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Diffstat (limited to 'source/know/concept/bell-state')
| -rw-r--r-- | source/know/concept/bell-state/index.md | 41 |
1 files changed, 21 insertions, 20 deletions
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md index f454264..fa289de 100644 --- a/source/know/concept/bell-state/index.md +++ b/source/know/concept/bell-state/index.md @@ -16,16 +16,16 @@ $$\begin{aligned} \boxed{ \begin{aligned} \ket{\Phi^{\pm}} - &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big) + &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{0}_B \pm \ket{1}_A \ket{1}_B \Big) \\ \ket{\Psi^{\pm}} - &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big) + &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{1}_B \pm \ket{1}_A \ket{0}_B \Big) \end{aligned} } \end{aligned}$$ -Where e.g. $$\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$$ -is the tensor product of qubit $$A$$ in state $$\Ket{0}$$ and $$B$$ in $$\Ket{1}$$. +Where e.g. $$\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$$ +is the tensor product of qubit $$A$$ in state $$\ket{0}$$ and $$B$$ in $$\ket{1}$$. These states form an orthonormal basis for the two-qubit [Hilbert space](/know/concept/hilbert-space/). @@ -37,7 +37,7 @@ Consider the following pure [density operator](/know/concept/density-operator/): $$\begin{aligned} \hat{\rho} = \ket{\Phi^{+}} \bra{\Phi^{+}} - &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big) + &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big) \end{aligned}$$ The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows: @@ -45,12 +45,12 @@ The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated $$\begin{aligned} \hat{\rho}_A &= \Tr_B(\hat{\rho}) - = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B + = \sum_{b = 0, 1} \bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \ket{b}_B \\ - &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big) - \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big) + &= \sum_{b = 0, 1} \Big( \ket{0}_A \inprod{b}{0}_B + \ket{1}_A \inprod{b}{1}_B \Big) + \Big( \bra{0}_A \inprod{0}{b}_B + \bra{1}_A \inprod{1}{b}_B \Big) \\ - &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big) + &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big) = \frac{1}{2} \hat{I} \end{aligned}$$ @@ -59,35 +59,36 @@ The same holds for the other three Bell states, and is equally true for qubit $$B$$. This means that a measurement of qubit $$A$$ -has a 50-50 chance to yield $$\Ket{0}$$ or $$\Ket{1}$$. +has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$. However, due to the entanglement, measuring $$A$$ also has consequences for qubit $$B$$: $$\begin{aligned} - \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2 + \big| \bra{0}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{0}{0}_B + \inprod{0}{1}_A \inprod{0}{1}_B \Big)^2 = \frac{1}{2} \\ - \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2 + \big| \bra{0}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{1}{0}_B + \inprod{0}{1}_A \inprod{1}{1}_B \Big)^2 = 0 \\ - \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2 + \big| \bra{1}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{0}{0}_B + \inprod{1}{1}_A \inprod{0}{1}_B \Big)^2 = 0 \\ - \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2 + \big| \bra{1}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{1}{0}_B + \inprod{1}{1}_A \inprod{1}{1}_B \Big)^2 = \frac{1}{2} \end{aligned}$$ -As an example, if $$A$$ collapses into $$\Ket{0}$$ due to a measurement, -then $$B$$ instantly also collapses into $$\Ket{0}$$, never $$\Ket{1}$$, +As an example, if $$A$$ collapses into $$\ket{0}$$ due to a measurement, +then $$B$$ instantly also collapses into $$\ket{0}$$, never $$\ket{1}$$, even if it was not measured. This was a specific example for $$\ket{\Phi^{+}}$$, but analogous results can be found for the other Bell states. + ## References 1. J.B. Brask, *Quantum information: lecture notes*, |