The Classical Channel - 2019-11-21

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1:34 PM

@glS I've been overloaded with worrrrkkk ... will be a while before I correct my answer ... I think it would be best for the (corrected) math to speak for itself :/

3 hours later…

4:29 PM

@MoreAnonymous who isn't =)

3 hours later…

7:08 PM

@glS thanks for ur patience

7:30 PM

@Mithrandir24601 any idea of this one?

1

Q: Born rule and the UV divergence?

A typical use of cutoffs is to prevent singularities from appearing during calculation. If some quantities are computed as integrals over energy or another physical quantity, these cutoffs determine the limits of integration. The exact physics is reproduced when the appropriate cutoffs...

quantum-field-theory renormalization quantum-measurement

Had enough time to pen down a quick question.

7:44 PM

no idea, I know next to knowing about cutoffs and such, but it seems like a good question. A couple of suggestions to improve it might be: 1) the title still isn't the question, but only a generic statement related to topic, 2) it's probably a good idea to highlight what part of the quoted text exactly is related to the question

3 hours later…

ChainedSymmetry

10:19 PM

@glS It might be better to continue the discussion about measurement basis in chat that comments.

One of the fundamental tenets of QM is that the state you measure becomes the state of the system post measurement.

And the post measurement state can never be a superposition of states, because measurements always find a particle in a definite state.

This is one of the most enigmatic and problematic features of QM, but it's been verified experimentally without exception.

@SanchayanDutta Is the above not clear to you either? This is usually covered in the first few chapters of any introductory QM textbook.

ChainedSymmetry

10:50 PM

If a state is $\vert \psi_0 \rangle = \vert 0 \rangle$, a measurement returns $\vert 0 \rangle$ with probability 1, and the post measurement state is $\vert 0 \rangle$.

Mithrandir24601

@ChainedSymmetry What if you start with $\left| 0\right>$ and measure in the X basis to get (say) $\frac{1}{\sqrt{2}}\left(\left|0\right> + \left|1\right>\right)$? (I'm missing context here, to be fair)

ChainedSymmetry

If a state is $\vert \psi_+ \rangle = \vert + \rangle$, a measurement returns $\vert 0 \rangle$ with probability 0.5 and $\vert 1 \rangle$ with probability 0.5. The post-measurement state is either $\vert 0 \rangle$ or $\vert 1 \rangle$, it is never $\vert + \rangle$.

Mithrandir24601

@ChainedSymmetry It is if you measure in that basis... Also, what makes you say that $\left|+\right> + \left|-\right>$ isn't a superposition?

ChainedSymmetry

@Mithrandir24601 I am saying that those are superpositions.

@Mithrandir24601 My point is that a measurement can't return a superpostion. It always finds a particle in a definite state.

And the state is always in the definite measured state post measurement.

Mithrandir24601

@ChainedSymmetry But the actual state $\left|\psi\right> = \frac{1}{\sqrt{2}}\left(\left|+\right> + \left|-\right>\right)$?

@ChainedSymmetry You need to be clearer here - there are types of measurement that (depending on how you define 'measurement') do exactly that

ChainedSymmetry

11:00 PM

The context was a measurement of a single qubit.

@Mithrandir24601 My statement was that this returns $\vert 0 \rangle$ or $\vert 1 \rangle$.

Mithrandir24601

@ChainedSymmetry So... What sort of qubit? Do you know the physical details of how it can be measured?

ChainedSymmetry

@Mithrandir24601 Is there a measurement of a qubit that returns something other than $\vert 0 \rangle$ or $\vert 1 \rangle$? To be clear I mean the measurement itself, not what can be inferred from the measurement based on a priori knowledge.

Mithrandir24601

@ChainedSymmetry For superconducting schemes, you can measure qubits using weak measurements, such as here. for single photon detectors, the measurement outcome isn't $\left|0\right>$ or $\left|1\right>$ as you need to include that the detector might not 'click' when it should

11:23 PM

@ChainedSymmetry so the main problem with this statement is that, at least in the way you are attaching meaning to it, it's tautological. A measurement will give you a definite result. What you are doing is defining the measurement basis as the "computational basis" denoting it with "$|i\rangle$". Then, of course, any measurement will give results in the computational basis, but this is a tautological statement

you cannot say that "$|+\rangle$ is a superposition while $|0\rangle$ is not", that doesn't mean much. Every single state is a "superposition" of other states. Being a superposition of other states is simply a statement about how you would find the state if measured in some other basis.

Mithrandir24601

@glS I've got the problem that, depending on how 'measurement' is defined, this isn't even necessarily true when it comes down to how the measurement is physically happening

ChainedSymmetry

@Mithrandir24601 I'm not very familiar with weak measurements, but this paper arxiv.org/ftp/arxiv/papers/1702/1702.04021.pdf (4th sentence of abstract) suggests it's entirely predicted by standard QM.

@Mithrandir24601 which part?

Mithrandir24601

@glS e.g. SPDs in photonics or weak measurements in superconducting

Oh sorry - "any measurement will give results in the computational basis"

@ChainedSymmetry a "weak measurement" is nothing but a process in which only part of the information about the state is "collapsed". For example, if you have $\sum_{k=1}^4 c_k|k\rangle$, you can "ask the question" is the state in the $\{|1\rangle,|2\rangle\}$ subspace or in the $\{|3\rangle,|4\rangle\}$ one?. Such a measurement will still give states as outcomes, and is thus "weak"

@Mithrandir24601 well I am saying that it is incorrect to say that one only measures in the computational basis. Unless one means the statement in a tautological way, that is, by defining "computational basis" as whatever basis is being used in a given experimental setup.

