Superdense coding: Difference between revisions
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[[File:Superdense coding.png|right|frame|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.]] |
[[File:Superdense coding.png|right|frame|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.]] |
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In [[quantum information theory]], '''superdense coding''' is |
In [[quantum information theory]], '''superdense coding''' is a protocol that transmits 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, as long as Alice and Bob already hold one pre-shared [[Bell pair]] (a specific two-qubit [[quantum entanglement|entangled]] state).<ref name="bennett1992communication">{{cite journal |last1=Bennett |first1=C. |last2=Wiesner |first2=S. |year=1992 |title=Communication via one- and two-particle operators on Einstein-Podolsky-Rosen states |journal=Physical Review Letters |doi=10.1103/PhysRevLett.69.2881 |volume=69 |issue=20 |pages=2881 |pmid=10046665}}</ref><ref name="NielsenChuang2010">{{cite book |last1=Nielsen |first1=Michael A. |last2=Chuang |first2=Isaac L. |date=9 December 2010 |title=Quantum Computation and Quantum Information: 10th Anniversary Edition |publisher=Cambridge University Press |isbn=978-1-139-49548-6 |section=2.3 Application: superdense coding |page=97 |url=https://books.google.com/books?id=-s4DEy7o-a0C}}</ref> 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]]. |
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It can be thought of as the |
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.<ref name="Wilde2013">{{cite book |last=Wilde |first=Mark |date=18 April 2013 |title=Quantum Information Theory |publisher=Cambridge University Press |isbn=978-1-107-03425-9 |page=181 |url=https://books.google.com/books?id=T36v2Sp7DnIC&pg=PA181}}</ref> |
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Both superdense coding and quantum teleportation make use of pre-shared entangled state (e.g. a [[Bell state|Bell state]]) between the sender and the receiver. |
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== Overview == |
== Overview == |
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In quantum information theory, superdense coding is a protocol that transmits 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, as long as Alice and Bob already hold one pre-shared Bell pair (a specific two-qubit 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
where denotes the tensor product, is prepared (e.g. using a Bell circuit or gate). Note: we can omit the tensor product symbol , so that the Bell state above can simply be written as
- .
Sharing
After the preparation of the Bell state , 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 can be very distant from each other).
The preparation and sharing of the entangled state between Alice and Bob, , can happen a lot before Alice wants to send any classical two-bit string to Bob.
Encoding
By only manipulating (using quantum gates) her qubit locally, Alice can transform the (shared) entangled state, , into any one of the 4 Bell states (including, of course, ). 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, , to her qubit, so that her qubit (of the shared entangled state ) remains unchanged. The resultant (shared) entangled state is then
In other words, the entangled state shared between Alice and Bob has not changed, i.e. it is still . The notation 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, , to her qubit, so that the resulting (shared) entangled quantum state becomes
3. If Alice wants to send (to Bob) the (classical) two-bit string 10, then she applies the quantum phase-flip gate to her qubit, so the resultant (shared) entangled state becomes
4. If, instead, Alice wants to send (to Bob) the (classical) two-bit string 11, then she applies the quantum gate to her (entangled) qubit, so that the resultant (shared) entangled state is
The matrices and are two of the Pauli matrices. The quantum states , , and (or, respectively, and ) 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[how?] her entangled qubit (of the entangled state shared between her and Bob) to Bob.
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 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 then after the application of the above unitary operations the entangled state will become
- If the resultant entangled state was then after the application of the above unitary operations the entangled state will become
- If the resultant entangled state was then after the application of the above unitary operations the entangled state will become
- If the resultant entangled state was then after the application of the above unitary operations the entangled state will become
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 or (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 , then a CNOT with A as control bit and B as target bit will change to become . Now, the Hadamard gate is applied only to A, to obtain
For simplicity, let's get rid of the subscripts, so we have
Now, Bob has the basis state , 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 has also 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
on subsystem 1. On the combined system, this is effected by
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 Fy. The probability that Bob's measuring apparatus registers the message y is
Therefore, to achieve the desired transmission, we require that
where δxy 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.