The Classical Channel

7:00 AM

@Blue: Maybe it's not so much the emphasis itself, but the attitude you have when you do it. It's as if you're on a high horse and speaking down to people. For example your comments to my answer here: quantumcomputing.stackexchange.com/questions/2443/…

The emphasis was on the word "discrete", which on its own would not be so bad, but not when you're speaking down at me as if you know more quantum than me. Can you explain why "boundary conditions" need to be present for energy levels to be discrete?

Anonymous

@user1271772 I'm sorry but I do not see how trying to correct an incorrect answer makes me seem as if I'm sitting on a high horse.

Fine Blue, maybe you will never see it, but can you not take the feedback and try to improve?

I'm telling you that what you are doing is making me feel uncomfortable. You should care to improve that whether or not you understand why I feel that way. Also, the question is not incorrect. Can you explain why "boundary conditions" need to be present for energy levels to be discrete?

Anonymous

@user1271772 Sure, I will explain it in detail. I don't have access to a PC now (on mobile), but will explain it once I reach home.

Sorry I meant to say, "the answer is not incorrect" not the "question is not incorrect". As a sophomore undergrad in electrical engineering, perhaps you can understand that someone that got a PhD in quantum computing almost a decade ago is not likely to be dumb enough to make a mistake as huge as you're claiming. Semiclassical told you that boundary conditions are what ensure states are discretized, and you didn't stop to think, maybe userXYZ is right too.. let's ask him a question

Rather than trying to act like you know some basic quantum that I failed to learn properly in my undergrad class.

Anonymous

7:16 AM

I really do not care what degree(s) you have. If you want to have a serious conversation maybe first stop pointing out who has what degrees. Let's have a fact based conversation. I'd be happy to be corrected if you find any mistake in my logic.

Blue, the "fact" is that you're still an undergrad student so it's not appropriate for you to tell a professor that they're wrong, in the way that you often do here.

Anonymous

@user1271772 This isn't an university. If you are under the impression that SE is designed to provide a higher platform to "professors" than students, I'm sorry to inform that that's not the case. SE is a democracy and a meritocracy. We only and only care about accurate questions and answers without caring about who is posting them. Moreover, it's pretty impossible to verify a person's credentials on SE. Escpecially an anonymous user's.

7:32 AM

You are right about a lot of these things, but it's the way that you say it that comes across rude. For example, you are telling me things as if it's you teaching me. I do find that rude. Keep in mind that we are all humans outside of the SE world and still can find things insulting the same way we would as humans do outside the SE world.

I'm still waiting to hear what was "incorrect" in that answer I gave you.

Anonymous

@user1271772 If I find an answer incorrect it's my (and other's) duty to point it out (to keep SE free from factual errors). If it comes across to you as rude, feel free to flag it. I've already explained what exactly is incorrect in your answer in the two comments but since you want elaboration, you'll have to wait till I get access to a PC. It's difficult to type from mobile.

7:45 AM

@user1271772 when you say things like people telling you that you're wrong makes you uncomfortable, why do you say things like "You have no clue what you're talking about." when you don't like people saying that to you?

Nelimee

@Blue github.com/nelimee/quantum-tools/blob/master/HHL/optim_hamil.py

Mith: When did I say "people telling me that I'm wrong makes me uncomfortable"

45 mins ago, by Blue

@user1271772 I'm sorry but I do not see how trying to correct an incorrect answer makes me seem as if I'm sitting on a high horse.

44 mins ago, by user1271772

I'm telling you that what you are doing is making me feel uncomfortable. You should care to improve that whether or not you understand why I feel that way. Also, the question is not incorrect. Can you explain why "boundary conditions" need to be present for energy levels to be discrete?

Specifically, the "I'm telling you that what you are doing is making me feel uncomfortable." bit

@user1271772 for everyone to get along and constructive and enjoyable communication

Anonymous

If it's wasting you're time, as I said earlier you are free to leave and come back later at your convenience. Telling a moderator that they're wasting your time whereas they're trying their best to be reasonable with you is unacceptable.

Anonymous

→ 4 messages moved to trash

Anonymous

7:59 AM

→ 1 message moved to trash

Anonymous

→ 1 message moved to trash

Anonymous

@Nelimee Could you explain that program a bit?

