The (Uhlmann-Josza) fidelity of quantum states $\rho$ and $\sigma$ is defined to be $$F(\rho, \sigma) := \mathrm{tr} \left[\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}} \right]^2.$$ Every article I've ever seen that discusses the fidelity uses this same definition. However, as discussed here, the cyclic property of the trace extends to arbitrary analytic functions: $$\mathrm{tr}[f(AB)] \equiv \mathrm{tr}[f(BA)]$$ for any analytic function $f$ whenever either side is well-defined. Letting $f$ be the square root function, this implies that $$F(\rho, \sigma) \equiv \mathrm{tr} \left[\sqrt{\rho \sigma} \right]^2$$ (I don't think the branch point at the origin of the square root function is an issue, because the function is still continuous there). This latter expression (a) is much easier to deal with, (b) requires fewer matrix diagonalizations to compute, and (c) clearly demonstrates the similarities and differences with the Hilbert-Schmidt inner product. Numerical experimentation seems to confirm that these two expressions are equivalent, but the latter one is much faster to compute.

Why is the much clunkier former expression with nested square roots always given in the literature? The only benefit of the original definition that I can see is that the operator inside the (outer) square root is obviously positive-semidefinite, so it's easier to see why the fidelity is a nonnegative real number, while this might be less obvious in the second form.

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    $\begingroup$ Perhaps it has to do with having a simpler expression in terms of the 1-norm? $\text{Tr}\left(\sqrt{\sqrt{\sigma}\rho\sqrt{\sigma}}\right) = |\sqrt{\rho}\sqrt{\sigma}|_1$ $\endgroup$
    – rnva
    Feb 21, 2020 at 14:11
  • $\begingroup$ I've always wondered this too. +1 for a great and comprehensive answer! $\endgroup$
    – knzhou
    Feb 22, 2020 at 6:40
  • $\begingroup$ Note that there are references which discuss that the fidelity is the sum of the square root of the eigenvalues of $\rho\sigma$. $\endgroup$ Feb 25, 2020 at 15:38

2 Answers 2


Okay, this is a rather subtle situation, but I think I've figured it out. The key is to be very careful about which mathematical results about Hermitian operators do and do not hold for generic operators. Let $H$ represent an arbitrary Hermitian matrix, $N$ an arbitrary normal one, and $M$ an arbitrary matrix, all acting on an $n$-dimensional Hilbert space.

Subtlety 1: For normal $N$, the numerical range $$\left \{ \frac{\langle \psi | N | \psi \rangle}{\langle \psi|\psi\rangle} \right \}$$ for all nonzero $\psi$ in the Hilbert space is the convex hull of the eigenvalues of $N$. (So for $H$ Hermitian, it's the real interval $[\min \lambda, \max \lambda]$.) For generic $M$, the numerical range is still convex and contains the eigenvalues, but is not necessarily a hull for them.

Subtlety 2: The term "positive (semi-)definite" is usually only defined for Hermitian matrices $H$. There, it can be defined in one of two equivalent ways:

PD1: A Hermitian matrix is positive (semi-)definite (PD1) if all of its eigenvalues are positive (nonnegative real).

PD2: A Hermitian matrix is positive (semi-)definite (PD2) if $\langle \psi | H | \psi \rangle > (\geq)\ 0$ for all nonzero $|\psi\rangle$ in the Hilbert space.

PD1 and PD2 are equivalent for Hermitian $H$, but their generalizations for generic (not necessarily Hermitian) $M$ are no longer equivalent: in the generic case, PD2 is strictly stronger than PD1. Indeed, PD2 (i.e. $\langle \psi | H | \psi \rangle \ge\ 0\ \forall\,|\psi\rangle$) implies that the matrix has to be Hermitian (which is why it is often considered the "standard" definition); on the other hand, there are clearly non-hermitian matrices which satisfy PD1 (i.e. have positive eigenvalues) -- see here for an example of a non-Hermitian $M$ that satisfies PD1 but not PD2.

