Generalized uncertainty relation between thermodynamic variables in quantum thermodynamics

Abstract

Macroscopic thermodynamics, via the weak coupling approximation, assumes that the equilibrium properties of a system are not affected by interactions with its environment. However, this assumption may not hold for quantum systems, where the strength of interaction between the system and the environment may become non-negligible in a strong coupling regime. In such a regime, the equilibrium properties of the system depend on the interaction energy and the system state is no longer of the Gibbs form. Regarding such interactions, using tools from the quantum estimation theory, we derive the thermodynamic uncertainty relation between intensive and extensive variables valid at all coupling regimes through the generalized Gibbs ensemble. Where we demonstrate the lower bound on the uncertainty of intensive variables increases in presence of quantum fluctuations. Also, we calculate the general uncertainty relations for several ensembles to corroborate the literature results, thus showing the versatility of our method.

Appendices

Appendix A. Partition function

The generalized Gibbs ensemble (GGE) is given by [16,17,18]

$$\begin{aligned} \rho = \frac{e^{-\sum _i\lambda _iA_i}}{Z(\{\lambda _i\})}~, \end{aligned}$$

(58)

where $[A_i,A_j]=0$ for all i, j, and the partition function is

$$\begin{aligned} Z(\{\lambda _i\})=\textrm{tr}\left( e^{-\sum _i \lambda _iA_i}\right) ~, \end{aligned}$$

(59)

a situation in which the ensemble is in thermodynamic equilibrium. It is important to note that

$$\begin{aligned} \langle A_i \rangle&= \textrm{tr}\left( \rho A_i\right) =\frac{1}{Z}\textrm{tr}\left( e^{-\sum _i\lambda _iA_i}A_i\right) \nonumber \\&=-\frac{1}{Z}\frac{\partial }{\partial \lambda _i}\textrm{tr}\left( e^{-\sum _i \lambda _i A_i}\right) ~, \end{aligned}$$

(60)

consequently

$$\begin{aligned} \langle A_i \rangle =-\frac{\partial }{\partial \lambda _i}\ln Z(\{\lambda _i\})=-\frac{1}{Z}\frac{\partial Z}{\partial \lambda _i}~. \end{aligned}$$

(61)

Moreover, the statistical fluctuations for the constants of motion $A_i$ can be found via the partition function Z as follows

$$\begin{aligned} \frac{\partial ^2 Z}{\partial \lambda _i \, \partial \lambda _j}=\textrm{tr}\left( e^{-\sum _k\lambda _kA_k}A_i A_j\right) ~. \end{aligned}$$

(62)

However, the calculations become easier when Z is replaced by $\ln Z$ in Eq.  (62), then we have

$$\begin{aligned} \frac{\partial ^2 \ln Z}{\partial \lambda _i \, \partial \lambda _j}&=\frac{\partial }{\partial \lambda _i}\frac{1}{Z}\textrm{tr}\left( e^{-\sum _k\lambda _kA_k}A_j\right) =\frac{1}{Z^2}\frac{\partial Z}{\partial \lambda _i}Z\langle A_j\rangle +\frac{1}{Z}\textrm{tr}\left( e^{-\sum _k\lambda _kA_k}A_iA_j\right) \nonumber \\&=\langle A_iA_j\rangle - \langle A_i\rangle \langle A_j\rangle ~. \end{aligned}$$

(63)

Appendix B. Equilibrium entropy

It is straightforward to obtain the entropy of a system in thermodynamic equilibrium via the partition function in the following path

$$\begin{aligned} \frac{S}{k}=-\langle \ln \rho \rangle&= - \textrm{tr}(\rho \ln \rho )\nonumber \\&=-\textrm{tr}\left\{ \frac{e^{-\sum _i\lambda _iA_i}}{Z}\ln \left( \frac{e^{-\sum _i\lambda _iA_i}}{Z}\right) \right\} \nonumber \\&=\ln Z \,\textrm{tr}\left\{ \frac{e^{-\sum _i\lambda _iA_i}}{Z}\right\} +\sum _i\lambda _i\,\textrm{tr}\left\{ \frac{e^{-\sum _i\lambda _iA_i}}{Z}A_i\right\} \nonumber \\&=\ln Z - \sum _i\lambda _i\frac{\partial }{\partial \lambda _i}\ln Z ~. \end{aligned}$$

(64)