Anonymous

I couldn't really understand what you did there

Mith: If you want everyone to get along and for there to be constructive communication, why not tell Blue not to talk down to others the way s/he always does? As if s/he knows so much more than everyone else? Especially when s/he is a sophomore student with still a long way to go before reaching the level of QM experience that some of us have?

8:19 AM

@user1271772 telling someone an answer is wrong is constructive for someone, whether it's the person who wrote the answer, the person claiming the answer is wrong or even both. I don't see Blue talking down to others, so it's hard to say such a thing but please do flag anything that's inappropriate. I also don't care about what degrees people do or do not have

As well as what's already been said above, I remember being told in undergrad that having a degree doesn't automatically make anyone better or smarter than someone who doesn't

Mith: It's the way that he does it. When you didn't understand something in your QM class, did you tell your professor that s/he was incorrect the way Blue does it? Or did you ask "what is it that I'm mis-understanding?" Anyway Blue is not going to have a good life outside of SE if s/he keeps acting this way in "real life" where by "real life" I mean life outside of SE. You do have to treat people with respect, and perhaps more when they're older or have more degrees.

"I also don't care about what degrees people do or do not have" you are in for a lot of trouble if that is true. I hope you don't treat the post-docs in your lab the way you treat the undergrads in your lab.

Nelimee

@Blue not now, maybe later or tomorrow :)

Quickly: I simulated my circuit (HHL) and computed the unitary performed by the circuit with the unitary_backend. Then, I computed the error between this unitary matrix and the expected one. This error is the quantity to minimise and the parameters are the 5 floating point parameters that are fixed in the article :)

@user1271772 Why do you find being taught something that you don't know rude?

@user1271772 actually yeah, I have gone up to lecturer after a lecture before and outright said "this is wrong". In this case, there's already been a discussion about this, so it's more like the time I was told by one prof that what a different prof had said was wrong, which makes things more complicated from that perspective

8:34 AM

One of the core tenants of SE is that everyone should be treated with respect, regardless of whether they're a professor with a PhD or a elementary school student, yes.

Anonymous

@Nelimee No worries! BTW could you give me your mail if you don't mind? I have something to discuss with you (it's better if said by private mail)

Anonymous

@Nelimee Oh, I see

Anonymous

I'll have to read the code a bit more thoroughly

You yourself have said that they are often correct when pointing out a technical inaccuracy. Why, then, do you prefer that those technical inaccuracies by perpetuated, by disallowing a correction because you feel that the person who (correctly) points out these inaccuracies is not "experienced" enough?

Anonymous

→ 1 message moved to trash

8:38 AM

also @blue moving messages like this in the middle of a conversation is not a good idea, and will only serve to exacerbate the situation.

Anonymous

@Mithrandir The "please don't message me" part was outright rude. However, if you can handle the situation well, I'm okay with it

@Mithrandir: I have no time to talk to you. Can you please not ask me questions? You are failing to see that my problem is with the rudeness with which Blue has been treating me. I never said that any "correction" should be disallowed.

Anonymous

@Nelimee I'm reading your code and jotting down my confusions. Ping me when you're free tomorrow :)

Anonymous

@user1271772 Please provide examples of messages or comments where I was rude instead of accusing without proof.

Also, I'm going to suggest something radical... @Blue, if you're aware that you're being perceived as rude, even if you think it's unwarranted, it's worth it to just try to word your comments a bit more proactively polite instead of passively not rude. Even if you think the offense is unwarranted, can you take the high road and just make an effort?

Because honestly, what I'm seeing here is user1271772 overreacting and Blue being stubborn. Can you please both make an effort?

I happen to be sick of the tangling here and I know your moderators are too. Would it really be such an effort to assume good faith on both sides and to make an effort to work together?

You're both knowledgeable people and your contributions to the site are valuable. I'd really appreciate if you can work out a way to realize that neither of you really wants to have these arguments and that you can work together, in the name of science.