Subtlety 3: There are two inequivalent definitions of a square root of a matrix.

SR1: $R$ is a square root (SR1) of a generic matrix $M$ if $R^2 = M$. Under this definition, a matrix has a finite number of square roots (e.g. $2^n$ if its eigenvalues are all distinct). This definition makes sense for any matrix. I'm not sure whether or not there's generically a natural choice of "principal" square root in this situation (e.g. if $M$ is defective), so the notation $\sqrt{M}$ is not (as far as I know) well defined.

SR2: $R$ is a square root (SR2) of a positive definite Hermitian matrix $P$ if $R^\dagger R = P$. (Since $P$ is Hermitian, we don't need to specify whether we mean PD1 or PD2 for "positive definite".) Under this definition, the set of square roots of a matrix $P$ is isomorphic to the Lie group $U(n)$, because if $R_1$ is a square root of $P$, then $R_2$ is a square root of $P$ iff $R_2 = U R_1$ for some unitary matrix $U$. $R$ is not necessarily Hermitian. But under this definition, we can define the natural "principal" square root of $P$, which we denote by $L = \sqrt{P}$, as the unique square root that is also Hermitian and positive (semi-)definite (again, no need to distinguish PD1 from PD2 here).

Since the principal square root $L$ is Hermitian by definition, it respects both definitions SR1 and SR2, as $L^\dagger L = L^2 = P$. But a generic square root (SR1) of $P$ will not be a square root (SR2) of $P$ or vice versa.

For a Hermitian $H$, the usual power series expansion of the square root function $$\sum_{n=0}^\infty \frac{(-1)^n (2n)!}{(1-2n)(n!)^2 4^n} (H - I)^n$$ will converge to $\sqrt{H}$ iff all the eigenvalues of $H$ lie in the interval $[0,2]$. For a generic $M$, this series will converge to a square root of $M$ iff all of the eigenvalues of $M$ lie in the disk in the complex plane that has that interval as a diameter. (If I recall correctly, the boundary points 0 and 2 are included, but the boundary points with nonzero imaginary part are all excluded.)

Subtlety 4: If either of two generic matrices $A$ or $B$ is invertible, then $AB$ and $BA$ are similar, but if both $A$ and $B$ are singular, then $AB$ and $BA$ are not necessarily similar (see here for a counterexample). But even in this case, $AB$ and $BA$ always have the same eigenvalues and in fact characteristic polynomials, so (for example) their traces and regions of convergence for any formal power series will be the same.

Subtlety 5: If we have two Hermitian positive-definite matrices $P_1$ and $P_2$, then their non-Hermitian product $P_1 P_2$ will satisfy PD1 but not necessarily PD2 (see the first link above for a counterexample), so it may or may not necessarily be "positive definite", depending on your definition.

Now we can finally try to answer my question. The standard definition of the fidelity is unambiguous, because only Hermitian positive-semidefinite operators are ever getting square rooted. Since $\rho \sigma$ is non-Hermitian, its numerical range is generically complex and it does not satisfy PD2. Moreover, we can't talk about its square roots using definition SR2. And generically, the notation $\sqrt{M}$ may not be meaningful for a non-Hermitian $M$ because it implies some natural principal branch.

But we can talk about the square roots (plural) of $\rho \sigma$ under definition SR1, as with any matrix. Moreover, $\rho \sigma$ is a highly non-generic matrix. It satisfies PD1 by Subtlety 5. In fact, by Subtlety 4, $\rho \sigma$ has the same characteristic polynomial (with all roots lying in $[0,1]$) as $\sqrt{\rho} \sigma \sqrt{\rho}$. So because the eigenvalues all lie in this interval (which is obviously not generically true), there is a natural choice of principal square root: the one given by the usual power series expansion above, which is guaranteed to converge in light of its eigenvalue spectrum. So in this particular case, we can get away with defining $\sqrt{\rho \sigma}$ by the formal power series above. Then by the logic outlined in my question, we can indeed cycle the operators inside the square root and always get the right answer.

