Real quantum operations and state transformations

First we formally introduce the resource theory of imaginarity [1, 17]. Given a reference basis $\{|i\rangle\}$, the free states in this theory are those states that have real density matrices. Specifically a state $\rho$ is called a free state when $\langle i|\rho|j\rangle\in\mathbb{R}$ for all $i$ and $j$. The free operations are those which can be characterised by real Kraus operators. A operation $\Lambda$ characterised by Kraus operators $\{K_{m}\}$ is free if $\langle i|K_{m}|j\rangle\in\mathbb{R}$ for all $i$, $j$ and $m$. In this section, we will explore state transformations under covariant operations. Here, covariant means, covariant with respect to transpose. A CP map ($\mathcal{E}$) is said to be “$\rho$-covariant” iff $(\mathcal{E}(\rho))^{T}=\mathcal{E}(\rho^{T})$. Similarly, a CP map ($\Lambda$) is said to be “covariant” iff $(\Lambda(\rho))^{T}=\Lambda(\rho^{T})$ for all $\rho$. But before presenting the results we introduce the following lemma which will be used later. Note that, in this letter by CP maps we mean completely positive trace non-increasing maps. We will now use these definitions to prove some interesting properties of real CP maps.

Using the above lemma, we now prove that, all covariant CP maps can be represented by real Kraus operators.

It is easy to see that any CP map with real Kraus operators is covariant. Using this fact, along with the above theorem, one can say that a CP map is real iff it is covariant. $\rho$-covariant maps correspond to a larger set than covariant maps. But when considering state transformations of $\rho$, we will see that, covariant maps are as powerful as $\rho$-covariant maps.

Next, we show that Theorem 2 can be used to derive necessary and sufficient conditions for state transformations under real quantum operations. In [31], the authors derived necessary and sufficient conditions for the existence of a CPTP map ($\mathcal{E}$), such that

$$\mathcal{E}(\rho_{i})=\sigma_{i}\,\,\forall i\in\{1,\cdots,n\},$$

(12)

where $\{\rho_{i}\}$ and $\{\sigma_{i}\}$ are two sets composed of $n$ density matrices each. It’s easy to see that, a transformation $\rho\rightarrow\sigma$ is possible via $\rho$-covariant operations iff there exists a CPTP map tranforming the set $\{\rho,\rho^{T}\}$ into $\{\sigma,\sigma^{T}\}$. Therefore, the transformation conditions for Eq. (12), along with Theorem 2, allow us to give necessary and sufficient conditions for the state transformations under real quantum operations, by substituting $\{\rho,\rho^{T}\}$ and $\{\sigma,\sigma^{T}\}$ instead of $\{\rho_{i}\}$ and $\{\sigma_{i}\}$. For the sake of completeness we provide these conditions below. We refer to [31] for a more detailed discussion.

Let $\{\rho,\sigma\}$ be density matrices acting on $\{\mathcal{H}_{\mathrm{B}},\mathcal{H}_{\mathrm{C}}\}$ and $\left|\phi_{+}^{\mathrm{AC}}\right\rangle$ denotes the maximally entangled state on $\mathcal{H}_{\mathrm{A}}\otimes\mathcal{H}_{\mathrm{C}}$. Additionally we define $d_{A}=d_{C}$, where $d_{A},d_{B}$ and $d_{C}$ are the dimensions of the Hilbert spaces $\mathcal{H}_{\mathrm{A}},\mathcal{H}_{\mathrm{B}}$ and $\mathcal{H}_{\mathrm{C}}$ respectively.

$$\Omega^{ABC}=\frac{1}{2}\omega_{1}\otimes\rho\otimes\sigma+\frac{1}{2}\omega_{2}\otimes\rho^{T}\otimes\sigma^{T}.$$

(13)

Then the following statements are equivalent:

1. There exists a real CPTP map $\mathcal{E}_{r}$, such that $\mathcal{E}_{r}(\rho)=\sigma$.

2. For any two states $\omega_{1}$ and $\omega_{2}$, we have

$$2^{-H_{\min}(A\mid B)_{\Omega}}\geqslant d_{\mathrm{C}}\left\langle\phi_{+}^{\mathrm{AC}}\left|\Omega^{\mathrm{AC}}\right|\phi_{+}^{\mathrm{AC}}\right\rangle.$$

