LAUSR

Eylee Jung et.al[1] had conjectured that Pmax=12\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_{max}=\frac{1}{2}$$\end{document} is a necessary and sufficient condition for the perfect two-party teleportation, and consequently, the Groverian measure of entanglement for the entanglement resource must be 12\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{1}{\sqrt{2}}$$\end{document}. It is also known that prototype W state is not useful for standard teleportation. Agrawal and Pati[2] have successfully executed perfect (standard) teleportation with non-prototype W state. Aligned with the protocol mentioned in[2], we have considered here Star type tripartite states and have shown that perfect teleportation is suitable with such states. Moreover, we have taken the linear superposition of non-prototype W state and its spin-flipped version and shown that it belongs to Star class. Also, standard teleportation is possible with these states. It is observed that genuine tripartite entanglement is not necessary requirement for a state to be used as a channel for successful standard teleportation. We have also shown that these Star class states are Pmax=14\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_{max}=\frac{1}{4}$$\end{document} states and their Groverian entanglement is 32\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{\sqrt{3}}{2}$$\end{document}, thus concluding that Jung conjecture is not a necessary condition.