The Classical Channel

@Blue Yes actually, although I'll need to make breakfast

Anonymous

@Mithrandir24601 No worries. If you get some time just see if you answer this one

Anonymous

cc @glS ^

Anonymous

I'm not completely sure if any quantum subroutine would be able to offer a speedup in that matrix multiplication and addition part...but worth a try I suppose

Anonymous

When there are lots of training samples the generation of W gets difficult and time consuming

Anonymous

6:26 AM

The time complexity is sort of around $\mathcal{O}(N^2)$ in the current situation if I'm not wrong, considering d,M ~ N

@Blue I'll have a think. d is the number of dimensions, not qubits, so this is $M\times$logarithmic in the number of qubits, which is a bit better than it initially looks like

Anonymous

@Mithrandir24601 Yeah, but they aren't using qubits as far as generating W is concerned

Anonymous

(If I haven't missed anything)

Anonymous

They're doing plain and simple matrix multiplication of a row and column vector and summing it over all the training samples, which is sort of costly

Q: Is it possible to speed up the generation of the weighting matrix using a quantum algorithm?

@Blue They are? Hmm.. I'll have a look at the paper in a few minutes

Anonymous

6:32 AM

Anonymous

In this part they are providing an alternate method to generate $\rho$, which is fine and all, but they didn't use that in the algorithm it seems :P It's sort of confusing

Anonymous

Their main speed-up seems to be coming in this part:

Quantum Computing

0

In this[1] paper, on page 2, they mention that they are generating the weighting matrix as follows: $$W = \frac{1}{Md}[\sum_{m=1}^{m=M} \mathbf{x}^{(m)}\left(\mathbf{x}^{(m)}\right)^{T}] - \frac{\Bbb I_d}{d}$$ where $\mathbf{x}^{(m)}$'s are the $d$-dimensional training samples (i.e. $\mathbf{x}...

Anonymous

6:42 AM

The operation $$\rho := \frac{1}{M}\sum_{m=1}^{M}|x^{(m)}\rangle \langle x^{(m)}|$$ is also $\mathcal{O}(Md)$, right?

@Blue I don't think this would actually matter - if you want to compare with e.g. Solovay-Kitaev, you'd need to compare like with like

Anonymous

@Mithrandir24601 Could you elaborate a bit?

Anonymous

Not getting

Anonymous

Solovay–Kitaev theorem

In the mathematical theory of computation, the Solovay–Kitaev theorem says, roughly, that if a set of single-qubit quantum gates generates a dense subset of SU(2) then that set is guaranteed to fill SU(2) quickly, which means good approximations to any desired gate can be created using fairly short sequences of gates from the generating set. It is one of the most important fundamental results in the field of quantum computation. Robert M. Solovay and Alexei Kitaev jointly came up with and proved the theorem. == References... ==

Anonymous

Haven't heard of that before

6:48 AM

@Blue If you use a different method to get the matrix that is described in terms of the number of qubits, comparing the efficiency of one algorithm in terms of number of dimensions with another in terms of qubits is distinctly unfair :P

(although I'm not sure if Solovay-Kitaev applies here)

Anonymous

@Mithrandir24601 Are you saying that if we manage to use qubits (around $\log_2 (d)$) to represent the $d$-dimensional training samples, then the generation time of the qubit states would have to be taken into account? (I mean for creating the qubit system also we need a large gate model circuit, I guess)

Anonymous

Obviously if we use qubits for the matrix multiplication we can't directly say the time complexity will be $\mathcal{O}(M\log(d))$

Anonymous

Because we have to create the qubit states too

@Blue I wrote an answer on this once

Anonymous

i.e. $\mathbf{x}\to |\mathbf{x}||x\rangle$

6:55 AM

@Blue No, I'm saying that $n$ qubits automatically 'have' $2^n$ dimensions. Most algorithms deal with qubit complexity, so you have to compare like with like

It looks like this is creating the qubit states

You could also think of this in optics, where it's the number of dimensions that matter, sure, but the way that (currently) works is that you just dial up the unitary you want to implement, so having a complexity in terms of dimensions is a bit odd to start with for that

Anonymous

@Mithrandir24601 What does "dial up the unitary" mean?

