The Wikipedia page discussing the Bures metric introduces it as the Hermitian 1-form operator $G$ defined implicitly by $\rho G+G\rho = \mathrm d\rho$, and which induces the corresponding Bures distance, which reads $$[d(\rho,\rho+\mathrm d\rho)]^2 = \frac12 \operatorname{Tr}(\mathrm d\rho \,G).$$ Shortly thereafter, they give the corresponding finite version of this metric, which turns out to have the form $$[D(\rho_1,\rho_2)]^2 = 2\Big(1 - \operatorname{Tr}\left|\sqrt{\rho_1}\sqrt{\rho_2}\right|\Big).$$ These two expression do not look particularly similar. How does one go from one to the other? I suppose this should involve taking the geodesic curve connecting $\rho_1$ and $\rho_2$, and then integrating the infinitesimal form over it, but I'm not sure how to actually perform such a calculation.

  • $\begingroup$ I just realised a possible derivation of this is (I think, I still have to work through it) in the second part of this great answer by @David Bar Moshe. I still think it might be worth having a more dedicated post here about this though, to make the information easier to retrieve. $\endgroup$
    – glS
    Commented Jul 2, 2021 at 9:16

1 Answer 1


The Bures metric is the limit of the Bures distance for two infinitesimally close density matrices $\rho$ and $\rho+d\rho$. The Bures distance however, is not unique. It depends on the space on which we integrate the Bures metric. The Bures distance for positive matrices differs from that of density matrices (with a unit trace), however they have the same limit when the separation becomes infinitesimal.

First, let me please remark that the calculation (by means of the Hamilton-Jacobi equation) in the previous answer for the case of the Fubini-Study distance of the space of pure states, is a much easier task than the general Bures distance evaluation for the space of density matrices. I haven't seen anywhere an evaluation of the Bures distance by integration of the Hamilton-Jacobi equation for the Bures metric. It should be a very nice exercise. It should be a case where the Hamilton- Jacobi equation has an exact solution which is not obtained by means of separation of variables.

The evaluation of the Bures distances in Bengtsson and Życzkowski (page 234 equations (9.28) to (9.32)), mentioned in the previous answer, was performed indirectly by minimizing the Euclidean or the spherical distance on the space of purifications, then showing that the infinitesimal version of which is the Bures metric by evaluating it on a horizontal vector (which by definition minimizes the distance).

Here, I'll perform the computation of the second part (the easier part), i.e., start from the Bures distance and obtain the Bures metric for the case of two infinitesimally separated density matrices by a direct computation: For the space of density matrices.

The Bures distance is given by: $$D_{\text{Bures}}(\rho_1, \rho_2) = \arccos(\sqrt{F}( \rho_1, \rho_2))$$ Where: $$\sqrt{F}( \rho_1, \rho _2) = \text{tr}\left(\sqrt{\sqrt{\rho _1}\rho _2\sqrt{\rho _1})}\right)$$

We need to evaluate this expression for

$$\rho_1 = \rho$$


$$\rho_2 = \rho+d\rho = \rho+ \frac{\partial\rho}{\partial \theta^a} d\theta^a := \rho + \partial_a \rho d\theta^a$$

($\theta_a$ are real coordinates locally parametrizing the space of density matrices). (Einstein's summation convention is implied for repeated indices).

When the density matrices are infinitesimally apart, the Uhlmnann's fidelity $(\sqrt{F}$ becomes close to unity and its arccosine can be approximated as:

$$D^2_{\text{Bures}}(\rho, \rho+d\rho) \approx 2- 2\sqrt{F}( \rho, \rho+d\rho)$$

Taking the Taylor series to the second order of the Uhlmann's fidelity, we obtain:

$$X \equiv\sqrt{\rho }(\rho+d\rho)\sqrt{\rho} = \rho^2 + \sqrt{\rho}\partial_a \rho \sqrt{\rho}d\theta^a + \frac{1}{2} \sqrt{\rho}\partial_a \partial_b \rho \sqrt{\rho}d\theta^a d\theta^b$$

We need to take the square root of the above expression. Suppose it has the form: $$\sqrt{X} = \sqrt{\sqrt{\rho }(\rho+d\rho)\sqrt{\rho}} = \rho + B_a d\theta^a + C_{ab} d\theta^a d\theta^b$$ To prove the uniqueness of the solution, let us assume that $\rho$ is invertible, thus none of its eigenvalues is equal to zero. Thus, for small enough $|d\theta^a|$ the square root will be positive definite. If we manage to find Hermitian solutions for $B_a$ and $C_{ab}$, then the result will be the unique Hermitian square root.

