as $\rho=\sum_{mn} (\langle r_m |\rho |r_n\rangle) |r_m\rangle\langle r_n|$. We then directly compute \begin{aligned} \mathrm{Tr}(A\rho)=&\sum_{mn} (\langle r_m |\rho |r_n\rangle) \mathrm{Tr}(A|r_m\rangle\langle r_n|)\\ =&\sum_{mn} (\langle r_m |\rho |r_n\rangle) \mathrm{Tr}(\lambda_m|r_m\rangle\langle r_n|)\\ =&\sum_{mn} (\langle r_m |\rho |r_n\rangle) \lambda_m\delta_{mn}\\ =&\sum_{n} (\langle r_n |\rho |r_n\rangle) \lambda_n. \end{aligned} Again, this only depends on the diagonal elements of $\rho$ in $R$'s eigenbasis.

I can give you a flavour of what happens when there are degeneracies. Say two eigenvalues of $R$ are the same. Call them $R_1=R_2=r$. Then the expectation value of that operator looks like $$\langle R\rangle=r(\langle r_1|\rho|r_1\rangle+\langle r_2|\rho|r_2\rangle)+\sum_{n>2} R_n\langle r_n|\rho|r_n\rangle.$$ Lo and behold, $$\langle r_1|\rho|r_1\rangle+\langle r_2|\rho|r_2\rangle=\langle \psi_1|\rho|\psi_1\rangle+\langle \psi_2|\rho|\psi_2\rangle$$ for transformations $$\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\end{pmatrix}=\mathsf{U}\begin{pmatrix}|r_1\rangle\\|r_2\rangle\end{pmatrix}$$ with $2\times 2$ unitary matrix $\mathsf{U}$. So it looks like we can learn about $\rho$ in some other $|\psi\rangle$ basis related to the $|r\rangle$ basis by a unitary transformation if $R$ or $A$ have degeneracy, but that's never giving us any new information because it will always be able to be rewritten as diagonal components in the $|r\rangle$ basis. For example, choose $\mathsf{U}=1/\sqrt{2}\begin{pmatrix}1&1\\ 1&-1\end{pmatrix}$ and observe that the expansion of $\langle \psi_1|\rho|\psi_1\rangle+\langle \psi_2|\rho|\psi_2\rangle$ has all of the cross-terms $\langle r_1|\rho|r_2\rangle$ cancel.

  • $\begingroup$ But you don't have to use the common eigenbasis, no? So e.g. what happens if $R$ and $A$ have some degeneracies... $\endgroup$ Commented Aug 4, 2023 at 21:02
  • $\begingroup$ Sorry I had to edit the answer @TobiasFünke $\endgroup$ Commented Aug 4, 2023 at 21:04
  • $\begingroup$ @TobiasFünke nothing in the edited answer changes with degeneracies. The operators can be simultaneously diagonalized, so we write them in that eigenbasis, and then the state can only be determined up to its projections onto that eigenbasis $\endgroup$ Commented Aug 4, 2023 at 21:06
  • $\begingroup$ My question/objection is that it might be possible to determine the matrix of $\rho$ in a basis which is not the common eigenbasis... I might miss something obvious here?! $\endgroup$ Commented Aug 4, 2023 at 21:10
  • $\begingroup$ @TobiasFünke Agree, this was my concern alluded to at the end of the OP. $\endgroup$
    – EE18
    Commented Aug 4, 2023 at 21:12

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