Semi device independence of the BB84 protocol

In the worst-case scenario, Alice, Bob, and the adversary Eve share a purification $| {\rm{\Psi }}\rangle \in {{ \mathcal H }}_{{\rm{A}}}\otimes {{ \mathcal H }}_{{\rm{B}}}\otimes {{ \mathcal H }}_{{\rm{E}}}$, prepared by Eve, of the state ${\rho }_{\mathrm{AB}}$ responsible for the observed correlations according to (2). When Alice measures u=z, the system in ${{ \mathcal H }}_{{\rm{B}}}\otimes {{ \mathcal H }}_{{\rm{E}}}$ is projected to the (unnormalised) state

$$\begin{eqnarray}&&\rho ={\mathrm{Tr}}_{{\rm{A}}}[({M}_{0}^{({\rm{z}})}\otimes {{\mathbb{1}}}_{\mathrm{BE}}){\rm{\Psi }}]\end{eqnarray} \tag{ 17 }$$

or

$$\begin{eqnarray}&&{\rho }^{\prime }={\mathrm{Tr}}_{{\rm{A}}}[({M}_{1}^{({\rm{z}})}\otimes {{\mathbb{1}}}_{\mathrm{BE}}){\rm{\Psi }}],\end{eqnarray} \tag{ 18 }$$

depending, respectively, on whether Alice gets the result a = 0 or a = 1. (We will in general write, e.g., Ψ as a shorthand for the density operator $| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}|$ associated to some pure state $| {\rm{\Psi }}\rangle$.) The normalisations of these states are related to the probabilities with which they are prepared according to $\mathrm{Tr}[\rho ]={P}_{{\rm{A}}}(0| {\rm{z}})$ and $\mathrm{Tr}[{\rho }^{\prime }]={P}_{{\rm{A}}}(1| {\rm{z}})$. The correlation between Alice's result a and the state available to Eve is summarised by the classical-quantum state

$$\begin{eqnarray}&&{\tau }_{A{\rm{E}}}=| 0\rangle \langle 0| \otimes {\rho }_{{\rm{E}}}^{\phantom{^{\prime} }}+| 1\rangle \langle 1| \otimes {\rho }_{{\rm{E}}}^{\prime },\end{eqnarray} \tag{ 19 }$$

in terms of Eve's parts ${\rho }_{{\rm{E}}}^{\phantom{^{\prime} }}={\mathrm{Tr}}_{{\rm{B}}}[\rho ]$ and ${\rho }_{{\rm{E}}}^{\prime }={\mathrm{Tr}}_{{\rm{B}}}[{\rho }^{\prime }]$ of the possible density operators ρ and ${\rho }^{\prime }$.

We consider the case where the key is extracted from the $u=v={\rm{z}}$ measurement results. In this case, the one-way asymptotic key rate secure against collective attacks is lower bounded by the Devetak–Winter rate [29], which can be expressed as the difference of two entropies

$$\begin{eqnarray}&&r=H(A| {\rm{E}})-H(A| B).\end{eqnarray} \tag{ 20 }$$

In (20), $H(A| B)$ is the Shannon entropy of Alice's outcome conditioned on Bob's and can either be computed directly or approximated by $H(A| B)\leqslant h({\delta }_{{\rm{z}}})=\phi ({E}_{{\rm{zz}}})$. The main problem, and the main goal of this section, is to derive a lower bound for the conditional von Neumann entropy $H(A| {\rm{E}})$, which is given by

$$\begin{eqnarray}H(A| {\rm{E}}) & = & S({\tau }_{A{\rm{E}}})-S({\tau }_{{\rm{E}}})\\ & = & S({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }})+S({\rho }_{{\rm{E}}}^{\prime })-S({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }}+{\rho }_{{\rm{E}}}^{\prime }),\end{eqnarray} \tag{ 21 }$$

where $S(\rho )=-\mathrm{Tr}[\rho \;{\mathrm{log}}_{2}(\rho )]$, when computed on the classical-quantum state (19).

The derivation followed in the remainder of this section uses a few mathematical tools (two of which are minor restatements of results in [30]) which are presented here as lemmas. Proofs for these are supplied as appendices to this article.

The starting point is the following relation for the conditional von Neumann entropy, which simplifies the problem to that of lower bounding the fidelity between the marginal states available to Eve.

Lemma 1. The conditional von Neumann entropy, computed on the classical-quantum state $| 0\rangle \langle 0| \otimes {\rho }_{{\rm{E}}}^{\phantom{^{\prime} }}+| 1\rangle \langle 1| \otimes {\rho }_{{\rm{E}}}^{\prime }$, is lower bounded by

$$\begin{eqnarray}&&H(A| {\rm{E}})\geqslant \phi ({A}_{{\rm{z}}})-\phi \left({\sqrt{{A}_{{\rm{z}}}^{\phantom{z}2}+4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}}}^{}\right)\end{eqnarray} \tag{ 22 }$$

in terms of the fidelity $F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$ between ${\rho }_{{\rm{E}}}^{\phantom{^{\prime} }}$ and ${\rho }_{{\rm{E}}}^{\prime }$. Furthermore, for fixed $F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$, the right-hand side of (22) is convex in ${A}_{{\rm{z}}}$ and is minimised with ${A}_{{\rm{z}}}=0$.

Here, we take the fidelity to be defined by $F(\rho ,\sigma )={\parallel \sqrt{\rho }\sqrt{\sigma }\parallel }_{1}$, where ${\parallel A\parallel }_{1}=\mathrm{Tr}[| A| ]=\mathrm{Tr}[\sqrt{{A}^{\dagger }A}]$ denotes the trace norm of an operator A, for (generally unnormalised) density operators ρ and σ. Note that the minimisation of (22) at ${A}_{{\rm{z}}}=0$ allows the bound for the von Neumann entropy to be simplified to

$$\begin{eqnarray}&&H(A| {\rm{E}})\geqslant 1-\phi (2F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })),\end{eqnarray} \tag{ 23 }$$

though this step is optional, since ${A}_{{\rm{z}}}$ is an observed parameter.