8:47 AM

@Mithrandir: that is sort of what I was trying to explain to Blue. If you think your professor has said something wrong, you don't flat out tell them "this is incorrect", you ask them what you've misunderstood. More often than not the professor is not wrong. I do think Blue is being a bit stubborn. Perhaps you can say I'm over-reacting, but I think I am just "reacting".

Anonymous

@Mithrandir I'm sorry Mithradir. It really isn't my intention to be rude to anyone. But user1271772 is now taking this to the point of no return: It's pretty much impossible to make an effort unless I'm provided with actual examples of where I'm being rude.

I do agree with you Mithrandir that I do not like to have these arguments. I do not appreciate Blue commenting on so many answers telling me that I'm wrong about stuff that isn't actually wrong but rather Blue is just trying to learn more by telling me it's wrong so that I explain it more clearly until s/he understands.

"But user1271772 is now taking this to the point of no return" ~ semantically null.

"It's pretty much impossible to make an effort unless I'm provided with actual examples of where I'm being rude." .... I don't have time to give you examples of every little thing. Why don't you go through your comments on my answers (for example the one that started this conversation) and reflect on it yourself. Why don't you try behaving the way you did in your comments to me, in the world outside of SE and see what happens?

For example tell your professor s/he is wrong about something s/he has a PhD in, and see what happens. Try it in real life and see what happens. I bet you won't even try that because you know it will turn out crappy for you. Treat people the way you would like to be treated. Treat them the way you'd treat people if you were talking to them face-to-face.

@Blue Looking at this situation it seems to me that one of the best ways to improve this is for you to drop the grudge, and don't get into an argument. Arguing whether or not you were actually rude is not going to help anything AT ALL now. It's just going to make things worse. To channel my inner Shog: Drop it.

Anonymous

@Mithrandir That's actually a good advice. Thanks.

I'm not asking you to sit down a write a textbook together. I'm asking you to be a bit more considerate to each other and recognize that in the end, it's all so that misinformation isn't perpetuated. You're working towards the same goal; you'll be better off if you don't waste your energy arguing with each other and just focus on actually making sure that the information is correct.

I'm frustrated here.

I'm frustrated because I see two users who could potentially be a great asset to the site instead having mostly petty arguments.

You're spending your limited energy on the wrong things, and it's harming the site and users around you.

Anonymous

9:12 AM

@Mithrandir I can completely understand and sympathize with your point of view. However, if conflicts were that easy to resolve the world would have been a much better place to live in. Fortunately, or unfortunately, it's much easier to simply use the ignore button and go on with our individual lives. On that note, maybe it's time to change the topic!

Q: What methods exist for cooling superconducting quantum computers?

@Blue See... conflicts can be that easy to resolve, if both participants realize the nature of what they're doing. Don't take a defeatist point of view; do what you can to resolve your end of the problem. It takes two to tango; don't let there be two willing to tango.

Anonymous

9:25 AM

@Mithrandir Well, it's a matter of debate whether refraining from a conflict classifies as the resolution of the conflict. ;)

Anonymous

9:47 AM

→ 1 message moved to trash

Anonymous

4

Existing super conducting quantum computers need to be cooled near absolute zero. For example, some of D-Wave's machines are cooled to about 20 mK. Their design uses a dilution refrigerator. Are there any other cooling methods for superconducting quantum computers besides a dilution refrigerat...

d-wave superconducting-quantum-computing dilution-refrigerator

Anonymous

This is an interesting question. I requested Daniel Sank to provide an answer. Let's see

12:36 PM

4 hours ago, by user1271772

For example tell your professor s/he is wrong about something s/he has a PhD in, and see what happens. Try it in real life and see what happens. I bet you won't even try that because you know it will turn out crappy for you. Treat people the way you would like to be treated. Treat them the way you'd treat people if you were talking to them face-to-face.

I find this wrong on so many levels. A qualification does not mean you cannot be corrected by someone without such qualification. "Professors" and such are wrong all the time, and that is perfectly normal and should be pointed out as necessary.

I'd... advise not poking sleeping bears.

At this point, it might be a good idea to just let it rest. We don't really need the argument starting again.