Mithrandir24601

11:28 PM

@glS Ahh, sure, yeah

ChainedSymmetry

@glS Is that not exactly what the computational basis is?

the bottom line is that it doesn't make much sense to say that any basis is privileged or special wrt others. All that matters is how bases interact with each other.

@ChainedSymmetry well, it depends on how you define it. Usually people just use "computational basis" to refer to some "reference basis". I agree that it is often, if not most of the time, the basis that is going to be used for the measurements, but this is a statement about notation, not physics

ChainedSymmetry

@glS The computational basis is the columns of the identity matrix. This is an extremely privileged basis.

but if you consider e.g. a Bell scenario, in which different measurement bases are used, then the whole point of the protocol is that you measure in different bases. Then sure, you could say that the measurements in the other bases are measurements in the computational + unitaries, but what's the point in that?

@ChainedSymmetry the way you describe an operator as a matrix is also purely a matter of notation

Mithrandir24601

@ChainedSymmetry I'm confused here - the identity matrix is entirely basis independent

11:33 PM

the meaning attached to the computational basis changes drastically between different physical scenarios. This is why I'm saying that it holds no physical meaning

ah, by the way, there is also this question about this exact topic

12

Q: What is meant by the term "computational basis"?

What is meant by the term "computational basis" in the context of quantum computing and quantum algorithms?

quantum-state terminology

and funnily enough, I did write there what I'm saying here. I didn't even remember that

ChainedSymmetry

@Mithrandir24601 That's exactly my point. In the same sense the computational basis is basis independent.

Mithrandir24601

@ChainedSymmetry The computational basis is a basis

@ChainedSymmetry there is no content in that statement. You could just as well define the "computational basis" as any other basis and do the same exact calculations

ChainedSymmetry

@glS That's like saying we can define the identity to be any other operator just because we have a linear map to the identity...

Mithrandir24601

@ChainedSymmetry Nooo, the identity is totally unique - in exactly the same way the '0' of addition or '1' of multiplication is

it's the 'do nothing' operator (regardless of what the input is)

11:41 PM

@ChainedSymmetry it's like saying that in a vector space no basis is privileged. Some might be easier to work with in some contexts, but that's all there is to it. If you define the "computational basis" as the columns of an arbitrary unitary $U$, nothing changes in the physics (of course, you also need to change the description of the other object so that the mantain the correct relations with the new "computational basis")

I mean at the end of the day I'm pretty sure you could do any calculation without ever working with the matrix/vector representations of your vectors, and then the whole matter of what is the "correct vector representation of a given basis" is meaningless

ChainedSymmetry

@Mithrandir24601 Yes, and the computational basis consists of the two columns of that totally unique operator.

As a sincere question, can either of you point to a quantum circuit that measures in something other than the computational basis?

Mithrandir24601

@ChainedSymmetry So... This is another way of looking at how the identity matrix is unique - it can be formed from any eigenbasis

@ChainedSymmetry the whole point is that there is no special way to understand the words "computational basis". Given an arbitrary circuit, it is just as correct to say that all measurements are done in, say, the $|\pm\rangle$ basis. It's just a matter of how you describe the process in the math

bit if you want, again, consider a Bell experiments, or any other context in which more than one measurement basis is used

or the examples in my answer here quantumcomputing.stackexchange.com/a/1425/55

or as another one, consider a photon's polarisation. Say the photon is in the horizontal polarisation, which is your computational basis. Measure with a polariser at 45 degrees wrt the polarisation of the photon. You just measured not in the computational basis

Mithrandir24601

@ChainedSymmetry One one level, any photonic circuit, a number of superconducting circuits (and I don't know what else), on another level, e.g. any circuit with single qubit unitaries before the measurement is performed

ChainedSymmetry

@glS The polariser at 45 degrees in that case is the computational basis.

11:51 PM

@Mithrandir24601 but I think even saying this is misleading. Any measurement is "done in the computational basis" or not, depending on how you describe it. What "computational basis" means is not about the physics, only about how you describe the process

@ChainedSymmetry sure, again, if you define it that way. I didn't, so it's not. That's the point. It's a matter of how you describe something, it's got nothing to do with the physics.

Mithrandir24601

@glS Yeah, this is just a definition thing

there is nothing wrong in using $(1,1)$ and $(1,-1)$ rather than $(1,0)$ and $(0,1)$ to describe the outcomes of a given measurement.

ChainedSymmetry

@glS No, in that case the direction of the polarizer is very privileged relative to other angles.

You have to at least agree with that correct?

Mithrandir24601

@ChainedSymmetry So... Have you read anything on measurement based quantum computing? - where different qubits are measured with different 'angles'? Or what about having two photons where one is measured in the horizontal basis and the other in the diagonal basis?

@ChainedSymmetry that only depends on what you are interested in. I don't agree that it makes any fundamental difference what basis you use. Again, if I'm studying how measuring with different polarizer's angles affect smy state, then what is the computational basis? If you change what you mean by the term every time you rotate the polarizer, then that makes things more convoluted, not easier, which is why that's not what you do

anyway, it's late and I've got to finish stuff before tomorrow =). See ya!

Mithrandir24601

11:57 PM

Yeah, I'm off to bed - goodnight!

ChainedSymmetry

@Mithrandir24601 Yes, my contention is directly from the GHZ paper.

here: arxiv.org/ftp/arxiv/papers/0712/0712.0921.pdf

But it is interesting that Bell’s results say nothing in the special case covered directly
by the EPR argument, namely the case where a measurement on one particle allows
one to predict what happens to the other particle with 100% certainty. This is the
case where one measures the spin on one particle, and then measures the other
either in the same or opposite direction. Not only does this case yield certainty in its
measurement, but in fact one can arrive at a classical model of the system which

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Nov '1921

The Classical Channel

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