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for some quantum state $\rho$ (a normalized non-negative matrix). We want to check under which conditions $\chi_{i,j}=\chi(t_1-t_j)$ is non-negative definite as a matrix.

for some quantum state $\rho$ (a normalized non-negative matrix). We want to check under which conditions $\chi_{i,j}=\chi(t_i-t_j)$ is non-negative definite as a matrix.

Theorem (Bochner) (Univariate case) $\chi (t)$ is the Fourier transform of a probability density $P(\omega)$ ($t,\omega \in\mathbb{R}$) if and only if for any $n$-tuple $t_1,t_2,\ldots t_n$ ($t_k \in \mathbb{R}, \ k=1,2,\ldots,n$) the $n\times n$ matrix with entries $\chi_{i,j}:= \chi(t_i-t_j)$ is non-negative definite (and hermitian).

 

Note: for the Multivariate $d$-dimensional generalization simply consider the obvious rephrasing with $t,\omega, t_k \in \mathbb{R}^d$.

Theorem (Bochner) (Univariate case) $\chi (t)$ is the Fourier transform of a probability density $P(\omega)$ ($t,\omega \in\mathbb{R}$) if and only if for any $n$-tuple $t_1,t_2,\ldots t_n$ ($t_k \in \mathbb{R}, \ k=1,2,\ldots,n$) the $n\times n$ matrix with entries $\chi_{i,j}:= \chi(t_i-t_j)$ is non-negative definite (and hermitian).

Note: for the Multivariate $d$-dimensional generalization simply consider the obvious rephrasing with $t,\omega, t_k \in \mathbb{R}^d$.

which is of the form $BB^\dagger$ and proves that $\chi_{i,j}$ is non-negative definite. Obviously this whole construction breaks down if $X_k$ are not mutually commuting. Since the condition in Bochner's theorem is an if and only if, in this case $\chi(t)$ is not the Fourier transform of a probability distribution}

Strictly speaking (as pointed out correctly by @AcuriousMind) this does not prove that the matrix $\chi_{i,j}$ is not non-negative definite for non-mutually commuting observables. For that one should find a counterexample, i.e. show that $\chi$ has a negative eigenvalue. However it does show where the argument breaks down.

which is of the form $BB^\dagger$ and proves that $\chi_{i,j}$ is non-negative definite. Obviously this whole construction breaks down if $X_k$ are not mutually commuting.

Strictly speaking (as pointed out correctly by @AcuriousMind) this does not prove that the matrix $\chi_{i,j}$ is not non-negative definite for non-mutually commuting observables. For that one should find a counterexample, i.e. show that $\chi$ has a negative eigenvalue. However it does show where the argument breaks down.

Added edit

Here I present a counterexample for a single qubit. It can be shown that for $n=2$ the $\chi_{i,j}$ matrix is always non-negative definite. So to look for the first counterexample we must take $n=3$.

Consider the following problem with incompatible (non-commuting) observables given by $\sigma^x$ and $\sigma^z$. As for the state we pick $\rho = | 0\rangle\langle 0|$. Hence the putative characteristic function is

\begin{align} \chi(t_x,t_z) &:= \langle 0| e^{i (t_x \sigma^x +t_z \sigma^z)}| 0 \rangle \\ & = \cos \left (\sqrt{t_x^2 +t_z^2}\right ) + i\frac{t_z}{\sqrt{t_x^2 +t_z^2}} \sin \left ( \sqrt{t_x^2 +t_z^2}\right ) . \end{align}

Now form the matrix $\chi_{i,j}$ for $n=3$. It can be shown that $\chi$ has the form

$$ \chi = 1\!\mathrm{l} + \Gamma $$

where the matrix $\Gamma$ is hermitian and has zero on the diagonal. Since $\Gamma$ is traceless $\chi$ fails to be non-negative definite if $\Gamma$ has an eigenvalue smaller than $-1$.

For simplicity let's call $a_{ij} = t_x^i-t_x^j$ and $b_{ij} = t_z^i-t_z^j$. Now simply pick random $a_{ij}, b_{ij}$:

\begin{align} a_{12} & = 1 \ \ b_{12} = 0.5 \\ a_{13} & = 2 \ \ b_{13} = 1.3 \\ a_{23} & = 0.4 \ \ b_{23} = 0.9 \\ \end{align}

The eigenvalues of $\Gamma$ turn out to be $\{1.499, \ -1.221, \ -0.278\}$, which implies $\chi$ is not non-negative definite. This shows that the Fourier transform of $\chi(t_x,t_z)$ is not a (joint) probability distribution. $\square$