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Diffstat (limited to 'source/know/concept/superdense-coding')
| -rw-r--r-- | source/know/concept/superdense-coding/index.md | 14 |
1 files changed, 8 insertions, 6 deletions
diff --git a/source/know/concept/superdense-coding/index.md b/source/know/concept/superdense-coding/index.md index 4338205..0ad8e9e 100644 --- a/source/know/concept/superdense-coding/index.md +++ b/source/know/concept/superdense-coding/index.md @@ -25,16 +25,17 @@ where $$A$$ and $$B$$ are qubits belonging to Alice and Bob, respectively. Based on the values of the two classical bits $$(a_1, a_2)$$, Alice performs the following operations on her side $$A$$ -of the Bell state: +of the Bell state, where $$\hat{\sigma}_x$$ and $$\hat{\sigma}_z$$ +are Pauli matrices (see [quantum gate](/know/concept/quantum-gate/)): | $$(a_1, a_2)$$ | **Operator** | **Result** | | :-: | :-: | :-: | -| $$00$$ | $$\hat{I}$$ | $$\displaystyle \ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big)$$ | -| $$01$$ | $$\hat{\sigma}_z$$ | $$\displaystyle \ket{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B - \Ket{1}_A \Ket{1}_B \Big)$$ | -| $$10$$ | $$\hat{\sigma}_x$$ | $$\displaystyle \ket{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B + \Ket{1}_A \Ket{0}_B \Big)$$ | -| $$11$$ | $$\hat{\sigma}_x \hat{\sigma}_z$$ | $$\displaystyle \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B - \Ket{1}_A \Ket{0}_B \Big)$$ | +| $$00$$ | $$\hat{I}$$ | $$\displaystyle \ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big( \:\:\: \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big)$$ | +| $$01$$ | $$\hat{\sigma}_x$$ | $$\displaystyle \ket{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big( \:\:\: \Ket{1}_A \Ket{0}_B + \Ket{0}_A \Ket{1}_B \Big)$$ | +| $$10$$ | $$\hat{\sigma}_z$$ | $$\displaystyle \ket{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big( \:\:\: \Ket{0}_A \Ket{0}_B - \Ket{1}_A \Ket{1}_B \Big)$$ | +| $$11$$ | $$\hat{\sigma}_x \hat{\sigma}_z$$ | $$\displaystyle \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big( \!-\! \Ket{1}_A \Ket{0}_B + \Ket{0}_A \Ket{1}_B \Big)$$ | -Her actions affect the state on Bob's side $$B$$ due to entanglement. +Her actions indirectly affect the state on Bob's side $$B$$ due to entanglement. Alice then sends her qubit $$A$$ to Bob over the quantum channel, so he has both sides of the entangled pair. @@ -45,6 +46,7 @@ In the end, Alice only sent a single qubit, and the rest of the information transfer was via entanglement. + ## References 1. J.B. Brask, *Quantum information: lecture notes*, |