 12:00 AM
@ChainedSymmetry I don't understand what the problem is... But I really am tired, so I'm off to bed, sorry! @Mithrandir24601 Good night!

8 hours later… 8:16 AM
@Mithrandir24601 and glS, I appreciate the discussion. I'll try to make my statement precise in hopes of understanding where the disagreement is.
The specific question (originating here) regards the probability that a single qubit measurement returns $\frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}$. 8:43 AM
As a mixed state $\frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}$ is a statistical ensemble of pure states, and cannot be described by a single ket.
When the mixed state $\frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}$ is measured, the wavefunction collapses to $\vert 0 \rangle$ or $\vert 1 \rangle$.
I went back to some of the basic literature to see if I was loosing my mind. Nielsen and Chuang states this almost verbatim.
"Recall that measurement of a qubit will give only either 0 or 1. Furthermore, measurement changes the state of a qubit, collapsing it from its superposition of $\vert 0 \rangle$ and $\vert 1 \rangle$ to the specific state consistent with the measurement result. For example, if measurement of $\vert + \rangle$ gives 0, then the post-measurement state of the qubit will be $\vert 0 \rangle$.
Why does this type of collapse occur? Nobody knows. As discussed in Chapter 2, this behavior is simply one of the fundamental postulates of quantum mechanics." (section 1.2)
It goes without saying that $\vert 0 \rangle$ and $\vert 1 \rangle$ can represent any number of different physical states depending on the experimental setup. But for a given computational circuit, this basis is well defined (perhaps up to permutation of basis states).
Where is the disagreement? 8:58 AM
@ChainedSymmetry this is confusing notation at best. What do you mean "$|\psi\rangle$ as a mixed state?" (where $|\psi\rangle=|0\rangle+|1\rangle$ here)
@ChainedSymmetry this is also a very weird statement. Is it verbatim from N&C? Can you tell me the pages? If you measure in a basis $\{|u_k\rangle\}_k$ the post-measurement result is one of the $|u_k\rangle$. If you measure in the $\{|+\rangle,|-\rangle\}$, the post-measurement result is either $|+\rangle$ or $|-\rangle$. There really isn't any doubt about this @glS Yes, this is copied verbatim (between the two quotation marks). Page 15 in 10th Anniversary Ed. @ChainedSymmetry right, I see it. But he means that you are measuring the state $|+\rangle$, as in, the state being measured is $|+\rangle$. In other words you are measuring $|+\rangle$ in the computational basis $|0\rangle,|1\rangle$.
then of course the post-measurement states will be either $|0\rangle$ or $|1\rangle$, no one is objecting that @glS But that's the extent of what I've been saying...
and people seem vehemently opposed to it, and are downvoting my answer, when it basically just repeats basic material from N&C @ChainedSymmetry this is the problem. But then how do you define the computational basis in the situation pointed out by @Mithrandir24601 before? Or if you have a photon measured first in one basis and then in another one
@ChainedSymmetry it doesn't. I do not agree with a statement such as "a single measurement can never return $|+\rangle$", unless you say ""a single measurement can never return $|+\rangle$ when you measure in the computational basis." Because if you measure in the $|\pm\rangle$ then no measurement can return $|0\rangle,|1\rangle$ instead @glS It seems the cleanest way to define computational basis in any context is by the pure states that can exist after measurement. 9:38 AM
Also, clearly it was a mistake to call $\frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}$ a mixed state above. My apologies. 10:03 AM
@ChainedSymmetry so if you are studying say Bell's inequalities you change what you mean $|0\rangle,|1\rangle$ in the middle of the derivations? That would be a bit awkward wouldn't it? More generally, there are many scenarios in which you want to use different measurement bases and how the results in those different bases relate to each other. You cannot define univocally a computational basis in these cases 10:18 AM
@glS The specific domain I'm concerned with is QIS/QIT. In that scenario, every qubit measurement results in a classical bit. The states $\vert 0 \rangle$ and $\vert 1 \rangle$ are defined by the eigenvectors of the system that return 0 and 1, respectively, with certainty.
Is that not what N&C are saying? Bell-type inequalities are fundamental to QI... @Mithrandir24601 So N&C is wrong?