Moreover, one can redefine S from $\ln Z$ by a Legendre transformation [3,4,5,6]. Let us consider the infinitesimal changes $d\langle A_i \rangle$ and $d\lambda _i$ of $\langle A_i \rangle$ and $\lambda _i$ in equilibrium, respectively. Accordingly,

$$\begin{aligned} d(\ln Z)= \sum _i\frac{\partial \ln Z}{\partial \lambda _i}d\lambda _i = - \sum _i \langle A_i \rangle d\lambda _i ~, \end{aligned}$$

(65)

and

$$\begin{aligned} \frac{dS}{k}=d(\ln Z) + \sum _id\lambda _i\langle A_i \rangle + \sum _i\lambda _id\langle A_i\rangle ~, \end{aligned}$$

(66)

then subsequently

$$\begin{aligned} dS = k \sum _i\lambda _id\langle A_i \rangle ~. \end{aligned}$$

(67)

It can be seen that Eqs. (65) and (67) capture the natural variables for S and $\ln Z$. Since S is a function of $\langle A_i \rangle$, and $\ln Z$ is also useful as a function of $\lambda _i$, one can write

$$\begin{aligned} \lambda _i = \frac{1}{k}\frac{\partial S}{\partial \langle A_i\rangle } ~, \end{aligned}$$

(68)

and

$$\begin{aligned} \ln Z = \frac{S}{k} - \frac{1}{k}\sum _i\langle A_i \rangle \frac{\partial S}{\partial \langle A_i \rangle } ~. \end{aligned}$$

(69)

Therefore, $\lambda _i$ is the conjugate variable of $\langle A_i \rangle$ concerning S/k, and $\langle A_i \rangle$ is the conjugate variable of $\lambda _i$ concerning $-\ln Z$.

Appendix C. Derivation of Eqs. (14) and (15)

In the context of quantum-enhanced parameter estimation, a suitable measure to determine the ultimate limit of estimation precision is the Cramér-Rao (CR) bound [37,38,39]

$$\begin{aligned} \Delta {\varvec{\theta }}\ge {\frac{1}{\sqrt{n\mathcal {F}(\theta )}}}~, \end{aligned}$$

(70)

with $\mathcal {F}(\theta )$ being the QFI and $\theta$ standing for the unknown parameter encoded in the initial state. By considering $\rho = e^G$ and $G=-\sum _i\lambda _iA_i - \ln Z$, where $\lambda _i$ are the Lagrange coefficients, and $A_i\,|e_j\rangle = a_{ij}\,|e_j\rangle$, we can write the QFI in the following form [41,42,43]

$$\begin{aligned} \mathcal F(\lambda _p)=2\sum _{n,m}\frac{|\langle e_n|\partial _{\lambda _p}\rho |e_m\rangle |^2}{p_n+p_m}~, \end{aligned}$$

(71)

for the desired Lagrange coefficient $\lambda _p$. Regarding Wilcox’s formula for the derivative of an exponential operator [49]

$$\begin{aligned} \partial _{\lambda _{p}}e^{G} = \int _0^1 e^{\alpha G}\,\partial _{\lambda _{p}}G \,e^{(1-\alpha )G}\,d\alpha ~, \end{aligned}$$

(72)

and spectral decomposition $\rho =\sum _n p_n |e_n\rangle \langle e_n|$, the QFI becomes

$$\begin{aligned} \mathcal F(\lambda _p)= & {} 2\sum _{n,m}\frac{1}{p_n+p_m}\left| \langle e_n|\int _0^1e^{\alpha G} \partial _{\lambda _p}[G] e^{(1-\alpha )G}|e_m\rangle \right| ^2 \nonumber \\= & {} 2\sum _{n,m}\frac{1}{p_n+p_m}|\langle e_n|\partial _{\lambda _p} [G]|e_m\rangle |^2 \left| \int _0^1e^{\alpha g_n+(1-\alpha )g_m}\right| ^2 \nonumber \\= & {} \sum _{n} \frac{1}{p_n}|\langle e_n|\partial _{\lambda _p}[G]|e_n\rangle |^2 e^{2g_n}\nonumber \\{} & {} \quad +2\sum _{n\ne m}\frac{1}{p_n+p_m}|\langle e_n|\partial _{\lambda _p} [G]|e_m\rangle |^2 \left( \frac{e^{g_m}-e^{g_n}}{g_m-g_n}\right) ^2 ~. \end{aligned}$$

(73)

