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Photonics for Quantum 2 - Shared screen with speaker view
Paul Kwiat
52:22
Hi Jeff — is the essential reason the OPA helps is that it allow you to precise select out only the exact mode of the reflected entangled photon, thereby excluding more noise?
liu han
53:43
Hi Prof Kwiat,
liu han
55:01
To my understanding, classical homodyne measurement is mode selective too. I believe with entanglement one can do better than selecting a single mode, in a way that is similar to superdense coding
Jim Gable
01:08:28
Thank you!
Mike Garrett
01:08:51
Is it possible to overcome the limited time-bandwidth product problem at microwave frequencies by exploiting other degrees of freedom such as orbital angular momentum states?
Poolad Imany
01:10:24
Thank you for the nice talk! Have you considered multiphoton entangled states and their effects on quantum illumination?
Andrew Weiner
01:10:45
Thanks Jeff for a great talk! What about comparison with classical radars that can violate the low brightness requirement? That is, can one envision applications where you can't win with classical techniques where we just turn up the transmitter power?
Paul Kwiat
01:11:46
Better?
liu han
01:12:28
Thank you for the wonderful talk. Do you think it is possible to implement QI with time-frequency entanglement between two photons?
shihan sajeed
01:12:49
May be I didn’t understand it correctly. The two papers on microwave quantum illumination, that claims to use QUANTUM two-mode squeezed radar: you say they are actually Classical as the amplifiers break the entanglement. But an actual quantum entanglement radar going through the entanglement breaking channel also suffers entanglement breaking, isn’t it? So, why the two microwave illumination experiments are classical?
Don Figer
01:14:46
Bow tie looks excellent! Iridescent.
shihan sajeed
01:20:18
thanks for the answer. I understand now.
Paul Kwiat
01:20:21
Let’s say the problem is different — I want to identify a particular shape. I could imagine that spatially structured light would offer an advantage; do I get the same factor of 4 advantage using spatial entanglement, or something else?
Peter McMahon
01:22:25
What is the intuition for why if you start with an entangled state, have enough loss that you "break" the entanglement, that you can still see an advantage from the source producing entangled light to begin with?
liu han
01:24:21
Is this related to superdense coding?
Paul Kwiat
01:25:34
Thanks Jeff!
Mike Garrett
01:25:51
Thank you!
Paul Kwiat
01:25:59
Batman!
Peter McMahon
01:26:22
Thanks, Jeff- that was great!