The Classical Channel

12:03 AM

@heather another thing: "And the development of those laws is still not really complete" what do you mean by this exactly? "Quantum mechanics" is not really a fundamental law of its own, but more or a general recipe for how to make theories. I wouldn't say its development is "incomplete". Sure, better theories could come in the future to replace it, but that does not mean that the current theory is not complete

heather

12:18 AM

@glS true, poor phrasing there - i was trying to point towards field theories, etc

Anonymous

3:35 AM

@heather I added a few points in the comments to your answer.

Anonymous

4:01 AM

Btw this is an excellent thread which I found the other day:

Anonymous

120

Q: What makes a theory "Quantum"?

Say you cook up a model about a physical system. Such a model consists of, say, a system of differential equations. What criterion decides whether the model is classical or quantum-mechanical? None of the following criteria are valid: Partial differential equations: Both the Maxwell equations ...

Mithrandir24601

6:49 AM

@Blue I disagree with (an extended version of) AFT's point on realism/determinism - classical physics is local and deterministic. Quantum physics isn't

different interpretations disagree over whether it's non-local or non-deterministic but they agree that it's non-(local and deterministic), as far as I'm aware

Ah-ha - this answer, in other words

Anonymous

@Mithrandir24601 Classical physics is not necessarily deterministic

Mithrandir

@heather Nice! I'm not knowledgeable about it enough to say anything about the accuracy, or Blue's comments, but it's understandable :)

Anonymous

25

A: What situations in classical physics are non-deterministic?

There are two famous cases in classical mechanics that fail to be deterministic. The first, and most famous, is Norton's Dome, which corresponds to a system with a force of the form $$F = \sqrt{r} $$ There are more details on the Wikipedia article (it's usually described as the result of a rea...

Anonymous

0celo and Slereah were having an extensive discussion regarding that issue a couple of months back

Anonymous

Locality is another very subtle issue and I wouldn't really take it for granted

Mithrandir24601

7:06 AM

@Blue 1. is it physical? (I mean, the second example is pretty much defined as non-physical) 2. More importantly, is it correlated?

Anonymous

@Mithrandir24601 I guess it depends on your definition of what constitutes "physical"

Mithrandir24601

Also, it's maybe possible to make the argument that that's a different type of non-determinism, if you want to be technical and pedantic about it :P

Anonymous

Maybe, but I have an issue with physicists brushing off legitimate mathematical solutions as "unphysical" simply because we haven't yet observed them

Anonymous

From a practical viewpoint that's fine perhaps

Mithrandir24601

@Blue well, the second example is unphysical due to an infinite acceleration

Anonymous

7:21 AM

@Mithrandir24601 And what prevents that? I don't think relativity puts a limit on acceleration (I'm not very knowledgeable on this, so genuine question)

Anonymous

There's definitely no limit as far as classical mechanics is concerned (afaik)

Mithrandir24601

@Blue Aside from having to use relativistic physics to treat this properly, infinite force implies the system has an infinite energy after any negligibly small amount of time (or has infinite mass, which is a black hole, which does require GR+QM)

Anonymous

7:40 AM

@Mithrandir24601 Okay. So we're putting some observable physical constraints on it (apparently): 1) Point particles don't exist 2) The universe is relativistic and some others. While that's very reasonable, I find it philosophically unsatisfactory that we should restrict ourselves to the laws of physics which are observable for our universe and neglect the other mathematically legitimate possibilities. Anyhow, thanks for the explanation!

Anonymous

At some point I'd like to learn the exact mathematical conditions for these things though, they're probably quite subtle

Anonymous

I don't know if there exist legitimate models of non-relavistic or non-quantum mechanical universes though

Mithrandir24601

8:13 AM

@Blue I'm ignoring point 2 for simplicity. If you want something classical with infinite mass, then that's much worse in that the gravitational force on everything in the universe will be infinite, so everything will either have infinite energy or will be contained in some infinitely massive object somewhere

@Blue Sounds like a question for Worldbuilding!

Anonymous

@Mithrandir24601 Come over to hbar, we were discussing that

Anonymous

@Mithrandir24601 Alternatively we could consider only massless particles (I guess)

Anonymous

@Nelimee Hey, you around?

Anonymous

I had a confusion regarding your answer

Nelimee

@Blue Hi! I'm here :)

This one: quantumcomputing.stackexchange.com/questions/2393/… ?

