1. Introduction
2. Preliminaries
3. Tighter monogamy relations for RαE
[37] For λ≥2 and ς≥0, one has
For 0 ≤ θ ≤ 1 and λ≥2, we see that
If θ = 0, the inequality holds. If θ ≠ 0, the core idea is to reformulate the desired inequality. Observing that equation (
For any tripartite quantum state ϱABC, we have for α ≥ 2 and η ≥ 2,
(1) If Rα(ϱA∣B) ≥ Rα(ϱA∣C), then
If Rα(ϱA∣B) ≥ Rα(ϱA∣C), we have
From lemma
For any N-qubit quantum state ${\varrho }_{A{B}_{1}\cdots {B}_{N-1}}\in {H}_{A}\otimes {H}_{{B}_{1}}\otimes \cdots \otimes {H}_{{B}_{N-1}}$, if ${R}_{\alpha }({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }({\varrho }_{A| {B}_{l}})$ for i = 1, 2, ⋯ , z, and ${R}_{\alpha }({\varrho }_{A| {B}_{j}})\,\leqslant \,\displaystyle {\sum }_{l=j+1}^{N-1}{R}_{\alpha }({\varrho }_{A| {B}_{l}})$ for j = z + 1, ⋯ , N − 2, ∀ 1 ≤ z ≤ N − 3, N ≥ 4, then
When ${R}_{\alpha }({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }({\varrho }_{A| {B}_{l}})$ for i = 1, 2, ⋯ , z, set ${\theta }_{i}=\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }({\varrho }_{A| {B}_{l}})/{R}_{\alpha }({\varrho }_{A| {B}_{i}})$. Substituting θi into the inequality (
When ${R}_{\alpha }({\varrho }_{A| {B}_{j}})\,\leqslant \,\displaystyle {\sum }_{l=j+1}^{N-1}{R}_{\alpha }({\varrho }_{A| {B}_{l}})$ for j = z + 1, ⋯ , N − 2, we use a strengthened version of lemma
For any N-qubit quantum state ${\varrho }_{A{B}_{1}\cdots {B}_{N-1}}$ $\in {H}_{A}\otimes {H}_{{B}_{1}}\otimes $ $\cdots \otimes {H}_{{B}_{N-1}}$, if ${R}_{\alpha }({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }({\varrho }_{A| {B}_{l}})$ for i = 1, 2, ⋯ , N − 2, then
Under local unitary operations, any three-qubit pure state can be written as [38, 39],
Figure 1. The y axis is the lower bound of Rα(ψA∣BC). Solid black line denotes R2(ψA∣BC) for the state given in equation ( |
For any tripartite quantum state ϱABC, we have when $\frac{\sqrt{7}-1}{2}\,\leqslant \,\alpha \lt 2$.
(1) If Rα(ϱA∣B) ≥ Rα(ϱA∣C), then
(2) If Rα(ϱA∣B) ≤ Rα(ϱA∣C), then
If Rα(ϱA∣B) ≥ Rα(ϱA∣C), we have
For any N-qubit quantum state ${\varrho }_{A{B}_{1}\cdots {B}_{N-1}}\in {H}_{A}\otimes {H}_{{B}_{1}}\otimes \cdots \otimes {H}_{{B}_{N-1}}$, if ${R}_{\alpha }^{2}({\varrho }_{A| {B}_{i}})\,\geqslant \displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }^{2}({\varrho }_{A| {B}_{l}})$ for i = 1, 2, ⋯ , z, and ${R}_{\alpha }^{2}({\varrho }_{A| {B}_{j}})\,\leqslant \,\displaystyle {\sum }_{l=j+1}^{N-1}{R}_{\alpha }^{2}({\varrho }_{A| {B}_{l}})$ for j = z + 1, ⋯ , N − 2, ∀ 1 ≤ z ≤ N − 3 and N ≥ 4, we see that
When ${R}_{\alpha }^{2}({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }^{2}({\varrho }_{A| {B}_{l}})$ for i = 1, 2, ⋯ , z, set ${\theta }_{i}^{{\prime} }=\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }^{2}({\varrho }_{A| {B}_{l}})/{R}_{\alpha }^{2}({\varrho }_{A| {B}_{i}})$. Substituting ${\theta }_{i}^{{\prime} }$ into the inequality (
