The quantum relative entropy (QRE) between two states $\rho$ and $\sigma$ is given by $$ S(\rho\|\sigma)=\operatorname{Tr}(\rho\ln\rho)-\operatorname{Tr}(\rho\ln\sigma) $$ Now if $\rho$ and $\sigma$ are infinitesimally related i.e, $\sigma=\rho+\delta\rho$, and we assumed a differential parametrization of the density matrix $\rho(\lambda)$ by a vector $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_n)$ such that $\delta\rho=\partial_i\rho~d\lambda_i$.

I know that the QRE is related to the quantum Fisher information metric $(F_{ij})$ by $$ S(\rho\mid\mid\rho+\delta\rho)\simeq\frac{1}{2}F_{ij}d\lambda_id\lambda_j+(\dots) $$ My calculation is leading me to the following result $$ S(\rho\|\rho+\delta\rho)\simeq-\operatorname{Tr}(\delta\rho)+\frac{1}{2}\operatorname{Tr}(\delta\rho\rho^{-1}\delta\rho). $$ where the first term is zero since $\operatorname{Tr}\rho=1$.

I am however unable to resolve the second term and obtain the standard form $$ F_{ij}=2~\sum_{kl}\frac{\operatorname{Re}(\langle k|\partial_i\rho| l\rangle\langle l|\partial_j\rho| k\rangle)}{\theta_k+\theta_l}, $$ where the eigenvalues $\theta_k$ and eigenvectors $| k\rangle$ of the density matrix potentially depend on $\lambda$.

  • $\begingroup$ The discrepancy here is mildly related to the problems of consistently discretizing the Fisher information (researchgate.net/publication/…). There, the equivalent ways (due to derivative/integral properties) to define the FI for a continuous parameter lead to different FIs when discretized; here, equivalences that hold for classical probability distributions (again due to derivative/integral properties) change when different things are quantized $\endgroup$ Feb 14 at 15:06

1 Answer 1


Expressing the derivative $\partial_i\rho$ in terms of its eigenvalues and eigenvectors will show us that these two are not equal. I will assume a full-rank density matrix $\rho$ to streamline the derivation.

Our first hint comes from writing $\rho^{-1}=\sum_k \theta_k^{-1}|k\rangle\langle k|$; then $$\mathrm{Tr}(\delta\rho \rho^{-1}\delta\rho)=\sum_{kl}\frac{\langle l|\partial_i \rho|k\rangle\langle k|\partial_j\rho|l\rangle}{\theta_k}d\lambda_i d\lambda_j.$$ We need to change the numerator into the real part and the denominator into $\theta_k+\theta_l$ to recover the expression for quantum Fisher information.

We can decompose the derivatives into $$\partial_i\rho=\sum_{m}\partial_i \theta_m|m\rangle\langle m|+\theta_m|\partial_i m\rangle\langle m|+ \theta_m|m\rangle\langle \partial_i m|.$$ Taking a matrix element yields \begin{align} \langle l|\partial_i \rho|k\rangle= \partial_i \theta_l \delta_{lk}+ \theta_k\langle l|\partial_i k\rangle+ \theta_l\langle \partial_i l|k\rangle. \end{align} Before we proceed, we use the trivial identity $0=\partial_i(\delta_{lk})=\partial_i\langle l|k\rangle=\langle \partial_i l|k\rangle+\langle l|\partial_i k\rangle$ to separate our matrix elements into terms with $k=l$ and terms with $k \neq l$: \begin{align} \langle l|\partial_i \rho|k\rangle= \partial_i \theta_l \delta_{lk}+ (1-\delta_{lk})(\theta_k-\theta_l)\langle l|\partial_i k\rangle. \end{align} The numerator now becomes \begin{align} \langle l|\partial_i \rho|k\rangle \langle k|\partial_j \rho|l\rangle= \partial_i \theta_l \partial_j \theta_l \delta_{lk}+ (1-\delta_{lk})(\theta_k-\theta_l)^2 \langle l|\partial_i k\rangle\langle k|\partial_j l\rangle. \end{align}