Anonymous

8:24 AM

@Nelimee Yeah, just one minute. I'm writing it out:

Anonymous

@Nelimee After the Phase Estimation step this is state of the system:

Anonymous

$$\left(\sum_{j=1}^{j=N}\beta_j|u_j\rangle\otimes |\tilde\lambda_j\rangle\right)\otimes |0\rangle_{\text{ancilla}}$$

Anonymous

But from that we can't separate out $|b\rangle$ in terms of its eigenvectors, right?

Nelimee

Yup, it seems correct

At least the "But from that we can't separate out $|b\rangle$ in terms of its eigenvectors, right?" is correct

I think the state is correct too, but I did not check enough to be 100% sure

Anonymous

@Nelimee And just to be sure, what exactly are we using the Hamiltonian simulation for?

Anonymous

8:32 AM

Anonymous

This is Schematic, I understand

Nelimee

At the beginning you have a system Ax=b. If a is Hamiltonian (its self adjoint) then you know that exp(iAt) is unitary and so can be simulated by a quantum circuit. Hamiltonian simulation is just the application of a quantum circuit that implements/approximates the effect of the unitary matrix exp(iAt) on the quantum state

Anonymous

@Nelimee I sort of get that part. However, I do not understand how we'd create a control on several qubits (which are in the control register). Do we have to break up $e^{iAt}$ into controlled-by-single-qubit gates?

Nelimee

And we use Hamiltonian simulation because the exp(iAt) matrix has eigenvalues that are related to A's eigenvalues (if L_i are the eigenvalues of A, exp(itL_i) are the eigenvalues of exp(iAt)), and from the shape of the eigenvalues of exp(iAt) we can see that the phase estimation algorithm will allow us to recover the eigenvalue of A from the simulation of exp(iAt) (Hamiltonian simulation)

Ha I see your point

The illustration is highly simplified

Anonymous

@Nelimee You can use MathJax in chat too: math.ucla.edu/~robjohn/math/mathjax.html

Nelimee

8:36 AM

Matjax does not render on my computer in chat :o

Do you see this rendered: $\lambda$ ?

Anonymous

@Nelimee See the link I gave

Anonymous

You need to drag it to your bookmarks bar

Anonymous

@Nelimee Yes

Nelimee

Ha ok, I'll look at it after :)

Anonymous

8:38 AM

Just takes 10 seconds ;)

Anonymous

Anyhow:

Nelimee

In the last picture (from the Wikipedia article on QPE), the $C-U^{2^n}$ operations are (in HHL) controlled $e^{iAt2^n}$ gates.

Anonymous

@Nelimee $e^{iAt2^n}$ means $(e^{iAt})^{2^n}$, yes? That is, the $e^{iAt}$ gate being applied $2^n$ times?

Nelimee

Nice! It works perfectly! Thanks for the link ;)

Anonymous

Yay :)

Nelimee

8:42 AM

And yes for your question, it means $e^{iAt}$ applied $2^n$ times

But this is not the most efficient way to implement it, of course :p

Anonymous

@Nelimee And the action of all those individual single-qubit controlled unitary gates is equivalent to $\sum_{\tau = 0}^{T-1}|\tau\rangle \langle \tau|^{C}\otimes e^{iA\tau t_0/T}$? I not very sure how to show that they're equivalent

Nelimee

Re-applying $e^{iAt}$ an exponential number of time will not improve that much the complexity of HHL ;)

Anonymous

Firstly, I forgot what $|\tau\rangle$ is, lol

Nelimee

$\left| \tau\right> = \left| i\right>$, $\tau$ is just a summation index

And I know that it works like that (for the sum), but I don't feel confident enough with it to explain it to someone else :/ I'll read a little more in a book I have, and see if I can have a clear explanation on that part

But the $\left|\tau\right>\left<\tau\right|$ is the "control" part, i.e. this means that the operation will be controlled by the register $C$.

Anonymous

@Nelimee Ah, and what does $T$ represent? What is $\sum_{0}^{T-1}|\tau\rangle$ the state of?

Anonymous

8:47 AM

This part is a bit confusing

Anonymous

The controlled gate thingy

Nelimee

It's just the state obtained after the application of an Hadamard gate to all the qubits of the register

Don't be fooled by the $T$ and $\tau$, you can replace them by $N$ and $n$ (or $i$ for summation index)

They just call it this way because the register is a "quantum clock". I'm sorry, but I don't understand this part for the moment, I don't know what is a quantum clock :/

Anonymous

So let's go a bit slow: The state of the system just after the Hadamard transform is: $(\frac{|0\rangle + |1\rangle}{\sqrt{2}})^{\otimes n}\otimes |b\rangle$, yes?