When ${R}_{\alpha }^{2}({\varrho }_{A| {B}_{j}})\,\leqslant \,\displaystyle {\sum }_{l=j+1}^{N-1}{R}_{\alpha }^{2}({\varrho }_{A| {B}_{l}})$ for j = z + 1, ⋯ , N − 2, we use a strengthened version of lemma
For any N-qubit quantum state ${\varrho }_{A{B}_{1}\cdots {B}_{N-1}}\in {H}_{A}\otimes {H}_{{B}_{1}}\otimes \cdots \otimes {H}_{{B}_{N-1}}$, if ${R}_{\alpha }^{2}({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{l=i+1}^{N-1}{R}_{\alpha }^{2}({\varrho }_{A| {B}_{l}})$ for i = 1, 2, ⋯ , N − 2, then
For state (
Figure 2. The y axis is the lower bound of Rα(ψA∣BC). Solid black line denotes Rα(ψA∣BC) of the state ( |
4. Tighter polygamy relations for RαEoA
For any 0 ≤ θ ≤ 1 and 0 ≤ v ≤ 1, then
If θ = 0, the inequality holds. If θ ≠ 0, the core idea is to reformulate the desired inequality. Observing that equation (
For any tripartite quantum state ϱABC, if ${R}_{\alpha }^{a}({\varrho }_{A| B})\,\geqslant \,{R}_{\alpha }^{a}({\varrho }_{A| C})$, we see that
It has been shown that ${R}_{\alpha }^{a}({\varrho }_{A| BC})\,\leqslant \,{R}_{\alpha }^{a}({\varrho }_{A| B})\,+{R}_{\alpha }^{a}({\varrho }_{A| C})$ in (
From lemma
For any N-qubit mixed state ${\varrho }_{A{B}_{1}\cdots {B}_{N-1}}\in {H}_{A}\otimes {H}_{{B}_{1}}\otimes \cdots \otimes {H}_{{B}_{N-1}}$, if ${R}_{\alpha }^{a}({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{k=i+1}^{N-1}{R}_{\alpha }^{a}({\varrho }_{A| {B}_{k}})$ for i = 1, 2, ⋯ , N − 2, we obtain
According to (
For any N-qubit mixed state ${\varrho }_{A{B}_{1}\cdots {B}_{N-1}}\in {H}_{A}\otimes {H}_{{B}_{1}}\otimes \cdots \otimes {H}_{{B}_{N-1}}$, if ${R}_{\alpha }^{a}({\varrho }_{A| {B}_{i}})\,\geqslant \,\displaystyle {\sum }_{k=i+1}^{N-1}{R}_{\alpha }^{a}({\varrho }_{A| {B}_{k}})$ for i = 1, 2, ⋯ , z, and ${R}_{\alpha }^{a}({\varrho }_{A| {B}_{j}})\,\leqslant \,\displaystyle {\sum }_{k=j+1}^{N-1}{R}_{\alpha }^{a}({\varrho }_{A| {B}_{k}})$ for j = z + 1, ⋯ , N − 2, ∀ 1 ≤ z ≤ N − 3 and N ≥ 4, then
From the proof of theorem
In addition, since ${R}_{\alpha }^{a}({\varrho }_{A| {B}_{j}})\,\leqslant \,\displaystyle \sum _{k=j+1}^{N-1}{R}_{\alpha }^{a}({\varrho }_{A| {B}_{k}})$ for j = z + 1, ⋯ , N − 2, we obtain
By the denotations of ${N}_{A{B}_{j}}$ and ${\overline{N}}_{A{B}_{j}}$, from the inequalities (
Consider a three-qubit generalised W-class state,
Equation (
Figure 3. The y axis is the upper bound of ${R}_{\alpha }^{a}({\psi }_{A| BC})$. Solid black line denotes ${R}_{\alpha }^{a}({\psi }_{A| BC})$ for the state ( |
5. Residual entanglement of RαE
For a three-qubit pure state ∣ψ〉ABC, we see that
For α ≥ 2 and μ ≥ 1 we get
Based on the definition of ${\tau }_{\mu }^{R}(| \psi {\rangle }_{ABC})$, we have
According to equation (
For the superpositions of the Greenberger–Horne–Zeilinger-state and the W-state,
Figure 4. The purple surface represents the lower bound of ${R}_{\alpha }^{a}({\varrho }_{AB})$ for the state ( |
Figure 5. The green line represents the lower bound of ${R}_{\alpha }^{a}({\varrho }_{AB})$ for the state ( |


