DBB vs EPR & GHZ: Questions & Answers

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In summary: It is the dimension of the Hilbert space. The authors show that violations of local realism are stronger for two maximally entangled quNits (3 # N # 9) than for two qubits and that they increase with N. Therefore, it can be said that EPR(N+1) is always "stronger" than EPR(N).
  • #1
Dmitry67
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dBB, as a theory with hidden variables, must be non-local.
So there is some magic how 'particles' affect each other with superluminal speed.
I don't care about the details, I know that somehow these rules are adjusted to satisfy Bell.

My questions:
1. is dBB also 'compatible' with GHZ?
2. does it need any extra assumptions to give correct predictions for GHZ?

Then, assuming some generalization - EPR with 2 observers is EPR(2), GHZ with 3 observers is EPR(3) - and I know that EPR(3) is "stronger" than EPR(2):

3. is EPR(N+1) always "stronger" than EPR(N)?
4. is dBB compatible with any EPR(N)?
 
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  • #2
I'm disappointed. Turns out this thread had nothing to do with Electron Paramagnetic Resonance in the Gigahertz range.. :)
 
  • #3
1. Yes
2. No
3. What do you mean by "stronger"?
4. Yes
 
  • #4
3. In a sense that GHZ is "stronger" than EPR
For example:

http://iftia9.univ.gda.pl/~pg/prace/2000PhRvL..85.4418K.pdf

We investigate the general case of two entangled quantum systems defined in N-dimensional Hilbert spaces, or “quNits.” Via a numerical linear optimization method we show that violations of local realism are stronger for two maximally entangled quNits (3 # N # 9) than for two qubits and that they increase with N.

Now when you answer "Yes" to the other questions you mean:
1. Non-relativistic dBB
2. Your version of relativistic dBB?
 
  • #5
Dmitry67 said:
Now when you answer "Yes" to the other questions you mean:
1. Non-relativistic dBB
2. Your version of relativistic dBB?
Both.
 
  • #6
Dmitry67 said:
Then, assuming some generalization - EPR with 2 observers is EPR(2), GHZ with 3 observers is EPR(3) - and I know that EPR(3) is "stronger" than EPR(2):

3. is EPR(N+1) always "stronger" than EPR(N)?
Dmitry67 said:
3. In a sense that GHZ is "stronger" than EPR
For example:

http://iftia9.univ.gda.pl/~pg/prace/2000PhRvL..85.4418K.pdf
In the paper above, N is not the number of observers.
 

1. What is the difference between DBB, EPR, and GHZ?

DBB (Deutsch-Brody-Braun) refers to entangled states of two qubits, while EPR (Einstein-Podolsky-Rosen) and GHZ (Greenberger-Horne-Zeilinger) refer to entangled states of multiple qubits. DBB and EPR are two-qubit entangled states, while GHZ involves three or more qubits.

2. How are DBB, EPR, and GHZ used in quantum computing?

DBB, EPR, and GHZ are all important resources for quantum computing. They can be used for performing certain quantum algorithms, such as quantum teleportation and superdense coding. They can also be used for quantum error correction and for creating larger entangled states.

3. What are the advantages of using DBB, EPR, and GHZ in quantum computing?

The use of entangled states like DBB, EPR, and GHZ allows for the manipulation of multiple qubits simultaneously, which can lead to faster and more efficient quantum computations. It also allows for the implementation of certain quantum protocols that are not possible with classical computing.

4. Are DBB, EPR, and GHZ the only types of entangled states used in quantum computing?

No, there are many other types of entangled states used in quantum computing, such as cluster states, cat states, and squeezed states. Each type of entangled state has its own unique properties and potential applications in quantum computing.

5. Can DBB, EPR, and GHZ states be created and manipulated reliably in quantum systems?

Yes, with advancements in quantum technology, DBB, EPR, and GHZ states can now be created and manipulated with high levels of reliability and precision. However, maintaining and controlling entanglement is still a major challenge in quantum computing and ongoing research is focused on improving and optimizing these processes.

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