Superdense coding: Difference between revisions

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==== Example ====
 
For example, if the resultant entangled state (after the operations performed by Alice) was <math>B_{01} = \fractfrac{1}{\sqrt{2}}(|1_A0_B\rangle + |0_A1_B\rangle)</math>, then a CNOT with A as control bit and B as target bit will change <math>B_{01}</math> to become <math>B_{01}' = \fractfrac{1}{\sqrt{2}}(|1_A1_B\rangle + |0_A1_B\rangle)</math>. Now, the Hadamard gate is applied only to A, to obtain
 
: <math>
<math>B_{01}'' = \frac{1}{\sqrt{2}} \left({\left(\frac{1}{\sqrt{2}}(|0 \rangle - |1 \rangle) \right) }_A \otimes
|1_B\rangle B_{01}'' += {\left(\fractfrac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle) \right) }_A \otimesleft[
{\left(\fractfrac{1}{\sqrt{2}}(|0 \rangle +- |1 \rangle)\right)}_A \otimes |11_B\rangle +
|1_B\rangle\right)</math>
<math>B_{01}'' = \frac{1}{\sqrt{2}} \left({\left(\fractfrac{1}{\sqrt{2}}(|0 \rangle -+ |1 \rangle) \right) }_A \otimes |1_B\rangle
\right)].
</math>
 
For simplicity, subscripts may be removed:
: <math>B_{01}'' =
\begin{align}
\frac{1}{\sqrt{2}}
B_{01}'' &=
\left(
\fractfrac{1}{\sqrt{2}}(|0 \rangle - |1 \rangle) \otimes |1\rangle +
\left(
\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle) \otimes |1\rangle
\fractfrac{1}{\sqrt{2}}(|01 0\rangle +- |11 1\rangle) \otimes |1\rangle +
\right)
\tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |1\rangle
=
\right) \\
\frac{1}{\sqrt{2}}
&=
\left(
\fractfrac{1}{\sqrt{2}}(|01 \rangle - |11 \rangle) +
\left(
\frac{1}{\sqrt{2}}(|01 \rangle + |11 \rangle)
\fractfrac{1}{\sqrt{2}}(|01 \rangle - \frac{1}{2} |11 \rangle) +
\right)
\tfrac{1}{\sqrt{2}}(|01 \rangle + |11\rangle)
=
\right) =
\frac{1}{2}|01 \rangle - \frac{1}{2} |11 \rangle +
\fractfrac{1}{2}|01 \rangle +- \fractfrac{1}{2}|11 \rangle +
= \tfrac{1}{2}|01\rangle + \tfrac{1}{2}|11\rangle
= |01\rangle.
\end{align}
</math>