Anonymous

To be clear, this is what I was thinking:

Anonymous

Generating $\rho$ and generating $W$ should take about the same time, if we are considering plain and simple classical matrix multiplication (no qubits involved here)

@Blue You've got so many phase shifters and (potentially variable) beam splitters. You 'twiddle some knobs' and that implements a unitary over everything at once, which is the entire circuit

(i.e. none of this 'using gates to build up to a circuit' or having to worry about which qubit can be used to control which)

Having said that, it's not the most scalable of methods, so measurement based looks much more promising in the longer term

Anonymous

@Mithrandir24601 I see. Makes sense! But where are they using that in this paper? (i.e. to generate $W$)

7:07 AM

@Blue It's highly unlikely. That's an implementation dependent thing, which people working on algorithms rarely worry about

"The $d$-level quantum system can be implemented by a register of $N=\lceil\log_2 d\rceil$ qubits"

Anonymous

@Mithrandir24601 Umm, are you saying the authors don't really care about how they'll generate $W$? :P Or, are you saying this ($\rho := \frac{1}{M}\sum_{m=1}^{M}|x^{(m)}\rangle \langle x^{(m)}|$) operation can be performed using some (controlled perhaps) unitary gates?

Anonymous

@Mithrandir24601 Yup, that's true. But I don't exactly understand how they're using that here for $W$

@Blue I'm reading it now

Anonymous

Okay!

7:30 AM

@Blue So they're saying that they have a channel or something that outputs $\rho$, not that they create it as part of the circuit

Anonymous

@Mithrandir24601 Huh! Where's that written?

Anonymous

I thought they're making the $U_t$ circuit just to find the eigenvalues and eigenvectors of $\rho$

Anonymous

But $\rho$ has to pre-made I thought

@Blue Page 2: "One can alternatively adapt a fully quantum perspective and take the activation patterns $\left|x\right>$ directly from a quantum device or as the output of a quantum channel", page 3: " If $\rho$ is the direct output of an unknown quantum device", so you've already got in in memory or something

@Blue yeah

Anonymous

@Mithrandir24601 Lol, so they're completely neglecting the creation of $\rho$ part although they're using $W$ later on in the algorithm on page 4

Anonymous

7:39 AM

I guess....that's......something worth addressing?

@Blue Looks like it. Also, it looks like a mixed state, which isn't something you can just make with a unitary matrix - it generally requires some CPTP map, which is a channel, so...

Anonymous

@Mithrandir24601 I'll have to read a bit thoroughly about CPTP maps it seems. Anyhow, any idea what the time complexity of that would be?

8:04 AM

@Blue I'm writing a slightly more detailed answer

Anonymous

@Mithrandir24601 Thanks :)

9:08 AM

@Blue Done :) First answer in a while

Anonymous

Thanks, reading!

9:36 AM

@Mithrandir24601 I'm not 100% sure, but generating that kind of mixed state shouldn't be hard given access to a qRAM. After all, the main feature of a qRAM device is that it allows to perform the operation $\sum_i \psi_i|i\rangle\mapsto\sum_i\psi_i|i\rangle|x^{(i)}\rangle$, so if you just "forget" to uncompute the index dof you should automatically get a mixture of the states of the data

@glS So, essentially, transfer data from qRAM to QPU, then decohere the two systems? That makes sense...

I'll make an edit

Anonymous

9:58 AM

@Mithrandir24601 I read your answer, but something is confusing me: You're saying that if the channel is simulated then the time complexity is around $\mathcal O\left(27n^34^{3n}\right)$ which is way worse than $\mathcal{O}(M2^n)$

Anonymous

I would rather calculate $\rho$ classically by matrix multiplication in that case

@Blue What's the complexity of multiplying 2 d times d matrices though?

(in any case, see updated version for something more reasonable)

Anonymous

@Mithrandir24601 I need to multiply two $d\times 1$ matrices I guess?

Anonymous

Not $d\times d$

@Blue OK, that's true enough

(although you then have to sum them as well)

Anonymous

10:05 AM

So that's around $\mathcal{O}(M2^n)$

Anonymous

@Mithrandir24601 I'm checking

@Blue You then have to physically create this in some way

Anonymous

@Mithrandir24601 Why would I need to "physically" create it? Can't I just code it up? (Given the training samples as input)

Anonymous

I'm not doing any quantum stuff

@Blue You'll still have to do 'Hamiltonian simulation'

It might(?) be more efficient just multiplying things classically anyway but I can't guarantee it without looking into it in more detail, which I couldn't really be bothered to do as we've now got a more efficient way anyway

Anonymous

10:13 AM

@Mithrandir24601 Yeah, but for that I don't think we need to physically "create" $H$. In Hamiltonian simulation given a matrix $H$ we try to simulate $e^{iHt}$ as closely as possible (if I'm not wrong)

Anonymous

I can even generate the matrix H I want by classical coding

Anonymous

@Mithrandir24601 But well, I need to check that qRAM method

Anonymous

Could you recommend me any source to read up the qRAM stuff?

