Superdense coding

When the sender and receiver share a Bell state, two classical bits can be packed into one qubit. See the section below "The protocol" for more details regarding this picture.

In quantum information theory, superdense coding (or dense coding) is a quantum communication protocol to transmit two classical bits of information (i.e., either 00, 01, 10 or 11) from a sender (often called Alice) to a receiver (often called Bob), by sending only one qubit from Alice to Bob, under the assumption of Alice and Bob pre-sharing an entangled state. This protocol was first proposed by Bennett and Wiesner in 1992 and experimentally actualized in 1996 by Mattle, Weinfurter, Kwiat and Zeilinger using entangled photon pairs. By performing one of four quantum gate operations on the (entangled) qubit she possesses, Alice can prearrange the measurement Bob makes. After receiving Alice's qubit, operating on the pair and measuring both, Bob has two classical bits of information. If Alice and Bob do not already share entanglement before the protocol begins, then it is impossible to send two classical bits using 1 qubit, as this would violate Holevo's theorem.

Superdense coding is the underlying principle of secure quantum secret coding. The necessity of having both qubits to decode the information being sent eliminates the risk of eavesdroppers intercepting messages.

It can be thought of as the opposite of quantum teleportation, in which one transfers one qubit from Alice to Bob by communicating two classical bits, as long as Alice and Bob have a pre-shared Bell pair.

Overview

Suppose Alice wants to send two classical bits of information (00, 01, 10, or 11) to Bob using qubits (instead of classical bits). To do this, an entangled state (e.g. a Bell state) is prepared using a Bell circuit or gate by Charlie, a third person. Charlie then sends one of these qubits (in the Bell state) to Alice and the other to Bob. Once Alice obtains her qubit in the entangled state, she applies a certain quantum gate to her qubit depending on which two-bit message (00, 01, 10 or 11) she wants to send to Bob. Her entangled qubit is then sent to Bob who, after applying the appropriate quantum gate and making a measurement, can retrieve the classical two-bit message. Observe that Alice does not need to communicate to Bob which gate to apply in order to obtain the correct classical bits from his projective measurement.

The protocol

The protocol can be split into five different steps: preparation, sharing, encoding, sending, and decoding.

Preparation

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

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

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

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

Sharing

After the preparation of the Bell state $|\Phi ^{+}\rangle$, 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 $|\Phi ^{+}\rangle$ and the rest of the steps in the procedure.

Encoding

By applying a quantum gate to her qubit locally, Alice can transform the entangled state $|\Phi ^{+}\rangle$ into any of the four Bell states (including, of course, $|\Phi ^{+}\rangle$). 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, $\mathbb {I} ={\begin{bmatrix}1&0\\0&1\end{bmatrix}}$, to her qubit, so that it remains unchanged. The resultant entangled state is then

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

In other words, the entangled state shared between Alice and Bob has not changed, i.e. it is still $|\Phi ^{+}\rangle$. The notation $|B_{00}\rangle$ 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, $X={\begin{bmatrix}0&1\\1&0\end{bmatrix}}$, to her qubit, so that the resultant entangled quantum state becomes

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

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

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

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

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

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

Sending

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.

Decoding

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

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 $|00\rangle ,|01\rangle ,|10\rangle$ or $|11\rangle$ (as you can see from the outcomes and the example below).

Example

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

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

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

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

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

Security

Superdense coding is a form of secure quantum communication. If an eavesdropper, commonly called Eve, intercepts Alice's qubit en route to Bob, all that is obtained by Eve is part of an entangled state. Without access to Bob's qubit, Eve is unable to get any information from Alice's qubit. A third party is unable to eavesdrop on information being communicated through superdense coding and an attempt to measure either qubit would collapse the state of that qubit and alert Bob and Alice.

General dense coding scheme

General dense coding schemes can be formulated in the language used to describe quantum channels. Alice and Bob share a maximally entangled state ω. Let the subsystems initially possessed by Alice and Bob be labeled 1 and 2, respectively. To transmit the message x, Alice applies an appropriate channel

$\;\Phi _{x}$

on subsystem 1. On the combined system, this is effected by

$\omega \rightarrow (\Phi _{x}\otimes I)(\omega )$

where I denotes the identity map on subsystem 2. Alice then sends her subsystem to Bob, who performs a measurement on the combined system to recover the message. Let the effects of Bob's measurement be F_y. The probability that Bob's measuring apparatus registers the message y is

$\operatorname {Tr} \;(\Phi _{x}\otimes I)(\omega )\cdot F_{y}.$

Therefore, to achieve the desired transmission, we require that

$\operatorname {Tr} \;(\Phi _{x}\otimes I)(\omega )\cdot F_{y}=\delta _{xy}$

where δ_xy is the Kronecker delta.

Experimental

The protocol of superdense coding has been actualized in several experiments using different systems to varying levels of channel capacity and fidelities. In 2004, trapped beryllium 9 ions were used in a maximally entangled state to achieve a channel capacity of 1.16 with a fidelity of 0.85. In 2017, a channel capacity of 1.665 was achieved with a fidelity of 0.87 through optical fibers. High dimensional ququarts (states formed in photon pairs by non-degenerate spontaneous parametric down-conversion) were used to reach a channel capacity of 2.09 (with a limit of 2.32) with a fidelity of 0.98. Nuclear Magnetic Resonance (NMR) has also been used to share among three parties.

References

• Wilde, Mark M., 2017, Quantum Information Theory, Cambridge University Press, Also available at eprint arXiv:1106.1145

External links

• Qubits, Quantum Mechanics and Computers Course Notes
• Superdense coding: how to send two bits using one qubit on YouTube by Michael Nielsen