@Mithrandir24601 I don't mean that flippantly... that's the source of my confusion. The unambiguous statements in section 1.2 of Nielsen and Chuang contradict what both of you are saying. Are the statements incorrect?
@Mithrandir24601 Also, I'm talking about measurement of a single qubit (stated from the start). You both keep replying with comments about bell states, where one is measuring correlation vs. anti-correlation entanglement states. @ChainedSymmetry It's on page 15 - this is still very much the introductory stuff, so there's a lot of stuff implicitly assumed, so it's a really vague and not-well defined statement @Mithrandir24601 The statements are precisely defined at that point. I'm open to accepting they are wrong, but there is no ambiguity in them. @ChainedSymmetry OK, I'm going to turn on my super-pedantic brain: What exactly does "Recall that measurement of a qubit will give only either 0 or 1" mean?
Because to my super-pedantic brain, that statement alone is vague and ambiguous 10:33 AM
@Mithrandir24601 It means measurement of a qubit returns exactly one classical bit, either 0 or 1. @ChainedSymmetry What's a measurement? How does it return this classical bit? At this level of abstraction - indifferent to physical realization
in other words, if there is a physical realization where the statements that follow are not true, those would be incorrect statements 10:52 AM
@ChainedSymmetry That doesn't really answer the question :/ The term 'measurement' itself is so not-well-defined. You can define the computational basis to be the measurement basis, in which case, what you're saying is a tautology (as glS said yesterday - and not even necessarily true physically) but you could also define it as the basis the state is input in, in which case, there is absolutely no physical, computational (or otherwise) reason the measurement has to be done in the computational basis
You could measure in the X basis and return a '1' if you measured the state $\left|+\right>$ and a '0' if you measured the state $\left|-\right>$ and this still satisfies "measurement of a qubit will give only either 0 or 1" 11:02 AM
@Mithrandir24601 If it's the case that measurement of the physical state $\vert + \rangle$ returns the classical state 1, with probability 1, then the physical state $\vert + \rangle$ is the computational basis state $\vert 1 \rangle$. This is not tautological, it's relied on for the binary decimal and/or integer representations of the state in quantum algorithms.

1 hour later… 12:11 PM
@ChainedSymmetry no, all that ever gets returned is e.g. an electrical signal, either there or not, or in one place or another. We then convert that signal to classical binary, it's not at all like you measure a qubit and a classical bit is mysteriously returned @Mithrandir24601 It's not mysterious, it's probabilistic. A state that returns classical 0 with probability 1 is $\vert 0 \rangle$, a state that returns classical 1 with probability 1 is $\vert 1 \rangle$. N&C state this explicitly, every quantum algorithm I've worked through relies on this implicitly.
@Mithrandir24601 I don't see any freedom (except possibly permutation of 0 and 1) in that definition of computational basis. If you change that definition the mathematics falls apart in the algorithms.
The only quantum hardware/software I use regularly is IBM/qiskit... it's beyond question that qiskit relies on this convention. 1:05 PM
@ChainedSymmetry That's so abstracted and not at all how the physics works - whoever makes the hardware/software decides these sorts of things and even then, weak measurements are often used
@ChainedSymmetry the point is that these signals get returned regardless of measurement basis, which doesn't necessarily have to be the Z basis
Also, measurement based quantum computing is a thing @glS I think you have a typo in your answer. Second to last paragraph, I think you meant to say $\{|+,0\rangle, |+,1\rangle\}$. @ChainedSymmetry as in, there's a physical process between a state getting measured and a bit getting returned @Mithrandir24601 I get that, but it seems to me the "computational basis" can only be defined in a meaningful way at level of abstraction that is agnostic to the underlying physical process.
@Mithrandir24601 Otherwise, even very high level algorithms would be hardware dependent, so to speak.
@glS wrt your approach, when you trace out the second qubit, the reduced density operator on the first qubit is a mixed state (is that agreed?). If so, it's not possible to calculate the probability amplitude that the first qubit is in state $\vert + \rangle$, right? 1:32 PM
@ChainedSymmetry I'd say the opposite - the computational basis is picked to be the easiest physical basis to deal with - usually that $\left|0\right>$ is the ground state of the physical system 1:42 PM
@Mithrandir24601 It's funny you say that. N&C identify that intuition explicitly as a dissonance between physicists and computer scientists/mathematicians in the notation section (page xxix). @ChainedSymmetry Then set $\left|0\right>$ as the excited state if you like - the basis hasn't changed at all, you've just swapped $\left|0\right>$ and $\left|1\right>$... @Mithrandir24601 Yeah, I've tried to state each time that this permutation is generally okay. It may just add an insignificant extra step to the calculations.
