In the context of Bures metric and quantum Fisher information, an important object is the symmetric logarithmic derivative (SLD). This is usually introduced as a way to express the derivative of a parametrised state as a superoperator acting on the state itself.

More precisely, let $\theta\mapsto\rho_\theta$ be some (differentiable) parametrisation of a set of states. The SLD is then the operator $L$ such that $$\partial_\theta\rho_\theta = \frac12 (L\rho_\theta+\rho_\theta L)\equiv \frac12 \{L,\rho_\theta\}.$$ This is discussed e.g. here in (Toth and Apellaniz 2014) in Eq. (83), or (Paris 2008) in Eq. (3). The Wikipedia page on the Bures metric also introduces the basic relevant definitions.

It seems clear that the SLD operator has some geometric significance. For one thing, it appears prominently on the expression for the metric induced by the Bures distance: $$[d(\rho,\rho+d\rho)]^2 = \frac12 \operatorname{Tr}(\partial_{\mu}\rho L_\nu)d\theta^\mu d\theta^\nu.$$

Is there a more direct way to see where this operator comes from? A way to justify why an operator whose anticommutator with $\rho_\theta$ produces $\partial_\theta\rho_\theta$ should be interesting? Or a way to justify why the SLD, in particular, gives the correct expression for the Bures metric?


2 Answers 2


First, the classical correspondence, explaining why the SLD should be present. The Fisher information is the expectation value of the score, where the score is the logarithmic derivative of the probability distribution: $$ L_{\mathrm{classical}}=\frac{\partial [\ln p(\pmb{x}|\theta)]}{\partial \theta}, $$ which leads to the relation $$ \frac{\partial [ p(\pmb{x}|\theta)]}{\partial \theta}= p(\pmb{x}|\theta)L_{\mathrm{classical}}=L_{\mathrm{classical}} \,p(\pmb{x}|\theta). $$ This clearly mirrors the expression for the SLD $L$, where the anticommutator is required to resolve the operator ordering ambiguity - it justifies the origin of the operator, but as of yet does not lend us too much geometric insight.

Now for some geometry. This is mostly taken from the older geometry paper and the recent geometrical review (the latter is available on arXiv but I'm not sure about the former). Various probability distributions $p(\pmb{x}|\theta)$ for a given parameter $\theta$ form a probability simplex and we are looking for a metric to know the distance between two distributions. A good candidate is the Fisher-Rao metric (here written for discrete variables) $$ ds^2=\sum_{jk}g_{jk} dp^j dp^k=\sum_j\frac{(d p^j)^2}{p_j}=\sum_j p_j (d\ln p^j)^2. $$ The metric makes sense for classical variables with expectation values $$\langle A\rangle=\sum_j A_j p^j,\quad \langle AB\rangle=\sum_{jk}A_j B_k g^{jk}=\sum_j A_j B_j p^j$$ because those expectation values constrain the metric tensor to obey $$g_{jk}=\frac{\delta_{jk}}{p^j}.$$ Displacement along a line segment $d\theta$ then leads us to the classical Fisher information $$ F=\frac{ds^2}{d\theta^2}=\sum_j p_j\left(\frac{\partial \ln p^j}{\partial \theta}\right)^2. $$ This already looks very similar to the metric induced by the Bures distance, if we recall the correspondence between the score and the SLD.