@Blue No but that's not the same as saying that if you've physically got $H$, it's not any faster (I haven't read this far, so I don't actually know, but I wouldn't go round making assumptions like this either)

@Blue I don't actually know of any beyond what I used in the answer :P (I've never remotely looked at RAM)

Anonymous

@Mithrandir24601 I see. I might sound stupid, but can a qRAM be simulated or is it a physical RAM sort of thing? I'm actually thinking if I could simulate the algorithm given in Lloyd's paper, that be good

10:20 AM

@Blue I don't know what you mean by simulated here

Anonymous

@Mithrandir24601 On something like the IBM Q computer, using the MNIST database for training the neural network and then using it for handwriting recognition

@Blue yes (at least, as far as you'd need it here, there's no reason why not as you don't want things like long coherence times), but you'd probably need a fair number of qubits to do the RAM part, so I doubt we have the capabilities yet

Anonymous

@Mithrandir24601 Umm, any idea how the authors of the Hopfield paper simulated their neural network on the RNA sequences? They seem to claim:

Anonymous

It seems like an actual experimental result from what they describe but they don't mention how they did that experiment. The whole qRAM thing seems like a black box to me.

10:33 AM

@Blue I didn't follow the whole argument, but the way I understand it the advantage of having a qRAM is that it would allow you to access in superposition data that is stored in a "classical" memory. So these works usually rely on the assumption that the data vectors (here the $x^{(i)}$) are already stored in some classical (but I guess "special" in some way) memory, similarly to what is the case in the classical case.

this means that it is not clear what it would mean to "simulate" its operation. You can simulate the output of a quantum memory giving the same output, but the point of a qRAM is that it doesn't require you to have a quantum memory

(actually, maybe I'll ask this last point on the main site, to see whether people agree with my understanding)

@Blue there isn't really a lot to read about it. If you look at the questions asked about it in this site you should find references to nearly all of the relevant papers.

Anonymous

@glS That makes sense. But Lloyd et al. seem to have actually run their neural network (on either an actual quantum computer or a simulated one). No idea how they did that

Anonymous

I can surely store the data vectors in classical arrays

Anonymous

But then...

Anonymous

Did they simulate the output of the qRAM classically?

Anonymous

Can't connect the dots

Anonymous

10:38 AM

"Author contributions:
P.R. and T.R.B developed the quantum Hopfield network
and quantum Hebbian learning, **performed simulations**
and wrote the manuscript. P.R. and S.L. conceived the
initial idea. C.W. supervised the project"

11:05 AM

From how they wrote the paper, I'd assume that they started with $\rho$ and just did a classical simulation of the quantum circuit

Anonymous

11:17 AM

@Mithrandir24601 Exactly, I thought so too. Classically it should be easy to code up the $\rho$ matrix given the training samples (although time consuming)

Anonymous

It's pretty improbable that they used any qRAM sort of architecture

Anonymous

Or even any CPTP channel

@Blue bearing in mind that they can only simulate a few qubits, this makes the most sense for what they did. That's also not the same as saying that that's what they would do if they had access to everything they could ever want

Also, it really doesn't look like the point of the paper was to come up with some way of getting $\rho$, but what to do once you've got it

That bit looked very much like 'here are a couple of concepts to show this isn't unreasonable'

So I wouldn't worry about it that much

Anonymous

11:49 AM

@Mithrandir24601 Agreed, yeah. Our prof was telling us to simulate the whole thing on the IBM Q and I was worried about the preparation of $\rho$ part. Looks like classical matrix multiplication is the way to go

Anonymous

Although I guess I could send a mail to Rebentrost and Lloyd and company asking what they actually did :P

12:10 PM

@Blue I don't think you can "simulate" the preparation of $\rho$, at least not in the efficient way they mention in the paper. I guess to replicate that algorithm on the IBMQ you would have to start from the assumption of having prepared $\rho$ somehow, which practically means that you will have to implement a state preparation procedure in the IBMQ generating a given input state.

this basically amounts to hard-coding the input into the algorithm itself, but with the idea that in a more realistic scenario this would be done via qRAM/memory or something like that

(not that even just simulating the rest of the algorithm, state preparation aside, looks simple, mind you)

in fact, doesn't implementing that Hopfield network stuff require to implement also HHL09 as a subroutine? Are there even implementations of HHL09 with IBMQ around?

also, off topic comment: am I the only one for which the writing panel is completely unreadable when editing a message?

@glS it works fine for me. Is it ok on other SE chats?

@Mithrandir24601 no it's the same. It's been doing this for a while to be honest, I just never bothered to actually investigate. But it is quite annoying. It's probably got something to do with some dark theme background I guess but I don't understand what exactly. Are there even settings to change the chat theme?

12:26 PM

@glS I don't think so, although @Mithrandir might know? Failing that, it sounds like a bug report for Meta Stack Exchange

hum, one moment

Do you have any userscripts enabled?

It doesn't do that for me, although it's reminiscent of a thing that was bugging me earlier - closing and opening it again a couple times fixed it in that case.