3. For any $\{\omega_{1},\omega_{2}\}\in\mathcal{B}\left(\mathcal{H}_{\mathrm{A}}\right)$ :

$$H_{\min}(A\mid B)_{\Omega}\leqslant H_{\min}(A\mid C)_{\Omega}.$$

4. $\alpha=1$, where

	$\displaystyle\alpha\equiv\min\operatorname{Tr}[Z]$
	subject to	$\displaystyle\quad I^{\mathrm{A}}\otimes Z\geqslant X_{1}\otimes\rho+X_{2}\otimes\rho^{T}$			(14)
		$\displaystyle\operatorname{Tr}\left(\sigma^{\mathrm{T}}X_{1}+\sigma X_{2}\right)=1;X_{1},X_{2}\geqslant 0.$

Here, $H_{\min}(R\mid A)_{\Omega}$ is the quantum conditional min-entropy of the state $\Omega^{RA}$, defined as

$$H_{\min}(R\mid A)_{\Omega}:=-\log\inf_{X_{A}\geq 0}\left\{\operatorname{tr}\left[X_{A}\right]:\mathbb{I}_{R}\otimes X_{A}\geq\Omega^{RA}\right\}.$$

(15)

The above statements provides necessary and sufficient conditions to check whether a transformation $\rho\rightarrow\sigma$ is possible under real operations. Note that the optimisation problem in point 4, is a semidefinite program.

In this section, we study the nature of real operations in spatially separated bipartite scenarios. We introduce a new class of monotones under local real operations and classical communication (LRCC) [17, 1] which are independent to local operation and classical communication (LOCC) monotones. Let’s assume a bipartite state is shared between Alice and Bob, where Alice and Bob share a classical channel and can perform LRCC operations. Note that, for any two real CP maps $\Lambda_{1}$ and $\Lambda_{2}$

$$(\Lambda_{1}\otimes\Lambda_{2}(\rho^{AB}))^{T_{B}}=\Lambda_{1}\otimes\Lambda_{2}((\rho^{AB})^{T_{B}})\,\,\textrm{for all}\,\,\rho^{AB}.$$

(16)

Here, $T_{B}$ is the partial transpose on Bob’s system. This is easy to see, as $\Lambda_{2}$ has only real Kraus operators. This implies that for any LRCC map ($\Lambda$),

$$(\Lambda(\rho^{AB}))^{T_{B}}=\Lambda((\rho^{AB})^{T_{B}})\,\,\textrm{for all}\,\,\rho^{AB}.$$

(17)

Using data-processing inequality of trace distance and Eq. (17), we get

	$\displaystyle\lVert\rho^{AB}-(\rho^{AB})^{T_{B}}\rVert_{1}$	$\displaystyle\geq\lVert\Lambda(\rho^{AB})-\Lambda((\rho^{AB})^{T_{B}}))\rVert_{1}$
		$\displaystyle\geq\lVert\Lambda(\rho^{AB})-\Lambda(\rho^{AB})^{T_{B}})\rVert_{1}$		(18)

Note that, the above inequality holds for all bipartite states $\rho^{AB}$ and all LRCC operations $\Lambda$. This shows that,

$$D(\rho^{AB})=\lVert\rho^{AB}-(\rho^{AB})^{T_{B}}\rVert_{1}$$

(19)

is a real entanglement monotone (as it does not increase under LRCC operations). Here, instead of trace distance, one can also choose any arbitrary distinguishability measure obeying data processing inequality under CPTP maps. An example of such a distinguishability measure would be the infidelity ($1-F$) between $\rho^{AB}$ and $(\rho^{AB})^{T_{B}}$. In this case, the infidelity ($1-F$) between $\rho^{AB}$ and $(\rho^{AB})^{T_{B}}$ can act as a LRCC monotone. Note that in Eq. (II.2) one can replace $T_{B}$ with $T_{A}$ (partial transpose on Alice’s system) and get a separate set of monotones.

As a special case, for any real bipartite state $\rho^{AB}$ with $\lVert\rho^{AB}-(\rho^{AB})^{T_{B}}\rVert_{1}$>0, Eq. (II.2) implies that there cannot exist an LRCC operation transforming $|00\rangle\langle 00|^{AB}$ into $\rho^{AB}$. Note that here $\rho^{AB}$ can also be a separable state. For example, lets take a real state which is “separable”: $\eta^{AB}=\frac{1}{4}(\openone+\sigma_{y}\otimes\sigma_{y})$, where $\sigma_{y}$ is the Pauli-y matrix. It is easy to see that $||\eta^{AB}-(\eta^{AB})^{T_{B}}||_{1}>0$. This shows that, even though this state is separable, it is “entangled” in real quantum theory. Therefore, this shows that invariance under partial transpose is a necessary condition for separable states in “real” quantum theory, which was shown in [32, 33].