Anonymous

Excluding the ancilla for now as we don't need it in the first step

Nelimee

Problem with the $\sqrt{2}$ but yes

Anonymous

8:50 AM

Typo :P Okay, and after that we apply the operator $\sum_{\tau = 0}^{T-1}|\tau\rangle \langle \tau|^{C}\otimes e^{iA\tau t_0/T}$ on it

Anonymous

And we need to prove that we get:

Anonymous

$\left(\sum_{j=1}^{j=N}\beta_j|u_j\rangle\otimes |\tilde\lambda_j\rangle\right)$

Nelimee

$\sum_{i=0}^{2^n-1} \left| i \right> = \left( \frac{\left|0\right> + \left|1\right>}{\sqrt{2}} \right)^n$ is OK for you?

Anonymous

@Nelimee Ah, yes, makes sense!

Nelimee

Ok, so this is clear

And no for your previous question, it's false

Let me write something to make it clear :)

From the image I linked in my answer (quantumcomputing.stackexchange.com/questions/2393/…), the first thing to do is a Quantum Phase Estimation (QPE).

After the QPE, the state will be:
1. Exactly the state you gave $\left( \sum_{j=1}^N \beta_j \left|u_j\right> \otimes \left|\tilde\lambda_j\right>\right)$ if the QPE had enough qubits to encode the eigenvalues in the register.
2. A state "close" to $\left( \sum_{j=1}^N \beta_j \left|u_j\right> \otimes \left|\tilde\lambda_j\right>\right)$ if the QPE hadn't enough qubits to encode the eigenvalues in the register.

Anonymous

8:55 AM

So basically our state is $\sum_{I=0}^{2^n-1} \left|I \right> \otimes |b\rangle$ and our operator is $\sum_{I = 0}^{2^n-1}|I\rangle \langle I|^{C}\otimes e^{iA I t_0/T}$

Anonymous

(you could replace $I$ with $\tau$ for convenience)

Nelimee

The exact state is $$\frac{1}{2^{n}}\sum_{x=0}^{2^n - 1} \sum_{k=0}^{2^n - 1} e^{-\frac{2\pi i k}{2^n} \left ( x - 2^n \theta \right )} |x\rangle \otimes |\psi\rangle.$$

Anonymous

Ah, interesting. Since all the $|I\rangle$'s are orthogonal, in case the indices don't match the terms will become $0$

Nelimee

And the QPE is:
1. Hadamard gate on each qubit
2. $$\sum_{\tau=0}^{T-1} \left|\tau\right>\left<\tau\right|^C \otimes e^{iA\tau t_0/T}$$
3. Inverse QFT

So the state you wrote is the state obtained after the QPE (with the QFT), not after only the controlled-operations

Anonymous

9:01 AM

@Nelimee Yup, I had written the result of this whole thing ^

Nelimee

Yes, then it should be approximately this.

Anonymous

$$\left(\sum_{j=1}^{j=N}\beta_j|u_j\rangle\otimes |\tilde\lambda_j\rangle\right)\otimes |0\rangle_{\text{ancilla}}$$

Nelimee

Proving it formally will be hard due to the fact that QPE is not always exact

You will have horrible summations like on the Wikipedia page (en.wikipedia.org/wiki/Quantum_phase_estimation_algorithm)

But if you consider that the QPE has enought qubits to encode the eigenvalues, then everything should be OK

Anonymous

@Nelimee Ah, true true

Anonymous

That part is a bit confusing to write out formally

Nelimee

9:03 AM

Yup

But if you just write the maths and blindly follow the rules of calculus, I think it should be possible to obtain something :)

Anonymous

@Nelimee How do we construct the gates for the $\text{FT}^{\dagger}$ though?

Anonymous

I sort of get it mathematically (for at least step (a) now)

Nelimee

It relatively easy: en.wikipedia.org/wiki/…

Anonymous

@Nelimee Oh, nice

Nelimee

If you can construct the gates for the QFT, then you can inverse them one by one (easy to do with a computer) to have the inverse QFT

Anonymous

9:05 AM

Nielsen and Chuang didn't cover it well unfortunately (the QFT part)

Nelimee

I have an implementation of the QFT in QISKit if you want, but the license is quite strange :p

And by strange I mean unusual

Not with silly requirements

Anonymous

@Nelimee What do you mean by inverse them one-by-one? How exactly does the circuit get altered ?

Anonymous

@Nelimee That might be useful for me, since I'm trying to implement the HHL09 for $3\times 3$ or larger matrices

Nelimee

You can look at the QISKit implementation but if you have a gate $G$ composed of $n$ simpler gates $g_i$, then the inverse of $G$ ($G^\dagger$) is obtained by taking each gate composing $G$ in reverse order and inverse them one by one

Anonymous

9:09 AM

I haven't seen any implementation of HHL09 for larger matrices (>$2\times 2$), for reasons I don't know.....strange

Nelimee

@Blue You want to implement it in a general framework (i.e. for a given matrix, which is not hardcoded) or not?

