
Chinese Researchers Achieve Quantum Entanglement Record - 0xbxd
https://www.scientificamerican.com/article/chinese-researchers-achieve-stunning-quantum-entanglement-record/
======
foxes
I visited USTC several years ago, (2014) and in this time they have basically
doubled the number of qubits they can entangle, so this is a great step
forward. At around ~100 qubits quantum computing becomes very useful, so maybe
we won't have to wait too long.

~~~
bscphil
What's the size needed for cracking large-primes based encryption? If not at
100 qubits, what are some useful calculations that we _can_ do at that size?

~~~
deepGem
So the number of raw qubits are in the order of a million for large prime
factorization or to make Shor's algorithm a reality. This is because the
physical qubits are highly unstable and prone to errors. The error correction
methodologies will result in a fault-tolerant logical qubit. Depending on the
error correction algorithm, the ratio of logical: data/physical qubit is
1:1000 at the minimum. You need 4000 logical qubits at least for large prime
factorization give or take 4 million raw qubits.

~~~
bscphil
This is probably a dumb question, but could you spread out the error
correction in time? I know it's hard to create very large entangled systems,
so would it be possible/easier to do the calculation with 4000 real qubits
1000 times and use the repetition for error correction? (Checking for correct
factorization is easy, after all.)

~~~
simcop2387
That's been one strategy I've seen mentioned in regards to Shor's algorithm
many times. Since there's the probability of an error, you still have to check
it and rerun it a few times until you confirm the right answer.

That's still many times faster than you'd get conventionally, so it's a
reasonable trade off.

But it doesn't address the possibility that you can't entangle enough qubits
to begin with, or keep them stable long enough to actually perform the
calculations, which is where the error correction stuff comes in. I don't know
enough about the challenges there to comment on how many you'd need (though
1000:1 seems high to my naive intuition, i'd guess maybe 10-100 for magnitude
myself).

~~~
sir_kin
For more info on that 1000:1 ratio, see [1].

[1]
[https://en.wikipedia.org/wiki/Quantum_threshold_theorem](https://en.wikipedia.org/wiki/Quantum_threshold_theorem)

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ArtWomb
Congrats! US House science committee also set to pass national initiative in
quantum tech with on order of $1B funding:

National Quantum Initiative - Action Plan

[https://www.lightourfuture.org/getattachment/85484dca-465a-4...](https://www.lightourfuture.org/getattachment/85484dca-465a-46f4-8c8c-090aeb845d09/FINAL-
Action-Plan-for-a-NQI-Apr-3-2018.pdf)

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ahelwer
Crash course in entanglement:

The state of each qbit is represented by a state vector of two complex numbers
[a, b] where |a|^2 + |b|^2 = 1. There are two special qbit values called the
_classical basis_ : [1, 0] which is the classical bit 0, and [0, 1] which is
the classical bit 1. If a qbit is not in one of the two classical states, we
say it is in _superposition_. When a qbit is in superposition, we can measure
it[0] and it will collapse probabilistically to 0 or 1; for a qbit [a, b], the
probability that it collapses to 0 is |a|^2 and the probability that it
collapses to 1 is |b|^2.

Things get more interesting when we have multiple qbits. If we have two qbits
[a, b] and [c, d], we define their _product state_ as their tensor product
[ac, ad, bc, bd]. For example, if we have two qbits both in state [1/sqrt(2),
1/sqrt(2)], their product state would be [1/2, 1/2, 1/2, 1/2]. We use the
product state to calculate the action of a quantum logic gate that operates on
multiple qbits - for a gate which operates on two qbits, we can always
represent its action as a 4x4 matrix.

Usually we can move back and forth between the product state representation
and writing out the individual qbits states. However, in certain scenarios
something very special happens: we cannot factor the product state back into
the individual state representation! Consider the product state [1/sqrt(2), 0,
0, 1/sqrt(2)]. If you try to write this as a tensor product of two states [a,
b] and [c, d], you cannot! It cannot be factored; the qbits have no individual
value, and we say they are _entangled_.

Well, what does this mean? It means when you measure one qbit, even if the
qbits are very far apart, you instantly know the value of the other qbit. So
if I entangled two qbits in the state [1/sqrt(2), 0, 0, 1/sqrt(2)], give you
one, then we go to opposite ends of the universe, if I measure my qbit and see
a 0 I'll know your qbit instantly also collapsed to 0 (or collapsed to 1 if I
measured 1). This phenomenon has been experimentally-verified to occur faster
than light. It is instantaneous, as far as we can tell. So, local realism is
wrong! Spooky action at a distance is real!

There is an important caveat: while the qbits seem to coordinate in some
faster-than-light way, you cannot use this to _communicate_ in a faster-than-
light way. All we have is a shared random number generator. I can't send some
chosen bit from my reference frame to yours. This is called the no-
communication theorem.

If you found this interesting, I have a full video on quantum computing for
computer scientists here:
[https://youtu.be/F_Riqjdh2oM](https://youtu.be/F_Riqjdh2oM)

[0] IDGAF about your chosen quantum mechanics interpretation, don't @ me