@Mithrandir I know, I kind of couldn't resist. I don't mean this as directly related to the discussion at end though, more as a general comment about academia

@Mithrandir feel free to delete it if you like

Anonymous

Fwiw, I personally have directly pointed out several times to professors when they were wrong. Sometimes they turned out to be actually wrong and sometimes it turned out that I was wrong (which I was happy to accept since I learnt something new). I developed very great respect for those who accepted their mistakes in a dignified manner and a lifelong hate for those who simply shut me down. Respect, really, isn't something a degree can earn for someone. Respect is earned by actions.

1:06 PM

@Blue I agree. By the way, what do you not find satisfying in this answer to your question (if anything)?

(I'm not saying you should be satisfied by it, just asking, if you are not, what you think should be added to it)

Anonymous

@glS Oh. He uses determinism as an argument. But we're not really sure if QM is deterministic or not (the notion depends on the interpretation we use).

Anonymous

Of course Copenhagen interpretation postulates that QM is non-deterministic

Anonymous

BTW I was trying to learn a bit of Bohm interpretation from SemiC

Anonymous

Afaik, Bohm and Many-Worlds are still in the development process and might even turn out to be wrong

Anonymous

But until I learn about them thoroughly, can't make any comments

Anonymous

1:12 PM

Bohmian interpretation has some pitfalls like there's no established explanation for the delayed choice experiment (afaik)

Anonymous

SemiC linked a nice paper about the topic, but I didn't yet get time to read it

@Blue eh, that's a good point. I myself found me thinking a lot about these things. I really don't see how the whole collapse thing can make any sense at all. It seems to me that the "collapse" has to arise from something like decoherence into the environment, but I don't find the explanations in this direction given by Zurek and co. at all sufficient to resolve the issue.

as for many-world.. I don't know, I never really got to study the papers deeply, but I don't see how something of that sort can be useful for the description of a physical system

@Blue was this in this chat?

Anonymous

@glS "It seems to me that the "collapse" has to arise from something like decoherence into the environment," Actually, that seems much more reasonable to me. I personally would prefer "entanglement with the large environment (say a billion atoms)" implying the observed collapse

Anonymous

But well, physics shouldn't depend on what I prefer :P

Anonymous

However, yes, I'd like to properly learn the foundations of quantum mechanics someday, thoroughly

Anonymous

1:20 PM

These are the actual juicy parts of physics ;)

Anonymous

@glS Nope, in hbar

Anonymous

Linking in a bit

Anonymous

in The h Bar, Jun 27 at 5:30, by Semiclassical

but Bohm, at least as far as non-relativistic QM goes, is experimentally equivalent

Anonymous

@glS ^

Anonymous

Follow the conversation

1:22 PM

@Blue anyway, I was wondering about how to answer the question. I believe at the end of the day the question is really about the difference between $|0\rangle\langle 0|+|1\rangle\langle1|$ and $|0\rangle+|1\rangle$, because a "classical spinning coin" simply means for our purposes something with a definite, if unknown, state. An answer could then be to point out that in the coherent case, you can apply an operation that destroys the uncertainty in the result (...)

(...) this is not something that you can do in the classical case: if the uncertainty is classical, then no matter what operations you do, some degree of uncertainty in the final outcome should remain. In the quantum case you can instead, e.g., apply $H$ to |0>+|1> and end up with a certain result. This indicates that what before appeared as uncertainty was really a completely well-defined state

in other words, classical uncertainty necessarily implies a lack of knowledge about the system, and this lack of knowledge cannot be decreased no matter what you do. In the quantum case, on the other hand, the "lack of knowledge" is "fake", because it can be removed by properly manipulating the state

Anonymous

@glS Umm. Yes, supposing the qubit is in a state $|0\rangle+|1\rangle$, and you apply a gate like $H$, the qubit will be in a certain definite state. But we cannot know for certain if the qubit ever actually started in the state $|0\rangle + |1\rangle$, in the first place, isn't it? That classical uncertainty is present even for a qubit, isn't it?

Anonymous

Actually, I'm not sure how you're defining "classical uncertainty"

@Blue well, you can say that you prepare the state always with the same apparatus, and measuring in the computational basis you see the outcome sometimes being 0 and sometimes being 1. If this was "classical uncertainty", it would mean that every time the state actually was either 0 or 1, it was just your fault in not being able to prepare it always in the same way. If this is the case, then even if you apply other operations afterwards, this uncertainty will always remain (...)