Anonymous

@Nelimee Yup, general framework

Nelimee

I linked you a 4x4 implementation in one of my answers: arxiv.org/abs/1110.2232v2 (v2, the v3 deleted many things)

How much time do you have to implement this?

Anonymous

@Nelimee There's no hurry as such but maybe within this month would be good. I'm still in the learning phase

Nelimee

Because that was my idea at the beginning, but Hamiltonian simulation for a general matrix is not even done in theoretical papers, and Hamiltonian simulation for specific matrices (sparse for example) is REALLY challenging

Anonymous

9:12 AM

I want to see how large we can go (in terms of matrix size)

Nelimee

Except if you are a genius, you won't be able to implement a general HHL within a month in my opinion

Even 6 months seem short to implement what you want

Anonymous

@Nelimee Uh....is there any existing example of Hamiltonian simulation on say IBM Q?

Nelimee

No, there is not. And efficient Hamiltonian simulation is really hard to understand.

Did you looked at some articles about this?

Anonymous

@Nelimee Lol, I don't really know the difficulties involved

Anonymous

As I said, I don't have any hurry as such

Nelimee

9:14 AM

All the other parts are trivial, QFT or even QPE for $n$ qubits is quite easy, but Hamiltonian simulation seems hard

Anonymous

@Nelimee I did, but not many examples of actual simulations

Nelimee

How much mathematical background do you have?

I have plenty of references on HHL, I'll give you a list of interesting papers if you want :)

Anonymous

@Nelimee Well, I'm an undergrad in EE (completed with freshman year)...don't know much information that conveys though

Nelimee

EE?

Anonymous

I guess I understand most of the math used in Nielsen and Chuang at least

Anonymous

9:17 AM

@Nelimee Electronics engineering

Nelimee

I don't have access to the book :/ Do you have a PDF link or something free?

Anonymous

@Nelimee www-reynal.ensea.fr/docs/iq/QC10th.pdf

Nelimee

Nice thanks!

Anonymous

@Nelimee I'd be very interested! If they contain an actual implementation on something like the IBM Q, even better

glS

@Blue @Nelimee I haven't read all of the messages. Any progress on HHL?

Anonymous

9:20 AM

@glS Sort of. At least the first step makes some sense to me know (The phase estimation)

Nelimee

I am not aware of any implementation of HHL on IBM Q except mine :/ And I'm not sure I can share it :/

I successfully implemented the algorithm on QISKit for the 4x4 matrix presented in arxiv.org/abs/1110.2232v2

Anonymous

@Nelimee License problem? Perhaps only the Hamiltonian simulation part would be very useful for me, for a start

Anonymous

You don't necessarily need to share the whole thing

Anonymous

Anyhow, it's not necessary to share it at all

Nelimee

I need to be sure that the license is OK, and also that I can share it :p

Anonymous

9:21 AM

You could just guide me a bit :P

glS

@Nelimee you are working on that? nice! Can I ask what part of the world do you work in (with as much precision as you like)?

Anonymous

I'm willing to work through it myself

Nelimee

@glS South of France

@Blue In the article I gave you, they provide an explicit decomposition of the Hamiltonian simulation.

You will have to decompose even more some gates (the square root of NOT gave me some headaches), but it's feasible :p

Anonymous

@Nelimee Okay, thanks. I'll read through it! :D (Good to know that it is feasible at least :P)

Nelimee

@gls I saw a pop up about one of your message but I did not have the time to read it... Did you message me in private or something like that? (I can't find any private chat, does it exist in SO?)

Yes it feasible! But there is not that much documentation :/

glS

9:27 AM

@Nelimee it might be an incomplete version of the message above that I sent by mistake and then deleted

Nelimee

@Blue For the square root of NOT, I started with quantumcomputing.stackexchange.com/questions/2228/… and changed a little the circuit of the sqrt(c-NOT)

Ok =) You work on HHL also? @glS

glS

@Nelimee not directly. I'm currently more on the side of classical ML applied to QM now, but I am trying to get a deeper understanding of QC and in particular QML algorithms. I would like to also go in that direction in the future. I started looking at those papers when I started my PhD but didn't get much out of it. I find kind of appalling how cryptic that whole area is

though it's probably just because no one understands it well enough to clear the waters

anyway, fully understanding HHL is kind of a dream of mine at this point lol

Nelimee

Yeah that's probably the problem :p

Anonymous

@glS I mean is it that hard? We could probably get through the details with around 10+ posts on the main site

Anonymous

Theoretically, it doesn't seem that hard. However, the implementation looks hard

Nelimee

9:35 AM

HHL is easy when you use black-boxes, especially for Hamiltonian simulation

BUT, Hamiltonian simulation is hard

Anonymous

So the main problem is really implementing Hamiltonian simulation rather than HHL as a whole :P

Anonymous

I see

Quantum Computing

0

Q: How many operations can a quantum computer perform per second?