~~~
bdamm
The part I don't understand is the "spooky action at a distance." Isn't this
just the same as if the qbits were already in whatever their final state was
as soon as they were entangled? Or in other words, is there any experimental
basis for determining that the "cat" stayed alive at all? It seems a lot less
magical when we just determine that in fact the value is the same no matter
when or where you measure it, although it may not be predictable. And since
the "entangled" qbit always had the same value, no matter when you measured
it, then there is no "spooky action at a distance". The action occurred when
d=0, and what you actually have is two copies of the same ROM.

~~~
ahelwer
The idea that the qbits choose which value they'll collapse to at time of
entanglement is called local hidden variable theory, which John Bell disproved
in 1964:
[https://en.wikipedia.org/wiki/Local_hidden_variable_theory](https://en.wikipedia.org/wiki/Local_hidden_variable_theory)

In more practical terms, the theory also falls apart when you start doing more
complicated things with entangled qbits than just measuring them, like quantum
teleportation or error correction.

~~~
moosinho
Can someone tell me if I understand this correctly. So Bell famously proved
that it is not the case that the correlation we see between 2 entangled
particles A and B is because of some common cause C. Therefore we concluded
that it must be that A => B or B => A and we called it a spooky action at a
distance. But aren't we forgetting that there's a one more way to get a
correlation between A and B without resorting to spooky action at a distance -
conditioning on a common effect, aka collider:
[http://www.the100.ci/2017/03/14/that-one-weird-third-
variabl...](http://www.the100.ci/2017/03/14/that-one-weird-third-variable-
problem-nobody-ever-mentions-conditioning-on-a-collider/) Has anyone proven
that this is not the case?

~~~
avip
Bell inequality proves (or is used to show) that the shared/correlated state
of "entangled" particles cannot be fixed before the act of measurement. It has
nothing to say about "A => B" or the likes. If there exists a third "C" it'll
_still_ have to set the state of A and B, at the moment of measurement,
regardless of distance.

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angrygoat
Here's a pre-print of the USTC team's paper:

18-qubit entanglement with photon's three degrees of freedom

[https://arxiv.org/abs/1801.04043](https://arxiv.org/abs/1801.04043)

------
xtrapolate
Is entanglement an actual physical phenomenon? What exactly needs to happen
in-order for a photon to "become" entangled?