(...) no matter what you do, you will always end up in one of two states (one arising from the evolution of the 0 state, and the other from the evolution of the 1 state). But this is not what happens: in the quantum case, if after having observed 0 50% of the times and 1 50% of the time, you apply an H gate and measure again, you will see that all uncertainty magically disappeared

Anonymous

1:39 PM

I guess the real question is for both your qubit and coin if you assume that you know the initial conditions and the time evolution with infinite precision, can you still distinguish them, on basis of outcome statistics? I think all the outcome statistics which can be produced by a qubit can be produced by a coin too

the only way to explain this is that it must have been the case that even if you before observed an uncertain result, the state was actually well-defined and always prepared in the same way

@Blue if it is a coin, then even after applying H you will still have uncertainty in the outcomes

Anonymous

@glS Why?

Anonymous

That I do not think is uncertainty but rather outcome probabilities

Anonymous

You can get exactly the same transition matrix for both the qubit and the coin

Anonymous

(but we need to keep in mind that for qubits we use 2-norm whereas for coins we use 1-norm)

1:42 PM

@Blue mathematically it can be seen because the "coin" is a mixed state, which you can represent as the two-dimensional identity. Then, no matter what operation you do to it, it will keep being an identity, and therefore produce uncertain outcomes: $U\rho U^\dagger = \rho$ if $\rho=I$ and $U$ is unitary

@Blue well the uncertainty is always in the outcome probabilities, isn't it?

@Blue in other words, no matter what question you ask the coin, you will get 50% of the times one answer and 50% of the times a different answer. On the other hand, for the qubit, you will get a deterministic answer if you choose the right question

"question" here meaning "measurement"

Anonymous

@glS Umm? If you know there's a 50% probability of outcome being 1 and 50% probability of outcome being 0 (since you know precisely the initial conditions and time evolution), I don't think that's called uncertainty. It would be uncertainty if you weren't sure whether it is 50% 1 and 50% 0 or 49% 1 and 51% 0

Anonymous

I guess you're using a different definition of "uncertainty" altogether

@Blue eh, define uncertainty then. In the classical world, every measurement has one and only one possible outcome. Uncertainty only arises from the lack of knowledge in the initial conditions

within QM it's trickier, but I wouldn't say totally different. Uncertainty can be "apparent" when measuring a pure state in some way, or still being "classical", that is, due to our lack of knowledge of the state being measured

I mean, knowing the probabilities doesn't mean that there is no uncertainty, does it?

Anonymous

@glS And what prevents us from saying that every measurement outcome in QM also has a definite outcome? (if I use a deterministic interpretation)

Anonymous

I'm trying to separate the notions of exact knowledge of density matrix or transition matrix, with measurement outcomes

Anonymous

1:51 PM

As far as I'm concerned, I would consider that if you have full knowledge of the density matrix there's no uncertainty (in the quantum sense) involved

Anonymous

But you're talking about classical uncertainity in the density matrix sense

A: How is a single qubit fundamentally different from a classical coin spinning in the air?

@Blue well, pragmatically speaking, because that's not what is observed. Imagine you have a black box that produces "states", that are then sent into another black box which "measures" the state in some way. There are different buttons in the second black box, which allow you to switch between different measurements to perform. When the input is a coin, you see 50/50 outputs no matter the button you chose on the second box. When the input is a qubit, some buttons give you 50/50 outcomes (..

..) but others give you always the same outcome

@Blue yes, what I mean is that pure states are completely well-defined states of nature. There is nothing more to know about them, and the uncertainty in outcomes is only "emergent". On the other hand, mixtures are different, because in principle there is information that you could add to your description of the state that will make it pure

Anonymous

@glS Oh. Good point. We were assuming that the outcomes of the coin can be continuous. That is, it can even land on an edge or we could stop it mid-air. So every outcome statistics of the qubit can be replicated with a coin

Anonymous

I need to make some correction to the question

Anonymous

3

How is a single qubit in a Bell state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ any different from a classical coin spinning in the air (on being tossed)? For both of them, the probability of getting heads is 1/2 and getting tails is also 1/2 (we can assume that heads$\equiv|1\rangle$ a...