Now we can go in two directions: either try to show how this looks like the expression for quantum Fisher information, or take the real part of this and try to show how it becomes the trace expression. The latter is easier, so let's do it: \begin{align} 2\sum_{kl}\frac{\Re (\langle k|\partial_i \rho|l\rangle \langle l|\partial_j \rho|k\rangle)}{\theta_k+\theta_l}&=2\sum_{kl}\frac{\Re (\langle l|\partial_i \rho|k\rangle \langle k|\partial_j \rho|l\rangle)}{\theta_k+\theta_l}\\ &=2\sum_{k=l}\frac{\partial_i\theta_k\partial_j\theta_l }{\theta_k+\theta_l}+2\sum_{k\neq l}\frac{(\theta_k-\theta_l)^2\Re (\langle l|\partial_i k\rangle\langle k|\partial_j l\rangle)}{\theta_k+\theta_l} \\ &=\sum_{k}\frac{\partial_i\theta_k\partial_j\theta_k }{\theta_k}+\sum_{k\neq l}(\theta_k-\theta_l)^2\frac{\langle l|\partial_i k\rangle\langle k|\partial_j l\rangle + \langle \partial_i k|l\rangle\langle \partial_j l|k\rangle}{\theta_k+\theta_l} \\ &=\sum_{k}\frac{\partial_i\theta_k\partial_j\theta_k }{\theta_k}+\sum_{k\neq l}(\theta_k-\theta_l)^2\frac{\langle l|\partial_i k\rangle\langle k|\partial_j l\rangle + \langle k|\partial_i l\rangle\langle l|\partial_j k\rangle}{\theta_k+\theta_l} \\ &=\sum_{k}\frac{\partial_i\theta_k\partial_j\theta_k }{\theta_k}+2\sum_{k\neq l}(\theta_k-\theta_l)^2\frac{\langle l|\partial_i k\rangle\langle k|\partial_j l\rangle }{\theta_k+\theta_l} \end{align}

As for the trace, it looks like \begin{align} \sum_{kl}\frac{\langle l|\partial_i \rho|k\rangle\langle k|\partial_j\rho|l\rangle}{\theta_k}&=\sum_{k}\frac{\partial_i\theta_k\partial_j\theta_k }{\theta_k}+\sum_{k\neq l}(\theta_k-\theta_l)^2\frac{\langle l|\partial_i k\rangle\langle k|\partial_j l\rangle }{\theta_k} \end{align}

These results are manifestly different (for example, the second term can be complex in the expression for the trace but is manifestly real in the expression for the Fisher information).

Why are these expressions different? The derivation of the infinitesimal version of the relative entropy is much more tricky than the Bures metric, or infinitesimal fidelity, etc., because of the logarithms. One has to write \begin{align} \mathrm{Tr}[\rho \ln \rho-\rho\ln(\rho+\delta\rho)]=\mathrm{Tr}[\rho \ln \rho-\rho\ln(\rho(\mathbb{I}+\rho^{-1}\delta\rho))] \end{align} and try to do a series expansion of $\ln(\rho(\mathbb{I}+\rho^{-1}\delta\rho))$. The problem is that one may not, in general, write $\ln(\rho(\mathbb{I}+\rho^{-1}\delta\rho))=\ln(\rho)+\ln(\mathbb{I}+\rho^{-1}\delta\rho)$ unless the commutator between $\rho$ and $\rho^{-1}\delta\rho$ vanishes, which it does not. This is why, even if the expressions may match for classical probability distributions, they do not in general for quantum states.

One needs to invoke the symmetric logarithmic derivatives to generally talk about a metric. See for example Eqs. (58-59) and the preceeding discussions here.

To do a proper derivation we need to be able to take the derivative of the logarithm of a matrix. See this math-SE post and this math overflow post and references therein; it's tricky! The components of the metric induced by the QRE can be defined by derivatives of the QRE with respect to the underlying parameters (this is how a metric is found classically from the relative entropy, see e.g. this paper eq. 2.20) $$g_{ij}=-\frac{\partial^2}{\partial \lambda_i \partial \lambda^\prime_j}\mathrm{Tr}(\rho\ln\rho-\rho\ln\rho^\prime)\big|_{\rho=\rho^\prime},$$ where $\rho$ depends on $\lambda$ and $\rho^\prime$ depends on $\rho^\prime$ (I have switched from the $\sigma$ notation to make it easier to see that $\sigma\to\rho$ when $\lambda^\prime\to\lambda$). Product rule follows normally for matrix functions, derivatives commute with traces, so we are left with only contributions from the second term $$g_{ij}=\mathrm{Tr}\left(\frac{\partial \rho}{\partial \lambda_i }\frac{\partial \ln\rho^\prime}{ \partial \lambda^\prime_j}\right)\bigg|_{\rho=\rho^\prime}=\mathrm{Tr}\left(\frac{\partial \rho}{\partial \lambda_i }\frac{\partial \ln\rho}{ \partial \lambda_j}\right).$$ This is exactly the expression for the QFI if one chooses the "symmetric logarithmic derivative" as $\frac{\partial \ln\rho}{ \partial \lambda_j}$ in this expression.