The approach we follow involves reducing the problem to considering pure states. To this end, we introduce orthonormal bases $\{| {0}_{u}\rangle ,| {1}_{u}\rangle \}$, $u\in \{{\rm{z}},{\rm{x}}\}$, in which Alice's (qubit Hermitian) POVM elements ${M}_{a}^{(u)}$ are diagonal. In these bases, Alice's POVMs can be expressed as convex sums

$$\begin{eqnarray}&&\{{M}_{0}^{(u)},{M}_{1}^{(u)}\}={m}_{1}^{(u)}\{{0}_{u},{1}_{u}\}+{m}_{2}^{(u)}\{{1}_{u},{0}_{u}\}+{m}_{3}^{(u)}\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}+{m}_{4}^{(u)}\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}\end{eqnarray} \tag{ 24 }$$

of the four projective measurements $\{{0}_{u},{1}_{u}\}$, $\{{1}_{u},{0}_{u}\}$, $\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}$, and $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$ for convex coefficients satisfying ${m}_{i}^{(u)}\geqslant 0$ and ${\displaystyle \sum }_{i}{m}_{i}^{(u)}=1$. (Here, ${0}_{u}$ and ${1}_{u}$ are shorthand for $| {0}_{u}\rangle \langle {0}_{u}|$ and $| {1}_{u}\rangle \langle {1}_{u}|$, and ${{\mathbb{1}}}_{{\rm{A}}}$ and ${{\mathbb{0}}}_{{\rm{A}}}$ denote the identity and null operators on ${{ \mathcal H }}_{{\rm{A}}}$.)

Concentrating on the z measurement, we can express the entangled state as

$$\begin{eqnarray}&&| {\rm{\Psi }}\rangle =| {0}_{{\rm{z}}}\rangle | \alpha \rangle +| {1}_{{\rm{z}}}\rangle | {\alpha }^{\prime }\rangle \end{eqnarray} \tag{ 25 }$$

for (unnormalised and not necessarily orthogonal) states $| \alpha \rangle ,| {\alpha }^{\prime }\rangle \in {{ \mathcal H }}_{{\rm{B}}}\otimes {{ \mathcal H }}_{{\rm{E}}}$. The fidelity between Eve's parts ${\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }}$ and ${\alpha }_{{\rm{E}}}^{\prime }$ of the states $| \alpha \rangle$ and $| {\alpha }^{\prime }\rangle$ introduced this way can, according to the following relation, be bounded in terms of an operator ${W}_{{\rm{B}}}$ on Bob's Hilbert space.

Lemma 2. The fidelity between Eve's partial traces ${\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }}$ and ${\alpha }_{{\rm{E}}}^{\prime }$ of the pure states $| \alpha \rangle$ and $| {\alpha }^{\prime }\rangle$ satisfies

$$\begin{eqnarray}&&2F({\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }},{\alpha }_{{\rm{E}}}^{\prime })\geqslant {\parallel {\text{}}{W}_{{\rm{B}}}\parallel }_{1},\end{eqnarray} \tag{ 26 }$$

where ${W}_{{\rm{B}}}={\mathrm{Tr}}_{{\rm{E}}}[{\text{}}W]$ and $W=| \alpha \rangle \langle {\alpha }^{\prime }| +| {\alpha }^{\prime }\rangle \langle \alpha |$.

We approach the problem of lower bounding ${\parallel {W}_{{\rm{B}}}\parallel }_{1}$ in the following way. Similar to (25), we express the entangled state as

$$\begin{eqnarray}&&| {\rm{\Psi }}\rangle =| {0}_{{\rm{x}}}\rangle | \beta \rangle +| {1}_{{\rm{x}}}\rangle | {\beta }^{\prime }\rangle \end{eqnarray} \tag{ 27 }$$

for the $u={\rm{x}}$ measurement. In an appropriate phase convention, the diagonalising bases are related by

$$\begin{eqnarray}&&\quad | {0}_{{\rm{z}}}\rangle =\mathrm{cos}\left(\tfrac{\varphi }{2}\right)| {0}_{{\rm{x}}}\rangle -\mathrm{sin}\left(\tfrac{\varphi }{2}\right)| {1}_{{\rm{x}}}\rangle ,\end{eqnarray} \tag{ 28 }$$

$$\begin{eqnarray}&&| {1}_{{\rm{z}}}\rangle =\mathrm{sin}\left(\tfrac{\varphi }{2}\right)| {0}_{{\rm{x}}}\rangle +\mathrm{cos}\left(\tfrac{\varphi }{2}\right)| {1}_{{\rm{x}}}\rangle \end{eqnarray} \tag{ 29 }$$

for some angle φ. From this and requiring that (25) and (27) are the same state, we extract

$$\begin{eqnarray}&&| \beta \rangle =\mathrm{cos}\left(\tfrac{\varphi }{2}\right)| \alpha \rangle +\mathrm{sin}\left(\tfrac{\varphi }{2}\right)| \alpha ^{\prime} \rangle ,\end{eqnarray} \tag{ 30 }$$

$$\begin{eqnarray}&&\quad | \beta ^{\prime} \rangle =-\mathrm{sin}\left(\tfrac{\varphi }{2}\right)| \alpha \rangle +\mathrm{cos}\left(\tfrac{\varphi }{2}\right)| \alpha ^{\prime} \rangle .\end{eqnarray} \tag{ 31 }$$