Anonymous

1:54 PM

See my comments under DW's answer

@glS it doesn't have to be 'into the environment' - decoherence can also arise in other ways, such as degradation of reference frames

Anonymous

> For both of them,
> the probability of getting heads is $\frac{1}{2}$ and getting tails is
> also $\frac{1}{2}$ (we can assume that $\text{heads} \equiv |1\rangle$ and
> $\text{tails} \equiv |0\rangle$ and that we are "measuring" in the heads-tails
> basis).

@Blue I don't think allowing continuous outcomes makes a difference. It only additionally murks the waters. I agree that a coin can replicate whatever outcome statistics is produced by a qubit. The difference is in the way you can change the outcome statistics when operating on them. Qubits allow you to produce 100/0 outcome statistics from a 50/50 one, coins do not.

(more precisely, I would say that a qubit allows you to go from 50/50 to 100/0 and then back to 50/50. A coin could still go from 50/50 to 100/0, but then not go back to 50/50)

mh, actually I'm not 100% sure about my last sentence

Anonymous

@glS If I define the time evolution of the coin to be something like - lands on edge (50-50 probability now) ----> hit it on the 1 face so that it falls down as 1 (100 % of the time) ----> re-toss it such that it falls on an edge (this time with no external hit). Now, measure the final outcome after the coin falls. That replicates what you want

Q: What makes a theory "Quantum"?

@Blue I mean, sure. I can simulate on my computer whatever outcome statistics of whatever state of a qubit. Or equivalently, I can simulate it with a "coin flipping computer".

Anonymous

2:07 PM

Yup. So it's not possible to distinguish between a qubit and a coin in this manner. But maybe there is some other way - we were discussing this in the hBar a few days ago

Anonymous

BTW interesting thread:

Anonymous

122

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 ...

quantum-mechanics classical-mechanics foundations

Anonymous

"All of this is to say that it seems difficult (and maybe impossible) to try to find a single distinguishing feature between classical and quantum mechanics without considering composite systems, so if that is what you want, I'm not sure I have an answer.

Anonymous

If you do allow for composite systems though, it is a pretty unambiguous distinction. Given this, it is perhaps not surprising that all the experimental tests we have which demonstrate that the world is quantum and not classical are based on entanglement."

Anonymous

Things become much simpler when you consider two or more qubits

Anonymous

2:09 PM

Entanglement produces certain statistics which a system of coins can never produce

@Blue yea I know the post. If you consider correlations then yes the answer is simpler, as you then do things in the quantum case that you straight-out cannot do classically. Although I do have half a feeling that you are cheating even then. These results are, as far as I understand, due to locality assumption. In other words, quantum correlations are more powerful in that case only because of nonlocality

but then again, even in those cases you can still simulate all the outcome statistics on your computer. You just cannot do it if you have two spatially separated devices

and then my question is, is this really different than the single qubit case? After all, the state of a single qubit can still be thought of as an entangled state between its two possible modes (0 and 1).

what I'm saying here is that $|0\rangle+|1\rangle$ is in some sense the same as $|10\rangle+|01\rangle$, where in the second notation I'm writing the state through the occupation number of the two different modes. Then the only difference between a Bell scenario and a "single qubit" scenario is that the first and second modes are spatially separated in one case, and not in the other

damn, I really have to go write a draft. I'll think about this and maybe we can continue the discussion later.

@Mithrandir24601 what do you mean by degradation of reference frames?

@Mithrandir24601 also, are you by any chance going to be in Nottingham in the next week?

Anonymous

2:26 PM

@glS I think you have a valid point. Can you simulate the outcome statistics of a Bell pair using two classical coins, on a computer, without creating a correlation between the two coins? I guess no. But if we allow for artificially induced correlations, then yes

Anonymous

@glS Okay, no worries. See you!