Here, we have used the fact that $G\,|e_j\rangle = g_j\,|e_j\rangle$, where $g_j = - \lambda _{j} a_{j} - \ln Z$. Also, it is substantial to keep the following in mind

$$\begin{aligned} \begin{aligned} \frac{\partial G}{\partial \lambda _p}&=\frac{\partial }{\partial \lambda _p}\left( -\sum _l \lambda _l A_l -\ln Z\right) \\&=\left( -\sum _l\frac{\partial \lambda _l}{\partial \lambda _p}A_l - \frac{\partial \ln Z}{\partial \lambda _p}\right) \\&=-A_p +\langle A_p\rangle =-\delta A_p ~. \end{aligned} \end{aligned}$$

(74)

In the second line of Eq. (74), we assumed that ${\lambda _l}$’s with $l\ne p$ and $A_p$ are independent of $\lambda _p$. Clearly, this assumption cannot be used for the strong coupling regime in which the energy operator has a dependence on temperature (its own Lagrange coefficient). By rewriting the GGE state in Eq. (9) as

$$\begin{aligned} \rho =\sum _ne^{G}|e_n\rangle \langle e_n|&= \sum _n e^{-\sum _i\lambda _iA_i-\ln Z}|e_n\rangle \langle e_n|\nonumber \\&=\sum _n e^{-\sum _i\lambda _ia_{ni}-\ln Z}|e_n\rangle \langle e_n|\nonumber \\ {}&=\sum _ne^{g_n}|e_n\rangle \langle e_n|=\sum _n p_n|e_n\rangle \langle e_n| ~, \end{aligned}$$

(75)

and substituting $g_n=\ln (p_n)$ together with Eq. (74) and by regarding $K_\alpha (\rho ,G)= \textrm{Var}[\rho ,G]=\sum _n\langle e_n| e^G (\delta G)^2 |e_n\rangle$, it turns out that

$$\begin{aligned} \mathcal F(\lambda _p)=\sum _n p_n|\langle e_n|\delta A|e_n\rangle |^2+2\sum _{n\ne m}\frac{(p_m-p_n)^2}{(p_n+p_m)(\ln (p_m/p_n))^2}|\langle e_n|\delta A_p |e_m\rangle |^2 , \end{aligned}$$

(76)

where

$$\langle e_n|\delta A_p |e_m\rangle = a_{pm}\delta _{nm}-\langle A_p \rangle \delta _{nm}~,$$

therefore, one comes to

$$\begin{aligned} \mathcal F(\lambda _p)=\sum _np_n( a_{pn}^2-2a_{pn}\langle A_p\rangle + \langle A_p\rangle ^2) = \sum _n p_n (a_{pn}-\langle A_p\rangle )^2 ~. \end{aligned}$$

(77)

At this point, let us consider the definition of classical uncertainty contribution

$$\begin{aligned} K_\alpha (\rho ,G)= \textrm{Var}(\rho ,G)=\langle (\delta G)^2\rangle = \sum _n\langle e_n| e^G (\delta G)^2 |e_n\rangle ~. \end{aligned}$$

(78)

Var() can be written as

$$\begin{aligned} \textrm{Var}(\rho ,G)&= \sum _{m,n,p}\langle e_n|e^{g_p}|e_p\rangle \langle e_p| \delta G |e_m\rangle \langle e_m|\delta G |e_n \rangle \nonumber \\&=\sum _{m,n}p_n \langle e_n|\delta G \vert e_m\rangle \langle e_m|\delta G|e_n\rangle \nonumber \\&=\sum _n p_n |\langle e_n|\delta G|e_n\rangle |^2 + \sum _{n\ne m}p_n |\langle e_n|\delta G|e_m\rangle |^2 ~, \end{aligned}$$

(79)

where $\delta G= -\sum _i \lambda _i \delta A_i$, hence

$$\begin{aligned} \textrm{Var}(\rho ,G)&= \sum _{n,i}p_n |\langle e_n | \lambda _i \delta A_i|e_n\rangle |^2 +\sum _{n\ne m,i}p_n |\langle e_n | \lambda _i \delta A_i|e_n\rangle |^2\nonumber \\&=\sum _{n,i}p_n\lambda _i^2(a^2_{ni}-2a_{pn}\langle A_i\rangle +\langle A_i\rangle ^2)\nonumber \\&=\sum _{n,i}p_n \lambda _i^2(a_{ni}-\langle A_i \rangle )^2 ~. \end{aligned}$$