@Mithrandir that was the first thing I thought, but it seems not. Tampermonkey reports no scripts enabled. I also have the Stylish extension which I used to use SE with dark theme, but that is also disabled.

also, it happens even if I change browser, from chrome to firefox, so it cannot be just the extensions (I basically don't have extensions on firefox)

what colors do you see for the rest of the bottom toolbar? do you see it like in my screenshot?

@Mithrandir ok, so it's really just the chat window's colors that are inverted... weird

ah, I found the problem

the only extension I had in both browsers: grammarly (it's a spell-checking extension). I have no idea why it should invert the colors of anything though, that's pretty weird (it doesn't do it anywhere else)

@glS ah. Now you've mentioned the name, I've realised that I've heard about this problem before

Anonymous

1:26 PM

@glS Yeah, that's the problem with Grammarly. I face it too. There was a workaround mentioned on Meta SE somewhere

Anonymous

@glS You mean the code for matrix inversion (as in solution for linear systems of equations? )

@Blue yes, the Harrow Hassidim Lloyd paper

Anonymous

Umm, it does seem to be quite popular but can't say if someone did it using IBM Q

Anonymous

Apparently someone our prof has a related paper: arxiv.org/abs/1801.00778 but not exactly the same thing

@Blue I don't remember having seen it, but people do all sort of stuff on the IBMQ lately so it's hard to keep track. I do remember having seen a proof of principle experimental implementation though, I believe it was with superconducting qubits by some group in china

@Blue yes, similar problem but different algorithm. I don't know if you could use that for the Hopfield network. But am I right that you need HHL09 to implement the Hopfield network? (you know the paper better than me)

1:55 PM

@Blue this perspective piece on QML came out recently, in case you are interested: medium.com/xanaduai/quantum-machine-learning-1-0-76a525c8cf69

it seems well-written

Anonymous

@glS I don't it's strictly necessary. They are simply estimating $A^{-1}|w\rangle$ by putting a restriction on the eigenvalues. I could very well use the previous algorithm I linked too for this purpose I think (the Grover's search one)

Anonymous

I haven't read that paper I linked though. Just spotted it while searching :P

Anonymous

Gotta ask the seniors who wrote that paper

Anonymous

@glS Oh, thanks I'll read :)

Anonymous

2:14 PM

While there does not yet exist a quantum computer that can truly offer a speedup over a classical computer, implementation of a "proof of concept" remains an important milestone in the development of a new quantum algorithm. Demonstrating the quantum algorithm for linear systems of equations remained a challenge for years after its proposal until 2013 when it was demonstrated by Cai et al., Barz et al. and Pan et al. in parallel.

Anonymous

en.m.wikipedia.org/wiki/…

Anonymous

Phew...the first implementation was in 2013

Anonymous

I don't think we can expect any previous implementation on IBM Q then

Anonymous

arxiv.org/abs/1302.4310

@Blue well they give the circuit they used, so it shouldn't be so hard to replicate that on the IBMQ

Anonymous

2:24 PM

@glS I'm worried that it will require way too many gates. They only gave the implementation for a system of two linear equations

Anonymous

The matrix A we have has much larger dimensions

Anonymous

Of the order $O(d^2)$

Anonymous

Can't be used for practical purposes then like taking input from 784 pixels as given in the MNIST database

2:41 PM

@Blue so, 1 more qubit? :P

Anonymous

log(784)~10 qubits. Am I missing something ? @Mithrandir24601

3:02 PM

@Blue I have no context for the current conversation, so I'm making a joke about how the dimensions scale with the number of qubits :)

(Hence the ':P')

hmm, what do y'all think of "basic"/beginner/very simple questions on main here?

Anonymous

@Mithrandir They're welcome

Anonymous

@Mithrandir24601 Okay lol

...like, "what is a qubit" might expand your SEO :)

@Mithrandir hopefully there's a tag wiki for that...

quantumcomputing.stackexchange.com/tags/qubit/info

3:09 PM

"What is a "qubit"? Google tells me that it's another term for a "quantum bit". What is a quantum bit? How is it quantum? What purpose does it serve?"

@Mithrandir if you feel the tag wiki isn't good enough, ask away!

@Mithrandir24601 that's a bit too technobabbly for a curious Googler

@Mithrandir oops... Probably my bad

And tag wikis don't have very good SEO in the first place.

This is true

Anonymous

3:14 PM

It's quite difficult to explain what a qubit physically is unless you already have an idea of (say) electron spin or light polarization

Anonymous

Mathematically it's quite easy

well. i guess a complete answer would explain what those are.

I asked it. I have nothing to lose, and knowledge to gain ;)

Anonymous

Ideally one would have to write a >2000 word essay to make it sensible for a beginner without any background in quantum mechanics or complex numbers ;) Anyhow, I'd say you should make it a bit more focused by asking what is a qubit "physically"? Later you could ask about its mathematical representation.