Introducing the correlators

$$\begin{eqnarray}&&\quad {\bar{\bar{E}}}_{{\rm{zx}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}({\alpha }_{{\rm{B}}}^{\phantom{^{\prime} }}-{\alpha }_{{\rm{B}}}^{\prime })],\end{eqnarray} \tag{ 32 }$$

$$\begin{eqnarray}&&{\bar{\bar{E}}}_{{\rm{xx}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}({\beta }_{{\rm{B}}}^{\phantom{^{\prime} }}-{\beta }_{{\rm{B}}}^{\prime })]\end{eqnarray} \tag{ 33 }$$

for the pure states and

$$\begin{eqnarray}&&{\bar{\bar{E}}}_{{\rm{wx}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}{\text{}}{W}_{{\rm{B}}}]\end{eqnarray} \tag{ 34 }$$

for the operator W appearing in lemma 2, the relations (30) and (31) imply

$$\begin{eqnarray}&&{\bar{\bar{E}}}_{{\rm{xx}}}=\mathrm{cos}(\varphi ){\bar{\bar{E}}}_{{\rm{zx}}}+\mathrm{sin}(\varphi ){\bar{\bar{E}}}_{{\rm{wx}}},\end{eqnarray} \tag{ 35 }$$

and applying the Cauchy–Schwarz inequality and rearranging, we obtain

$$\begin{eqnarray}&&{\bar{\bar{E}}}_{{\rm{wx}}}^{\phantom{{wx}}2}\geqslant {\bar{\bar{E}}}_{{\rm{xx}}}^{\phantom{{xx}}2}-{\bar{\bar{E}}}_{{\rm{zx}}}^{\phantom{{zx}}2},\end{eqnarray} \tag{ 36 }$$

similar the outline of the previous section. Finally, since ${\hat{B}}_{{\rm{x}}}$ is the difference of two POVM elements, it satisfies the operator inequalities $-{{\mathbb{1}}}_{{\rm{B}}}\leqslant {\hat{B}}_{{\rm{x}}}\leqslant {{\mathbb{1}}}_{{\rm{B}}};$ this allows ${\bar{\bar{E}}}_{{\rm{wx}}}$ to be used as a lower bound on the trace norm ${\parallel {W}_{{\rm{B}}}\parallel }_{1}$ of ${W}_{{\rm{B}}}$:

$$\begin{eqnarray}&&{\bar{\bar{E}}}_{{\rm{wx}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}{\text{}}{W}_{{\rm{B}}}]\leqslant {\parallel {\text{}}{W}_{{\rm{B}}}\parallel }_{1}{\parallel {\hat{B}}_{{\rm{x}}}\parallel }_{\infty }\leqslant {\parallel {\text{}}{W}_{{\rm{B}}}\parallel }_{1},\end{eqnarray} \tag{ 37 }$$

from which we finally obtain

$$\begin{eqnarray}&&4F{({\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }},{\alpha }_{{\rm{E}}}^{\prime })}^{2}\geqslant {\bar{\bar{E}}}_{{\rm{xx}}}^{\phantom{{xx}}2}-{\bar{\bar{E}}}_{{\rm{zx}}}^{\phantom{{zx}}2}.\end{eqnarray} \tag{ 38 }$$

The remaining problem is to convert (38) into a lower bound on $F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$ depending on the observed parameters A_u, B_v, and E_uv which can be used in lemma 1 (or (23)). Part of the problem is to relate these parameters to the pure-state versions ${\bar{\bar{E}}}_{{\rm{xx}}}$ and ${\bar{\bar{E}}}_{{\rm{zx}}}$ appearing in (38). From the POVM decomposition (24) we can deduce

$$\begin{eqnarray}&&{E}_{{uv}}=({m}_{1}^{(u)}-{m}_{2}^{(u)}){\bar{\bar{E}}}_{{uv}}+({m}_{3}^{(u)}-{m}_{4}^{(u)}){B}_{v},\end{eqnarray} \tag{ 39 }$$

which will allow the ${\bar{\bar{E}}}_{{uv}}{\rm{s}}$ to be related to the E_uvs and B_vs. For the z measurement, we will also need to be able to relate the fidelity $F({\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }},{\alpha }_{{\rm{E}}}^{\prime })$ in (38) to $F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$. For this, we will need the following general bound for the fidelity between mixtures of two states.

Lemma 3. Let $\rho$, $\sigma$, ${\tau }_{0}$, and ${\tau }_{1}$ be (not necessarily normalised) density operators related by

$$\begin{eqnarray}&&\quad \rho ={p}_{0}{\tau }_{0}+{p}_{1}{\tau }_{1},\end{eqnarray} \tag{ 40 }$$

$$\begin{eqnarray}&&\sigma ={q}_{0}{\tau }_{0}+{q}_{1}{\tau }_{1}\end{eqnarray} \tag{ 41 }$$

for parameters ${p}_{0},{p}_{1},{q}_{0},{q}_{1}\geqslant 0$. Then,

$$\begin{eqnarray}&&F{(\rho ,\sigma )}^{2}\geqslant {\left(\sqrt{{p}_{0}{q}_{0}}{\parallel {\tau }_{0}\parallel }_{1}+\sqrt{{p}_{1}{q}_{1}}{\parallel {\tau }_{1}\parallel }_{1}\right)}^{2}+{\left(\sqrt{{p}_{0}{q}_{1}}-\sqrt{{p}_{1}{q}_{0}}\right)}^{2}F{({\tau }_{0},{\tau }_{1})}^{2}.\end{eqnarray} \tag{ 42 }$$

The $u={\rm{x}}$ measurement is the simplest to handle, since it is not used for key generation, so we deal with it first. Rewriting the decomposition (39) for ${E}_{{\rm{xx}}}$ as

$$\begin{eqnarray}&&{E}_{{\rm{xx}}}=\lambda {\bar{\bar{E}}}_{{\rm{xx}}}+\mu {B}_{{\rm{x}}},\end{eqnarray} \tag{ 43 }$$

with $\lambda ={m}_{1}^{({\rm{x}})}-{m}_{2}^{({\rm{x}})}$ and $\mu ={m}_{3}^{({\rm{x}})}-{m}_{4}^{({\rm{x}})}$, the triangle inequality and the constraint $| \mu | \leqslant 1-| \lambda |$ together imply

$$\begin{eqnarray}&&| {E}_{{\rm{xx}}}| \leqslant | \lambda | | {\bar{\bar{E}}}_{{\rm{xx}}}| +(1-| \lambda | )| {B}_{{\rm{x}}}| ,\end{eqnarray} \tag{ 44 }$$

which rearranges to

$$\begin{eqnarray}&&| \lambda | (| {\bar{\bar{E}}}_{{\rm{xx}}}| -| {E}_{{\rm{xx}}}| )\geqslant (1-| \lambda | )(| {E}_{{\rm{xx}}}| -| {B}_{{\rm{x}}}| ).\end{eqnarray} \tag{ 45 }$$