I want to know what time complexity is considered efficient/inefficient for quantum computers. For this, I need to know how many operations a quantum computer can perform per second. ჩan anyone tell me how to calculate it and what factors it depends on (implementation details or number of qubits ...

quantum-computer quantum-algorithms asymptotic-complexity time-complexity quantum-complexity-theory

glS

@Blue it's probably not that hard to go through the details, no. Although getting through and understanding the details still doesn't fully qualify as "fully understanding" for me. I feel confident in saying I "fully understand" something only when it feels "totally obvious" and intuitive. Indeed, I have yet to understand anything in QC

Anonymous

@glS I see, makes sense. We'll probably get to that stage if we stick with this for a few months (years?) :P

Quantum Computing

9:54 AM

0

Q: When can we expect the first (universal) quantum computer being able to do something useful?

The governments, big companies (list of quantum processors) and smaller ones are in the competition of building bigger and bigger quantum computers. Not unexpectedly the number of qubits of those quantum computers seem to double every year but those qubits are noisy qubits. What is a more meani...

quantum-computer

glS

damn airport internet connection

Anonymous

@glS lol

Anonymous

I know the feeling

glS

are you kidding? I've tried to send like 10 messages and none got through, and then that is the only message that gets sent lol

Anonymous

hehehe

Anonymous

10:05 AM

It's biased ;)

Anonymous

I always keep a light-weight browser on my phone for such situations

Anonymous

Something like Opera mini

glS

@Blue what parts did you manage to get in this question? quantumcomputing.stackexchange.com/q/2393/55

@Blue anyway, I was just saying that I've got time!

Anonymous

@glS Just before the $R(\lambda^{-1})$ gate part

Anonymous

3

A: Quantum algorithm for linear systems of equations (HHL09): Step 2 - What is $|\Psi_0\rangle$?

1. Definitions Names and symbols used in this answer follow the ones defined in Quantum linear systems algorithms: a primer (Dervovic, Herbster, Mountney, Severini, Usher & Wossnig, 2018). A recall is done below. 1.1 Register names Register names are defined in Figure 5. of Quantum linear syst...

Anonymous

10:13 AM

I'm still reading through the references in that answer

Anonymous

@Nelimee How do you decompose the $R(\lambda^{-1})$ into single-qubit controlled gates though?

glS

@Blue I think that is basically what I was also wondering the other day. You can write that gate explicitly, but its decomposition is not obvious

Anonymous

@glS Exactly. I am currently in the process of framing a question for that part. But maybe @Nelimee could give us some useful inputs before that

Anonymous

@glS See page 29 of arxiv.org/pdf/1802.08227.pdf

Anonymous

They're saying something related but I don't completely understand it

Anonymous

10:22 AM

Ah, but it's basically the same thing as what you said

Anonymous

They don't seem to mention the multi control part

glS

@Blue by the way, I still think you are trying to cram too much information and too many questions into a single thread

Anonymous

@glS Even at this rate I think I'll cross 10+ questions on HHL itself :P But okay, I'll spread it out a bit more

glS

I think, generally, the more bite-sized the questions (and thus likely the answers) the better

@Blue which would be very telling as to how compressed the information in that paper is! But really I don't see any problem with that (the many question, not the compressed paper)

Anonymous

@glS Okaies...I'll do that :)

Anonymous

10:29 AM

@glS I think there's another issue with the $R(\lambda^{-1})$ i.e. how will we know the estimated eigenvalues beforehand (in order to construct the gates)?

Anonymous

To create the gate I guess we actually have to know $\tilde{\lambda}_j$?

glS

I don't understand very well the second question in quantumcomputing.stackexchange.com/q/2393/55. Isn't the procedure the other way around? They "magically" have $|\Psi_0\rangle|b\rangle$, and they apply that conditional operation to make it into the state that you say is after the phase estimation step

Anonymous

@glS Actually in that paper they didn't even create the exact state $\left(\sum_{j=1}^{j=N}\beta_j|u_j\rangle\otimes |\tilde\lambda_j\rangle\right)\otimes |0\rangle_{\text{ancilla}}$ anywhere

Anonymous

There are apparently two different versions of the algorithm

Anonymous

10:41 AM

arxiv.org/pdf/0811.3171.pdf and arxiv.org/pdf/1802.08227.pdf are a bit different

glS

Ok I give up sorry, this connection is really killing me. Let's continue this evening when I can use a decent Wi-Fi!