@Mithrandir24601 But if you try to identify the partially excited state with the classical output 0 or 1, you seem to run into intractable problems.
@Mithrandir24601 If that's not physically precise, by partially excited, I mean whatever N&C means by the "halfway" state on page 14.
@Mithrandir24601 After talking, I understand better why people are uncomfortable with my original statement. But doesn't the fact remain that the probability amplitude of the first qubit can't be measured? I.e. isn't this the correct answer?
As a side note, I really appreciate you and glS talking this through with me. I know it was frustrating for both of you. 2:14 PM
@ChainedSymmetry What's a 'partially excited state'? @Mithrandir24601 Whatever N&C means by the "halfway" state on page 14. @ChainedSymmetry Oooh, $\left|+\right>$, OK, fair
@ChainedSymmetry It's not about identifying, it's about a bit appearing on a screen - how that's interpreted depends very much on the basis used - you may as well say 'orange' and 'blue'
@ChainedSymmetry No, no, it's fine. There are other things going on in life at the minute that are way beyond 'frustrating', so I don't mind being here discussing stuff @Mithrandir24601 Sorry to hear that :(
@Mithrandir24601 Regardless of basis definition though, isn't it fundamentally true that we can't measure the probability amplitude of the state of the first qubit. I.e. is this not a correct answer? @ChainedSymmetry Thanks :) I'm looking forward to next year, for sure
@ChainedSymmetry Sure you can - give me a moment @Mithrandir24601 Of course, no rush. 2:25 PM
@ChainedSymmetry What does the above circuit do? @Mithrandir24601 Not sure, I can only see the first angle in the U3 gates. Are those Hadamard gates? @ChainedSymmetry They're arbitrary - the point of the U3s and CNOT is just as an arbitrary 2-qubit gate (I haven't specified angles or anything) @ChainedSymmetry well, that depends. If the two qubits are separable than tracing out the second qubit still gives you a pure state on the first one. Otherwise sure, if the qubits are not separable then tracing the second one leaves you with a mixed state on the first one, for which you therefore cannot talk of probability amplitudes anymore
but that doesn't mean that you cannot talk of probabilities of different outcomes on the reduced state, which is what the question is about
@ChainedSymmetry yes that was a typo, thanks 2:41 PM
@Mithrandir24601 it returns $\vert 01 \rangle$ @ChainedSymmetry by the way, would you be able to summarise the misconception that led to this discussion in a way fit for a question? It seems to be a more or less widespread confusion, so it might be good to have a question dedicated to it. The reason I wouldn't know how to do it is because I don't really understand why quantumcomputing.stackexchange.com/questions/1410 doesn't already address the confusion @glS I think the core of the confusion is how are $\vert 0 \rangle$ and $\vert 1 \rangle$ "special"? From N&C (p. 13): "The special states $\vert 0 \rangle$ and $\vert 1 \rangle$ are known as computational basis states, and form an orthonormal basis for this vector space."
@glS Your post directly contradicts N&C that this basis is "special." So one is left very uncertain who to believe. 2:57 PM
@ChainedSymmetry right, I think that's a matter of taking the sentence out of context. They are special in the sense that they are essentially used to denote the natural choice of basis in a given context (as per my answer in answer above..). But they are not special in a sense that they have special physical meaning. That would like saying that describing vectors in a specific basis is preferred over using any other basis, which is not true @glS The comments in N&C reinforce my background from Lie algebra that diagonal elements are very special in a representation theory setting.
@glS See I see it as saying diagonal elements have a special relationship to eigenvalues, as compared to off diagonal elements. @ChainedSymmetry but I mean even in representation theory they are not. What's to prevent you from saying that a diagonal operator $A$ is one that is diagonal in some basis $(v_k)_k$?