If $| {E}_{{\rm{xx}}}| \gt | {B}_{{\rm{x}}}|$ then the only way that (45) can be satisfied is if $| \lambda | \gt 0$ and if $| {\bar{\bar{E}}}_{{\rm{xx}}}| \geqslant | {E}_{{\rm{xx}}}|$. In this case ${E}_{{\rm{xx}}}$ can safely be substituted in place of ${\bar{\bar{E}}}_{{\rm{xx}}}$ in the pure-state fidelity bound (38). Otherwise, it is perfectly possible for the ${\rm{x}}$ measurement POVM decomposition (43) to be satisfied with ${\bar{\bar{E}}}_{{\rm{xx}}}=0$. In the following, we will assume that $| {E}_{{\rm{xx}}}| \gt | {B}_{{\rm{x}}}|$, since (38) becomes trivial otherwise.

The POVM decomposition (24) implies that the states ρ and ${\rho }^{\prime }$ prepared on ${{ \mathcal H }}_{{\rm{B}}}\otimes {{ \mathcal H }}_{{\rm{E}}}$ are related to α and ${\alpha }^{\prime }$ by

$$\begin{eqnarray}&&\;\rho ={m}_{1}^{({\rm{z}})}\alpha +{m}_{2}^{({\rm{z}})}\alpha ^{\prime} +{m}_{3}^{({\rm{z}})}(\alpha +\alpha ^{\prime} ),\end{eqnarray} \tag{ 46 }$$

$$\begin{eqnarray}&&\rho ^{\prime} ={m}_{1}^{({\rm{z}})}\alpha ^{\prime} +{m}_{2}^{({\rm{z}})}\alpha +{m}_{4}^{({\rm{z}})}(\alpha +\alpha ^{\prime} ).\end{eqnarray} \tag{ 47 }$$

In general, the decomposition (24) for POVMs is not unique, so we have some freedom to choose a decomposition which will simplify the problem of turning the fidelity bound

$$\begin{eqnarray}&&4F{({\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }},{\alpha }_{{\rm{E}}}^{\prime })}^{2}\geqslant {E}_{{\rm{xx}}}^{\phantom{{\rm{xx}}}2}-{\bar{\bar{E}}}_{{\rm{zx}}}^{\phantom{{zx}}2}\end{eqnarray} \tag{ 48 }$$

into a lower bound for $F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$ depending on observed parameters A_u, B_v, and E_uv. Specifically, the identity

$$\begin{eqnarray}&&\{{0}_{{\rm{z}}},{1}_{{\rm{z}}}\}+\{{1}_{{\rm{z}}},{0}_{{\rm{z}}}\}=\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}+\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}\end{eqnarray} \tag{ 49 }$$

implies that one of the POVMs $\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}$ or $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$ can always be eliminated, meaning we can assume that one of ${m}_{3}^{{\rm{z}}}$ and ${m}_{4}^{{\rm{z}}}$ in (24) is zero without loss of generality.

We proceed in two steps, first considering mixtures of the measurements $\{{0}_{{\rm{z}}},{1}_{{\rm{z}}}\}$ and $\{{1}_{{\rm{z}}},{0}_{{\rm{z}}}\}$, before accounting for a contribution from one of the measurements $\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}$ or $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$. In anticipation, and assuming a contribution from $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$ for example, we re-express (46) and (47) as

$$\begin{eqnarray}&&\rho =p(q\alpha +q^{\prime} \alpha ^{\prime} ),\end{eqnarray} \tag{ 50 }$$

$$\begin{eqnarray}&&\quad \rho ^{\prime} =p(q^{\prime} \alpha +q\alpha ^{\prime} )+p^{\prime} (\alpha +\alpha ^{\prime} ),\end{eqnarray} \tag{ 51 }$$

where the nonnegative parameters p, ${p}^{\prime }$, q, ${q}^{\prime }$ are related to the ${m}_{i}^{({\rm{z}})}{\rm{s}}$ by $p={m}_{1}^{({\rm{z}})}+{m}_{2}^{({\rm{z}})}$, ${p}^{\prime }={m}_{4}^{({\rm{z}})}$, ${pq}={m}_{1}^{({\rm{z}})}$, and ${{pq}}^{\prime }={m}_{2}^{({\rm{z}})}$ and satisfy $p+{p}^{\prime }=q+{q}^{\prime }=1$.

For the contribution from $\{{0}_{{\rm{z}}},{1}_{{\rm{z}}}\}$ and $\{{1}_{{\rm{z}}},{0}_{{\rm{z}}}\}$, we set

$$\begin{eqnarray}&&\quad \bar{\rho }=q\alpha +q^{\prime} \alpha ^{\prime} ,\end{eqnarray} \tag{ 52 }$$

$$\begin{eqnarray}&&\quad \bar{\rho }^{\prime} =q^{\prime} \alpha +q\alpha ^{\prime} ,\end{eqnarray} \tag{ 53 }$$

and, applying lemma 3 and the pure-state fidelity bound (48), we have

$$\begin{eqnarray}4F{({\bar{\rho }}_{{\rm{E}}}^{\phantom{^{\prime} }},{\bar{\rho }}_{{\rm{E}}}^{\prime })}^{2} & \geqslant & 4{qq}^{\prime} +{(q-q^{\prime} )}^{2}4F{({\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }},{\alpha }_{{\rm{E}}}^{\prime })}^{2}\\ & \geqslant & 4{qq}^{\prime} +{(q-q^{\prime} )}^{2}({E}_{{\rm{xx}}}^{\phantom{{\rm{xx}}}2}-{\bar{\bar{E}}}_{{\rm{zx}}}^{\phantom{{\rm{zx}}}2}).\end{eqnarray} \tag{ 54 }$$

