12:09 AM
@glS huh, yeah, that's weird. (arxiv version here: arxiv.org/abs/1211.7132)

12:21 AM
@glS looking at it briefly, it involves two d-level systems (for odd prime d). The protocol is along the lines of: Alice prepares a specific two-qudit state, and sends her second particle off to Bob. Bob performs a certain measurement on this particle to encode his message, and sends it back to Alice. Alice then makes a measurement on the two qudits and uses this to deduce Bob's message with probability 1-1/d. (Otherwise the outcome is inconclusive.)
So that's a good deal more involved than Alice and Bob sharing an entangled Bell state
But the fact that it works at all is deeply strange to me

12:51 AM
(should be possible to understand the d=3 case tho)

4 hours later…
5:03 AM
@Semiclassical thank you for clarifications Semi

2 hours later…
6:49 AM
Hi @JohnRennie

@cOnnectOrTR12 Hi :-)
I drew a diagram to show what the solution to that problem was doing. Give me a moment and I'll upload it.
At the top I have drawn two separate plates and their electric fields.
The problem is using σ to mean the total area charge density i.e. adding the two surfaces together, so I have done the same.
That means the field either side of the two separated plates is E = σ/2ε
OK so far?

I think they are using sigma because they are assuming it’s a sheet and in the book of resnik Halliday there is no mention of total area charge density. Sigma is defined as surface charge density.
See they have mentioned “sheet”

OK
So are you happy with my description of the problem so far?

Now we bring the two sheets together at some distance d. When we bring the sheets together their electric fields overlap. In that case the total electric field is the sum of the fields of the two sheets.

7:04 AM
Yeah!

I've shown this in the bottom diagram. I've drawn the field of the positive sheet in red and the field of the negative sheet in blue.

Note that the electric field is a vector, so when we are adding the fields the direction matters. Fields in the same direction add and fields in opposite directions subtract. OK so far?

Yeah

In between the sheets the two fields are in the same direction so they add. Is this clear from my diagram?

7:08 AM
Yeah

So can you see how the solution works now?

Every thing is clear but this. What will be the field of a sheet of conductor?( Please take sigma as surface charge density)
You say first there is a plate with sigma as surface charge density on both sides and then if we decrease it’s thickness we will get 2sigma. What if it’s a sheet already.

You are making a distinction between a plate and a sheet that I don't think is justified.
In this particular case the result of the calculation is the same whether we use a plate with a non-zero thickness or a sheet with zero thickness.

So a plate and sheet is same ?

We just need to make sure we use σ to mean the total area charge density in both cases.

7:23 AM
Why have they used E=sigma/2e0 for separate plates and added. Are you saying they have taken sigma as total charge density?

@Semiclassical The event of a measurement without knowing the result changes nothing - isn't that exactly what the quantum eraser is about?

@JohnRennie ?

@cOnnectOrTR12 Yes, they have taken σ to be the total area charge density
This applies whether we are considering a plate with a non-zero thickness or a sheet of zero thickness.

So a sheet will also have two surfaces with total charge density as sigma and surface charge density as sigma/2.
Like a plate?

We keep getting stuck on this point.
How we define σ is entirely up to us. We can define it any way we want.
In this case the person who wrote the question has defined σ to be the total area charge density.
And we know that if the total area charge density is σ then the field either side is σ/2ε. Yes?

7:40 AM
@Semiclassical ah, right. The state being sent to the other party after the measurement means it's just a different setup

My confusion is whether the sheet is a thin film that charges on it have effect on both sides. Not two layers of charges but only one layer of charge.

@cOnnectOrTR12 Why does it matter?

I'd say it's different here though. Whatever happened to the other side of a (possibly entangled) bipartite state cannot affect your side, sure. In this case some information *is* passed between the two parties, so it seems conceivable that Bob might be able to predict what happens to his side better because of it (which corresponds to saying his description of his partial state should be updated).

I mean, it turns out this is not the case as per calculations above, but I'm not seeing why that should obviously be the case, yet

@glS But how is it different (from Bob's viewpoint) from the eraser situation, where we also know that a measurement happened but not what it resulted in?
In the abstract I agree that it is not obviously absurd that the information might change something, but the concrete situation seems to me to be exactly like the eraser, where it is well-known to not change anything

Because if there is two layer ,total charge density will be sigma. Surface charge density will be sigma/2. But if it’s one layer then surface charge density and total charge density will be same.

7:45 AM
@cOnnectOrTR12 But it makes no difference to our calculation.

For a plate it will be sigma/e0. For a sheet it will be sigma/2e0
@JohnRennie yes?
O yes! You are right!
It will be same
If we take sigma as total charge density:)

8:06 AM
:-)

5 hours later…
12:46 PM
Hey
I don't really understand the concept of gauge
What does gauge transformation even mean in qcd? Does it represent anything physical?

