The (Uhlmann-Jozsa) fidelity of quantum states $\rho$ and $\sigma$ is defined to be $$F(\rho, \sigma) := \left(\mathrm{tr} \left[\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}} \right]\right)^2.$$ However, as discussed here, the cyclical 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 seems to imply that $$F(\rho, \sigma) \equiv \big(\mathrm{tr} \left[\sqrt{\rho \sigma} \right]\big)^2,$$ which is much easier to deal with. (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.)

  1. Am I correct that these two expressions are equivalent?

  2. If so, is there any reason why the much clunkier former expression with nested square roots is always given?

The only benefit of the original definition that I can see is that it makes it clear that the operator inside the trace is Hermitian and positive-semidefinite, so that the resulting fidelity is a non-negative real number.

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    $\begingroup$ This is actually very interesting. The math seems to be correct, and also checks out numerically (tried with random states and many dimensions). I wonder if this is easier to evaluate numerically. The original expression involves computing the sum of the singular values of $\sqrt\rho\sqrt\sigma$, or the sum of sqrt of eigvals of $\sqrt\rho\sigma\sqrt\rho$. This one the sum of the sqrt of eigvals of $\rho\sigma$, which all turn out to be real positive despite the matrix itself not being even normal. In other words, there is one less diagonalisation involved $\endgroup$
    – glS
    Feb 19, 2020 at 10:15
  • 1
    $\begingroup$ I did a quick test with MMA and I get that your expression takes less than half the time to evaluate. Granted, this is using possibly naive implementations: fidelityOld[a_, b_] := With[{sqrtA = MatrixFunction[Sqrt, a]}, Total@Sqrt@Re@Eigenvalues@Dot[sqrtA, b, sqrtA] ]; fidelityNew[a_, b_] := Total@Sqrt@Re@Eigenvalues@Dot[a, b]; Note that the first method is the one also used in qutip, see the relevant lines on GitHub. $\endgroup$
    – glS
    Feb 19, 2020 at 10:33
  • $\begingroup$ @glS Did you try with any singular density matrices? Maybe there is some subtlety about the branch point of the square root function at the origin that messes things up? $\endgroup$
    – tparker
    Feb 19, 2020 at 13:58
  • $\begingroup$ is that even a problem here? $\rho\sigma$ has positive eigvals, as it equals, up to permutation, $\sqrt\rho\sigma\sqrt\rho\ge0$. This means that the square root ever only operates on positive reals. But yes, numerically I've tried with for example pure states and it works fine. You can see easily that the expressions are equal from the fact that $\sqrt\rho\sigma\sqrt\rho$ and $\rho\sigma$ have the same eigenvalues, which then immediately implies the rest $\endgroup$
    – glS
    Feb 19, 2020 at 17:07
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    $\begingroup$ @glS I don't think that this is a problem in this case, but $\rho \sigma$ doesn't necessarily have only positive eigenvalues - it can also have zero eigenvalues if one or both density operators are singular. So the square root can operate on $0$. $\endgroup$
    – tparker
    Feb 19, 2020 at 22:27

2 Answers 2

  1. Square root is not differentiable at 0, so that cyclic property cannot be applied
  2. While $\rho\sigma$ has the same non-negative eigenvalues as $\sqrt\rho\sigma\sqrt\rho$, it's not self-adjoint. Non self-adjoint matrices are not diagonalizable in general, so the square root $\sqrt{\rho\sigma}$ can be not well-defined (see edit below).
  3. Anyway, $\text{Tr}(\sqrt{\sqrt\rho\sigma\sqrt\rho})$ is equal to the sum of square roots of eigenvalues of $\rho\sigma$.

It turns out that $\rho\sigma$ is always diagonalizable https://www.sciencedirect.com/science/article/pii/002437959190239S

So, taking principal square root of it is a correct operation. And it is indeed possible to write this shorter formula. Though this is not very well known and not conventional, since $\rho\sigma$ is not self-adjoint.