Introducing the correlator

$$\begin{eqnarray}&&{\bar{E}}_{{\rm{zx}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}({\bar{\rho }}_{{\rm{B}}}^{\phantom{^{\prime} }}-{\bar{\rho }}_{{\rm{B}}}^{\prime })],\end{eqnarray} \tag{ 55 }$$

related to ${\bar{\bar{E}}}_{{\rm{zx}}}$ by ${\bar{E}}_{{\rm{zx}}}=(q-{q}^{\prime }){\bar{\bar{E}}}_{{\rm{zx}}}$, and using that $4{{qq}}^{\prime }\geqslant 4{{qq}}^{\prime }{E}_{{\rm{xx}}}^{\phantom{{\rm{xx}}}2}$,

$$\begin{eqnarray}4F{({\bar{\rho }}_{{\rm{E}}}^{\phantom{^{\prime} }},{\bar{\rho }}_{{\rm{E}}}^{\prime })}^{2} & \geqslant & (4{qq}^{\prime} +{(q-q^{\prime} )}^{2}){E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{\bar{E}}_{{\rm{zx}}}^{\phantom{{zx}}2}\\ & = & {(q+q^{\prime} )}^{2}{E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{\bar{E}}_{{\rm{zx}}}^{\phantom{{\rm{zx}}}2}\end{eqnarray} \tag{ 56 }$$

or

$$\begin{eqnarray}&&4F{({\bar{\rho }}_{{\rm{E}}}^{\phantom{^{\prime} }},{\bar{\rho }}_{{\rm{E}}}^{\prime })}^{2}\geqslant {E}_{{\rm{xx}}}^{\phantom{{\rm{xx}}}2}-{\bar{E}}_{{\rm{zx}}}^{\phantom{{zx}}2},\end{eqnarray} \tag{ 57 }$$

which shows that allowing mixtures of the measurements $\{{0}_{{\rm{z}}},{1}_{{\rm{z}}}\}$ and $\{{1}_{{\rm{z}}},{0}_{{\rm{z}}}\}$ alone will not affect the key-rate formula.

Finally, we account for the effect of a contribution from one of the degenerate measurements $\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}$ or $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$. Assuming first a contribution from $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$, according to (50) and (51) and using that $\bar{\rho }+{\bar{\rho }}^{\prime }=\alpha +{\alpha }^{\prime }$, ρ and ${\rho }^{\prime }$ are related to the states $\bar{\rho }$ and ${\bar{\rho }}^{\prime }$ defined above by

$$\begin{eqnarray}&&\rho =p\bar{\rho },\end{eqnarray} \tag{ 58 }$$

$$\begin{eqnarray}&&\quad \rho ^{\prime} =p^{\prime} \bar{\rho }+\bar{\rho }^{\prime} .\end{eqnarray} \tag{ 59 }$$

Applying lemma 3 again,

$$\begin{eqnarray}&&F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}\geqslant {{pp}}^{\prime }{\parallel \bar{\rho }\parallel }_{1}^{\phantom{1}2}+{pF}{({\bar{\rho }}_{{\rm{E}}}^{\phantom{^{\prime} }},{\bar{\rho }}_{{\rm{E}}}^{\prime })}^{2}.\end{eqnarray} \tag{ 60 }$$

Inserting the lower bound (57) for $F({\bar{\rho }}_{{\rm{E}}}^{\phantom{^{\prime} }},{\bar{\rho }}_{{\rm{E}}}^{\prime })$ and recognising that

$$\begin{eqnarray}&&p{\parallel \bar{\rho }\parallel }_{1}={\parallel \rho \parallel }_{1}={P}_{{\rm{A}}}(0| {\rm{z}})=(1+{A}_{{\rm{z}}})/2,\end{eqnarray} \tag{ 61 }$$

the lower bound for $F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$ becomes

$$\begin{eqnarray}&&4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}\geqslant \left(\tfrac{1}{p}-1\right){(1+{A}_{{\rm{z}}})}^{2}+{{pE}}_{{\rm{xx}}}^{\phantom{{xx}}2}-p{\bar{E}}_{{\rm{zx}}}^{\phantom{{zx}}2}.\end{eqnarray} \tag{ 62 }$$

The observed parameters

$$\begin{eqnarray}&&{E}_{{\rm{zx}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}({\rho }_{{\rm{B}}}^{\phantom{^{\prime} }}-{\rho }_{{\rm{B}}}^{\prime })]\end{eqnarray} \tag{ 63 }$$

and

$$\begin{eqnarray}&&{B}_{{\rm{x}}}=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}({\rho }_{{\rm{B}}}^{\phantom{^{\prime} }}+{\rho }_{{\rm{B}}}^{\prime })]=\mathrm{Tr}[{\hat{B}}_{{\rm{x}}}({\bar{\rho }}_{{\rm{B}}}^{\phantom{^{\prime} }}+{\bar{\rho }}_{{\rm{B}}}^{\prime })]\end{eqnarray} \tag{ 64 }$$

are related to ${\bar{E}}_{{\rm{zx}}}$ by

$$\begin{eqnarray}&&{E}_{{\rm{zx}}}=p{\bar{E}}_{{\rm{zx}}}-{p}^{\prime }{B}_{{\rm{x}}}.\end{eqnarray} \tag{ 65 }$$