@ChainedSymmetry but that's again tautological. You are just saying that if the elements of a basis are eigenvectors of the operator, then they have a special relationship with the operator via its eigenvalues. Of course that's true
what you represent in vector notation as $(1,0,0)$, $(0,1,0)$ etc depends on the kind of description you want to use, it doesn't affect anything about the physics, nor the structure of an operator in a vector space It's more subtle than that, in the representation theory of $\mathfrak{sl}_2 \mathbb{C}$, H is the diagonal element, X and Y are the off-diagonal elements. The "correct" interpretation of the commutation relations defining $\mathfrak{sl}_2 \mathbb{C}$ is "X and Y are eigenvectors for the adjoint action of H on $\mathfrak{sl}_2 \mathbb{C}$." (Fulton and Harris p. 162)
There's an incredible amount of information that leads up to that "correct interpretation" thought.
@glS One of the early pieces of advice that I got was to learn group/representation theory from mathematicians, not physicists. Physicists are generally pretty sloppy when it comes to group theory, and many computer scientists are oblivious to it, so it's not easy to port that knowledge into QIT.
@glS Also I reread it, and my interpretation isn't out of context. There's nothing in the chapter (or surrounding chapters to my memory) that suggests the "specialness" of $\vert 0 \rangle$ and $\vert 1 \rangle$ should be qualified to non-physical specialness. Quite the opposite, in the same section they use it to discuss ground states vs. excited states of atoms. 3:24 PM
@ChainedSymmetry it might be that they just take it for granted that there is nothing physical about the choice. Which I understand really, I still don't really get what exactly the issue is. Can you tell me what is your definition of "computational basis" again?
@ChainedSymmetry so what? You can describe the adjoints of the operators in any basis, not description/basis is inherently "better", apart from some making it easier for us to see some interesting properties 3:48 PM
@glS My interpretation of the computational basis is measurement of state $\vert 0 \rangle$ maps to classical output 0, with probability 1, and measurement of $\vert 1 \rangle$ maps to classical output 1 with probability 1, when no entanglement is present.
In other words on measurement the wave function collapses into that basis.
As discussed with Mithrandir earlier, this definition only makes sense at a "hardware agnostic" level of abstraction, which is how I interpret N&C's comments regarding "specialness". @ChainedSymmetry What is your definition of the state $|0\rangle$? 4:15 PM
@ChainedSymmetry as @SanchayanDutta is pointing out (I think) that doesn't answer the question, only transforms it into "what is the definition of $|0\rangle$" @SanchayanDutta I would say that the stated relationship defines $\vert 0 \rangle$ wrt the computational basis.
Is that problematic?
It seems well defined, up to phase. Maybe another relationship is necessary to address relative phase? @ChainedSymmetry That's just circular. You invoked $|0\rangle$ while defining the computational basis here, but when I ask you to define $|0\rangle$, you're invoking the computational basis. So let me ask again, how do you define $|0\rangle$ by being "hardware-agnostic"? 4:38 PM
@SanchayanDutta For any physical realization of qubits, only two states can be mapped to probability 1 measurements (up to phase). So it's defined up to phase and permutation of 0 and 1. @ChainedSymmetry That's false. Take electron spin (it's a qubit); I can measure it along the z-axis as well as the x-axis. If $S_z=+1/2$, then I could call that the $|0\rangle$ state. But if $S_x = +1/2$, I might call that the $|0\rangle$ state too. Any two orthogonal states of a electron can be labelled $|0\rangle$ and $|1\rangle$ respectively.
Yes, qubits have only two basis states. But those basis states are not unique. What basis will be chosen for a particular experiment, depends.  @SanchayanDutta That's just hardware implementation. Hardware designer has to pick a direction to measure. One is $\vert 0 \rangle$, opposite direction is $\vert 1 \rangle$. Analogous to calling 0 volts 0 and 0.5 volts 1 on a charge storage cell.
Let me ask the converse, how do you all interpret the computational basis when designing a circuit in, say, qiskit. @ChainedSymmetry Sure, and the hardware designer could choose to call your $|+\rangle$ as their $|0\rangle$. So it's not correct to say that the measured state cannot be $|+\rangle$, when it's merely a matter of choice. @SanchayanDutta No, it's not arbitrary. Algorithm design depends on the numeric representation of the physical system states. Algorithms interpret system states as binary decimals, integers, etc.
