Superdense coding

The protocol starts with the preparation of an entangled state, which is later shared between Alice and Bob. Suppose the following Bell state

${\displaystyle |\Phi ^{+}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |0\rangle _{B}+|1\rangle _{A}\otimes |1\rangle _{B})}$

where ${\displaystyle \otimes }$ denotes the tensor product, is prepared. Note: we can omit the tensor product symbol ${\displaystyle \otimes }$ and write the Bell state as

${\displaystyle |\Phi ^{+}\rangle ={\frac {1}{\sqrt {2}}}(|0_{A}0_{B}\rangle +|1_{A}1_{B}\rangle )}$.

After the preparation of the Bell state ${\displaystyle |\Phi ^{+}\rangle }$, the qubit denoted by subscript A is sent to Alice and the qubit denoted by subscript B is sent to Bob (note: this is the reason these states have subscripts). At this point, Alice and Bob may be in completely different locations (which might be very distant from each other).

There may be a long period of time between the preparation and sharing of the entangled state ${\displaystyle |\Phi ^{+}\rangle }$ and the rest of the steps in the procedure.

By applying a quantum gate to her qubit locally, Alice can transform the entangled state ${\displaystyle |\Phi ^{+}\rangle }$ into any of the four Bell states (including, of course, ${\displaystyle |\Phi ^{+}\rangle }$). Note that this process cannot "break" the entanglement between the two qubits.

Let's now describe which operations Alice needs to perform on her entangled qubit, depending on which classical two-bit message she wants to send to Bob. We'll later see why these specific operations are performed. There are four cases, which correspond to the four possible two-bit strings that Alice may want to send.

1. If Alice wants to send the classical two-bit string 00 to Bob, then she applies the identity quantum gate, ${\displaystyle \mathbb {I} ={\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$, to her qubit, so that it remains unchanged. The resultant entangled state is then

${\displaystyle |\Phi ^{+}\rangle$ :=|B_{00}\rangle ={\frac {1}{\sqrt {2}}}(|0_{A}0_{B}\rangle +|1_{A}1_{B}\rangle )}

In other words, the entangled state shared between Alice and Bob has not changed, i.e. it is still ${\displaystyle |\Phi ^{+}\rangle }$. The notation ${\displaystyle |B_{00}\rangle }$ is also used to remind us of the fact that Alice wants to send the two-bit string 00.

2. If Alice wants to send the classical two-bit string 01 to Bob, then she applies the quantum NOT (or bit-flip) gate, ${\displaystyle X={\begin{bmatrix}0&1\\1&0\end{bmatrix}}}$, to her qubit, so that the resultant entangled quantum state becomes

${\displaystyle |\Psi ^{+}\rangle$ :=|B_{01}\rangle ={\frac {1}{\sqrt {2}}}(|0_{A}1_{B}\rangle +|1_{A}0_{B}\rangle )}

3. If Alice wants to send the classical two-bit string 10 to Bob, then she applies the quantum phase-flip gate ${\displaystyle Z={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}}$ to her qubit, so the resultant entangled state becomes

${\displaystyle |\Phi ^{-}\rangle$ :=|B_{10}\rangle ={\frac {1}{\sqrt {2}}}(|0_{A}0_{B}\rangle -|1_{A}1_{B}\rangle )}

4. If, instead, Alice wants to send the classical two-bit string 11 to Bob, then she applies the quantum gate ${\displaystyle Z*X}$ to her qubit, so that the resultant entangled state becomes

${\displaystyle |\Psi ^{-}\rangle$ :=|B_{11}\rangle ={\frac {1}{\sqrt {2}}}(|0_{A}1_{B}\rangle -|1_{A}0_{B}\rangle )}

The matrices ${\displaystyle X}$ and ${\displaystyle Z}$ are two of the Pauli matrices. The quantum states ${\displaystyle |\Phi ^{+}\rangle }$, ${\displaystyle |\Psi ^{+}\rangle }$, ${\displaystyle |\Phi ^{-}\rangle }$ and ${\displaystyle |\Psi ^{-}\rangle }$ (or, respectively, ${\displaystyle B_{00},B_{01},B_{10}}$ and ${\displaystyle B_{11}}$) are the Bell states.

After having performed one of the operations described above, Alice can send her entangled qubit to Bob using a quantum network through some conventional physical medium.

In order for Bob to find out which classical bits Alice sent he will perform the CNOT unitary operation, with A as control qubit and B as target qubit. Then, he will perform ${\displaystyle H\otimes I}$ unitary operation on the entangled qubit A. In other words, the Hadamard quantum gate H is only applied to A (see the figure above).

• If the resultant entangled state was ${\displaystyle B_{00}}$ then after the application of the above unitary operations the entangled state will become ${\displaystyle |00\rangle }$
• If the resultant entangled state was ${\displaystyle B_{01}}$ then after the application of the above unitary operations the entangled state will become ${\displaystyle |01\rangle }$
• If the resultant entangled state was ${\displaystyle B_{10}}$ then after the application of the above unitary operations the entangled state will become ${\displaystyle |10\rangle }$
• If the resultant entangled state was ${\displaystyle B_{11}}$ then after the application of the above unitary operations the entangled state will become ${\displaystyle |11\rangle }$

These operations performed by Bob can be seen as a measurement which projects the entangled state onto one of the four two-qubit basis vectors ${\displaystyle |00\rangle ,|01\rangle ,|10\rangle }$ or ${\displaystyle |11\rangle }$ (as you can see from the outcomes and the example below).

For example, if the resultant entangled state (after the operations performed by Alice) was ${\displaystyle |\Psi ^{+}\rangle$ :=B_{01}={\frac {1}{\sqrt {2}}}(|0_{A}1_{B}\rangle +|1_{A}0_{B}\rangle )} , then a CNOT with A as control bit and B as target bit will change ${\displaystyle B_{01}}$ to become ${\displaystyle B_{01}'={\frac {1}{\sqrt {2}}}(|0_{A}1_{B}\rangle +|1_{A}1_{B}\rangle )}$. Now, the Hadamard gate is applied only to A, to obtain

${\displaystyle B_{01}''={\frac {1}{\sqrt {2}}}\left({\left({\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )\right)}_{A}\otimes |1_{B}\rangle +{\left({\frac {1}{\sqrt {2}}}(|0\rangle -|1\rangle )\right)}_{A}\otimes |1_{B}\rangle \right)}$

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

${\displaystyle B_{01}''={\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )\otimes |1\rangle +{\frac {1}{\sqrt {2}}}(|0\rangle -|1\rangle )\otimes |1\rangle \right)={\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {2}}}(|01\rangle +|11\rangle )+{\frac {1}{\sqrt {2}}}(|01\rangle -|11\rangle )\right)={\frac {1}{2}}|01\rangle +{\frac {1}{2}}|11\rangle +{\frac {1}{2}}|01\rangle -{\frac {1}{2}}|11\rangle =|01\rangle }$

Now, Bob has the basis state ${\displaystyle |01\rangle }$, so he knows that Alice wanted to send the two-bit string 01.