Rearranging for ${\bar{E}}_{{\rm{zx}}}$ and inserting in (62), we obtain

$$\begin{eqnarray}&&4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}\geqslant \left(\tfrac{1}{p}-1\right){(1+{A}_{{\rm{z}}})}^{2}+{{pE}}_{{\rm{xx}}}^{\phantom{{xx}}2}-p{\left(\tfrac{1}{p}{E}_{{\rm{zx}}}+\left(\tfrac{1}{p}-1\right){B}_{{\rm{x}}}\right)}^{2}\end{eqnarray} \tag{ 66 }$$

or, subtracting ${E}_{{\rm{xx}}}^{\phantom{{\rm{xx}}}2}-{E}_{{\rm{zx}}}^{\phantom{{\rm{zx}}}2}$ from both sides,

$$\begin{eqnarray}&&4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}-({E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{E}_{{\rm{zx}}}^{\phantom{{zx}}2})\geqslant \left(\tfrac{1}{p}-1\right)[{(1+{A}_{{\rm{z}}})}^{2}-p({E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{B}_{{\rm{x}}}^{\phantom{x}2})-{({E}_{{\rm{zx}}}+{B}_{{\rm{x}}})}^{2}].\end{eqnarray} \tag{ 67 }$$

By following similar reasoning starting from the decomposition

$$\begin{eqnarray}&&\quad \rho =\bar{\rho }+p^{\prime} \bar{\rho }^{\prime} ,\end{eqnarray} \tag{ 68 }$$

$$\begin{eqnarray}&&\rho ^{\prime} =p\bar{\rho }^{\prime} ,\end{eqnarray} \tag{ 69 }$$

assuming a contribution from $\{{{\mathbb{1}}}_{{\rm{A}}},{{\mathbb{0}}}_{{\rm{A}}}\}$ instead of $\{{{\mathbb{0}}}_{{\rm{A}}},{{\mathbb{1}}}_{{\rm{A}}}\}$, we obtain the same result as (67) except with the sign changes ${A}_{{\rm{z}}}\to -{A}_{{\rm{z}}}$ and ${B}_{{\rm{x}}}\to -{B}_{{\rm{x}}}$. The worst of the two bounds obtained this way is

$$\begin{eqnarray}&&4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}-({E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{E}_{{\rm{zx}}}^{\phantom{{zx}}2})\geqslant \left(\tfrac{1}{p}-1\right)[1-2| {A}_{{\rm{z}}}-{E}_{{\rm{zx}}}{B}_{{\rm{x}}}| +{A}_{z}^{\phantom{z}2}\\ &&\;p({E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{B}_{{\rm{x}}}^{\phantom{x}2})-({E}_{{\rm{zx}}}^{\phantom{{zx}}2}+{B}_{{\rm{x}}}^{\phantom{x}2})]\phantom{\rule{0ex}{2.16ex}}.\end{eqnarray} \tag{ 70 }$$

The multiplicative factor $1/p-1$ is nonnegative, so the right-hand side of (70) is nonnegative if

$$\begin{eqnarray}&&p({E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{B}_{{\rm{x}}}^{\phantom{x}2})+({E}_{{\rm{zx}}}^{\phantom{{zx}}2}+{B}_{{\rm{x}}}^{\phantom{x}2})\leqslant 1-2| {A}_{{\rm{z}}}-{E}_{{\rm{zx}}}{B}_{{\rm{x}}}| +{A}_{{\rm{z}}}^{\phantom{z}2}.\end{eqnarray} \tag{ 71 }$$

Finally, since we are assuming $| {E}_{{\rm{xx}}}| \gt | {B}_{{\rm{x}}}|$, the term $p({E}_{{\rm{xx}}}^{\phantom{{xx}}2}-{B}_{{\rm{x}}}^{\phantom{x}2})$ is nonnegative and is maximised with p = 1. This implies that (71) is satisfied for all $p\leqslant 1$ if it is satisfied for p = 1, i.e., if

$$\begin{eqnarray}&&{E}_{{\rm{xx}}}^{\phantom{{xx}}2}+{E}_{{\rm{zx}}}^{\phantom{{zx}}2}\leqslant 1-2| {A}_{{\rm{z}}}-{E}_{{\rm{zx}}}{B}_{{\rm{x}}}| +{A}_{{\rm{z}}}^{\phantom{z}2},\end{eqnarray} \tag{ 72 }$$

which is the condition given in the previous section. If this condition is met then the lower bound

$$\begin{eqnarray}&&4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}\geqslant {E}_{{\rm{xx}}}^{\phantom{{\rm{xx}}}2}-{E}_{{\rm{zx}}}^{\phantom{{zx}}2}\end{eqnarray} \tag{ 73 }$$

can be used for the fidelity in lemma 1.

The conditional von Neumann entropy satisfies $H(A| {\rm{E}})\geqslant H(A| {\mathrm{EE}}^{\prime })$ for any extension ${{ \mathcal H }}_{{\rm{E}}}\otimes {{ \mathcal H }}_{{{\rm{E}}}^{\prime }}$ of Eve's Hilbert space ${{ \mathcal H }}_{{\rm{E}}}$. We use this to replace the (unnormalised) density operators ${\rho }_{{\rm{E}}}^{\phantom{^{\prime} }}$ and ${\rho }_{{\rm{E}}}^{\prime }$ appearing in the classical-quantum state (19) with purifications $| \psi \rangle$ and $| {\psi }^{\prime }\rangle ;$ by Uhlmann's theorem (which still holds for unnormalised states), these can be chosen such that $\langle \psi | {\psi }^{\prime }\rangle =F({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })$. We this way obtain

$$\begin{eqnarray}H(A| {\rm{E}}) & \geqslant & S(\psi )+S(\psi ^{\prime} )-S(\psi +\psi ^{\prime} )\\ & = & h({P}_{{\rm{A}}}(0| {\rm{z}}))-h({\lambda }_{+}),\end{eqnarray} \tag{ 75 }$$

where

$$\begin{eqnarray}&&{\lambda }_{\pm }=\tfrac{1}{2}\pm \tfrac{1}{2}\sqrt{{({\parallel \psi \parallel }_{1}-{\parallel {\psi }^{\prime }\parallel }_{1})}^{2}+4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}}\end{eqnarray} \tag{ 76 }$$