Substituting in $\sqrt{X} \sqrt{X} = X$, we obtain: $$ \sqrt{\rho}\partial_a \rho \sqrt{\rho}= \rho B_a + B_a \rho \quad (1)$$ $$\frac{1}{2} \sqrt{\rho}\partial_a \partial_b\rho \sqrt{\rho}= \rho C_{ab} + C_{ab} \rho + \frac{1}{2} (B_a B_b + B_b B_a) \quad (2)$$ Using the definition of the symmetric logarithmic derivative: $$\partial_a \rho = \frac{1}{2}(\rho L_a + L_a \rho)$$ We can see that: $$ B_a = \frac{1}{2} \sqrt{\rho} L_a \sqrt{\rho}$$ is a solution of (1) In the second equation, writing: $$\partial_a \partial_b\rho = \frac{1}{2} (\partial_a \partial_b + \partial_b \partial_a)\rho$$ and expressing all the derivatives of the density matrix by means of the symmetric logarithmic derivative and substituting in (2), we see that: $$C_{ab} =\frac{1}{16}\sqrt{\rho} (\partial_a L_b + \partial_b L_a + L_a L_b + L_b L_a) \sqrt{\rho}$$
is a solution of (2).

Substituting of the Taylor approximation of the Uhlmann fidelity into the Bures distance, we get the infinitesimal Bures distance: $$D^2_{\text{Bures}}(\rho, \rho+d\rho) = -2\mathrm{tr}(B_a)d\theta^a - \mathrm{tr}(C_{ab}) d\theta^ad\theta^b$$

The first term: $$\mathrm{tr}(B_a) = \mathrm{tr}(\rho L_a) = \frac{1}{2} \mathrm{tr}(\rho L_a+ L_a \rho) = \frac{1}{2} \mathrm{tr}(\partial_a \rho) = \partial_a \mathrm{tr}(\rho) = 0 $$ The second term: $$\mathrm{tr}(C_{ab}) = \frac{1}{16}\mathrm{tr}\left(\rho(\partial_a L_b + \partial_b L_a + L_a L_b + L_b L_a) \right) $$ Using the identity $$\partial_a\partial_b\mathrm{tr}\rho = 0$$ We have: $$\mathrm{tr}\left(\rho(\partial_a L_b + \partial_b L_a)\right) = -2 \mathrm{tr}\left(\rho( L_a L_b + L_b L_a)\right)$$ Thus: $$\mathrm{tr}(C_{ab}) =- \frac{1}{16}\mathrm{tr}\left(\rho(L_a L_b + L_b L_a) \right) $$ Therefore: $$D^2_{\text{Bures}}(\rho, \rho+d\rho) = \frac{1}{8}\mathrm{tr}\left(\rho(L_a L_b + L_b L_a) \right) d\theta^ad\theta^b = \frac{1}{8}\mathrm{tr}\left(L_a (\rho L_b + L_b \rho) \right) d\theta^ad\theta^b = \frac{1}{4}\mathrm{tr}\left(L_a \partial_b\rho \right) d\theta^ad\theta^b $$

  • $\begingroup$ very nice, thanks. The nontrivial part seems to be computing the Taylor approximation of $\sqrt{F}(\rho,\rho+d\rho)$. I wonder if reasoning in terms of its expression with the trace norm, $\sqrt{F}(\rho,\sigma)=\|\sqrt\rho\sqrt\sigma\|_1$ might make for a more "elegant" computation. The nontrivial part would probably then be to find an expansion for $\sqrt{\rho+d\rho}$. A few posts I found related to this are math.stackexchange.com/a/1320527/173147, mathoverflow.net/a/193921/84108, and physics.stackexchange.com/a/196720/58382 $\endgroup$
    – glS
    Commented Jul 8, 2021 at 12:18
  • $\begingroup$ Solving these seems to eventually boil down to solving a Sylvester equation, which makes sense as that is what computing the SLD also boils down to, I think. $\endgroup$
    – glS
    Commented Jul 8, 2021 at 12:18

Your Answer

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

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