@Blue where is this from? I gladly see that the calculations I did the other day were on point! =)

Aand now it sends the old message of course

Anonymous

11:06 AM

@glS Here: arxiv.org/pdf/1802.08227.pdf

Anonymous

Page 29

Nelimee

11:20 AM

I can't help you on $R(\tilde \lambda^{-1})$ part, I did not understand it correctly myself :/ I just implemented it as they present it in the 4x4 paper, and it wasn't working until I magically changed a swap before the rotations. So for the moment, I don't even know why my implementation is working ><

Anonymous

@Nelimee Eh...okay, we'll figure it out

Nelimee

And @Blue, they don't know the eigenvalue beforehand. They just perform rotations of a fixed angle, controlled by the quantum register that encodes the eigenvalue

And thanks to this control, they seems to approximate a rotation that depends on the value of the eigenvalue

Anonymous

@Nelimee And how is the multi-qubit controlling done?

Anonymous

(theoretically)

Anonymous

I mean the circuit is there for 4*4 :P

Nelimee

11:24 AM

Yeah, take a look at the circuit

It's just controlled Ry gates

On QISKit you can implement them with cu3 (warning, there is a phase in the implementation of cu3 that should not be here, look at the issue I opened on the GitHub of QISKit)

github.com/QISKit/qiskit-core/issues/546 <- Implementation of controlled Rx, you can adapt it for controlled Ry

Anonymous

@Nelimee Which issue? Link?

Anonymous

@Nelimee Ah, okay, thanks!

Quantum Computing

11:52 AM

1

Q: Density matrices acting on $\mathcal{H}_A$

For a Hilbert space $\mathcal{H}_A$, I have seen the phrase density matrices acting of $\mathcal{H}_A$ multiple times, e.g. here. It is clear to me that if $\mathcal{H}_A$ is finite ($|A|=n$), then this makes sense mathematically, because a density matrix $\rho$ can be written as $\rho \i...

density-matrix

Quantum Computing

3:22 PM

0

Q: Are all [[n, k, d]] quantum codes equivalent to additive self-orthogonal GF(4)^n classical codes?

Theorem 2 of [1] states: Suppose $C$ is an additive self-orthogonal sub-code of $\textrm{GF}(4)^n$, containing $2^{n-k}$ vectors, such that there are no vectors of weight $<d$ in $C^\perp/C$. Then any eigenspace of $\phi^{-1}(C)$ is an additive quantum-error-correcting code with parameters $[...

quantum-error-correction stabilizer-code

Andrew O

3:42 PM

Sorry I missed the chat yesterday. What'd I miss?

Mithrandir24601

3:59 PM

@AndrewO not very much to be honest. We're thinking about giving the feeds to this room better names but that's about it. Also, there were issues with @Blue's chat account recently meaning the event got deleted and recreated, so you might want to resubscribe

Andrew O

4:20 PM

Ok thanks

heather

would you agree @Mithrandir24601 that the key difference between classical and quantum probabilities (or one of them) is the way via superposition quantum probabilities affect each other? along with measurement

Anonymous

The main difference really is quantum interference

Anonymous

It's the heart and soul of QM

Anonymous

In the classical case you do not have the cross terms involved

Anonymous

Professor Vazirani covers it well I think

Mithrandir24601

4:24 PM

@Blue One word: waves

Anonymous

@Mithrandir24601 Eh?

Mithrandir24601

@heather I would say that it's the violation of Bell/CHSH-type inequalities

Anonymous

en.m.wikipedia.org/wiki/Wave_interference#Quantum_interference

Anonymous

@heather ^

heather

@Mithrandir24601 hmm, good point

i'll have to think about how to write this part up.

Mithrandir24601

4:27 PM

@Blue waves interfere. I don't think you're justified in saying that 'quantum interference' is what makes quantum special as you're just moving the argument to 'what's special about quantum interference?'

heather

i kinda want to finish writing the "layman" portion of the answer and then go back through explaining and formalize everything in a second part, but it's going to be really long.

Anonymous

@Mithrandir24601 Wait a bit. Wave inference (of say light) isn't a classical phenomenon in the first place. Are you talking about say interference of sound ?

Anonymous

Thats isn't what I meant by quantum interference.