The geometric imaginarity of a pure state $|\psi\rangle$ can be expressed as [17]

$$\mathscr{I}_{g}(|\psi\rangle)=1-\max_{|\phi\rangle\in\mathscr{R}}|\langle\phi|\psi\rangle|^{2},$$

(20)

where the maximization is taken over all the real pure states. The definition can be extended to mixed states by considering the minimization of average imaginarity over all pure state decompositions as follows

$$\mathscr{I}_{g}(\rho)=\min_{\{p_{j},|\psi_{j}\rangle\}}\sum_{j}p_{j}\mathscr{I}_{g}(|\psi_{j}\rangle),$$

(21)

such that $\rho=\sum_{j}p_{j}|\psi_{j}\rangle\!\langle\psi_{j}|$. Note that the above definition of geometric imaginarity is equivalent to the following distance-based measure [34]

$$\mathscr{I}_{g}(\rho)=1-\max_{\sigma\in\mathcal{R}}F(\rho,\sigma),$$

(22)

where $\mathcal{R}$ represents set of all real states and $\sqrt{F(\rho,\sigma)}=\operatorname{Tr}\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}$ is the root fidelity. In [17], analytical expression of geometric measure of imaginarity has been derived for pure states. In the following theorem, we extend the previous result to find an analytical expression for the geometric measure of imaginarity for arbitrary states.

The proof is given in the Appendix. In the following, we show that geometric measure of imaginarity plays a key role in probabilistic and stochastic-approximate state transformations.

As described in the introduction, sometimes exact deterministic state transformations are not possible. Then, we consider the probabilistic scenario. Previously, probabilistic transformations between pure states have been studied [1, 17]. Here in the following theorem, we extend this to a scenario where the target state is mixed.

The above result gives an optimal probability of converting a pure state into a mixed state and it can be easily computed as geometric measure is easy to find.

In the above, we consider exact transformations. Here, we consider a more general scenario where we consider both the stochastic and approximate scenarios together. Stochastic-approximate state transformations have been studied previously for general resource theories [29, 25, 35, 30]. The probability for stochastic-approximate conversion is defined as the maximum probability of converting a state $\rho$ into $\sigma$ with fidelity at least $f$ [29]:

$$P_{f}(\rho\rightarrow\sigma)=\max_{\Lambda}\left\{\operatorname{Tr}\Lambda(\rho):F\left(\frac{\Lambda(\rho)}{\operatorname{Tr}[\Lambda(\rho)]},\sigma\right)\geq f\right\},$$

(30)

where $\Lambda$ corresponds to the set of all real operations. Similarly, the fidelity for stochastic-approximate conversion provides the maximal fidelity of transforming a state $\rho$ into $\sigma$ with a success probability at least $p$ and can be expressed as [29]

$$F_{p}(\rho\rightarrow\sigma)=\max_{\Lambda}\left\{F\left(\frac{\Lambda(\rho)}{\operatorname{Tr}[\Lambda(\rho)]},\sigma\right):\operatorname{Tr}\Lambda(\rho)\geq p\right\}.$$

(31)

Before, going into the result we introduce some important results, which will be used to find the above quantities.

For a given target fidelity $f$ to achieve a state $\rho$, the above theorem gives the least possible geometric measure among all states $\rho^{\prime}$ such that the fidelity between $\rho$ and $\rho^{\prime}$ is at least $f$. Note that, the case of maximum geometric measure is discussed only for pure states and the case for mixed states remains open. In order to prove this theorem, we use the following lemma. We refer to the Appendix for the proof of the following lemma.

Similar to the two-qubit entanglement theory [22, 36, 37], this lemma shows that there exists a pure state decomposition for any state $\rho$, such that all the pure states have same geometric measure as $\rho$. Using this lemma, we proove theorem 5 in the Appendix. Equipped with this we are now ready to discuss the stochastic-approximate transformations of an initial pure state.

