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8:00 PM
It's (very) early days yet, so we just haven't got to the point where they're properly useful - there are scientists at D-Wave doing actual science (completely unrelated to quantum computing) but we're a fair bit away from reaching the point where a quantum computer is unquestionably faster than any classical computer (known as quantum computational supremacy, although many people don't like that name)
 
glS
@Blue indeed, in the HHL paper they use the notation $|\tilde{\lambda}_k\rangle$, to express the fact that the output of the QPE gives you an approximation to the (state corresponding to the) true eigenvalue. I think you have to look at the error analysis to see the scaling in terms of number of qubits for a given target precision
 
@Mithrandir24601 oh the people here are doing research only ?
 
@Blue Wait, did someone mention non-Hermitian?
 
Anonymous
@Mithrandir24601 They did mention a work-around for non-Hermitian matrices
 
@Blue Where?
 
Anonymous
8:04 PM
 
Anonymous
Page 2
 
Anonymous
Equation (1)
 
Anonymous
In the Algorithm part
 
Anonymous
@glS Checking it
 
@O.Rares At the companies, practically everything is research - there are people making programming languages but that's also pretty much research. If you wanted to create an app, that would pretty much be research as well because no-one's (to my knowledge) done so before, so you'd probably need to come up with new ways/algorithms for things to work well. On this site, we do have researchers, but not everyone is a researcher
@Blue Thanks!! :D
Oh yeah, I've seen that before :P
:(
I got super excited there for a moment
 
Anonymous
8:07 PM
Just a trick XD
 
@Blue Yep, although a rather neat one
There was a preprint put on arXiv a week or so ago about making a quantum PT-symmetric photonic system would require adding additional thermal noise, which I was a bit sad about though
 
@Mithrandir24601 thanks ,I think that I should wait a while in order to study this in a way that can be a safe bet for me
Good look everyone
 
@O.Rares It depends on what you find interesting. Quantum cryptography is something that companies are now beginning to sell...
 
Anonymous
@O.Rares The main reason why theorists study quantum computing and quantum information isn't to make a quantum computer though. You can learn a lot of auxiliary stuff from CS and math too.
 
@Blue Also this: researchers research because it's interesting
 
Anonymous
8:12 PM
It really isn't like if you get a PhD in QIT you can't go ahead and become a pure mathematician or even a software developer
 
Anonymous
:P
 
glS
@Mithrandir24601 are you looking for ways to implement/simulate non-hermitian matrices with photonics?
 
Anonymous
@Mithrandir24601 True, true :)
 
Anonymous
I mainly love this subject because of the interdisciplinary nature (which I've mentioned several times already I guess :P)
 
what if the quantum computers will fail ? We are close to 6 nm at processors
 
8:13 PM
Then we'll do something else?
 
@glS Well, I'd be looking for better ways to do so...
 
so people know what to expect
 
Anonymous
@O.Rares Don't underestimate the amount of progress we make from failures.
 
If it's successful, people will keep doing it. If it's not, people will do something else.
 
@Blue I will remember this quote in future
 
8:15 PM
@O.Rares There's an interesting thing here in that people researching spintronics could (are!) finding applications for classical computing due to the small size of processors
 
glS
@Blue Scott Aaronson often says that if it turned out that quantum computation cannot do better than classical computation, than that would be even more interesting than if it can
 
Anonymous
@glS Nielsen and Chuang, page: 224; Thus to successfully obtain $\varphi$ accurate to $n$-bits with probability of success at least $1-\epsilon$ we choose $$t=n+\lceil{\log(2+\frac{1}{2\epsilon})\rceil}$$
 
Anonymous
$t$ is the number of qubits register which finally contains $|\tilde\varphi\rangle$
 
Anonymous
But then I'm a bit confused how many qubits we'd need to represent $(\sum_{i=1}^{N}\beta_i|u_i\rangle |\lambda_i\rangle)\otimes |0\rangle_{\text{ancilla}}$
 
Anonymous
to say $3$-bits of accuracy
 
Anonymous
8:24 PM
@glS That's very true! Would be a great news for the mathematicians, not so much for the experimentalists :P
 
@Mithrandir24601 interesting that Intel and AMD don't complain about patent theft
 
glS
@Blue a naive guess would be $\sim\log_2 N$ for the eigenvector register, plus that number you gave from N&C, plus 1 for the ancillae. The $\epsilon$ has to be chosen beforehand and determines the precision of the result
 
Anonymous
@glS The eigenvalue register would probably need $(3+\lceil{\log(2)\rceil})\times \log_2(N)$ qubits if we want 3-bits of accuracy, amirite?
 
