Imperfect Quantum Teleportation

Quantum Computing is part of Stack Overflow’s open communities: specialist spaces where curiosity is welcome, knowledge is shared freely, and the best answers rise to the top. Stack Overflow for Teams is now called Stack Internal. Bring the best of human thought and AI automation together at your work. Bring the best of human thought and AI automation together at your work. Learn more Bring the best of human thought and AI automation together at your work. Quantum Teleportation consists in transferring a qubit state from Alice to Bob : |m⟩A=α|0⟩+β|1⟩\begin{equation}\tag{1} \label{eq:1} |m \rangle_{A} = \alpha |0\rangle + \beta |1\rangle \end{equation} by means of Local Quantum Operations and Classical Communications. The helpful resource for this is the Bell Pair : |ϕ+⟩=1√2[|00⟩+|11⟩]\begin{equation} \label{eq:2} |\phi^+ \rangle = \frac{1}{\sqrt{2}} [|00\rangle + |11\rangle ] \end{equation} The steps for this are first to write the overall state |ϕ⟩|\phi\rangle composed of the message tensored with the Bell Pair : |ψ⟩=|m⟩A⊗|ϕ+⟩=1√2[α|000⟩+β|100⟩+α|011⟩+β|111⟩]=12[|ϕ+⟩⊗|m⟩+|ϕ−⟩⊗|mz⟩+|ψ+⟩⊗|mx⟩+|ψ−⟩⊗|mxz⟩](3)\begin{equation} \label{eq:3} \begin{split} |\psi\rangle &= |m\rangle_{A}\otimes |\phi^{ + }\rangle \\ & = \frac{1}{\sqrt{2}}\biggl[\alpha \lvert 000 \rangle + \beta\lvert 100 \rangle + \alpha \lvert 011 \rangle + \beta\lvert 111 \rangle\biggr]\\ & = \frac{1}{2}\biggl[ \lvert \phi^{ +} \rangle\otimes\lvert m \rangle + \lvert \phi^{ -} \rangle\otimes \lvert m_{z} \rangle + \lvert \psi^{ + } \rangle\otimes\lvert m_{x} \rangle +\lvert \psi^{ - } \rangle\otimes\lvert m_{xz} \rangle\biggr] \end{split}\qquad (3) \end{equation} with |mx⟩=X|m⟩,|mz⟩=Z|m⟩,|mxz⟩=XZ|m⟩\begin{equation} \label{eq:4} \lvert m_{x} \rangle = X\lvert m \rangle\;,\quad \lvert m_{z} \rangle = Z\lvert m \rangle\;,\quad \lvert m_{xz} \rangle = XZ\lvert m \rangle \end{equation} and |ϕ+⟩=1√2(|00⟩+|11⟩),|ϕ−⟩=1√2(|00⟩−|11⟩)|ψ+⟩=1√2(|01⟩+|10⟩),|ψ−⟩=1√2(|01⟩−|10⟩)\begin{equation} \label{eq:5} \begin{split} \lvert \phi^{ +} \rangle= \frac{1}{\sqrt{2}}(\lvert 00 \rangle+\lvert 11 \rangle)\;&,\quad \lvert \phi^{ -} \rangle= \frac{1}{\sqrt{2}}(\lvert 00 \rangle-\lvert 11 \rangle) \\ \lvert \psi^{+} \rangle= \frac{1}{\sqrt{2}}(\lvert 01 \rangle + \lvert 10 \rangle)\;&,\quad \lvert \psi^{-} \rangle= \frac{1}{\sqrt{2}}(\lvert 01 \rangle-\lvert 10 \rangle) \end{split} \end{equation} As a consequence, Bob is left with state |m⟩B|m\rangle_B which is either |m⟩\lvert m \rangle or |mz⟩\lvert m_z \rangle or |mx⟩\lvert m_x \rangle or |mxz⟩\lvert m_{xz} \rangle. One can check that for every case : Zx1Xx2|m⟩B=|m⟩A\begin{equation} \label{eq:6} Z^{x_{1}}X^{x_{2}}|m\rangle_B = |m\rangle_A \end{equation} However, this protocol relies on having one perfect Bell State Pair, which is not the case in reality. I wonder then how the protocol behaves if instead of having (1), one actually gets : |ϕ+⟩=cosθ|00⟩+sinθ|11⟩]\begin{equation} \label{eq:7} |\phi^+ \rangle = \cos\theta|00\rangle + \sin\theta|11\rangle ] \end{equation} with θ=π4+ε,|ε|≪π4\begin{equation} \label{eq:8} \theta = \frac{\pi}{4} + \varepsilon\;,\quad |\varepsilon| \ll \frac{\pi}{4} \end{equation} In the sequel, we use the fact that : cosθ=cos(π4+ε)=1√2(cosε−sinε)sinθ=sin(π4+ε)=1√2(cosε+sinε)\begin{equation} \label{eq:9} \begin{split} \cos\theta &= \cos\biggl(\frac{\pi}{4} + \varepsilon\biggr) = \frac{1}{\sqrt{2}}(\cos\varepsilon - \sin\varepsilon)\\ \sin\theta &= \sin\biggl(\frac{\pi}{4} + \varepsilon\biggr) = \frac{1}{\sqrt{2}}(\cos\varepsilon + \sin\varepsilon) \end{split} \end{equation} |ψ⟩=12|ϕ+⟩⊗(cosε|m⟩−sinε|mz⟩⏟|u+⟩)+12|ϕ−⟩⊗(cosε|mz⟩−sinε|m⟩⏟|u−⟩)+12|ψ+⟩⊗(cosε|mx⟩+sinε|mxz⟩⏟|v+⟩)+12|ψ−⟩⊗(cosε|mxz⟩+sinε|mx⟩⏟|v−⟩)=12|ϕ+⟩⊗|u+⟩+12|ϕ−⟩⊗|u−⟩+12|ψ+⟩⊗|v+⟩+12|ψ−⟩⊗|v−⟩\begin{equation} \label{eq:10} \begin{split} \lvert \psi \rangle &= \frac{1}{2} \lvert \phi^{ + } \rangle\otimes\biggl(\underbrace{\cos\varepsilon\,\lvert