I'm going to predict this will hit HNQ...

Anonymous

It always does :P

3:29 PM

More site hits! :) We did really well last week, so I want to do better this week :)

Anonymous

I personally do encourage beginner questions but I want this site to be more of a Math Overflow rather than Math SE. That is, you are expected to thoroughly read at least Wikipedia before asking a question

Anonymous

Physics SE is a site with lots of visitors too, but the level has dropped a lot over the years with newbies beginning to ask hw questions without showing any effort

Anonymous

What I mean to say is that, more site hits doesn't necessarily mean that the site is healthy

Anonymous

@Mithrandir I really encourage you to go through the Wikipedia page on qubits and make your question more focused. Trust me, you'll learn a lot too, in the process of making sense of a technobabbly page and filtering out the exact portions which you don't understand

Anonymous

Asking the right questions is a very important skill if you ever want to be a scientist

A: Making sense of principal component analysis, eigenvectors & eigenvalues

3:40 PM

@Mithrandir I'm actually surprised to not find a highly-voted version of that exact question on physics.SE

though it's possibly because it would get closed

maybe one could make it into a question like "How can you explain what is a qubit to a X year old?"? There is handful of very nice questions/answers of that kind on math.SE and stats.SE

I'm thinking for example of this: stats.stackexchange.com/a/140579/82418

874

Imagine a big family dinner, where everybody starts asking you about PCA. First you explain it to your great-grandmother; then to you grandmother; then to your mother; then to your spouse; finally, to your daughter (who is a mathematician). Each time the next person is less of a layman. Here is h...

@Blue but neither do few hits

@Mithrandir24601 what's HNQ?

@glS hot network questions

Anonymous

@Mithrandir24601 Sure. But the questions during beta phase serve as good indicators of what type of audience we want in the future

Anonymous

I think we are doing well currently though

3:52 PM

@Blue it is also true that if the site grows we can expect many variations of this kind of question to pop up. Having a well written question/answer can be useful as reference to close the others as duplicate

@Blue considering this is quantum computing, I hope that's going to fluidly change as people figure out what's going on an start using them

Also, there's nothing wrong with having questions and answers as alternatives to Wiki, although I do want more research questions, of course!

Quantum Computing

1

Q: What is a qubit?

What is a "qubit"? Google tells me that it's another term for a "quantum bit". What is a quantum bit? How is it quantum? What purpose does it serve in quantum computing? (I'd prefer an explanation that is easily understood by laypeople; terms specific to quantum computing should preferably be ex...

qubit

Anonymous

@glS Yes, but what do you think is a better question: "What is meant when they say a photon can behave as a two dimensional quantum mechanical system and can exist as a superposition of two different polarization states ?" or "what is a qubit"? I think the latter is a tad too broad and lacks research effort.

Anonymous

However, yes, if we are creating a set of canonical Q's with large essay type answers that might be suitable, however, personally I don't like such questions as models of what questions should be asked in the future. Anyhow that's a completely subjective opinion. :)

@Blue that's why you've got an option to flag and VTC

Anonymous

4:04 PM

@Mithrandir24601 Judging from the past community behaviour towards such questions I don't think sufficient people would agree with me anyway :P One VTC is useless and I don't want to hamper the process if you really want to create such canonical Q&A's

Anonymous

I'll perhaps just leave behind a comment mentioning possible improvements

@Blue I mean, sure, it does lack research effort (no offence Mithrandir =) ). I'm thinking of it more like a canonical question that can be handy to have. But it is also true that it shouldn't be seen as a model for others to follow. Maybe we can get the best of both worlds by adding a disclaimer stating something along the line of "this is a canonical question and not to be seen as model for other questions"?

Anonymous

Well, community wiki was meant for exactly this purpose earlier

Anonymous

We could use that maybe...But then again Robert might come rushing XD

@Blue also true. I don't know I'm conflicted about this

Anonymous

4:09 PM

What's the current use of CW though ? I don't really understand its scope

@Blue I'm a mod, so my purpose is to try and do what the community wants (within reason), so if you want a particular requirement, it's not up to me to say yes or no

Anonymous

@Mithrandir24601 I wasn't really telling you to say yes or no, nor was I asking for any mod action. Was simply casually expressing my views :P

Anonymous

Anyhow, discussing site policies is both boring and increases everyone's blood pressure. Let's change the topic ;)

@Blue ah, OK, I thought "if you really want to create such canonical Q&A's" was referring specifically to me :P

Anonymous

4:31 PM

To those who wish to answer this question: It would be great if you point out the difference between classical and quantum probabilities in your answers. That is, how is a quantum state like $\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$ different from a coin which when tossed in the air has a $50-50$ chance of turning out to be heads or tails. Why can't we say that a classical coin is a "qubit" or call a set of classical coins a system of qubits? — Blue 5 mins ago

Anonymous

I think this is an important point which not many introductory sources cover

Anonymous

@Mithrandir24601 Nope, I meant "you" collectively ;)

@Blue you meant yous!