Note that, if we want exact transformation, i.e., $f=1$, then we recover the result stated in Eq. (29). The optimal fidelity for a given probability can also be found easily:

$$F_{p}(|\psi\rangle\rightarrow\rho)=\begin{cases}1\,\,\,\mathrm{for}\,\,\,p\leq\frac{\mathscr{I}_{g}(|\psi\rangle)}{\mathscr{I}_{g}(\rho)},\\ \cos^{2}\left[\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\rho)}-\sin^{-1}\!\sqrt{\frac{\mathscr{I}_{g}(|\psi\rangle)}{p}}\right]\,\,\mathrm{otherwise}.\end{cases}$$

(35)

The details can be found in the appendix. The above result provides a complete solution for state transformations under real operations, when starting from a pure state. Note that for a special case when $\rho$ is a maximally imaginary state and $p=1$, we get back the result derived in [29].

In the following, we show that when starting from an arbitrary state $\rho_{A}=\sum_{i,j}c_{ij}|i\rangle\!\langle j|$, for a given probability of transformation $p$, the optimal achievable fidelity to reach a pure target state $|\psi_{B}\rangle$, can be computed by a semidefinite programe (SDP). Note that a CP map is real iff the choi matrix associated with it is real [38]. Let’s take a real CP map $\Lambda:A\rightarrow B$ and $\Sigma_{\Lambda}$ is the choi matrix of $\Lambda$, given by

$$\Sigma_{\Lambda}=\openone\otimes\Lambda\left(\sum_{i,j}|i\rangle\!\langle j|\otimes|i\rangle\!\langle j|\right).$$

(36)

Here, $d$ is the dimension of the Hilbert space of $A$. It is known that (see Eq. (4.2.12) of [39])

$$\Lambda(\rho_{A})=\operatorname{Tr}_{A}(\Sigma_{\Lambda}(\rho_{A}^{T}\otimes\openone_{B})),$$

(37)

where $T$ represents the transpose in above mentioned basis. Using Eq. (37), for any pure state $|\psi_{B}\rangle$

$\displaystyle\langle\psi_{B}|\Lambda(\rho_{A})|\psi_{B}\rangle$

$\displaystyle=\operatorname{Tr}(\Sigma_{\Lambda}(\rho_{A}^{T}\otimes|\psi_{B}\rangle\!\langle\psi_{B}|))$

(38)

Note that, $\Lambda^{\prime}$ can be considered as another real operation which completes the real operation $\Lambda$ (i.e $\Lambda+\Lambda^{\prime}$ is a real CPTP map). This leads us to the following theorem.

Using this SDP, we can compute the optimal achievable fidelity for transforming an arbitrary state into a pure state, for a given probability of transformation.

We note that, for pure states, geometric measure of imaginarity is given by [17]

$$\mathscr{I}_{g}(|\psi\rangle)=\frac{1-|\langle\psi^{*}|\psi\rangle|}{2}=\frac{1-\sqrt{F(|\psi\rangle\!\langle\psi|,|\psi^{*}\rangle\!\langle\psi^{*}|)}}{2}.$$

(40)

Let $\{p_{j},|\psi_{j}\rangle\}$ be the ensemble of $\rho$ which achieves the minimisation in Eq. (21). Using the joint concavity property of root-fidelity [40], we find

	$\displaystyle\sqrt{F(\rho,\rho^{T})}$	$\displaystyle\geq$	$\displaystyle\sum_{j}p_{j}\sqrt{F(|\psi_{j}\rangle\!\langle\psi_{j}|,|\psi_{j}^{*}\rangle\langle\psi_{j}^{*}|)}$
	$\displaystyle\frac{1-\sqrt{F(\rho,\rho^{T})}}{2}$	$\displaystyle\leq$	$\displaystyle\sum_{j}p_{j}\frac{1-\sqrt{F(|\psi_{j}\rangle\langle\psi_{j}|,|\psi^{*}_{j}\rangle\langle\psi^{*}_{j}|)}}{2}$		(41)
		$\displaystyle=$	$\displaystyle\mathscr{I}_{g}(\rho),$

where we consider the fact that $\sum_{j}p_{j}=1$ and in the last line we use the expression for pure states given in Eq. (40). In the above, we derive a lower bound for geometric imaginarity, given by,

$$\mathscr{I}_{g}(\rho)\geq\frac{1-\sqrt{F(\rho,\rho^{T})}}{2}.$$

(42)

Since, we know

$$\mathscr{I}_{g}(\rho)=\min_{\{p_{j},|\psi_{j}\rangle\}}\sum_{j}p_{j}\mathscr{I}_{g}(|\psi_{j}\rangle),$$