  • 1
    $\begingroup$ for 3), why not? $\sqrt{\rho\sigma}$ is perfectly well-defined by having the square root operate on the eigenvalues (which are guaranteed to be non-negative). I mean you can also try the formula numerically on your computer algebra system of choice and see that it gives the correct result $\endgroup$
    – glS
    Feb 21, 2020 at 20:15
  • $\begingroup$ In general, matrix could have no eigenvectors for some eigenvalues, i.e. it can be non-diagonalizable (en.wikipedia.org/wiki/Defective_matrix). Square root of such matrix is not well-defined. I can't find a concrete example for $\rho$ and $\sigma$ right now, but I think it exists. $\endgroup$
    – Danylo Y
    Feb 21, 2020 at 20:23
  • $\begingroup$ in general, sure. In this case, it doesn't look like it. What we do know is that $\sqrt\rho\sigma\sqrt\rho$ and $\sigma\rho$ have the same characteristic polynomial (see math.stackexchange.com/q/311342/173147), hence at least the generalised eigenvalues will always be the same, i.e. one could define the matrix function on the Jordan normal form, and be ensured to get a correct result. The minimal polynomials might be different, and thus $\rho\sigma$ not be diagonalisable, but this doesn't seem to ever happen, at least picking the states at random $\endgroup$
    – glS
    Feb 21, 2020 at 20:36
  • $\begingroup$ Well, if we pick $\rho$ and $\sigma$ at random then $\rho\sigma$ will have different eigenvalues, so it will be diagonalizable. The non-diagonalizable example should have at least two zero eigenvalues. $\endgroup$
    – Danylo Y
    Feb 21, 2020 at 20:49
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    $\begingroup$ @gIS Well, for Hermitian matrices $A,B \geq 0$ we can reduce the problem of finding Jordan form of $AB$ to the case when $A = I_k \oplus 0_{n-k}$. This is because there is always a matrix $S$ such that $\hat{A} = SAS^\dagger = I_k \oplus 0_{n-k}$, so for $\hat{B} = (S^{-1})^{\dagger}BS^{-1}$ we have $\hat{A}\hat{B} \sim AB$. This is only the first step, the following proof is a bit technical and I don't think there is a short version. $\endgroup$
    – Danylo Y
    Feb 24, 2020 at 17:11

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, $D$ be a generic diagonalizable matrix, 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: Therefore, there are two definitions of "positive (semi-)definite" that are equivalent for Hermitian $H$ but not for generic $M$:

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

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

PD1 and PD2 are equivalent for Hermitian $H$, but PD2 (which I think is the "real" meaning of positive-definite) is strictly stronger for generic $M$. 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 diagonalizable matrix $D$, this series will converge to a square root of $D$ iff all of the eigenvalues of $D$ 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.

  • $\begingroup$ Again, look at the $M = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ example. It has two 0 eigenvalues, so it's PD1 (by your definition). How that power series can converge to $\sqrt{M}$ if it doesn't exist? $\endgroup$
    – Danylo Y
    Feb 24, 2020 at 9:12
  • $\begingroup$ @DanyloY You're right, I added hte requirement that the matrix be diagonalizable. I think (but correct me if I'm wrong) that your matrix $M$ cannot be expressed as a product of two density matrices. $\endgroup$
    – tparker
    Feb 24, 2020 at 13:21
  • 2
    $\begingroup$ Yes, $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ is not diagonalizable since it's already in a Jordan normal form. And the paper I linked proves that the product of two density matrices is always diagonalizable (it's far from trivial). For diagonalizable matrices that power series indeed converges if the spectra lay in $[0,2]$. Though this a bit overkill because it's enough to define $\sqrt{M} = S\sqrt{D}S^{-1}$, where $M = SDS^{-1}$ with $D$ diagonal. $\endgroup$
    – Danylo Y
    Feb 24, 2020 at 17:29
  • $\begingroup$ @DanyloY Yes, I agree that the result that $\rho \sigma$ is always diagonalizable with nonnegative eigenvalues (which I didn't know when I wrote this answer) makes most of this discussion unnecessary. $\endgroup$
    – tparker
    Feb 25, 2020 at 3:44

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