~~~
mundo
Not sure what "actual physical phenomenon" means, but probably no. If you mean
can you look at a photon and observe something about it that tells you it's
entangled to some other photon, no. If you mean do we know what physical
process underlies entanglement, also no. If you mean does entanglement really
happen as opposed to being theoretical, yes. The link from the first paragraph
of the article may help: [https://www.scientificamerican.com/article/chinese-
researche...](https://www.scientificamerican.com/article/chinese-researchers-
achieve-stunning-quantum-entanglement-record/)

------
downrightmike
Even neater is that they are using quantum entanglement in radar to detect
stealth planes: [https://www.popsci.com/china-quantum-radar-detects-
stealth-p...](https://www.popsci.com/china-quantum-radar-detects-stealth-
planes-missiles#page-2)

~~~
madeuptempacct
Yea, you have to explain this. It made no sense when I first read it, and it
still makes no sense.

Here is how radar works: You shine a flashlight somewhere. If there is an
object within range, some of the light is reflected back at you and you know
there is an object. This is literally what radar does.

So, what part of this, and how, does this "quantum" magic affect?

~~~
nabla9
It works like normal radar, except that they send one entangled photon towards
the plane and keep the other.

When the reflection returns, they can use the entanglement to sort out signal
from noise much better and radar is safe from jamming and spoofing.

~~~
madeuptempacct
Thanks, this helps a lot, but just to clarify:

Originally "We 'shine' a bunch of photons at an object, and see if they come
back. Except there could be dragged decoy spamming photons back at us, and we
don't know if there is. Or it could be sending out so much light, our photons
coming back are lost in the mix."

With quantum: "We can verify that any photons that get reflected are the ones
that we sent out and check only our photons when flooded with photons from
enemy flashlights."

Right?

~~~
Filligree
Right. Oversimplifying a lot, it's putting a name tag on the photons. Then you
can sort out the tagged ones from all the others, drastically reducing the
practical noise floor.

~~~
renox
Except that photons don't have tags.. Each time a story talk about
entanglement, I'm stuck: what does it mean experimentally to verify that two
particles are entangled? I don't know..

------
tzakrajs
> “It’s as though you took six bits in your computer, but each bit tripled in
> how much information it could hold,” Schreppler said

"Each bit is tripled in how much information it could hold"? Normally bits
have two states, but Shreppler is claiming these qubits can have six states?
What am I missing?

~~~
barbegal
It's a confusing quote. They have achieved 18 qubits from 6 photons. But 18
qubits can store more information than 18 regular bits however when the 6
photon system is measured the quantum state collapses to 18 bits of data.

How do you get 18 qubits from 6 photons? You use multiple degrees of freedom.
Whereas a bit in traditional memory has a single degree of freedom (electric
charge), photons have many degrees of freedom such as polarity, direction of
travel and angular momentum.

~~~
21
Even in traditional memory, you can use a single degree of freedom to store
multiple bits, by quantizing the charge. Like in TLC flash memory, where you
can store 3 bit per cell.

This is because in traditional memory, a single "cell" stores many elementary
charges (electrons).

------
nabla9
Question:

Why superfluid 4He is not quantum-entangled? What differentiates macroscopic
many-body quantum states from many-body quantum entanglement?

~~~
barbegal
Superfluid 4He is quantum-entangled and non-local effects are observable.
[https://www.sciencedaily.com/releases/2017/03/170321110344.h...](https://www.sciencedaily.com/releases/2017/03/170321110344.htm)

~~~
nabla9
So why it's not taken into account when quantum-entanglement records are
broken?

You can have flasks of superfluid 4He with thousands of moles of entangled
atoms clearly visible.

~~~
barbegal
Quite correct but usually quantum-entanglement records are for qubits rather
than generic atoms. And qubits must be able to be measured in isolation e.g.
the polarisation of a single photon can be measured. With superfluid 4He you
can't isolate the qubits.

~~~
nabla9
Ah, that makes sense. Thank you.

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kelvinsilva
[https://www.reddit.com/r/QRL/](https://www.reddit.com/r/QRL/)

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xpuente
uhmmm ...

[https://www.quantamagazine.org/gil-kalais-argument-
against-q...](https://www.quantamagazine.org/gil-kalais-argument-against-
quantum-computers-20180207/)

~~~
Cobord
Take that more as a challenge and a caution to avoid overhype.