The standard procedure for deriving the quantum Fisher information is to write $$\partial_i \rho=\frac{L_i\rho+\rho L_i}{2}$$ for the symmetric logarithmic derivative $L$, which can be proven to always exist and be Hermitian and in general depends on the underlying parameters $\lambda$. We can write an explicit solution as $$L_i=2\sum_{kl}\frac{\langle k|\partial_i\rho|l\rangle}{\theta_k+\theta_l}|k\rangle\langle l|,$$ where a straightforward computation using orthogonality of the eigenstates and two resolutions of the identity verifies $$L_i\rho+\rho L_i=2\sum_{kl}\frac{\langle k|\partial_i\rho|l\rangle}{\theta_k+\theta_l}|k\rangle\langle l|(\theta_k+\theta_l)=2\partial_i \rho.$$ Then the quantum Fisher information takes the form $$F_{ij}=\mathrm{Tr}[\frac{\partial \rho}{\partial \lambda_i}L_j]=\mathrm{Tr}[\rho(L_iL_j+L_jL_i)]/2.$$ The expression in the question takes the form \begin{align}\mathrm{Tr}(\delta\rho \rho^{-1}\delta\rho)&=\sum_{ij}\mathrm{Tr}\left(\frac{L_i\rho+\rho L_i}{2}\rho^{-1} \frac{L_j\rho+\rho L_j}{2}\right)d\lambda_i d\lambda_j\\ &=\sum_{ij}\mathrm{Tr}\left(2\rho L_iL_j+\rho L_jL_i+\rho L_i \rho^{-1}L_j\rho\right)d\lambda_i d\lambda_j/4 \\ &=\sum_{ij}\mathrm{Tr}\left[F_{ij}-\mathrm{Tr}\left(\rho\frac{ L_jL_i- L_i \rho^{-1}L_j\rho}{4}\right)\right]d\lambda_i d\lambda_j. \end{align} This explicitly shows the error between the two expressions due to $\mathrm{Tr}\left(\rho\frac{ L_jL_i- L_i \rho^{-1}L_j\rho}{4}\right)\neq 0$. Again, if all of the states and their derivatives were just probability distributions, everything would commute and the latter term would vanish.

At this point I have not proven that $L_j$ can replace $\frac{\partial \ln\rho}{ \partial \lambda_j}$ in the trace expression for derivatives of the QRE; it is highly unlikely that $L_j=\frac{\partial \ln\rho}{ \partial \lambda_j}$ holds in general because that would make solving the matrix logarithm derivative problem too easy. This is the crucial missing step required to connect the QFI and the QRE and is probably why people don't often derive the QFI from the QRE. Yes, there are classical/quantum-correspondence reasons for choosing the symmetric logarithmic derivative here, while other choices for the logarithmic derivative lead to other quantum Fisher information matrices. One should always choose one that yields a proper metric on the space of quantum states but even that is insufficient to single out a particular constrution. See e.g., the above quoted article

The classical Fisher information is the unique Riemannian metric on the space of classical probability distribtions that has the property of contraction under coarse graining. However, in the quantum-mechanical case the possibility of non-commuting operators breaks this uniqueness, and instead we obtain a family of metrics. It is therefore interesting to explore some alternative quantum extensions of the Fisher information.

My conclusion is thus: the curvature of the quantum relative entropy is given by the quantum Fisher information if the matrix derivative of the logarithm in the former is given by the same logarithmic derivative used in the latter.

  • $\begingroup$ Thanks a lot for this beautifully presented answer. It couldn't be better than this. I just want to know whether there are some example of the choice of identifying matrix derivative of the logarithm of density matrix as the Logarithmic derivative in the literature. How is this choice compatible with the derivation of the Bures metric from the Fidelity. $\endgroup$
    – m1rohit
    May 15 at 15:54
  • $\begingroup$ @m1rohit unfortunately I'm not sure off the top of my head. I can point you to the article quoted in my answer for discussing right logarithmic derivative vs symmetric logarithmic derivative $\endgroup$ May 15 at 19:21
  • $\begingroup$ It will help if you can show that this choice of the Logarithmic derivative required in the relative entropy is a solution to the original defining equation of SLD? I am still trying to work it out. $\endgroup$
    – m1rohit
    May 17 at 2:50
  • $\begingroup$ @m1rohit not sure if this is your problem, but let's try. If $\rho$ and $L_i$ commute, $\rho^{-1}\partial \rho/\partial \theta=L_i$. That's a differential equation solved by $\rho\propto e^{\theta L}$ and is equivalent to $\partial \ln \rho/\partial \theta=L_i$. That is the connection between logarithmic derivative and SLD $\endgroup$ May 17 at 12:32
  • $\begingroup$ I was asking how to show that $$ L_i=\frac{\partial \ln\rho}{\partial\lambda_i}\neq \rho^{-1}\frac{\partial\rho}{\partial\lambda_i}\neq \frac{\partial\rho}{\partial\lambda_i}\rho^{-1}$$ satisfies $$ L_i\rho+\rho L_i=2\partial_i\rho$$ $\endgroup$
    – m1rohit
    May 18 at 15:46

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