Anonymous

@glS The plural of you is you too, you fools! :P

Anonymous

(justjokingXD)

4:44 PM

@Blue don't tell that to the Irish!

Anonymous

lol

Anonymous

You're Irish too?

@Blue no, but am currently living there (on the north), and I find yous to be a great invention. English really lacks a plural form of "you"!

Anonymous

Okay, I see. Hehe. I've heard a lot of good things about the great natural scenery in Northern Ireland :)

Anonymous

@glS I'd surely vouch for the addition of yous in the OED ;)

Queen's University Belfast

4:50 PM

@glS :) nice. I think you've mentioned Queens uni Belfast before?

Anonymous

Queen's University Belfast (informally Queen's or QUB) is a public research university in Belfast, Northern Ireland. The university was chartered in 1845, and opened in 1849 as "Queen's College, Belfast". The university forms the focal point of the Queen's Quarter area of the city, one of Belfast's six cultural districts. It offers academic degrees at various levels and across a broad subject range, with over 300 degree programmes available. Its acting President and Vice-Chancellor is James McElnay, and its Chancellor is Thomas Moran. The annual income of the institution for 2016–17 was £337.6...

:45214634 Northern Ireland, which may or may not be Ireland depending on who you're talking to :P

Anonymous

Lol, I'm so poor in Geography

Anonymous

::Hides face and runs away::

Anonymous

@Mithrandir24601 Haha, why though? :P

4:55 PM

@Blue yes they are quite nice indeed

@Mithrandir24601 yes, that's why I felt the need to specify on the north!

@Blue why is there a price tag on the bottom left of the symbol?

Anonymous

@glS Which one? The endowment amount?

Anonymous

I don't see that. Probably your browser configurations are faulty

@Blue historical reasons. Just maybe don't ask random people these questions if you come around these parts (it's one of the first things I learnt arriving here lol)

Anonymous

5:13 PM

@glS Good to know, lol

Anonymous

I'll probably visit Ireland someday. It sounds like a nice place to stay

Anonymous

I have a special liking for places which are low in population and away from populated cities (aka country-sides)

@Blue the character limit is 30,000 characters ;)

Anonymous

@Mithrandir That wasn't the point, but anyway...

Anonymous

5:30 PM

0

A: What is a qubit?

Google tells me that it's another term for a "quantum bit". That is correct, but the relationship between these words is even stronger than that: "qubit" is actually just short for "quantum bit" so the latter defines the former. What is a "quantum bit" physically? How is it "quantum"? ...

Anonymous

"The reason why the qubit examples I gave come in quanta are because they exist as solutions to something called the Schrödinger Equation."

Anonymous

ugh...

6:02 PM

@Blue I call writing an answer in <2 hours quick :P

long is when it takes more than about 2 days

Anonymous

@Mithrandir24601 Same. Btw that answer seems to be wrong (or at least ambiguous) in several places. It makes it sound as if the SE is something intrinsic/fundamental to quantum mechanics

Anonymous

"Two solutions to the Schrödinger equation (the 0 solution, and the 1 solution) can exist at the same time."

Anonymous

0

A: What is a qubit?

It's a quantum system that lives in a 2 dimensional complex vector space. However, more than that is required to do computations. There needs to exist two orthogonal basis vectors in that vector space, call them |0> and |1>, that are stable in the sense that you can set the system very precisely ...

@Blue I didn't write it, so the best idea would either be to leave a comment under the answer or make an edit that fixes the problem(s)

Anonymous

This is what happens when you ask non-focused questions :/

Anonymous

6:08 PM

@Mithrandir24601 I wouldn't dare to edit userXYZ's posts :'D

Anonymous

Are you crazy? Lol

Anonymous

I might leave a comment, but it seems (s)he'd tend to start an argument over it

Anonymous

I'll rather leave it and write my own answer later if and when I get time

Anonymous

My main objection is that the SE does not dictate that the states of a quantum system have to be discrete

The eigenstates of the free Hamiltonian on an infinite line being a notable exception to that rule.

6:15 PM

@Blue I'm a mod. I (try and) know what is and isn't allowed. This includes saying that you've got the privilege to edit and if you feel a post needs editing and know how to fix the problem(s) then (assuming you're not abusing the system in any way) you can do that if you so wish. No-one gets special permission to say 'you cannot edit my posts'

One needs boundary conditions to get discrete states.

Anonymous

@Semiclassical Exactly!