are the eigenvalues of $\psi +{\psi }^{\prime }$. Recognising that

$$\begin{eqnarray}{\parallel \psi \parallel }_{1}-{\parallel \psi ^{\prime} \parallel }_{1} & = & {\parallel \rho \parallel }_{1}-{\parallel \rho ^{\prime} \parallel }_{1}\\ & = & {P}_{{\rm{A}}}(0| {\rm{z}})-{P}_{{\rm{A}}}(1| {\rm{z}})\\ & = & {A}_{{\rm{z}}},\end{eqnarray} \tag{ 77 }$$

we obtain

$$\begin{eqnarray}&&H(A| {\rm{E}})\geqslant \phi ({A}_{{\rm{z}}})-\phi \left(\sqrt{{A}_{{\rm{z}}}^{\phantom{z}2}+4F{({\rho }_{{\rm{E}}}^{\phantom{^{\prime} }},{\rho }_{{\rm{E}}}^{\prime })}^{2}}\right),\end{eqnarray} \tag{ 78 }$$

which is the lower bound claimed in the statement of lemma 1.

The right-hand side of (78) has the form

$$\begin{eqnarray}&&f(x)=\phi (x)-\phi (\sqrt{{x}^{2}+{y}^{2}}),\end{eqnarray} \tag{ 79 }$$

where we treat y as a fixed parameter and x should satisfy ${x}^{2}+{y}^{2}\leqslant 1$. We show that this function is convex by lower bounding its second derivative. First, the first and second derivatives of ϕ are

$$\begin{eqnarray}&&\quad {\phi }^{\prime }(x)=-\displaystyle \frac{1}{2}\;{\mathrm{log}}_{2}\;\left(\displaystyle \frac{1+x}{1-x}\right)\end{eqnarray} \tag{ 80 }$$

and

$$\begin{eqnarray}&&{\phi }^{\prime\prime }(x)=-\displaystyle \frac{1}{\mathrm{ln}(2)}\displaystyle \frac{1}{1-{x}^{2}}.\end{eqnarray} \tag{ 81 }$$

Applying the product rule, the first and second derivatives of f are

$$\begin{eqnarray}&&{f}^{\prime }(x)={\phi }^{\prime }(x)-{\phi }^{\prime }\left(\sqrt{{x}^{2}+{y}^{2}}\right)\displaystyle \frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\end{eqnarray} \tag{ 82 }$$

and

$$\begin{eqnarray}&&{f}^{\prime\prime }(x)={\phi }^{\prime\prime }(x)-{\phi }^{\prime\prime }\left(\sqrt{{x}^{2}+{y}^{2}}\right)\displaystyle \frac{{x}^{2}}{{x}^{2}+{y}^{2}}-{\phi }^{\prime }\left(\sqrt{{x}^{2}+{y}^{2}}\right)\displaystyle \frac{{y}^{2}}{{({x}^{2}+{y}^{2})}^{3/2}}.\end{eqnarray} \tag{ 83 }$$

Using that $\mathrm{ln}\left(\tfrac{1+| x| }{1-| x| }\right)\geqslant 2| x|$, the last term can be replaced with

$$\begin{eqnarray}&&-{\phi }^{\prime }\left(\sqrt{{x}^{2}+{y}^{2}}\right)\displaystyle \frac{{y}^{2}}{{({x}^{2}+{y}^{2})}^{3/2}}\geqslant \displaystyle \frac{1}{\mathrm{ln}(2)}\displaystyle \frac{{y}^{2}}{{x}^{2}+{y}^{2}},\end{eqnarray} \tag{ 84 }$$

so that

$$\begin{eqnarray}f^{\prime \prime} (x) & \geqslant & \displaystyle \frac{1}{\mathrm{ln}(2)}\left[-\displaystyle \frac{1}{1-{x}^{2}}+\displaystyle \frac{1}{1-{x}^{2}-{y}^{2}}\displaystyle \frac{{x}^{2}}{{x}^{2}+{y}^{2}}+\displaystyle \frac{{y}^{2}}{{x}^{2}+{y}^{2}}\right]\\ & = & \displaystyle \frac{1}{\mathrm{ln}(2)}\left[-\displaystyle \frac{1}{1-{x}^{2}}+\displaystyle \frac{1-{y}^{2}}{1-{x}^{2}-{y}^{2}}\right]\\ & = & \displaystyle \frac{1}{\mathrm{ln}(2)}\displaystyle \frac{{x}^{2}{y}^{2}}{(1-{x}^{2})(1-{x}^{2}-{y}^{2})}\\ & \geqslant & 0,\end{eqnarray} \tag{ 85 }$$

which shows that f is convex. Noticing that ${f}^{\prime }(0)=0$ (or just that f is an even function) implies that x = 0 is the global minimum.

A basic property of the trace norm is that ${\parallel {W}_{{\rm{B}}}\parallel }_{1}=\mathrm{Tr}[{U}_{{\rm{B}}}{W}_{{\rm{B}}}]$ for some unitary operator ${U}_{{\rm{B}}}$; furthermore, since ${W}_{{\rm{B}}}$ is Hermitian, ${U}_{{\rm{B}}}$ can also be taken to be Hermitian. From here and using that $W=| \alpha \rangle \langle {\alpha }^{\prime }| +| {\alpha }^{\prime }\rangle \langle \alpha |$,

$$\begin{eqnarray}\parallel {\text{}}{W}_{{\rm{B}}}{\parallel }_{1} & = & \mathrm{Tr}[{U}_{{\rm{B}}}{W}_{{\rm{B}}}]\\ & = & \mathrm{Tr}[({U}_{{\rm{B}}}\otimes {{\mathbb{1}}}_{{\rm{E}}})W]\\ & = & 2\mathrm{Re}[\langle \alpha | {U}_{{\rm{B}}}\otimes {{\mathbb{1}}}_{{\rm{E}}}| \alpha ^{\prime} \rangle ]\\ & \leqslant & 2| \langle \alpha | {U}_{{\rm{B}}}\otimes {{\mathbb{1}}}_{{\rm{E}}}| \alpha ^{\prime} \rangle | \\ & \leqslant & 2F({\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }},{\alpha }_{{\rm{E}}}^{\prime }).\end{eqnarray} \tag{ 86 }$$

The final line follows, by Uhlmann's theorem, from noticing that $| \alpha \rangle$ and ${U}_{{\rm{B}}}\otimes {{\mathbb{1}}}_{{\rm{E}}}| {\alpha }^{\prime }\rangle$ are purifications of ${\alpha }_{{\rm{E}}}^{\phantom{^{\prime} }}$ and ${\alpha }_{{\rm{E}}}^{\prime }$.