(80)

Appendix D. Derivation of Eq. (22)

According to the previous methods

$$\begin{aligned} \begin{aligned} \mathcal F(\lambda _p)&=\sum _{n,m}\frac{2}{p_n+p_m}\left| \langle e_n|\int _0^1e^{\alpha G^*_S}\partial _{\lambda _p}G^*_S e^{(1-\alpha )G^*_S}|e_m\rangle \right| ^2\\&=\sum _{n,m,l,q}\frac{2}{p_n+p_m}\left| \langle e_n|\int _0^1e^{\alpha g_l}|e_l\rangle \langle e_l|\partial _{\lambda _p}G^*_S e^{(1-\alpha )g_q}|e_q\rangle \langle e_q|e_m\rangle \right| ^2\\&=2\sum _{n,m}\frac{1}{p_n+p_m}|\langle e_n|\partial _{\lambda _p}G^*_S |e_m\rangle |^2\left| \int _0^1e^{\alpha g_n+(1-\alpha )g_m}\right| ^2\\&=\sum _{n} \frac{1}{p_n}|\langle e_n|\partial _{\lambda _p}G^*_S |e_n\rangle |^2 e^{2g_n}\\&+2\sum _{n\ne m}\frac{1}{p_n+p_m}|\langle e_n|\partial _{\lambda _p}G^*_S |e_m\rangle |^2\left( \frac{e^{g_m}-e^{g_n}}{g_m-g_n}\right) ^2 ~, \end{aligned} \end{aligned}$$

(81)

where we have used the fact that $G^*_S|e_j\rangle = g_j|e_j\rangle$ and

$$\begin{aligned} \begin{aligned} \frac{\partial G^*_S}{\partial \lambda _p}&=\frac{\partial }{\partial \lambda _p}\left( -\sum _i \lambda _i A^*_i -\ln Z^*_S\right) \\&=-\frac{\partial }{\partial \lambda _p}\sum _i\lambda _iA^*_i +\left\langle \frac{\partial }{\partial \lambda _p}\left( \sum _i\lambda _iA^*_i\right) \right\rangle \\&=-E^*_S +\langle E^*_S\rangle =-\delta E^*_S ~. \end{aligned} \end{aligned}$$

(82)

It is also shown that

$$\begin{aligned} \rho _S&=\sum _ne^{G^*_S}|e_n\rangle \langle e_n| = e^{-\sum _i\lambda _iA^*_i-\ln Z^*_S}|e_n\rangle \langle e_n|\nonumber \\&=\sum _n e^{-\sum _i\lambda _ia_{ni}-\ln Z^*_S}|e_n\rangle \langle e_n|=\sum _ne^{g_n}|e_n\rangle \langle e_n|\nonumber \\ {}&=\sum _np_n|e_n\rangle \langle e_n|~. \end{aligned}$$

(83)

Implementing Eqs. (82) and (83) as well as $g_n=\ln (p_n)$, leads to

$$\begin{aligned} \mathcal F(\lambda _p)=\sum _n p_n|\langle e_n|\delta E^*_S|e_n\rangle |^2+2\sum _{n\ne m}\frac{(p_m-p_n)^2}{(p_n+p_m)(\ln (p_m/p_n))^2}|\langle e_n|\delta E^*_S |e_m\rangle |^2 ~, \end{aligned}$$

(84)

and by applying $|\langle e_n|\delta E^*_S |e_m\rangle | = |\langle e_n| E^*_S |e_m\rangle |$ for $n\ne m$, we have

$$\begin{aligned} \mathcal F(\lambda _p)=\sum _n p_n|\langle e_n|\delta E^*_S|e_n\rangle |^2+2\sum _{n\ne m}\frac{(p_m-p_n)^2}{(p_n+p_m)(\ln (p_m/p_n))^2}|\langle e_n| E^*_S |e_m\rangle |^2 ~. \end{aligned}$$

(85)

Given that

$$\begin{aligned} \textrm{Var}(\rho _S,E^*_S)=\langle (\delta E^*_S)^2\rangle = \sum _n\langle e_n|e^{G^*_S}(\delta E^*_S)^2|e_n\rangle ~, \end{aligned}$$