Anonymous

It was a good discussion

2:59 PM

@glS I'm on my phone, so it's kind of hard to explain but I can try and remember to send a couple of papers tomorrow? Terry Rudolph did a really good review

@glS no, sorry - I won't be outside of Bristol (as far as I know) until the 26th(?) July, when I'm heading to Dublin for a few days

Q: How is entanglement achieved between two particles in quantum computing?

0

Many texts (especially meant for public consumption) discussing quantum mechanics tend to skim over exactly how entanglement is achieved. Even the Wikipedia article on quantum entanglement describes the phenomenon as follows: "Quantum entanglement is a physical phenomenon which occurs when pairs...

quantum-computer quantum-entanglement

@Blue not without adding something artificial to make it effectively 'non-classical', which is part of why it's so important, really

Anonymous

@Mithrandir24601 Yeah, well, it comes down to our definition of "classical" and "quantum" again. In the usual definition, creating those artificial correlations would make the system "quantum", in the sense (we can say) that it is the main defining property of what separates quantum from classical

Anonymous

This point is unfortunately not well-explained in most textbooks

Anonymous

3:15 PM

I guess I should have said "one of the main", but it conveys the point anyway

Q: How the joint state of these qubits is derived?

0

Can someone show to me the steps to derive the joint state at the bottom of this image, please? I tried to follow his explanation but I didn't get the same results… This is taken from the lecture notes of Ronald de Wolf in case it may help

quantum-gate qubit-state quantum-state

Q: Currently, what are the best structured courses available online on quantum computing?

3:53 PM

0

As we make inroads into Machine Learning, there seem to be plenty of respectable courses available online via Coursera, edX, etc. on the topic. As quantum computing is still in its infancy, not to mention, incredibly daunting, it is vital that easy to understand, introductory courses are made ava...

quantum-computer

Q: Quantum Teleportation trough linear algebra

6:15 PM

0

I am learning to manipulate Qbits and recently I saw the teleportation algorithm. I read about it in two places: Wikipedia and Lecture Notes from Ronald de Wolf (Page 7, 1.5 Example: Quantum Teleportation). I'd like to understand how to operate with Qbits using linear algebra when an entaglement...

quantum-algorithms quantum-entanglement

7:39 PM

@user1271772 if you find something rude, the best course of action is usually to flag the offending post/comment/message

Mith this is becoming exhausting. You keep telling me that. But flagging is useless in terms of explaining to the person how I feel about what they wrote. They won't know that they were flagged unless a moderator tells them, and that only happens in the rarest of instances. You don't explain to someone that they're being too rude every single time something is perceived as rude.

Also the moderators are a high school student, a PhD student who is rather friendly with Blue and seems not to be neutral, and a post-doc who isn't very active here, so flagging isn't necessarily going to achieve what I want, which is for Blue and gIS to change their attitude in the way they are telling me that they think I'm wrong.

Can you just stay out of it and let Blue and gIS see what I wrote?

Anonymous

7:59 PM

@user1271772 Consider a particle in an infinite one-dimensional box system. The general solution to the SE for such a system would be $\psi_E = Ae^{ikx} + B e^{-ikx}$. If I use the standard boundary conditions $\psi(x=0)=0$ and $\psi(x=l)=0$ you'd get the quantized eigenfunction set and the energies $E_n$.

Anonymous

Now, you could say that the two lowest discrete orthogonal energy levels are $|0\rangle$ and $|1\rangle$. Notice that you couldn't have proven that the energy eigenvalues are discrete without considering the boundary conditions.

I'm going to be blunt now. You are constantly saying that they are being rude, but in the process of making such a statement, you are being rude yourself as well. You are being rude to your moderators, to the other chat members, and myself, on the grounds of someone else being rude. That's not how it works.

3

You've been told many times that the proper way to go about dealing with perceived rudeness is a flag, or raising a constructive, non-accusatory post on Quantum Computing Meta. You have been asked by several moderators to stop discussing this here. Why do you insist on starting arguments here when you know very well that this is not the proper place for that?

4

The age of a user, or the qualifications they have, is irrelevant. What matters is the way they act, and the information they present. Every user is on an even footing here. This is not a lecture hall or a classroom; it is an internet site.

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Anonymous

Let me know if you can give me any example of a qubit system where eigenvalues are discretize-able without considering the boundary conditions.