When we do the same thing with operators and probability distributions found via the Born rule, we see the need to somehow divide by the probability distribution. This is achieved via the superoperator $$ \mathcal{R}_{\rho}(O)=\frac{\left\{\rho,O\right\}}{2}, $$ with which the correlation between two observables can be written using $$ \left\langle \frac{\left\{A,B\right\}}{2}\right\rangle=\mathrm{Tr}\left[A \mathcal{R}_\rho (B)\right]=\mathrm{Tr}\left[B \mathcal{R}_\rho (A)\right]. $$ In this expression, $R_\rho(\cdot)$ plays the role of the metric tensor $g^{jk}$. To find the operator playing the role of the metric tensor with lowered indices, we want the inverse of this operator $\mathcal{L}_\rho(\cdot)=\mathcal{R}^{-1}_\rho(\cdot)$ (because $\sum_{k}g_{jk}g^{kl}=\delta_{kl}$), which must obey $$ \mathcal{L}_\rho[\mathcal{R}_\rho(B)]=B. $$ We can then use this to construct a scalar product between quantum states as $$ \langle\sigma_1,\sigma_2\rangle=\mathrm{Tr}\left[\sigma_1\mathcal{L}_\rho(\sigma_2)\right] $$ and with that define the infinitesimal distance $$ ds^2=\mathrm{Tr}\left[d\rho\mathcal{L}_\rho(d\rho)\right]. $$ The quantum Fisher information then takes the form $$ Q=\frac{ds^2}{d\theta^2}=\mathrm{Tr}\left[\frac{\partial\rho}{\partial \theta}\mathcal{L}_\rho\left(\frac{\partial \rho}{\partial \theta}\right)\right]. $$

The only remaining ingredient is to identify the SLD as $$L=\mathcal{L}_\rho\left(\frac{\partial \rho}{\partial \theta}\right),$$ which can be proven by expanding all of the operators in the eigenbasis of $\rho$. We then see how the SLD is related to the classical score, the distance between classical probabability distributions, and the distance between quantum states.

TL;DR: We're looking for an inner product between $\partial\rho/\partial \theta^\mu$ and $\partial\rho/\partial \theta^\nu$, and the metric for the inner product is related to the SLD $L_\nu=\mathcal{L}_\rho(\partial \rho/\partial \theta^\nu)$ because this metric makes classical expectation values like $\langle A\rangle$ and $\langle AB\rangle$ agree with each other.


The inverse operator satisfies $$ \mathrm{Tr}(AB)=\mathrm{Re}\left\{\mathrm{Tr}\left[\rho A\mathcal{L}_\rho(B)\right]\right\}. $$

  • $\begingroup$ thanks a lot, this is very useful, as is the linked review, I didn't know about that one. I'm still a bit unclear about how $R_\rho(O)$ is introduced though. What do you mean exactly with it "playing the role" of the metric tensor? $\endgroup$
    – glS
    Commented Jun 29, 2021 at 18:41
  • $\begingroup$ $R_\rho$ lets you define an inner product between two observables $A$ and $B$ via $\langle A,B\rangle=\mathrm{Tr}[A R_\rho(B)]$, or more generally $\langle \cdot,\cdot\rangle=\mathrm{Tr}[\cdot R_{\rho}(\cdot)]$. Inner products come in the form $\sum_{\mu\nu}A_\mu B_\nu g^{\mu\nu}$ or $\sum_{\mu\nu}A^\mu B^\nu g_{\mu\nu}$, depending whether $A$ and $B$ are to be treated as covariant or contravariant. In the classical expression, probabilities have superscripts and the variables have subscripts, so the inner product for quantum states needs the inverse tensor as that for observables $\endgroup$ Commented Jun 29, 2021 at 19:09
  • $\begingroup$ I understand that as saying that $R_\rho$ is just the metric, with respect to the $L_2$ inner product structure on the operators. But I suppose what I was trying to ask is how do you justify it having the specific form with the anticommutator? I mean, the classical metric tensor seems to be $g^{ij}=\delta_{ij} p_i$ here, so defining $R$ via anticommutator we do get something which gives the probability, reproducing the classical expression. But I'm still left wondering if there is a way to more specifically pinpoint this form of $R_\rho$ (other than: it works) $\endgroup$
    – glS
    Commented Jun 30, 2021 at 10:09
  • $\begingroup$ At this point I'm not sure I have a better answer for you - we know that $R$ works, so if we could prove that nothing else works then we'd be done, but I don't have that proof handy, I only suspect it to be true $\endgroup$ Commented Jun 30, 2021 at 13:39
  • $\begingroup$ that's fine, this is already quite useful as it stands $\endgroup$
    – glS
    Commented Jun 30, 2021 at 13:41

Although, the Bures metric, the Fisher tensor and the symmetric logarithmic derivative appear mainly in quantum estimation theory, and even though the original discovery by Helstrom was in this context; they have a deeper origin in the geometric formulation of quantum mechanics.