You can't just arbitrarily assign the binary states. 5:14 PM
You cannot simply make assertions without explanations. If it's not arbitrary, what's your explanation for that? I already gave you the electron spin example.
And you had mentioned previously that algorithms are hardware agnostic. So that makes your statements more vague. Done! :) and thanks ... Also any experience on MSE? Might have abused the conjecture tag there :/

https://math.stackexchange.com/questions/3446764/integral-form-of-work-during-an-irreversible-process @SanchayanDutta I already explained that. The computational basis has to map to the two probability 1 states. However those are implemented in the physical realization. In transmons, for example it's simple. Computational basis is excited/not excited, other states are superpositions of the two.
@SanchayanDutta In a single atom ground state/excited state map to computational basis, other states map to superpositions of excitation state.
@SanchayanDutta One of us is over complicating this, in most realizations this is very straightforward. @ChainedSymmetry Can I ask for the link to the question. (not that I plan on joining but I am curious) We're going in circles again. Sorry, I'll quit here. ? 5:28 PM
@SanchayanDutta Sorry, I don't know how to explain it more clearly. You never answered my question, how do you interpret the computational basis in qiskit?
@MoreAnonymous This question started it, but the conversation doesn't have much to do with this question any more.
Current topic started by glS asking my interpretation of the computational basis. @ChainedSymmetry Also I suspect what you are experiencing is a culture difference. I experienced one when I did my first masters in theoretical physics and did the second in mathematical physics :) ... @MoreAnonymous It was around here. Also I didn't see the (removed) message ... I am multitasking :/ (ping me and then remove) or I can stay for 10 seconds? 5:55 PM
@MoreAnonymous The (removed) just failed to tag you as a response. @ChainedSymmetry Ah ... @ChainedSymmetry they really, really don't - they can in some contexts but in e.g. measurement based quantum computing, they have to be able to measure in any basis. Also have you looked up projection value measures?
@ChainedSymmetry I don't understand what you mean by this :/ @Mithrandir24601 I remember you saying something about QED so I'd tagged you about a post. Any quick comments? \\chat.stackexchange.com/transcript/message/52648318#52648318 6:14 PM
@MoreAnonymous Whoops, sorry! my instinct (which is completely unfounded) is that it could simply be related to the fact that we don't know the physics at the highest of energy scales - there'll be an energy where quantum gravity plays an important part, after all! @Mithrandir24601 So which one do u wager on Born rule vs QFT?
:P @Mithrandir24601 I respect what you're saying, it just doesn't match what textbooks and literature say. So I'm not sure how else to respond.
@Mithrandir24601 Here's one of many papers saying the same thing as N&C (link Section 2.1). @ChainedSymmetry This actually matches how I like to think of physics. P.S I haven't read the entire (massive) conversation so I might be taking things outta context :/ @Mithrandir24601 "Qubits have two basis states |0⟩ and |1⟩, which are the analogues of the classical 0 and 1 states.... To obtain classical output, a qubit is measured, collapsing its state to either |0⟩ or |1⟩."
Authors are from Princeton, IBM, Univ. of Maryland. @ChainedSymmetry I'm curious about what you think of this question: quantumcomputing.stackexchange.com/questions/8597/…
? 6:29 PM
@Mithrandir24601 I know you know your stuff, which is why I've been trying to understand the disconnect between what you're saying here and what I'm reading there. I just can't seem to figure it out.
@MoreAnonymous You accepted my answer to that question. :) @ChainedSymmetry HAha just saw it was you .. lol
:P
@ChainedSymmetry Do you think it would be fruitful to create an algorithm encode it in the position operator and see how things differ in relativistic QM? (not QFT) @MoreAnonymous Isn't QFT just relativistic QM? I'm not aware of any other theories that reconcile SR and QM. @ChainedSymmetry QFT involves 2nd quantization
@ChainedSymmetry I read this book for fun
He's a famous german physicist 6:51 PM
@MoreAnonymous Looks really good, especially chapter 16. I'll probably pick it up just for that chapter. But this book is an intermediate step towards QFT/QED, which he covers in the next volume in his series. Note the Feynman diagram on the cover. @ChainedSymmetry I never got the chance to read the whole thing :/ ... never had enough time :/
@ChainedSymmetry I agree but he even discusses the how the dirac delta function of QM is actually a gaussian in RQM  