I am reading through "Direct Fidelity Estimation from Few Pauli Measurements" and it states that the measure of fidelity between a desired pure state $\rho$ and an arbitrary state $\sigma$ is $\mathrm{tr}(\rho\sigma)$. It then describes a 'characteristic function' $\chi_\rho(k) = \mathrm{tr}(\rho W_k/\sqrt d)$ where $W_k (k = 1, ... , d^2)$ are all possible Pauli operators (n-fold tensor products of $I$, $X$, $Y$ and $Z$). It then states that:

$$\mathrm{tr}(\rho\sigma) = \sum_k \chi_\rho(k) \chi_\sigma(k)$$

Which is where I get confused. How do the two equal each other? A proof is much appreciated.

Additionally, what do they mean by 'characteristic function'. Is it the type of characteristic function defined in probability theory. If so, I don't see how it is derived.


1 Answer 1



If $v_1, v_2, \dots, v_n$ is an orthonormal basis in the inner product space $V$, then any vector $u\in V$ can be expressed as a linear combination

$$ u = \alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_n v_n.\tag1 $$

Moreover, the coefficients can be computed using $\alpha_k=\langle v_k, u\rangle$, as can be seen by applying $\langle v_k, .\rangle$ to both sides of $(1)$.

Fidelity in terms of the characteristic function

The set $L(\mathcal{H})$ of linear operators on a $d$-dimensional Hilbert space $\mathcal{H}$ forms an inner product space with the inner product defined as

$$ \langle A, B\rangle = \mathrm{tr}(A^\dagger B). $$

It is easy to check that the normalized Pauli operators $B_k = W_k/\sqrt{d}$ form an orthonormal basis in $L(\mathcal{H})$ with respect to this inner product. Therefore, any operator $\rho \in L(\mathcal{H})$ can be written as

$$ \rho = \alpha_1 B_1 + \alpha_2 B_2 + \dots + \alpha_{d^2} B_{d^2}\tag{1'} $$

and the coefficients can be computed as

$$ \alpha_k = \langle B_k, \rho\rangle = \mathrm{tr}(B_k^\dagger \rho) = \mathrm{tr}(\rho B_k) = \chi_\rho(k).\tag2 $$

Finally, using $(1')$ and $(2)$, we find

$$ \begin{align} \rho = & \sum_{i=1}^{d^2}\chi_\rho(i)B_i \\ \rho\sigma = & \sum_{i=1}^{d^2}\chi_\rho(i)B_i\sigma \\ \mathrm{tr}(\rho\sigma) = & \mathrm{tr}\left(\sum_{i=1}^{d^2}\chi_\rho(i)B_i\sigma\right) \\ \mathrm{tr}(\rho\sigma) = & \sum_{i=1}^{d^2}\chi_\rho(i)\,\mathrm{tr}\left(B_i\sigma\right) \\ \mathrm{tr}(\rho\sigma) = & \sum_{i=1}^{d^2}\chi_\rho(i)\chi_\sigma(i) \end{align} $$

which is the desired equality.

I am not aware of any connection between the characteristic function defined in the paper and the characteristic function of a random variable.

  • $\begingroup$ that's awesome. thanks a bunch. from what I understand, the characteristic function is just something they define to use as a tool for neatly proving other things. it doesn't seem important how they derived it (if at all). would be interesting to ask them what their thought process was on defining it and using it to prove many other things. $\endgroup$ Jul 3, 2021 at 19:19
  • $\begingroup$ also tbh I wish these kind of papers were more explicit with their proofs. they don't include some proofs that at the bare minimum should have been included in their supplemental material/appendix. $\endgroup$ Jul 3, 2021 at 19:20
  • 1
    $\begingroup$ @QuantumGuy123 Well, to be honest, the characteristic function is simply the expansion coefficients in an orthonormal basis, here the normalised Pauli operators, so a pretty natural thing to look at if you're doing linear algebra. BTW the identity for the trace inner product is proven for any inner product and orthonormal basis in a standard linear algebra course ... So it's perfectly fine to leave that out in a research paper. $\endgroup$ Jul 5, 2021 at 10:29
  • $\begingroup$ ah ok, my bad. I didn't realize it was something commonly known in linear algebra. I took a first year university course in linear algebra many years ago, I must've forgot the proof or I wasn't taught it. I've been doing quite a bit of catching up in the past year so I can understand QC :) $\endgroup$ Jul 5, 2021 at 21:27
  • 2
    $\begingroup$ @QuantumGuy123 I think an important lesson to learn is that the space of linear operators also forms a Hilbert space with the trace/Hilbert-Schmidt inner product. Thus, the same "ket and bra" reasoning for states also applies to operators. $\endgroup$ Jul 6, 2021 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.