Anonymous
No no
 
Anonymous
There's that $1/\epsilon$ factor
 
Anonymous
8:35 PM
Say I want 0.90 probability
 
Anonymous
So that would need $(3+\lceil{\log(7)\rceil})\times \log_2(N)$ qubits
 
Anonymous
Where $A$ is a $N\times N$ Hermitian matrix
 
Anonymous
So basically, $(6)\log_2(N)$ for eigenvalue register. $\log_2(N)$ for eigenvector register and 1 for ancilla
 
Anonymous
That's quite a lot...
 
I bounced this story (well, a story like this) off of the MSE chat and wanted to get a reaction from here as well
Every week, two people (Alice and Bob) each pick one of three different shows (1,2,3) to watch on TV. They have similar enough tastes that: (1) if they watch the same show, then they'll both either dislike the show (+1) or both dislike the show (-1); if they were to swap which shows they watch, they'd get the same outcomes.
 
glS
8:47 PM
@Mithrandir24601 as in you're already doing it/have done it somehow? (I think I've already asked this question but don't remember the answer)
 
(So if A likes show 1 and B dislikes show 2, then B would have liked show 1 and A would have disliked show 2.)
In addition, none of the three shows are particularly good and so they're just as likely as not to enjoy a given show
 
glS
@Blue that looks correct. So $7\log_2 N+1$. For $N=4$ you get $15$. Just what you need to improve on the state of the art of $N=2$ right?
 
From the above, the conditional probabilities must satisfy:
P(++|11) = P(++|22) = P(++|33) = P(--|11) = P(--|22) = P(--|33) = 1/2,
P(+-|11) = P(+-|22) = P(+-|33) = P(-+|11) = P(-+|22) = P(-+|33) = 0,
P(++|12)  = 1/2 - P(-+|12) = 1/2 - P(-+|12)  = P(--|12),
 
Anonymous
 
Anonymous
@glS Yes, but it's still nowhere close to what they did. They seem to have simulated it for a $200\times 200$ sized matrix $A$
 
Anonymous
8:53 PM
From the tone in that paper it seems they actually did some simulation with the H1N1 data
 
Anonymous
$d=100, M=8$
 
Anonymous
So $W$ is of dimension $d\times d$
 
Anonymous
And $A$ has $W$ as one of its blocks
 
Anonymous
So it is $2d\times 2d$
 
bah, last line was incomplete
P(++|12)  = 1/2 - P(-+|12) = 1/2 - P(+-|12)  = P(--|12) = P(++|21) = ...,
P(++|13)  = 1/2 - P(-+|13) = 1/2 - P(+-|13)  = P(--|13) = P(++|31) = ...,
P(++|23)  = 1/2 - P(-+|23) = 1/2 - P(+-|23)  = P(--|23) = P(++|32) = ...,
With the upshot being that this scenario is dictated entirely by the probabilities P(++|12), P(++|13), P(++|23)
 
Anonymous
8:57 PM
@glS For their H1N1 simulation we'd need a minimum of 57 qubits it seems
 
The question: Given a particular choice of those three numbers, does this scenario make sense?
As an example, suppose P(++|12) = P(++|13) = P(++|23)=p. How small can p be?
(The surprise is that this is basically a Bell inequality in disguise.)
 
@glS How to answer this? In a fashion, yes
 
glS
@Mithrandir24601 as in, it's not published yet? =)
 
@glS You would be right on that :P
 
glS
I understand, no worries!
 
9:06 PM
It is on my poster, but that poster still hasn't been outside of Bristol yet
Oh! My secondary supervisor might have mentioned it at the PT-symmetry conference
"I will present experimental results for multi-particle correlations and quantum information across the PT transition, obtained by using a quantum photonic chip" :) Yay! It's now been at a conference
(although most/the vast majority of it wasn't my work, to be fair)
 
(If anyone is curious, I'll explain my above rambling. But I'm not sure how interesting I made it soooo yeah)
 
glS
@Mithrandir24601 haha congratulations. The abstract looks interesting
@Semiclassical sorry, that looks interesting but I don't have time atm to read it carefully (though it looks a lot like the stuff you get in quantum contextuality). Will read better later
 
mmkay
 
glS
@Blue that still sounds like a lot. I don't think they can "easily" simulate that many qubits without using tricks
 
There's two punchlines, one of which I think I fully comprehend and one which I'm not sure I do.
So I'm not without an ulterior motive.
 
Anonymous
9:43 PM
@glS True. I'm thinking of mailing the corresponding author tomorrow morning. It's a bit annoying when people don't explain the experimental procedure clearly
 
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