TLDR: The expression $\sqrt{M}$ is not uniquely defined for a generic matrix $M$ that is not Hermitian and positive semidefinite. But in this case, the special properties of the matrix $\rho \sigma$ guarantee that the formal power series above converges, so we can use that power series to (non-conventionally) define $\sqrt{\rho \sigma}$. If we use that convention, then we will indeed always get the same answer as the traditional definition. However, this is a bit of a hack, and the traditional definition's meaning is clear without needing to make any additional implicit definitions.

  • $\begingroup$ Do you have a trustworthy source for your definition PD1? I have always only ever seen PD2 (which has nice features such that A>0, B>0 => A+B>0 etc. - certain "normal" math still works), which I strongly doubt holds for PD1. $\endgroup$ Feb 22, 2020 at 14:29
  • $\begingroup$ To add to this, I'm not sure about all this fuzz (not only in this answer) about subtleties. Spectra or products (except for zeros) are invariant under cyclic permutations, so as you correctly observed functions like square root etc. are cyclic as well, as long as there are no issues with branches - which is clearly the case here. I think if there is an answer to your question, it is that hermitian objects are more familiar and "nice", and Uhlmann and Josza certainly didn't think about computing those numerically for large systems. $\endgroup$ Feb 22, 2020 at 14:34
  • $\begingroup$ (P.S.: Note that even if there were branches, if you define them in some specific way you are fine, since tr(f(X)) is defined on the eigenvalues, and those do not change. $\endgroup$ Feb 22, 2020 at 14:36
  • $\begingroup$ @NorbertSchuch (1) I've never seen definition PD1 in the literature, but I thought it was worth stating it separately (a) to show that it's distinct from the more usual definition PD2, and (b) because in this particular case (convergence of power series expansions), property PD1 is actually the more relevant one. $\endgroup$
    – tparker
    Feb 23, 2020 at 18:27
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    $\begingroup$ PD1 is not a definition of "positive semidefinite" for non-normal matrices. Apparently, it's also not "natural" to talk about "positive semitedefinite" in such contexts, see discussion here $\endgroup$
    – glS
    Feb 24, 2020 at 12:01

The square root function is not analytic, but has a branch cut (which is usually chosen along the negative real axis, but this is merely a choice).

To convince yourself that this is not a technical point but a real problem with your alternative definition, note that the square root function doesn't have a standard power series expansion around zero. When we want to define what is $\sqrt{\rho}$ we cannot use the series expansion properties and rather define it using the fact that $\rho$ can be diagonalized $\rho = \sum_n r_n |n\rangle \langle n|$ then $$ \sqrt{\rho} : = \sum_n \sqrt{r_n} |n\rangle \langle n|$$ and since $\rho$ is Hermitian and $r_n$ are real non-negatives so there's no ambiguity, and $\sqrt{\rho}$ is also Hermitian.

However, $\rho\sigma$ is in general not Hermitian (it is Hermitian only in the trivial case where they commute). Therefore, we have no safe way to define $\sqrt{\rho\sigma}$ (in contrast to $\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}$ which is a square root of an Hermitian operator).

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    $\begingroup$ $\rho \sigma$ isn't Hermitian, but it's still always diagonalizable and its spectrum lies in $[0, 1]$. So we can just expand the power series about the identity rather than the zero matrix, and it will converge even for zero eigenvalues. Or, we can calculate the square root using the Jordan block method. $\endgroup$
    – tparker
    Feb 21, 2020 at 12:08
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    $\begingroup$ I think that for finite matrices one can prove that there is no difference. However for infinite matrices things get more complicated (intuition from finite to infinite spaces can be tricky, as for example $\rm{tr}\{[A,B]\}=0$ in finite spaces but not in infinite ones). I guess that as we restrict ourselves to density operators, which are bounded and "nice", this would work also in infinite spaces. However, the usual definition is safer and nicer. $\endgroup$
    – user245141
    Feb 21, 2020 at 16:52

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