(43)

with $\sum_{j}p_{j}|\psi_{j}\rangle\!\langle\psi_{j}|=\rho$ and for each pure state $|\psi_{j}\rangle$

$$\mathscr{I}_{g}(|\psi_{j}\rangle)=\frac{1-|\langle\psi_{j}|\psi^{*}_{j}\rangle|}{2},$$

(44)

Eq. (42), is equivalent to

$$\max_{\{p_{j},|\psi_{j}\rangle\}}\sum_{j}p_{j}|\langle\psi_{j}|\psi^{*}_{j}\rangle|\leq\sqrt{F(\rho,\rho^{T})}.$$

(45)

Now, in the following we construct a special decomposition, which saturates the above bound. Note that any other ensemble of $\rho$ (let’s say $\{q_{i},|\phi_{i}\rangle\}$) is related to the ensemble $\{p_{j},|\psi_{j}\rangle\}$ in the following way

$$\sqrt{q_{i}}|\phi_{i}\rangle=\sum_{j}U^{*}_{ij}\sqrt{p_{j}}|\psi_{j}\rangle,$$

(46)

where $U_{ji}$ are the elements of a Unitary matrix $U$. Without loss of generality, we can choose $\{p_{j},|\psi_{j}\rangle\}$ to be the eigenvalues and eigenstates of $\rho$.

$$\sqrt{q_{i}q_{j}}\langle\phi_{i}|\phi^{*}_{j}\rangle=(UAU^{T})_{ij},$$

(47)

where $A_{ij}=\sqrt{p_{i}p_{j}}\langle\psi_{i}|\psi^{*}_{j}\rangle$. In [41], the authors show that for any symmetric matrix $S$, there exists a singular value decomposition of the form

$$S=Q\Sigma Q^{T},$$

(48)

where $\Sigma$ is a positive diagonal matrix and $Q$ is a unitary matrix. Eq. (48) also appears in [42] (P263, Corollary 4.4.4 (c)). The singular values of $A$, are the square roots of eigenvalues of $AA^{\dagger}(=AA^{*}$). Note that,

	$\displaystyle(AA^{*})_{ij}$	$\displaystyle=$	$\displaystyle\sum_{k}p_{k}\sqrt{p_{i}p_{j}}\langle\psi_{i}|\psi^{*}_{k}\rangle\langle\psi^{*}_{k}|\psi_{j}\rangle$		(49)
		$\displaystyle=$	$\displaystyle\sqrt{p_{i}p_{j}}\langle\psi_{i}|\rho^{T}|\psi_{j}\rangle$		(50)
		$\displaystyle=$	$\displaystyle\langle\psi_{i}|\sqrt{\rho}\rho^{T}\sqrt{\rho}|\psi_{j}\rangle.$		(51)

Therefore, $(AA^{*})_{ij}$ (matrix) is $\sqrt{\rho}\rho^{T}\sqrt{\rho}$ (operator), written (in matrix form) in the eigenbasis of $\rho$. This shows that the singular values of $(A)_{ij}$ (matrix) are the eigenvalues of $\sqrt{\sqrt{\rho}\rho^{T}\sqrt{\rho}}$. Eqs. (47) and (48) shows that one can find a decomposition (say $\{\lambda_{i},|\mu_{i}\rangle\}$), which is “conjugate-orthogonal”

$$\sqrt{\lambda_{i}\lambda_{j}}\langle\mu_{i}|\mu^{*}_{j}\rangle=\delta_{ij}D_{j}$$

(52)

where, $D_{j}$ is the $j^{th}$ eigenvalue of $\sqrt{\sqrt{\rho}\rho^{T}\sqrt{\rho}}$. This completes the proof. We found out that an independent proof of this theorem can be found in [43].

Let $\{\lambda_{j},|\mu_{j}\rangle\}$ be a pure state decomposition we described in Eq. (52). Therefore, $\mathscr{I}_{g}(\rho)=\sum_{j}\lambda_{j}\mathscr{I}_{g}(|\mu_{j}\rangle)$. We now show that we can achieve a decomposition (say $\{p_{i},|\psi_{i}\rangle\}$), where

$$\langle\psi_{i}|\psi^{*}_{i}\rangle=\langle\psi_{j}|\psi^{*}_{j}\rangle\,\,\forall\,\,i,j.$$

(53)

Firstly, note that

$$\langle\mu_{i}|\mu^{*}_{i}\rangle\geq 0\,\,\forall\,\,i.$$

(54)

We further define $\langle\mu_{i}|\mu^{*}_{i}\rangle$ as the “conjugate product” of $|\mu_{i}\rangle$. Therefore, the “average conjugate product” ($\sum_{i}\lambda_{i}\langle\mu_{i}|\mu^{*}_{i}\rangle$) of the decomposition $\{\lambda_{j},|\mu_{j}\rangle\}$ is equal to $1-\mathscr{I}_{g}(\rho)$. Now, we go into another decomposition such that the “average conjugate product” remains constant.

