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What is the implementation of quantum superdense coding protocol if the sender and receiver share the entangled state $ \frac{1}{\sqrt{2}} (\left| 01 \right> - \left| 10\right>) $?

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If you apply $\mathrm{X}$ on second qubit, you will get state $$ \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle) $$

After that you can apply $\mathrm{Z}$ on second qubit as well to get Bell state $\beta_{00}$: $$ \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) $$

Now you can employ superdense conding as usual.

Note: matrix description of above mentioned operations is $(\mathrm{I}\otimes\mathrm{Z})(\mathrm{I}\otimes\mathrm{X})$.

Here is a circuit for doing all mentioned above:

Circuit

A first part prepares state $|\psi_0\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$. This is an initial state you questioned about. Second part changes $|\psi_0\rangle$ to $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, i.e. Bell state $\beta_{00}$. Now you can apply superdense coding as usual - this is symbolized by gates $\mathrm{X}$ and $\mathrm{Z}$ (of course, application of these gates depends on two bits you want to encode, e.g. for encoding string 00 you will apply neither gate). Last part is measuring in Bell basis - again usual part of superdense coding.

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  • $\begingroup$ your answer in which case ? I need transmit bit in cases 00 ,01,10,11 and which gate I will apply for each one. @Martin Vesely $\endgroup$ – Ba. Taj Mar 4 '20 at 13:43
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    $\begingroup$ @Ba.Taj: My construction changes your state to $\beta_{00}=\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. After that you can apply $X$ and $Z$ gates to do coding as you are used to from "normal superdense coding". $\endgroup$ – Martin Vesely Mar 4 '20 at 13:45
  • $\begingroup$ could you please draw circuit for implementation of the above given state ,and I would appreciate if you could elaborate it extensively in advance thank you, @Martin Vesely $\endgroup$ – Ba. Taj Mar 4 '20 at 14:12
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    $\begingroup$ @Ba.Taj: See edited question. $\endgroup$ – Martin Vesely Mar 4 '20 at 14:48

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