It means you pick a different representation for the same field

1:31 PM
@ACuriousMind I do think this is a test of intuition vs concepts. In usual settings, a message consists of many bits (including some that are used to ensure the message is properly received)
So in that context it makes sense to think that an incomplete message would still carry some information
But in this context it’s either boom or bust, so to speak: an incomplete message from Alice doesn’t alter Bob’s expectations
(And Bob arguably has gained something if Alice states her measurement direction; the ability to interpret Alice saying “I got +1”, but Bob can’t exploit that until Alice actually sends that part of the message.)

2:25 PM
Gotta say, Kurzgesagt makes some surprisingly high quality videos

3:08 PM
@Semiclassical I'm not really sure I follow your intuition here: Let's take a purely classical example: Alice is measuring position and momentum of a particle and communicates the result to Bob in the format "$x$:position value" or "$p$:momentum value"
if I truncate one of these messages to just "$x$" or "$p$", Bob isn't getting any information anymore (because classical measurements don't alter the state anyway), even though I didn't cut all the bits from the message!
i.e. I think it is actually not at all intuitive to expect that a truncated message transports any information at all

3:52 PM
@ACuriousMind it might be similar, I'm not really sure. I never quite grokked the point of the eraser experiment. From a cursory read on the Wiki page, you have a two boson entangled state $\rho\sim |00\rangle+|11\rangle$. Thus measurements on B cannot detect interference, as $\rho_B=I/2$. But if A measures after applying an Hadamard $H$, then Bob's state becomes either $|+\rangle$ or $|-\rangle$. Thus conditionally to A telling their outcome, B can observe interference
but then again, without knowing A's outcome, B's state is always the same. So yes, in this sense I can see the analogy. Though in the context of the eraser the question is posed in a slightly different way
to me, this seems like an even stronger way to see how entanglement is really just a (stronger than "usual') form of correlation between measurement outcomes, rather than any ort of causal link between the physical states themselves.

2 hours later…
5:29 PM
@ACuriousMind Sure, that'd be another context where truncating a message would remove the info. I'm talking in the day-to-day context of 'information'. In that case it makes sense to think that a partial message still carries some info.
and thus why it's tempting to import that intuition into the quantum context
@glS i sorta feel like the way they describe it is misleading too. to me it's more appropriate to say there's three subsystems A,B,C. Alice can't do anything on B and Bob can't do anything on A, but both can do something on C
hmm, no, i'm mischaracterizing it. Alice can act on the entire system AB, Bob can only act on B.

6:21 PM
@Semiclassical the way I'm understanding it, we have a simple bipartite system. Bob doing "interference fringes" amounts to him measuring in the $|\pm\rangle$ basis. And Alice either measures in the computational, or in the $|\pm\rangle$ basis. If she measures in computational, Bob has uncertainty in outcome in any scenario. If she measures in $|\pm\rangle$, then Bob's result is determined from Alice, but only if she tells him the observed outcome. I don't think it's anything more than this

7:01 PM
hmm, interesting. so while "the choice of measurement is the signal" is a nice sounding title, it's really very specific to their scenario

7:47 PM
@Semiclassical are you referring to the eraser or the PRL?

@Semiclassical oh. Wait, my previous comment was about the eraser

ahh
i was talking about the PRL there too

@Semiclassical if this is about the PRL, then yes I think I agree with this way of seeing it

i still haven't understood how their protocol works, though

7:52 PM
in that paper the context is just completely different, indeed. Alice can act on the entire system, as you say, which also means they assume there is a post-measurement state after Bob's measurement. But then, I don't really understand what sort of state they're assuming is left after the measurement. I suppose a convex mixture of the possible measurement outcomes, weighted with the probabilities, given that Alice is not told the outcome?
so the question is something like, how much information is encoded in mixtures of the form $\sum_k |\langle k|\psi\rangle|^2 |k\rangle\!\langle k|$, with $|k\rangle$ some orthonormal basis (choice of (projective) measurement), and $|\psi\rangle$ initial Bob's state

yeah

Bob can choose which basis $|k\rangle$ to use, and Alice wants to somehow recover information on this choice from the convex mixture.

i'd need to see the d=3 case to really grok it i think

it makes sense that you can get something from it. If Bob e.g. uses a basis containing $|\psi\rangle$, Alice gets a pure state; if Bob chooses a basis mutually unbiased wrt $|\psi\rangle$, Alice gets a totally mixed state. So the thing might boil down to encoding information in the purity of the state Alice gets or something

neat
the language of mutually unbiased bases isn't one i'm familiar with
i imagine it's nothing too complicated
the conclusions paragraph in the Signal paper is sorta interesting
with the implication, i think, that ultimately one still does "report" a measurement result. but it's Alice's second measurement which does the reporting, not the first.
tricky

8:10 PM
is anyone familiar with mixed type surfaces
trying to understand the definition

8:38 PM
0

As a teacher of Mathematics and Physics at a high school I ask some questions closely related to the level of knowledge for students aged $14$ to $19$. On the Math.se website there are some tags such as education, soft-question, memorization-techniques, etc... . What are the best or existing tag...