Superdense coding

In quantum information theory, superdense 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.^[1]^[2] 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.

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.^[3]

Overview

Suppose Alice (the sender) wanted to send two (classical) bits of information (00, 01, 10 or 11) to Bob (the receiver) using qubits (instead of classical bits). To do that, an entangled state (e.g. a Bell state) is first prepared (e.g. using a Bell circuit or gate), e.g. by Eve, who later sends one of these qubits (in the Bell state) to Alice and the other to Bob. Once Alice obtains her qubit (which is in the entangled state), depending on which two-bit message (00, 01, 10 or 11) she wants to send to Bob, she applies a certain quantum gate to her (entangled) qubit. Her (entangled) qubit is then sent to Bob, who, after a measurement, can retrieve the classical two-bit message.

The protocol

The protocol can be split into different steps (as described below).

Preparation

The protocol starts with the preparation of an entangled state, which is later shared between Alice (the sender) and Bob (the receiver). So, 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 (e.g. using a Bell circuit or gate). Note: we can omit the tensor product symbol $\otimes$ , so that the Bell state above can simply be written as

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

.

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).

A long time might elapse between the preparation and sharing of the entangled state $|\Phi ^{+}\rangle$ and the rest of the steps in the procedure.

Encoding

By only manipulating (using quantum gates) her qubit locally, Alice can transform the (shared) entangled state, $|\Phi ^{+}\rangle$ , into any one of the 4 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 4 cases, which correspond to the 4 possible two-bit strings that Alice may want to send.

1. If Alice wants to send (to Bob) the (classical) two-bit string 00, then she applies the identity quantum gate, $\mathbb {I} ={\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ , to her qubit, so that her qubit (of the shared entangled state $|\Phi ^{+}\rangle$ ) remains unchanged. The resultant (shared) 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 (to Bob) the (classical) two-bit string 01, 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 resulting (shared) entangled quantum state becomes

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

3. If Alice wants to send (to Bob) the (classical) two-bit string 10, then she applies the quantum phase-flip gate $Z={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}$ to her qubit, so the resultant (shared) 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 (to Bob) the (classical) two-bit string 11, then she applies the quantum gate $Z*X$ to her (entangled) qubit, so that the resultant (shared) entangled state is

|\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 (depending on the message she wants to send), Alice can send, using a quantum network, her entangled qubit (of the entangled state shared between her and Bob) to Bob through some conventional physical medium.

Decoding

Now, if Bob wants to find which classical bits Alice wanted to send 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 bit 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 $|\Psi ^{+}\rangle :=B_{01}={\frac {1}{\sqrt {2}}}(|1_{A}0_{B}\rangle +|0_{A}1_{B}\rangle )$ , then a CNOT with A as control bit and B as target bit will change $B_{01}$ to become $B_{01}'={\frac {1}{\sqrt {2}}}(|1_{A}1_{B}\rangle +|0_{A}1_{B}\rangle )$ . 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 (classical) two-bit string 01.

Eavesdropper

If an eavesdropper, which we can call Eve, intercepts Alice's qubit en route to Bob, all that is obtained by Eve is part of an entangled state. Therefore, no useful information whatsoever is gained by Eve, unless she also has access to Bob's qubit.

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.

References

↑ Bennett, C.; Wiesner, S. (1992). "Communication via one- and two-particle operators on Einstein-Podolsky-Rosen states". Physical Review Letters. 69 (20): 2881. doi:10.1103/PhysRevLett.69.2881. PMID 10046665.
↑ Nielsen, Michael A.; Chuang, Isaac L. (9 December 2010). "2.3 Application: superdense coding". Quantum Computation and Quantum Information: 10th Anniversary Edition. Cambridge University Press. p. 97. ISBN 978-1-139-49548-6.
↑ Wilde, Mark (18 April 2013). Quantum Information Theory. Cambridge University Press. p. 181. ISBN 978-1-107-03425-9.

External links

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[bennett1992communication-1] Bennett, C.; Wiesner, S. (1992). "Communication via one- and two-particle operators on Einstein-Podolsky-Rosen states". Physical Review Letters. 69 (20): 2881. doi:10.1103/PhysRevLett.69.2881. PMID 10046665.

[NielsenChuang2010-2] Nielsen, Michael A.; Chuang, Isaac L. (9 December 2010). "2.3 Application: superdense coding". Quantum Computation and Quantum Information: 10th Anniversary Edition. Cambridge University Press. p. 97. ISBN 978-1-139-49548-6.

[Wilde2013-3] Wilde, Mark (18 April 2013). Quantum Information Theory. Cambridge University Press. p. 181. ISBN 978-1-107-03425-9.