(86)

one obtains

$$\begin{aligned} \begin{aligned} \textrm{Var}(\rho _S,E^*_S)&=\frac{1}{Z_S^*(\{\lambda _i\})}\sum _{m,n,p}\langle e_n|e^{-\sum _i\lambda _ia_{pi}}|e_p\rangle \langle e_p|\delta E^*_S|e_m\rangle \langle e_m|\delta E^*_S |e_n\rangle \\&=\sum _{m,n}p_n\langle e_n|\delta E^*_S|e_m\rangle \langle e_m| \delta E^*_S |e_n\rangle \\&=\sum _n p_n |\langle e_n|\delta E^*_S|e_n\rangle |^2+\sum _{m\ne n}p_n|\langle e_n| E^*_S|e_m\rangle |^2 . \end{aligned} \end{aligned}$$

(87)

Subsequently, the QFI is given as the following form

$$\begin{aligned} \mathcal F(\lambda _p)=\textrm{Var}(\rho _S,E^*_S)+\sum _{n\ne m}\left[ \frac{2(p_m-p_n)^2}{(p_n+p_m)(\ln (p_m/p_n))^2}-p_n\right] |\langle e_n| E^*_S |e_m\rangle |^2 ~. \end{aligned}$$

(88)

Appendix E. Derivation of Eq. (24)

On the one hand, the skew information can be derived in the following way

$$\begin{aligned} Q(\rho _S,E^*_S) = \int _0^1 d\alpha Q_\alpha (\rho _S,E^*_S)=-\frac{1}{2}\int _0^1 d\alpha \textrm{tr}\left\{ [E^*_S,\rho _S^\alpha ][E^*_S,\rho _S^{1-\alpha }]\right\} ~, \end{aligned}$$

(89)

where

$$\begin{aligned} \begin{aligned}&\textrm{tr}\left\{ [{E^*_S},\rho _S^\alpha ][E^*_S,\rho _S^{1-\alpha }]\right\} =\textrm{tr}\left\{ [{E^*_S}\rho _S^\alpha -\rho _S^\alpha E^*_S][E^*_S\rho _{S}^{(1-\alpha )}-\rho _{S}^{(1-\alpha )}E^*_S]\right\} \\&\quad =\textrm{tr}\left\{ E^*_S\rho _S^{\alpha }E^*_S\rho _S^{1-\alpha }\right\} -\textrm{tr}\left\{ E^*_S\rho _S^{\alpha }\rho _S^{1-\alpha }E^*_S\right\} -\textrm{tr}\left\{ \rho _S^{\alpha }E^*_SE^*_S\rho _S^{1-\alpha }\right\} \\&\qquad +\textrm{tr}\left\{ \rho _S^{\alpha }E^*_S\rho _S^{1-\alpha }E^*_S\right\} ~. \end{aligned} \end{aligned}$$

On the other hand, knowing that $\rho _S^\alpha \rho _S^{1-\alpha }=\rho _S$ and that a trace has the cyclic permutation property, we find

$$\begin{aligned} \textrm{tr}(E^*_S\rho _S^\alpha \rho _S^{1-\alpha }E^*_S)=\textrm{tr}({E^{*}_{S}}^{2}\rho _S) = \langle {E^*_S}^{2} \rangle ~. \end{aligned}$$

(90)

Also, by utilizing this fact

$$\begin{aligned} \textrm{tr}(E^*_S\rho ^{\alpha }_S E^*_S\rho ^{1-\alpha }_S)=\textrm{tr}(\rho _S^\alpha E^*_S\rho _S^{1-\alpha }E^*_S)~, \end{aligned}$$

(91)

we obtain

$$\begin{aligned}{} & {} \int _0^1 d\alpha \,\textrm{tr}(E^*_S\rho _S^\alpha E^*_S\rho _S^{1-\alpha }) =\sum _n\langle e_n|E^*_S \int _0^1 d\alpha \,e^{-G^*_S}E^*_S e^{-(1-\alpha )G^*_S}|e_n\rangle \nonumber \\{} & {} \quad =\sum _{n,m,l}\langle e_n|E^*_S\int _0^1dae^{-\alpha g_m}|e_m\rangle \langle e_m|E^*_S e^{-(1-\alpha )g_l}|e_l\rangle \langle e_l|e_n\rangle \nonumber \\{} & {} \quad =\sum _{n,m}\langle e_n|E^*_S|e_m\rangle \langle e_m|E^*_S|e_n\rangle \int _0^1 d\alpha \,e^{[\alpha g_m+(1-\alpha )g_n]}\nonumber \\{} & {} \quad =\sum _n e^{-g_n}|\langle e_n|E^*_S|e_n\rangle |^2 + \sum _{n\ne m} |\langle e_n|E^*_S|e_m\rangle |^2\left( \frac{e^{g_m}-e^{g_n}}{g_m-g_n}\right) \nonumber \\{} & {} \quad =\sum _n p_n|\langle e_n|E^*_S|e_n\rangle |^2+\sum _{n\ne m} \frac{p_m-p_n}{\ln (p_m/p_n)} |\langle e_n|E^*_S|e_m\rangle |^2 ~, \end{aligned}$$