As we know (from Eq.(46)), we can go into another decomposition by mixing the sub-normalised pure states of the decomposition with elements of a unitary matrix (For our case, we just need a orthogonal matrix). The procedure to achieve the desired decomposition goes as follows.

If $\mathscr{I}_{g}(|\mu_{i}\rangle)=\mathscr{I}_{g}(\rho)$, we already have the required decomposition. In the other case, there exist at least two pure states (without loss of generality say) $|\mu_{1}\rangle$ and $|\mu_{2}\rangle$, such that

$$\mathscr{I}_{g}(|\mu_{1}\rangle)>\mathscr{I}_{g}(\rho)\,\,\textrm{and}\,\,\mathscr{I}_{g}(|\mu_{2}\rangle)<\mathscr{I}_{g}(\rho).$$

(55)

Now we mix these two sub-normalised pure states in the following way to achieve a new decomposition.

$$\begin{split}\sqrt{\lambda_{1}^{\prime}}|\mu^{\prime}_{1}\rangle&=\cos{\alpha}\sqrt{\lambda_{1}}|\mu_{1}\rangle+\sin{\alpha}\sqrt{\lambda_{2}}|\mu_{2}\rangle\,\mbox{and}\\ \sqrt{\lambda_{2}^{\prime}}|\mu^{\prime}_{2}\rangle&=-\sin{\alpha}\sqrt{\lambda_{1}}|\mu_{1}\rangle+\cos{\alpha}\sqrt{\lambda_{2}}|\mu_{2}\rangle.\end{split}$$

(56)

Without loss of generality, we can take $\alpha\in[0,\pi/2]$. Since, $\mathscr{I}_{g}(|\mu^{\prime}_{1}\rangle)$ varies continuously with $\alpha$ and it is easy to see that, at $\alpha=0$

$$|\mu^{\prime}_{1}\rangle=|\mu_{1}\rangle\implies\mathscr{I}_{g}(|\mu^{\prime}_{1}\rangle)=\mathscr{I}_{g}(|\mu_{1}\rangle)$$

(57)

and at $\alpha$ = $\pi/2$

$$|\mu^{\prime}_{1}\rangle=|\mu_{2}\rangle\implies\mathscr{I}_{g}(|\mu^{\prime}_{1}\rangle)=\mathscr{I}_{g}(|\mu_{2}\rangle).$$

(58)

Therefore, there exists a $\alpha$ = $\alpha^{*}$, such that

	$\displaystyle\sqrt{\lambda^{\prime}_{1}}|\mu^{\prime}_{1}\rangle=\cos{(\alpha^{*})}\sqrt{\lambda_{1}}|\mu_{1}\rangle+\sin{(\alpha^{*})}\sqrt{\lambda_{2}}|\mu_{2}\rangle\,\mbox{and}$
	$\displaystyle\mathscr{I}_{g}(|\mu^{\prime}_{1}\rangle)=\mathscr{I}_{g}({\rho}).$		(59)

Now after this we have a new decomposition of $\rho$. Note, that the average conjugate product of the new decomposition is same as the previous one and $\{|\mu^{\prime}_{2}\rangle,|\mu_{j}\rangle\},j=3,4...$ are cojugate-orthonormal. Therefore, we can continue this procedure recursively for the remaining vectors until we reach the required decomposition.