m \rangle - \sin\varepsilon\, \lvert m_{z} \rangle}_{\lvert u^{ +} \rangle}\biggr) + \frac{1}{2} \lvert \phi^{ - } \rangle\otimes\biggl(\underbrace{\cos\varepsilon\, \lvert m_{z} \rangle -\sin\varepsilon \,\lvert m \rangle}_{\lvert u^{ -} \rangle}\biggr) \\ &\quad + \frac{1}{2} \lvert \psi^{ + } \rangle\otimes\biggl(\underbrace{\cos\varepsilon\,\lvert m_{x} \rangle +\sin\varepsilon\,\lvert m_{xz} \rangle}_{\lvert v^{ + } \rangle}\biggr) + \frac{1}{2} \lvert \psi^{ - } \rangle\otimes\biggl(\underbrace{\cos\varepsilon\,\lvert m_{xz} \rangle +\sin\varepsilon\, \lvert m_{x} \rangle}_{\lvert v^{ -} \rangle}\biggr)\\ & = \frac{1}{2} \lvert \phi^{ + } \rangle\otimes\lvert u^{ +} \rangle + \frac{1}{2} \lvert \phi^{ - } \rangle\otimes\lvert u^{ -} \rangle + \frac{1}{2} \lvert \psi^{ + } \rangle\otimes\lvert v^{ + } \rangle + \frac{1}{2} \lvert \psi^{ - } \rangle\otimes\lvert v^{ -} \rangle \end{split} \end{equation} One can check that this reduces to (3) for ε=0\varepsilon = 0. The protocol steps remain the same. The outcome (x1,x2)(x_{1},x_{2}) of joint measurement performed is : The state that was finally teleport is : |m′⟩=cosε|m⟩±sinε|mz⟩=α(cosε±sinε)|0⟩+β(cosε∓sinε)|1⟩\begin{equation} \label{eq:18} \lvert m^{\prime} \rangle = \cos\varepsilon \lvert m \rangle \pm \sin\varepsilon \lvert m_{z} \rangle = \alpha\,(\cos\varepsilon \pm \sin\varepsilon)\,\lvert 0 \rangle + \beta\,(\cos\varepsilon \mp \sin\varepsilon)\,\lvert 1 \rangle \end{equation} What is the error ? One can use the norm between state ‖\begin{equation} \label{eq:19} \begin{split} \lVert\lvert m^{\prime} \rangle - \lvert m\rangle \rVert^{2} &= \lVert(\cos\varepsilon -1)\lvert m \rangle \pm \sin\varepsilon \lvert m_{z}\rangle\rVert^{2}\\ & = (\,(\cos\varepsilon -1)\langle m \rvert \pm \sin\varepsilon \langle m_{z} \rvert\,)(\,(\cos\varepsilon -1)\lvert m \rangle \pm \sin\varepsilon \lvert m_{z} \rangle\,)\\ & = (\cos\varepsilon -1)^{2} + \sin^{2}\varepsilon \pm 2(\cos\varepsilon -1)\sin\varepsilon\,\langle m \rvert Z\lvert m \rangle \\ & = 2(1 -\cos\varepsilon) \pm 2(\cos\varepsilon -1)\sin\varepsilon\,\langle m \rvert Z\lvert m \rangle\\ & = 2(1-\cos\varepsilon)\,(1 \mp\sin\varepsilon)\,\langle m \rvert Z\lvert m \rangle\\ & = 2(1-\cos\varepsilon)\,(1 \mp\sin\varepsilon)\,(|\alpha|^{2} - |\beta|^{2}) \end{split}\qquad (19) \end{equation} Finally, for \varepsilon \ll 1\varepsilon \ll 1, the error behaves as : \begin{equation} \label{eq:20} \lVert \lvert m^{\prime} \rangle - \lvert m \rangle\rVert^{2} \approx (\varepsilon^{2})\,(1 \mp\varepsilon)\,(|\alpha|^{2} - |\beta|^{2}) = \mathcal{O}(\varepsilon^{2})\quad\Rightarrow\quad \lVert m^{\prime} \rangle - \lvert m \rangle \rVert = \mathcal{O}(\varepsilon) \end{equation}\begin{equation} \label{eq:20} \lVert \lvert m^{\prime} \rangle - \lvert m \rangle\rVert^{2} \approx (\varepsilon^{2})\,(1 \mp\varepsilon)\,(|\alpha|^{2} - |\beta|^{2}) = \mathcal{O}(\varepsilon^{2})\quad\Rightarrow\quad \lVert m^{\prime} \rangle - \lvert m \rangle \rVert = \mathcal{O}(\varepsilon) \end{equation} The error grows then linearly with the Bell State imperfectness, at least for \varepsilon\ll 1\varepsilon\ll 1. But from (19), one can note that the error depends also on the message itself. Indeed, if the qubit state sent is maximally mixed, i.e. |\alpha| = |\beta||\alpha| = |\beta|, then the error vanishes whatever \varepsilon\varepsilon. This sounds strange to me, is it normal ? and is (19) the best way to assess the error due to imperfect Bell Pair ? Thanks for contributing an answer to Quantum Computing Stack Exchange! Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. This comment attacks a person or group. Learn more in our Abusive behavior policy. This comment is rude or condescending. Learn more in our Code of Conduct. A problem not listed above. Try to be as specific as possible. 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