(92)

hence

$$\begin{aligned}{} & {} Q(\rho _S,E^*_S)= -\frac{1}{2}\left[ 2\sum _n p_n|\langle e_n|E^*_S|e_n\rangle |^2\right] \nonumber \\{} & {} \quad -\frac{1}{2}\left[ 2\sum _{n\ne m}\frac{p_m-p_n}{\ln (p_m/p_n)} \langle e_n|E^*_S|e_m\rangle |^2 - 2\langle {E^{*}_{S}}^2\rangle \right] ~. \end{aligned}$$

(93)

One can translate $\langle {E^{*}_{S}}^2\rangle$ by

$$\begin{aligned} \langle {E^{*}_{S}}^2\rangle&= \sum _{n,m,l} \langle e_l|e^{g_n}|e_n\rangle \langle e_n |E^*_S|e_m\rangle \langle e_m |E^*_S|e_n\rangle \nonumber \\&=\sum _{n,m}p_n|\langle e_n |E^*_S|e_m\rangle |^2\nonumber \\&=\sum _{n}p_n|\langle e_n|E^*_S|e_n\rangle |^2 +\sum _{n\ne m}p_n|\langle e_n|E^*_S|e_m\rangle |^2 ~. \end{aligned}$$

(94)

Therefore

$$\begin{aligned} Q(\rho _S,E^*_S)=\sum _{n\ne m}\left( p_n-\frac{p_m-p_n}{\ln (p_m/p_n)}\right) |\langle e_n|E^*_S|e_m\rangle |^2 ~. \end{aligned}$$

(95)

Appendix F. Derivation of Eqs. (25) and (31)

By adding the terms related to the skew information and the QFI, we achieve

$$\begin{aligned}&\mathcal F(\lambda _p)+Q(\rho _S,E^*_S) =\textrm{Var}(\rho _S,E^*_S)+\sum _{n\ne m}\nonumber \\&\left[ \frac{2(p_m-p_n)^2}{(p_n+p_m)(\ln (p_m/p_n))^2}-\frac{p_m-p_n}{\ln (p_m/p_n)}\right] |\langle e_n|E^*_S|e_m\rangle |^2 ~. \end{aligned}$$

(96)

Now, the following factorization can be considered

$$\begin{aligned} \left[ \frac{2(p_m-p_n)}{(p_n+p_m)(\ln (p_m/p_n))}-1\right] \left( \frac{p_m-p_n}{\ln (p_m/p_n)}\right) ~, \end{aligned}$$

(97)

where the second term in the multiplication is always positive, since $p_m,p_n\in [0,1]$. Thus, the condition for the expression (97) to be positive is assigned to the following inequality

$$\begin{aligned} \frac{(p_m-p_n)}{(p_n+p_m)}&>\frac{1}{2}\ln (p_m/p_n)~. \end{aligned}$$

(98)

One can follow the solution in two different situations: The first one is when

$$\begin{aligned} \frac{x-1}{x+1}>\frac{1}{2}\ln x, \qquad x=\frac{p_m}{p_n}~. \end{aligned}$$

(99)

The inequality in Eq. (99) is satisfied for the interval $0<x<1$, where $p_m<p_n$.