Quantum Computing Meta

1

Q: Listing all Nielsen & Chuang exercises as a resource for future generations of QC students

Like many quantum computation researchers, back when I was first learning the basics of the field, I relied heavily on Nielsen & Chuang's "Quantum computation and quantum information" textbook. However, one frustrating aspect of doing so was that no official set of solutions was ever released fo...

discussion feature-request

Mithrandir24601

@Blue interference of anything that's described as a wave

Anonymous

4:32 PM

@Mithrandir24601 Eh, there is considerable difference between classical and quantum interference. The latter means something completely different

Anonymous

Secondly "waves" isn't a part of QM in the first place. We just deal with "wavefunctions"

Mithrandir24601

@Blue at its heart, it's still a variety of wave interference. Double slit is wave interference. If you show me an experiment and go 'look, interference!', I'm not going to assume that you've got a quantum system

Anonymous

@Mithrandir24601 I'm very confused about how you defining a "wave" in the first place....

Mithrandir24601

@Blue and? If you say the difference is that one is quantum and the other is classical, you're just moving the question again

@Blue something that can be described as a function that evolves using some type of wave equation

Anonymous

@Mithrandir24601 I wouldn't say that, no. However I'll definitely say that if you want to explain the interference of waves on the surface of water using QM, you'll face some issues

Andrew O

4:40 PM

Looks like we'll be at 3 months of public beta in two weeks.

Any word on what comes next?

Anonymous

Yes, cross terms can occur classically too, but that's in a different context.

Anonymous

In this case we are comparing coins and qubits

Anonymous

While cross terms will exist for the latter it won't exist for the former

Mithrandir24601

@Blue It's definitely possible in principle as classical non-relativistic physics can in principle be described by QM

Anonymous

@Mithrandir24601 I really would like to see that explanation written out . That is, how quantum interference causes interference of water waves :P

Andrew O

4:42 PM

@heather I thought is was more that classical is probability and quantum is probability density.

At least that's the argument aaronson makes in his book

Mithrandir24601

@AndrewO I believe we keep going in public beta until we get things like >10 questions per day (not that there's a huge distinction between this site and a graduated version)

Anonymous

If that's not what you are pointing at, we're really talking about different things

Anonymous

Yes, all classical phenomena are explainable by QM but that doesn't mean classical interference = quantum interference

Mithrandir24601

@Blue I'm not saying that, I'm saying that QM can describe a water wave

Anonymous

@Mithrandir24601 Sure

Anonymous

4:45 PM

But that's not gonna be very easy either :P

Anonymous

Anyhow I think we're over thinking this one

Mithrandir24601

@Blue if what you mean by quantum interference is e.g. the thing that causes antibunching, then yeah, that's non-classical. It also violates an inequality describing the boundary between classical and quantum physics :P

heather

5:01 PM

@Blue I think with my latest edit I have answered all your points (except perhaps the vector one; but I'm not going to talk about isomorphisms and such).

I need to start working on adding pictures/diagrams and links now, and then maybe I can tackle a more formalized section. We'll see what I have time for.

hmm, i need to run through for grammar. and expand on cathode ray tubes.

Mithrandir24601

@Blue for your point 1 on @heather's post, this feels more like a point on interpretations

4 feels a tad too mathematical to me, to be honest. There's nothing wrong with analogies, especially one that is exact under some subset of all possibilities

Anonymous

@Mithrandir24601 Let's take something much simpler: Quantum interference is precisely what helps us explain YDSE, something that classical interference in wave optics cannot.

Mithrandir24601

Also, physicists tend to go to fair lengths to ensure that energy is bounded from below, but this isn't quite what you're referring to, I think

Anonymous

@Mithrandir24601 The infinite dimensional case is really very important and it isn't isomorphic to our notion of "arrows" in any way. Infinite dimensional spaces are extremely common too!

Mithrandir24601

@Blue The amplitude of a classical wave through a double slit is very strongly related to the probability of measuring an electron

Anonymous

5:16 PM

@Mithrandir24601 Sure

Mithrandir24601

As is that classical oil droplet experiment

@Blue again, and? An arrow is a vector, even if not all vectors are arrows, so it's a very good analogy that would give a much better intuition than throwing maths terminology in the answer

Everything is a rabbit hole, so in explaining anything you have to stop somewhere

Anonymous

@Mithrandir24601 I never said that's not good intuition but my point was it introduces misconceptions in people like "operators" are "infinite dimensional matrices" and "infinite dimensional vectors" are analogous to "infinite dimensional arrows". While my nitpicking is indeed irritating, I say this precisely because I struggled a lot due to these exact misconceptions (which various introductory level books seem to contain)

Anonymous

Heather and everybody else is surely free to write at whatever depth they wish to. However, I'm also free to mention my views. And so are you free to oppose them. ;)

heather

I think even just the mention that there's more to it than the analogy can be helpful.