Let us make use of the following distance measure:

$$D(\rho,\sigma)=\cos^{-1}\left(\sqrt{F(\rho,\sigma)}\right),$$

(60)

known as the Bures angle. It has the following properties: i) $D(\rho,\sigma)\geq 0$ for all $\rho$ and $\sigma$, ii) $D(\rho,\sigma)=0$ if an only if $\rho=\sigma$, iii) It satisfies the triangle inequality, i.e., $D(\rho,\sigma)\leq D(\rho,\eta)+D(\eta,\sigma)$. Let us note that, the geometric measure of imaginarity for any state lies between 0 to $1/2$. Consider two states $\rho$ and $\rho^{\prime}$ with $\rho^{\prime}\in S_{\rho,f}$. Further we assume that $\rho_{r}$ and $\rho_{r}^{\prime}$ are the closest real states to $\rho$ and $\rho^{\prime}$ (with respect to Bures angle) respectively. Using the fact that $\rho_{r}$ is the closest real state to $\rho$ and the triangle inequality, we find

	$\displaystyle\cos^{-1}\left(\sqrt{1-\mathscr{I}_{g}(\rho)}\right)$	$\displaystyle=$	$\displaystyle D(\rho,\rho_{r})\leq D(\rho,\rho^{\prime}_{r})$
		$\displaystyle\leq$	$\displaystyle D(\rho,\rho^{\prime})+D(\rho^{\prime},\rho^{\prime}_{r})$
		$\displaystyle\leq$	$\displaystyle\cos^{-1}(\sqrt{f})+\cos^{-1}\left(\sqrt{1-\mathscr{I}_{g}(\rho^{\prime})}\right).$

Simplifying this, we have

$$\mathscr{I}_{g}(\rho^{\prime})\geq\sin^{2}\left(\max\left\{\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\rho)}-\cos^{-1}\!\sqrt{f},0\right\}\right).$$

(62)

An upper bound on $\mathscr{I}_{g}(\rho^{\prime})$ can be derived in similar fashion. Again using the fact that $\rho^{\prime}_{r}$ is the closest real state to $\rho^{\prime}$ and the triangle inequality, we get

	$\displaystyle\cos^{-1}\left(\sqrt{1-\mathscr{I}_{g}(\rho^{\prime})}\right)$	$\displaystyle=$	$\displaystyle D(\rho^{\prime},\rho^{\prime}_{r})\leq D(\rho^{\prime},\rho_{r})$
		$\displaystyle\leq$	$\displaystyle D(\rho,\rho^{\prime})+D(\rho,\rho_{r})$
		$\displaystyle\leq$	$\displaystyle\cos^{-1}(\sqrt{f})+\cos^{-1}\left(\sqrt{1-\mathscr{I}_{g}(\rho)}\right).$

Simplifying this, we will end up with the following upper bound

$$\mathscr{I}_{g}(\rho^{\prime})\leq\sin^{2}\left(\min\left\{\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\rho)}+\cos^{-1}\!\sqrt{f},\frac{\pi}{4}\right\}\right).$$

(64)

Now we show that lower bound in Eq. (62) can be achieved. From lemma 2, we know that for any state $\rho$, we can find a pure state decomposition such that $\rho=\sum_{i}p_{i}|\psi_{i}\rangle\!\langle\psi_{i}|$ and $\mathscr{I}_{g}(\psi_{i})=\mathscr{I}_{g}(\rho)$ holds true for all states $|\psi_{i}\rangle$. This implies that each of the states $|\psi_{i}\rangle$ can be written as

$\displaystyle|\psi_{i}\rangle$

$\displaystyle=\cos{\alpha}|a_{i}\rangle+i\sin{\alpha}|a^{\perp}_{i}\rangle,$

(65)

where $\langle a_{i}|a^{\perp}_{i}\rangle=0$ and $|a_{i}\rangle$ and $|a^{\perp}_{i}\rangle$ are real states. Here, we assumed, $\alpha=\sin^{-1}\sqrt{\mathscr{I}_{g}(\rho)}$. Note that $\alpha\in\{0,\frac{\pi}{4}\}$. Let’s consider another state $\rho_{\min}$ to be

$\displaystyle\rho_{\min}=\sum_{i}q_{i}|\phi_{i}\rangle\!\langle\phi_{i}|$

(66)

with

$\displaystyle|\phi_{i}\rangle$

$\displaystyle=\cos{\tilde{\beta}}|a_{i}\rangle+\sin{\tilde{\beta}}|a^{\perp}_{i}\rangle,$

(67)

where $\tilde{\beta}\in[0,\pi/4]$. The probabilities are given by

$$q_{i}=\frac{p_{i}|\langle\psi_{i}|\phi_{i}\rangle|^{2}}{\sum_{k}p_{k}|\langle\psi_{k}|\phi_{k}\rangle|^{2}}.$$

(68)