This metric is the geometrized version of the Jordan structure of the observable algebra of quantum systems.

The algebra of observables $\mathcal{A}_{SA}$ of a quantum system has a Lie-Jordan structure: The commutator (divided by the imaginary unit) and the anti-commutator of two observables are observables. $$\frac{1}{i} [A, B] \in \mathcal{A}_{SA}$$ $$\{A, B\} \in \mathcal{A}_{SA}$$ Please see, for example, the following article by Clemente-Gallardo and Marmo.

It is worthwhile to mention that the Jordan algebra is commutative but not associative. Its non-associativity is one of the sources of quantumness and its anti-commutator appears in the Schrodinger's stronger version of the Heisenberg uncertainty relation.

Given a space of quantum states, (which, for the sake of clarity, is assumed to be parametrized by a manifold $\Theta$, i.e., $\hat{\rho}(\theta), \theta \in \Theta$, the Lie structure induces a Poisson structure, while the Jordan structure induces a Riemannian structure on the space of quantum states as follows:

Let the functions on the state space $e_A(\theta), e_B(\theta), …$ be the expectations of the operators $A, B, ...$. (In the finite $N$-dimensional case, by letting $A$ run on the $N^2-1$ basic observables, we can consider the $e_A$s as coordinate functions). $$ e_A(\theta) = \mathrm{tr}\left(\hat{\rho}(\theta) A\right)$$

Then the Lie-Jordan structure induces a Poisson structure and a metric as follows: $$\Lambda(de_A(\theta), de_B(\theta)) = e_{[A, B]}(\theta)$$ $$G (de_A(\theta), de_B(\theta)) = e_{\{A, B\}}(\theta)$$

It is not hard to verify that $\Lambda$ satisfies the properties of a Poisson vector and $G$ the properties of an inverse metric (the metric $G$ contracts into forms, thus it is an inverse metric)

(In the sequel, I'll be following the article by Ciagala and Jost with some changes and simplifications in the notation)

In order to take the inverse inverse $G$ and compute the metric $g$, we can introduce the gradient vector fields defined by their action on the "coordinate" functions: $$Y_A(e_B) = e_{\{A, B\}}$$ The metric is thus given by: $$g (Y_A, Y_B) = e_{\{A, B\}}(\theta)$$ The density matrix $\rho$ lives in the dual of the observable algebra $\mathcal{A}$, i.e., $$\langle \rho, A\rangle \equiv \rho(A) = \mathrm{tr}(\hat{\rho} A)$$

Observing that:

$$Y_A(e_B) = e_{\{A, B\}} = \mathrm{tr}(\{\hat{\rho}, A\}, B)$$ Thus $Y_A$ can be represented on the cotangent space of the density matrix $\hat{\rho }$ by the operator $$\hat{Y}_A = \{\hat{\rho}, A\}$$ Since: $$\hat{Y}_A(B) = Y_A(e_B) = e_{\{A, B\}}$$

Now, we would like to use the metric $g$ to find the length of the difference between two infinitesimally close density matrices $d\hat{\rho}$. We observe that $d\hat{\rho}$ is not yet in the form of a gradient vector represented by $\hat{Y}_A = \{\hat{\rho}, A\}$ (i.e., it is not of the form of an anti-commutator with the density matrix), but if we define: $$ d\hat{\rho}= \{\hat{\rho}, d_L\hat{\rho}\}$$

($d_L\hat{\rho}$ denotes the symmetric logarithmic derivative). (Of course this definition needs to be proven for existence and uniqueness). We get: $$g (d\hat{\rho}, d\hat{\rho}) = e_{\{ d_L\hat{\rho}, d_L\hat{\rho}\}} = \mathrm{tr}(\hat{\rho} d_L\hat{\rho} d_L\hat{\rho}) = \frac{1}{2}\left ((\hat{\rho} d_L\hat{\rho} + d_L\hat{\rho}\hat{\rho} ) d_L\hat{\rho}\right) = \frac{1}{2}\left (d\hat{\rho} d_L\hat{\rho}\right)$$


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