Superdense coding: Difference between revisions

Revision as of 15:41, 2 September 2021

Revision as of 15:49, 2 September 2021

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==== Example ====

For example, if the resultant entangled state (after the operations performed by Alice) was <math>|\Psi ^{+}\rangle := B_{0110} = \frac{1}{\sqrt{2}}(|0_A1_B1_A0_B\rangle + |1_A0_B0_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_{0110}' = \frac{1}{\sqrt{2}}(|0_A1_B1_A1_B\rangle + |1_A1_B0_A1_B\rangle)</math>. Now, the Hadamard gate is applied only to A, to obtain

<math>B_{0110}'' = \frac{1}{\sqrt{2}} \left({\left(\frac{1}{\sqrt{2}}(|0 \rangle +- |1 \rangle) \right) }_A \otimes

|1_B\rangle + {\left(\frac{1}{\sqrt{2}}(|0 \rangle -+ |1 \rangle) \right) }_A \otimes

|1_B\rangle\right)</math>

For simplicity, let's get rid of the subscripts, so we have

<math>B_{0110}'' =

\frac{1}{\sqrt{2}}

\left(

\frac{1}{\sqrt{2}}(|0 \rangle +- |1 \rangle) \otimes |1\rangle +

\frac{1}{\sqrt{2}}(|0 \rangle -+ |1 \rangle) \otimes |1\rangle

\right)

=

\frac{1}{\sqrt{2}}

\left(

\frac{1}{\sqrt{2}}(|01 \rangle +- |11 \rangle) +

\frac{1}{\sqrt{2}}(|01 \rangle -+ |11 \rangle)

\right)

=

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</math>

Now, Bob has the basis state <math>|01\rangle</math>, so he knows that Alice wanted to send the two-bit string 01.

== Security ==