We introduce purifications $| {\chi }_{0}\rangle$ and $| {\chi }_{1}\rangle$ of ${\tau }_{0}$ and ${\tau }_{1}$ such that $F({\tau }_{0},{\tau }_{1})=\langle {\chi }_{0}| {\chi }_{1}\rangle$. In terms of these, note that

$$\begin{eqnarray}&&\quad | \psi \rangle =\sqrt{{p}_{0}}| {\chi }_{0}\rangle | {\gamma }_{0}\rangle +\sqrt{{p}_{1}}| {\chi }_{1}\rangle | {\gamma }_{1}\rangle ,\end{eqnarray} \tag{ 87 }$$

$$\begin{eqnarray}&&| \phi \rangle =\sqrt{{q}_{0}}| {\chi }_{0}\rangle | {\delta }_{0}\rangle +\sqrt{{q}_{1}}| {\chi }_{1}\rangle | {\delta }_{1}\rangle ,\end{eqnarray} \tag{ 88 }$$

where $\{| {\gamma }_{0}\rangle ,| {\gamma }_{1}\rangle \}$ and $\{| {\delta }_{0}\rangle ,| {\delta }_{1}\rangle \}$ are orthonormal bases, are purifications of ρ and σ. Using Uhlmann's theorem and expanding, the fidelity between ρ and σ is lower bounded by

$$\begin{eqnarray}F(\rho ,\sigma ) & \geqslant & | \langle \psi | \phi \rangle | \\ & = & \left|\displaystyle \sum _{{ij}}\sqrt{{p}_{i}{q}_{j}}\langle {\chi }_{i}| {\chi }_{j}\rangle \langle {\gamma }_{i}| {\delta }_{j}\rangle \right|\\ & = & \left|\displaystyle \sum _{{ij}}\sqrt{{p}_{i}{q}_{j}}F({\tau }_{i},{\tau }_{j}){U}_{{ji}}\right|\\ & = & | \mathrm{Tr}[{UT}]| ,\end{eqnarray} \tag{ 89 }$$

where U and T are the matrices of elements ${U}_{{ji}}=\langle {\gamma }_{i}| {\delta }_{j}\rangle$ and ${T}_{{ij}}=\sqrt{{p}_{i}{q}_{j}}F({\tau }_{i},{\tau }_{j})$. By exploiting the freedom to choose the bases $\{| {\gamma }_{0}\rangle ,| {\gamma }_{1}\rangle \}$ and $\{| {\delta }_{0}\rangle ,| {\delta }_{1}\rangle \}$, U can be made to be any 2 × 2 unitary matrix. Maximising the right-hand side over U, we obtain

$$\begin{eqnarray}&&F(\rho ,\sigma )\geqslant {\parallel T\parallel }_{1},\end{eqnarray} \tag{ 90 }$$

with

$$\begin{eqnarray}T=\left[\begin{array}{cc}\sqrt{{p}_{0}{q}_{0}}{\parallel {\tau }_{0}\parallel }_{1} & \sqrt{{p}_{0}{q}_{1}}F({\tau }_{0},{\tau }_{1})\\ \sqrt{{p}_{1}{q}_{0}}F({\tau }_{0},{\tau }_{1}) & \sqrt{{p}_{1}{q}_{1}}{\parallel {\tau }_{1}\parallel }_{1}\end{array}\right],\end{eqnarray} \tag{ 91 }$$

in which we inserted that $F({\tau }_{i},{\tau }_{i})={\parallel {\tau }_{i}\parallel }_{1}$.

In general, the trace norm of a 2 × 2 matrix $M=\left[\begin{array}{ll}\alpha & \beta \\ \gamma & \delta \end{array}\right]$ is given by

$$\begin{eqnarray}&&{\parallel M\parallel }_{1}=\sqrt{T+2\sqrt{D}},\end{eqnarray} \tag{ 92 }$$

where

$$\begin{eqnarray}&&T=| \alpha {| }^{2}+| \beta {| }^{2}+| \gamma {| }^{2}+| \delta {| }^{2},\end{eqnarray} \tag{ 93 }$$

$$\begin{eqnarray}&&\sqrt{D}=| \alpha \delta -\beta \gamma | \end{eqnarray} \tag{ 94 }$$

are respectively the trace of $| M{| }^{2}={M}^{\dagger }M$ and the root of its determinant. Applying this to obtain an explicit expression for the trace norm of (91) and using that $F({\tau }_{0},{\tau }_{1})\leqslant \sqrt{{\parallel {\tau }_{0}\parallel }_{1}}\sqrt{{\parallel {\tau }_{1}\parallel }_{1}}$ produces the result

$$\begin{eqnarray}&&F{(\rho ,\sigma )}^{2}\geqslant {\left(\sqrt{{p}_{0}{q}_{0}}{\parallel {\tau }_{0}\parallel }_{1}+\sqrt{{p}_{1}{q}_{1}}{\parallel {\tau }_{1}\parallel }_{1}^{}\right)}^{2}+\left(\sqrt{{p}_{0}{q}_{1}}\right.-{\left.\sqrt{{p}_{1}{q}_{0}}\right)}^{2}F{({\tau }_{0},{\tau }_{1})}^{2}.\end{eqnarray} \tag{ 95 }$$