5
$\begingroup$

I just came across Bennett's laws and I wonder what the second law mean. It states that 1 qubit "can do the job" of 1 ebit. However, the definition of ebit (entanglement bit, wiki just refers it to the Bell state) and the notion of "can do the job" are unclear. Can you clarify more rigorously what the second Bennett's law says?

$\endgroup$
5
$\begingroup$

Without sacrificing any generality we can define an ebit as a Bell state $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ shared between two parties $A$ and $B$, and since we're concerned with communicating information each party gets to use any many local operations as they want.

Then $$ \text{1 qubit} \geq \text{1 ebit } $$

can be understood as "transmitting a qubit between $A$ and $B$ allows you to share an ebit over $AB$. The protocol is simple: $A$ prepares a bell state and transmits one of the two qubits to $B$. This consumes $\text{1 qubit}_\rightarrow$ and results in $\text{1 ebit}$ shared between the parties, where the subscript indicates that we only considered one-way transmission from $A \rightarrow B$ of the qubit. But since the same is true if we switch roles, we also have $\text{1 qubit}_\leftarrow \geq \text{1 ebit}$ and we can generally drop the arrows and recover law #2. This interpretation also applies to the other laws:

(1) $\text{1 qubit}_\rightarrow \geq \text{1 cbit}_\rightarrow$ means you can transmit one cbit in one direction by the following protocol: $A$ prepares a qubit in either $|0\rangle$ or $|1\rangle$, transmits it to $B$, and then $B$ performs a computational basis measurement.

(3) $\text{1 ebit} + \text{1 qubit}_\rightarrow \geq \text{2 cbits}_\rightarrow$ means shared entanglement plus one-way transmission of a qubit can be used to transmit two cbits in the same direction (superdense coding).

(4) $\text{1 ebit} + \text{2 cbits}_\rightarrow \geq \text{1 qubit}_\rightarrow$ means shared entanglement plus one-way transmission of two cbits can be used to transmit a qubit of information (quantum teleportation).

Note since all of these protocols can be performed in the opposite direction then the arrows indicating $A$ transmitting to $B$ aren't strictly necessary and they can be omitted. But they help clarify that in any of the above protocols information flows in a specific direction; for example it would be very strange to see a statement like $\text{1 qubit}_\rightarrow \geq \text{1 cbit}_\leftarrow$

$\endgroup$
4
  • $\begingroup$ Ok, I am trying to frame it in context of transmitting information via some quantum (or even any) channel. I have problem with the notion of "sharing" ebit (or anybit). More precisely, I wonder if $\text{1 anybit}\geqslant\text{1 anybit}_\rightarrow$ and if $\text{1 anybit}_\leftarrow+\text{1 anybit}_\rightarrow=\text{1 anybit}$. The first one is intuitive, if we share the same $\text{1 anybit}$ and if you send me your $\text{1 anybit}$ then I as well may look at my $\text{1 anybit}$. However, if we swap $\text{anybits}$ do we share $\text{1 anybit}$ or $\text{2 anybits}$? $\endgroup$ – Fallen Apart 2 days ago
  • $\begingroup$ Or does the sharing only have sense in the context of $\text{ebits}$ by entanglement? $\endgroup$ – Fallen Apart 2 days ago
  • $\begingroup$ The sharing only really makes sense in terms of an $\text{ebit}$, because its the only resource that we define as having been split between both systems - a bell state won't let us do quantum teleportation if $A$ holds both of its qubits. $\endgroup$ – forky40 2 days ago
  • $\begingroup$ maybe I don't understand the other statements; however a statement like $\text{1 qubit} \geq \text{1 qubit}_\rightarrow$ doesn't really parse without explicitly saying which direction the qubit on the LHS is transmitted; meanwhile $\text{1 qubit}_\rightarrow + \text{1 qubit}_\leftarrow$ actually requires two uses of a quantum channel so that equality you gave doesn't work. These kinds of statements are made in order to count the uses of a classical or quantum channel are required for a protocol (or composition of protocols) so the number of transmitted (qu)bits is the relevant quantity. $\endgroup$ – forky40 2 days ago
1
$\begingroup$

Similarities ebit and qubit:

An ebit is one unit of bipartite entanglement, the amount of entanglement that is contained in a maximally entangled two-qubit state (Bell state).

Requirement:

If a state is said to have X ebits of entanglement (quantified by some entanglement measure) it has the same amount of entanglement (in that measure) as X Bell states. If a task requires Y ebits, it can be done with Y or more Bell states, but not with fewer.

"Can do the job"

There must also always be at least as many ebits as there are qubits. This is what is meant by "can do the job".

New contributor
DEX7RA is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
2
  • $\begingroup$ "There must also be at least as many qubits as there are ebits". Actually, I would believe that is the case (this in fact motivated me to post the question), but the law actually states the converse. $\endgroup$ – Fallen Apart 2 days ago
  • $\begingroup$ Thanks for the hint, I got it mixed up. Let me correct that. $\endgroup$ – DEX7RA 2 days ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.