Superdense coding: Difference between revisions

Revision as of 11:54, 23 February 2014

Revision as of 07:21, 22 April 2014

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* Suppose, Alice wants to sent the classical bits 00

Then she will perform Identity unitary operation on her particle. Obviously, her entangled qbitqubit remains unchanged. The resultant tangled qbitqubit would be <math>|B_{00}\rangle = \frac{1}{\sqrt{2}}(|0_A0_B\rangle + |1_A1_B\rangle)</math>

* Suppose, Alice wants to sent 01 bits. Then she will perform <math>X</math> unitary operation.

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</math> are called Bell states.

Now, if Bob wants to find which classical bits did Alice wants to send he will perform the <math>CNOT</math> unitary operation followed by <math>H\otimes I</math>unitary operation on the entangled qbitqubit.

* If the resultant entangled qbitqubit was <math>B_{00}</math> then after the application of the above unitary operations the entangled qbitqubit will become <math>|00\rangle</math>

* If the resultant entangled qbitqubit was <math>B_{01}</math> then after the application of the above unitary operations the entangled qbitqubit will become <math>|01\rangle</math>

* If the resultant entangled qbitqubit was <math>B_{10}</math> then after the application of the above unitary operations the entangled qbitqubit will become <math>|10\rangle</math>

* If the resultant entangled qbitqubit was <math>B_{11}</math> then after the application of the above unitary operations the entangled qbitqubit will become <math>|11\rangle</math>

So, depending on the message she would like to send, Alice performs one of the four local operations given above and sends her qubit to Bob. By performing a projective measurement in the Bell basis on the two particle system, Bob decodes the desired message.