@Mithrandir: If find your last message rude.

@Blue: What about for a spin-1/2 particle where the two levels are +1/2 and -1/2?

ArtOfCode

lurks ominously

Anonymous

8:06 PM

@user1271772 Let me know when you write down the Schrodinger equation solutions for that.

@Blue: Why should I do that?

A somewhat problematic example is a particle on a ring. The only boundary condition there is that the wavefunction be a single-valued function

Anonymous

Eh, the ring one is pathological :P

i don't really think that's an example, though, since 1) that's still a boundary condition, and 2) it's still a restricted geometry

@Semiclassical I'd argue that's a periodic boundary condition though

8:08 PM

sure

@Semiclassical: you have a strange definition of "boundary condition". The fact that a wavefunction has to be a single-valued function, is part of the definition of a function.

I suppose the better way to put it is that the single-valuedness condition enforces periodic boundary conditions.

In any case, I don't think you generally get quantized energy levels from the Schrodinger equation without boundary conditions of some sort. But quantum mechanics is not only the Schrodinger equation.

But to talk about the origin of spin, one has to talk about the Dirac equation, and I don't really feel like thinking about that right nwo

What are the boundary conditions for a Morse potential @Semiclassical?

@Semiclassical: The Dirac equation is one way to understand the origin of spin. But how about H|psi> = E|psi> where H is a 2x2 matrix?

@user1271772 $\psi$ and its derivatives vanish sufficiently fast at infinity

same as usual for a bound state subject to a confining potential

@user1271772 In what sense does the latter tell you about the origin of spin? It describes the situation, sure, but it hardly explains why you should be using a 2-by-2 matrix in the first place.

I wouldn't call that an explanation of spin. Just a statement of how spin is typically presented formally in quantum mechanics.

I'm not saying it tells us about the "origin" of spin, the Dirac equation does

8:16 PM

Then I have no idea what your statement was trying to say.

JoErNanO

Folks please let's keep things civil in here.

H|psi> = E|psi> is the TISE.

3 mins ago, by user1271772

@Semiclassical: The Dirac equation is one way to understand the origin of spin. But how about H|psi> = E|psi> where H is a 2x2 matrix?

The latter is how you incorporate spin into the Schrodinger equation.

Sigma_z|psi> = E|psi> is a Schroedinger equation, what are the solutions to it?

The former is how you understand why spin should show up in this manner in the first place.

8:18 PM

What are the solutions to it @Semiclassical?

Depends on what basis you insist upon. But, certainly, once you allow yourself 2-by-2 operators, you can have quantized energy levels.

Precisely, and what is the Hamiltonian for a qubit?

If you're arguing that the matrix formalism allows you to get quantized energy levels without using boundary conditions, I don't disagree.

Thank you.

However, I'll point out that in that case you're also not really doing the schrodinger equation in the usual sense of $H=-p^2/2m+V$

even if you allow $V$ to be a spin-dependent potential

8:19 PM

That's not the only Hamiltonian in the world

the moment you include the kinetic term, you have to include boundary conditionso

So if the claim is that not all Hamiltonians of interest in QM involve a kinetic term, then sure

Since when do you describe the kinetic energy of qubits?

Depends on the system.

@Semiclassical I'd also make the point that such a qubit is either confined, giving boundary conditions, or a photon, which is somewhat complicated

Yes anyway I think Blue could have been more polite about the way s/he was trying to lecture me on basic year-1 QM.

8:22 PM

@user1271772 Drop it.

You've been asked many times to stop bringing it up here.

An observation. : an anagram of 'drop' is 'prod.'

maybe that explains why people so often fail to follow the advice to 'drop it'

8:43 PM

@Mithrandir24601 it's also worth noting that the parameters of a qubit usually arise as a result of said boundary conditions. for instance, if i've got a two-site system, then the single-site states will overlap and therefore one can expect an interaction term

the resulting cross-term in the two-site Hamiltonian should be something on the order of $e^{-\Delta E/\hbar}$ where $\Delta E$ is the barrier height

main point is that it depends on the geometry of the two-site system, and in that sense is very much in line with 'boundary conditions'