Of course, if you go around making wrong edits all the time (which I haven't noticed anyone doing, to be clear), then that's a different matter

The case of scattering states is probably the most practical example

Anonymous

@Semiclassical Does this sentence make sense to you: "The reason why the qubit examples I gave come in quanta are because they exist as solutions to something called the Schrödinger Equation. Two solutions to the Schrödinger equation (the 0 solution, and the 1 solution) can exist at the same time."?

6:18 PM

I don't mind the first one so terribly much because of how much 'boundary conditions' are taken for granted

And I'm guessing that the 0 and 1 solutions are not intended as $\psi=0,1$ but $|0\rangle, |1\rangle$

My more problem is that, when you do the Schrodinger equation, you don't get just two solutions even if you apply boundary conditions

you get an infinite dimensional Hilbert space of solutions

@Semiclassical I believe they're saying that you only take two of those solutions, but this is not clear

yeah, so you restrict yourself to 'low energy' states to get a simple Hilbert space

the reason I don't really like that, though, is that the simplest example I know of a two-state system is a spin-1/2 electron

and it has to be anharmonic etc. etc.

@Semiclassical Which, technically isn't well described by the SE, but requires the Dirac equation (just to be pedantic about things)

Right.

Within non-relativistic QM one just takes it granted: there is this system, and it behaves this way

which is fine imo

@Semiclassical Sure. The really pedantic point is then saying that the Dirac equation also allows for infinite solutions (with different momenta) and you're back to having to add constraints (or similar) in some way to have only 2 possible solutions (again)

Although, to be fair, the dimensions spanned by the eigenvectors is still only 4, only you've got this extra relativistic DOF on top, which gets confusing

6:28 PM

Ugh

6:45 PM

@Mithrandir24601 seconded

> If you are not comfortable with the idea of your contributions being collaboratively edited by other trusted users, this may not be the site for you.

(source)

Anonymous

@Mithrandir24601 @glS What does the $R(\lambda^{-1})$ gate do? (For example: Page 2 here)

Anonymous

Haven't come across it before

Anonymous

Any source to read it up?

@Blue I think they just mean a rotation by an angle $\lambda^{-1}$

that is, $R_z(\lambda^{-1})$

ok maybe not exactly. But it should just be a weird notation for a rotation around Z. You should be able to find out the $\theta$ such that $R_z(\theta)$ sends $|0\rangle$ to that state there

Anonymous

6:58 PM

also I remember that same equation in the HHL paper, maybe it's explained better there?

Anonymous

This part is a bit confusing i.e. how they are ending up with the circled state

Anonymous

@glS The HHL paper is even more cryptic...lol

Anonymous

It's here: arxiv.org/pdf/0811.3171.pdf

Anonymous

They are doing some Hamiltonian simulation stuff again

Anonymous

7:01 PM

Anonymous

I was just wondering whether this circuit which they implemented for $2\times 2$ system of linear equations can be applied for larger systems

Anonymous

Should I ask this on the main site? :/

this is what I guess they are doing?

I missed to add an $i$ in the top equation and in the last equation $j$ should be $k$

though I don't know that $C$ is

Anonymous

@glS What is the last word in that sentence: "they do the controlled sp"?

Anonymous

What is "sp"?

7:14 PM

sorry, op, as in operation

Anonymous

What gate are they using for that though? It's a bit hazy for me

Anonymous

I understood till before that sentence

Anonymous

What controlled operation is it?

@Blue what do you mean "what gate"? is the unitary in my last equation (the whole one, including the control space)

if you notice in my last equation, depending on the state of the first system ($|\lambda_k\rangle$), the second is rotated by a different angle

@Blue the answer to this question is almost always yes! I think it would be great to have a series of questions on these kinds of papers. It would be a great resource for anyone studying them (I know I would have loved to have it when I first started!)

@Blue for example, it would definitely be a worthy question how they think these $R(\lambda^{-1})$ controlled rotations can be implemented

Anonymous

@glS To be clear, there are three systems of qubits here: 1) The input $|b\rangle$ 2) The Eigenvalue qubits $(|0\rangle)^{\otimes n}$ 3) The single ancilla qubit state $|0\rangle$. I understand that after phase estimation the state generated is $$(\sum_{i=1}^{N}\beta_i|u_i\rangle |\lambda_i\rangle)\otimes |0\rangle_{\text{ancilla}}$$

Anonymous

7:23 PM

After that I'm not clear on how we're getting to this part:

Anonymous

Gotta think for a bit

Anonymous

Okay, the state of the eigenvalue qubit register along with the ancilla register after phase estimation step is:

Anonymous

$$(\sum_{i=1}^{N}\beta_i |\lambda_i\rangle)\otimes |0\rangle_{\text{ancilla}}$$ if I'm not wrong

@Blue yes, from this state they ignore the first register (with the $|u_i\rangle$), and apply a controlled rotation, which conditionally to the state of the second register (the $|lambda_i\rangle$) applies a rotation on the ancilla qubit. In particular, this rotation is a $R_x(\theta)$ rotation with an angle $\theta$ that depends on $|\lambda_i\rangle$

(where $R_x(\theta)\equiv e^{i\theta X}$)

Anonymous

7:28 PM

@glS And after that controlled rotation we end up with this:

Anonymous

?