The second one is when $p_m>p_n$, and we have

$$\begin{aligned} \frac{1-x}{1+x}>-\frac{1}{2}\ln (x)~, \end{aligned}$$

(100)

with $x=p_n/p_m$, which is exactly the same inequality as before with only the sign changed. Therefore, the positivity of the second term on the right side of Eq. (96) boils down to the inequalities (99) and (100), which were proven to be valid for any real values of $p_m$ and $p_n$ between 0 and 1. From these considerations, we can write an inequality as

$$\begin{aligned} \mathcal F(\lambda _p) \le \textrm{Var}(\rho _S,E^*_S)-Q(\rho _S,E^*_S). \end{aligned}$$

(101)

Furthermore, we express the uncertainty using the operators $A^*_i$’s by

$$\begin{aligned} E^{*p}_{S}=\sum _i\frac{\partial (\lambda _iA^*_i)}{\partial \lambda _p}~, \end{aligned}$$

(102)

then

$$\begin{aligned} K(\rho _S,E^{*p}_S)&=\int _0^1d\alpha \,\textrm{tr}\{\rho _S^\alpha \delta E^{*p}_S\rho _S^{1-\alpha }\delta E^{*p}_S\}~, \end{aligned}$$

(103)

where $\delta E^{*p}_S=E^{*p}_S-\langle E^{*p}_S\rangle$. By considering the dependence $\lambda _i$ and $A^*_p$ on $\lambda _p$, we have

$$\begin{aligned} E^{*p}_{S} = \sum _{i\ne p}(\dot{\lambda }_iA_i^*+\lambda _i\dot{\lambda }_i\frac{\partial A_i^*}{\partial \lambda _i})+\partial _{\lambda _p}[\lambda _p A_p^*], \qquad \textrm{with} \quad \dot{\lambda }_i=\frac{\partial \lambda _i}{\partial \lambda _p}~. \end{aligned}$$

(104)

Finally, given that $[A^*_p,\rho _S]=0~(i\ne p)$, we derive a general expression for classical uncertainty in the following form

$$\begin{aligned}&K(\rho _S, E^{*p}_{S}) = \int _0^1d\alpha \textrm{tr} \left\{ \rho _S^\alpha [\partial _{\lambda _p}[ {\lambda _p}A^*_p] \rho _S^{1-\alpha }\partial _{\lambda _p} [{\lambda _p}A^*_p]\right\} \nonumber \\&\quad - \langle \partial _{\lambda _p}[ \lambda _p A^*_p]\rangle ^2+\sum _{i\ne p}\dot{\lambda }_i\langle \partial _{\lambda _p}[ {\lambda _p}A^*_p][A^*_i+\lambda _i \frac{\partial A^*_i}{\partial \lambda _i}]\rangle \nonumber \\&+\sum _{j\ne p}\dot{\lambda }_j\langle \partial _{\lambda _p}[\lambda _p A^*_p][A^*_j+\lambda _j\frac{\partial A^*_j}{\partial \lambda _j}]\rangle -\sum _{i\ne p}\dot{\lambda }_i\langle \partial _{\lambda _p} [{\lambda _p}A^*_p]\rangle ~\langle A^*_i+\lambda _i \frac{\partial A^*_i}{\partial \lambda _i}\rangle \nonumber \\&-\sum _{j\ne p}\dot{\lambda }_j\langle \partial _{\lambda _p}[ {\lambda _p}A^*_p]\rangle ~\langle A^*_j+\lambda _j\frac{\partial A^*_j}{\partial \lambda _j}\rangle +\sum _{ij\ne p}\dot{\lambda }_i\dot{\lambda }_j \langle [A^*_i+\lambda _i \frac{\partial A^*_i}{\partial \lambda _i}][A^*_j+\lambda _j\frac{\partial A^*_j}{\partial \lambda _j}] \rangle \nonumber \\&-\sum _{ij\ne p}\dot{\lambda }_i\dot{\lambda }_j\langle A^*_i+\lambda _i \frac{\partial A^*_i}{\partial \lambda _i}\rangle ~ \langle A^*_j+\lambda _j\frac{\partial A^*_j}{\partial \lambda _j}\rangle \nonumber \\&=K(\rho _S,\partial _{\lambda _p}[\lambda _p A^*_p])+2\sum _{i\ne p} \dot{\lambda }_i~ \textrm{Cov}~(\partial _{\lambda _p}[\lambda _p A^*_p],~A^*_i+\lambda _i \frac{\partial A^*_i}{\partial \lambda _i})\nonumber \\ {}&+\sum _{ij\ne p}\dot{\lambda }_i\dot{\lambda }_j~ \textrm{Cov}~(A^*_i+\lambda _i \frac{\partial A^*_i}{\partial \lambda _i},~A^*_j+\lambda _j\frac{\partial A^*_j}{\partial \lambda _j})~. \end{aligned}$$

(105)

where $\textrm{Cov}(X,Y)=\langle X Y\rangle -\langle X\rangle \langle Y\rangle$.