# What is the implementation of the state in superdense coding?

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

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:

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.

• 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 – Ba. Taj Mar 4 '20 at 13:43
• @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". – Martin Vesely Mar 4 '20 at 13:45
• 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 – Ba. Taj Mar 4 '20 at 14:12
• @Ba.Taj: See edited question. – Martin Vesely Mar 4 '20 at 14:48