From the convexity of the geometric measure of imaginarity, we obtain

$$\mathscr{I}_{g}(\rho_{\min})\leq\sin^{2}\tilde{\beta}.$$

(69)

Now we use the properties of root-fidelity [40] to obtain

	$\displaystyle\sqrt{F(\rho,\rho_{\min})}$	$\displaystyle\geq\sum_{i}\sqrt{p_{i}q_{i}}|\langle\psi_{i}|\phi_{i}\rangle|$
		$\displaystyle=\sqrt{\sum_{i}p_{i}|\langle\psi_{i}|\phi_{i}\rangle|^{2}}=|\cos(\alpha-\tilde{\beta})|.$		(70)

Therefore,

$$F(\rho,\rho_{\min})\geq\cos^{2}(\alpha-\tilde{\beta}).$$

(71)

Now, let us assume $\cos^{-1}\sqrt{f}=k$ and set

$$\tilde{\beta}=\max\{\alpha-k,0\}.$$

(72)

For this choice of $\tilde{\beta}$, we get

$$F(\rho,\rho_{\min})\geq\cos^{2}\left(\min\{k,\alpha\}\right)\geq\cos^{2}k\geq f.$$

(73)

Here, we used the fact that $\cos^{2}$ is a monotonically decreasing function in $[0,\pi/4]$ and $\min\{k,\alpha\}$ is within $[0,\pi/4]$. This shows that, the state $\rho_{\min}$ is within $S_{\rho,f}$. Further, from Eq. (69) we get

	$\displaystyle\mathscr{I}_{g}(\rho_{\min})$	$\displaystyle\leq$	$\displaystyle\sin^{2}\left(\max\{\alpha-k,0\}\right)$		(74)
		$\displaystyle\leq$	$\displaystyle\sin^{2}\left(\max\left\{\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\rho)}-\cos^{-1}\!\sqrt{f},0\right\}\right).$

Comparing Eqs. (62)and (74), we obtain

$$\mathscr{I}_{g}(\rho_{\min})=\sin^{2}\left(\max\left\{\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\rho)}-\cos^{-1}\!\sqrt{f},0\right\}\right).$$

(75)

Hence, the minimum geometric measure of imaginarity within the set $S_{\rho,f}$ is given by

$$\min_{\rho^{\prime}\in S_{\rho,f}}\mathscr{I}_{g}(\rho^{\prime})=\sin^{2}\left(\max\left\{\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\rho)}-\cos^{-1}\!\sqrt{f},0\right\}\right).$$

(76)

Now, we focus on the case when $\rho$ is a pure state (let’s say $|\psi\rangle$):

$\displaystyle|\psi\rangle=\cos{\alpha}|a\rangle+i\sin{\alpha}|a^{\perp}\rangle.$

(77)

For this case, the upper bound in Eq. (64) is achievable as well. To show this, we consider the following pure state

	$\displaystyle|\psi_{\max}\rangle$	$\displaystyle=\cos(\min\{\alpha+k,\pi/4\})|a\rangle$		(78)
		$\displaystyle+i\sin(\min\{\alpha+k,\pi/4\})|a^{\perp}\rangle.$

Note that

$\displaystyle\mathscr{I}_{g}(|\psi_{\max}\rangle)$

$\displaystyle=\sin^{2}(\min\{\alpha+k,\pi/4\}).$

(79)

We also note that

$$F(\psi,\psi_{\max})=\cos^{2}\left(\min\left\{k,\frac{\pi}{4}-\alpha\right\}\right)\geq\cos^{2}k\geq f.$$

(80)

Here we use the fact that, $\cos^{2}$ is a monotonically decreasing function in $[0,\pi/4]$ and $\min\{k,\pi/4-\alpha\}$ is within $[0,\pi/4]$. Therefore, $|\psi_{\max}\rangle$ is inside $S_{\psi,f}$ and has the maximum possible geometric measure of imaginarity given by the following equation:

	$\displaystyle\max_{\rho^{\prime}\in S_{\psi,f}}\mathscr{I}_{g}(\rho^{\prime})$	$\displaystyle=$	$\displaystyle\sin^{2}\left(\min\{\alpha+k,\pi/4\}\right)$
		$\displaystyle=$	$\displaystyle\sin^{2}\left(\min\left\{\sin^{-1}\!\sqrt{\mathscr{I}_{g}(\psi)}+\cos^{-1}\!\sqrt{f},\pi/4\right\}\right).$

This completes the proof.