@Blue yes. Try and apply the operator I wrote my last equation in the figure to the state after phase estimation

you should "magically" get exactly the red-circled state

by the way, I just now thought about it, but did you have a look at this review? arxiv.org/abs/1804.03719

Anonymous

@glS I'll do that on pen-and-paper then ;) But, one min, you were saying that there's some problem with $R(\lambda^{-1})$ i.e. controlled operations are difficult to implement. Why is it so? Isn't there a standard gate-set for controlled rotations? (I'm a complete noob in this)

Anonymous

@glS I did see the abstract before, but didn't read it. Checking

7:34 PM

@Blue it should be "easy" to implement any controlled-$R_x(\theta)$ for any given $\theta$. However what they do there is a little bit more complicated, because they have many qubits as control, and they want to apply a rotation with an angle that depends on the exact multiqubit state of the controls register. I'm not sure how that can be done

Anonymous

@glS Oh, I see. I'll ask it on main, then! And one more thing: Any idea why all these experimental implementation papers are only solving this problem for $2\times 2$ systems? Haven't seen a single implementation of (say) $3\times 3$

@Blue it probably depends on the experimental constraints, that is, how many qubits they have available. What is the last implementation that has been done?

Anonymous

@glS The latest IBM Q implementations of this algorithm I could find are all from 2013

Anonymous

They all used the 5 qubit QC

Anonymous

Now they've released the 15 qubit one too I guess but haven't seen any implementation with that

7:38 PM

and how many qubits does HHL need?

for a given dimension I mean

Anonymous

Proportional to the dimension of A I guess

Anonymous

Checking it

Anonymous

So say you have a $N\times N$ matrix $A$

Anonymous

Which is Hermitian (assume for now)

Anonymous

$e^{iAt}$ will be unitary

Anonymous

7:43 PM

It will have $N$ eigenvalues and $N$ eigenvectors

Anonymous

So to represent the state $(\sum_{i=1}^{N}\beta_i|u_i\rangle |\lambda_i\rangle)\otimes |0\rangle_{\text{ancilla}}$ we'll need how many qubits?

@Blue I wouldn't be certain that the 15 qubit one is actually good enough to do much on

Hi

Anonymous

$2\lceil{\log_2(N)\rceil} + 1$ I guess? @glS

@O.Rares Hi! Welcome to our chat room :)

7:46 PM

is there a motivation to create quantum apps ? I mean can they be sold?

@Blue hint: use lceil and rceil

Anonymous

Ah, nice

Anonymous

@O.Rares Well, assuming people (will) have access to quantum computers, why not

@Blue and at this moment ? I try to find motivation to learn but idk if it is worth it

@O.Rares Depends what you mean by quantum apps - an app that accesses some quantum computer and allows you to use it easily would be quite nice. One that access a quantum computer is, at the minute, interesting but not overly useful due to the small scale things are at as well as (most people) often needing to wait to get results when they make a call to whichever chip they're using

Anonymous

7:49 PM

@O.Rares It's sort of like wanting to prepare Android apps in the 1950's

@O.Rares If you find it interesting, absolutely! If you want something that could potentially be useful sooner rather than later, quantum cryptography/key distribution would be the way to go

@Blue I hope we're a bit further ahead than that... But well, it's slow going, so... Maybe not

didn't Microsoft shared their quantum APIs for some marketing benefits?

@Blue I like that analogy! seems fitting

@O.Rares Are you thinking of the Q# language? They don't have an actual quantum computer (yet) but would probably like people to start thinking of higher-level ways of using them

@Mithrandir24601 yeah,Q# .I read that they have few quantum bits . So why are people studying it if they can't apply it in order to get a benefit ?

Anonymous

7:56 PM

@glS So it seems for a $2\times 2$ system we'd need $2+1=3$ qubits and for $3\times 3$ system we'd need $5$ qubits

Anonymous

Maybe we could go upto maximum $100\times 100$ for Hermitian A it seems with the 15 qubit computer

Anonymous

And $50\times 50$ for non-Hermitian

@O.Rares They'll have access to a few bits - IBM has free 5 qubit and 16(?) qubit chips (available online here), as well as commercial access to higher numbers. Google has some (unknown) number, Rigetti has 20-ish, D-Wave has a thing called a quantum annealer, I (at Bristol) have access to a photonic chip (which isn't referred to in terms of qubits)