Anonymous

Anyhow, I believe this isn't a productive usage of either of our time, so I'll stop

heather

5:29 PM

I do appreciate your commentary @Blue, it helped the answer a lot, I think.

Mithrandir24601

@Blue ahh, fair. So this is essentially the different paths up the same mountain thing. I eventually got used to being told that what I learned up to that point was wrong :P

@Blue yeah, this is a good discussion, which is why I mentioned it in chat and not in comments

@Blue but I had a kung fu analogy! :P

Anonymous

7:01 PM

@heather Honestly though, do you really understand what you wrote about Bell's inequality violation, as in, it makes sense to you? Because it doesn't make sense to me. It's one of the more subtle topics and I'd really like a discussion about that when you're free. It's one of those topics which I'm not confident about

Anonymous

@heather Thanks

Anonymous

2

A: What is a qubit?

This is a good question and in my view gets at the heart of a qubit. Like the comment by @blue its not that it can be a equal superposition as this is the same as a classical probability distribution. It is that it can have negative signs. Take this example. Imagine you have a bit in the $0$ st...

Anonymous

Upon reading carefully, this one makes more sense to me

Anonymous

Although I still have a few confusions looming

Anonymous

@heather Is it possible to relate your Bell inequality argument with the transition matrix argument (for any number of qubits) ?

Anonymous

7:15 PM

@Semiclassical Ah, you're here at the right time. Any views on the relation between Bell's inequalities and quantum interference (as in the transition matrix written in that answer)?

Semiclassical

I actually think it might be on point.

For instance, that configuration I think is the one which would maximize the violation of the CHSH inequality.

I can ramble on a bit about that if you want.

(I'd also note that, in terms of Pauli matrices, that transition matrix is of the form $M=\frac{1}{\sqrt{2}}(\sigma_x+\sigma_z)$

as such, one has $M^\top = M$ and $M^\top M=\frac{1}{2}(\sigma_x^2+\sigma_z^2+\sigma_x \sigma_z+\sigma_z\sigma_x) = \frac12(I+I)=I$

so that $M$ is a unitary matrix.

...which, I guess, is what they said already

derp

heather

7:37 PM

@Blue let me catch up a minute here

I clicked over to an SE tab and saw I had 9 messages =)

Anonymous

@heather Come over to hbar

Mithrandir24601

9:58 PM

@Blue There comes a point when you look at something and start going "hmm... I'm not sure about this... Is that right? I want more detail here". This is how things are - there's a fundamental point at which science goes "it could be this, this, this or even this but we're just not sure" and you kind of have to live with that, in much the same way that every year you'll have to live with getting told "what you knew last year was wrong" - this *never* stops.
Many answers and published papers have this 'problem'. It's just finding out where the 'problem' is. Like Siereah's link currently on th…

Although @heather have you considered making headings in your answer to make it easier to read?

(as it is getting quite long)

heather

10:28 PM

@Mithrandir24601 that might be a good idea =P i'll do that right now.

okay, done.

Mithrandir24601

10:55 PM

@heather Nice :) I'll have a read through it tomorrow morning

heather

i have so much more i want to say, that's the problem =) do you know what the character limit is on an SE post?

Mithrandir24601

@heather Long. Very long. I'll try and track down one of the really long ones on Worldbuilding

@heather this one's quite long

Nowhere near the longest

heather

dang

so i've got some room =)

Mithrandir24601

There's apparently a 30k character limit

Ah yes, this would be it

at 32418 characters, of course

@Anoplexian woow, nice)). Was forced to trim it, it is about max length to post on SE, less then 50chars left. I guess I trimmed it about 25% to be able to post it. — MolbOrg Dec 20 '16 at 18:58

:'D

SLesslyTall

11:11 PM

The first Nielsen & Chuang exercise question is up! I will try and add more in (with answers where I can) regularly, but others should feel free to contribute

The meta community wiki post listing them all will go live tomorrow morning. I'm not staying up 40 minutes just to post that tonight. Bon nuit all.

Mithrandir24601

I'm off to bed as well, so night all!

Quantum Computing

0

Q: Nielsen & Chaung Exercise 2.1 - "Linear dependence: example"

Reproduced from Exercise 2.1 of Nielsen & Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition): Show that (1, −1), (1, 2) and (2, 1) are linearly dependent.

quantum-information

heather

@Mithrandir24601 good night =)

Transcript for

The Classical Channel