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Category: Quantum Computing
Tags: ErrorCorrectionQuantumQubitsRydbergStabilizers
Entities: Eraser errorsQuantum repetition codesRydberg atomsStabilizersSuperconducting qubits
00:00
Thank you. Yeah.
So all the talk is going to be on the whiteboard. So it'll be very very slow.
Stop me anytime. Um and it's and it's it's you know it's
00:16
familiar topics that you have seen so far like error correction. I think Steve talked about the st quantum error correction with stabilizers.
So I'll talk a little bit of that. I think you learned about cross operators this morning with um Danny and then you also heard about erasers from Taka.
So it's
00:33
like a combination of all of that. Some of the things what I might be saying might be a repetition for you in which case just just tell me stop.
Uh or sometimes it's good to have a repetition. Um yeah.
So I I can talk about this stuff for like days on end.
00:49
So if you tell me you already know this, I can talk about you know something ahead. Great.
So um the official topic is uh rberg atoms uh and error correction. However, the main uh concept there is something
01:06
called as eraser errors. So we'll focus I would say like three4s of the talk on what eraser errors are, how we er correct them, why they are you know so or why they're considered better for uh
01:22
error correction and then how do we engineer uh a cubit that have erasers that only has eraser errors in practice in in different platforms. The first platform being the Rberg atoms.
Okay. So
01:39
um I think you may have heard this word erasia error today for sure. Um so maybe I can I can ask you guys a question.
What is an eraser error?
01:58
>> It's when a bit goes missing it gets lost in transit entirely. >> Perfect.
Yeah. So erasia errors like in the in the classical case is when a bit goes uh missing right.
So suppose I have a a classical
02:13
repetition code and you have you know so instead of sending a bit zero I send three bit zeros and one I said send three bit ones and so as I'm sending these bits say some of them go missing. to say the
02:29
middle bit got got I was sending this and the middle bit got missing so I don't receive anything here and so in this case uh I can I I know that you know since I didn't receive anything in the second bit I should not trust it um there's something called as a um
02:46
biased eraser error did you hear about this word so far okay bias eraser error is when this kind of missing error can only happen when the bit started in say the zero state or the one state. So in
03:02
this case for example uh you know you might suppose the uh eraser or the bit can go missing only when it started in the zero state but it will never go missing from the one state. This is
03:19
in fact in the classical case you don't need any error correction here because when you don't receive the bit you know that this was zero to begin with right so in this case it's like 100% correctable a biased eraser error um in
03:34
fact you don't even need any encoding here because like suppose a message you want to send is like 0 1 1 0 you will send the same message without any encoding and anytime you don't receive something you know that the user had
03:50
tried to send a zero. So this is like a kind of trivial almost like no errors uh case in the classical scenario but it turns out that in quantum scenario you can have quantum eraser errors but um you still need error correction uh for
04:06
that. So it's not one-on-one uh doesn't while in classical case you don't need any error correction in quantum case for bias quantum eraser you do need error correction.
So uh let's let's look at the the
04:22
quantum case for just a second. So let's look at the quantum repetition code and I'm going to write the repetition code for um uh zs.
So repetition
04:37
quantum repetition code for poly z errors. So I think this is something that Steve just mentioned just touched upon last time but he didn't go into the details.
Uh so I'm going to go a little bit more
04:54
into the details here. Um okay so suppose um so here I want to correct for z errors.
So my code words are so um one code word um I'm going to call this zero is so I'm
05:12
going to do a three cubit quantum repetition code. So I take three cubits um and this is one of the states and this is another state.
So let me call this bar represents the logical state and without the bar is the physical states. So I
05:29
prepared three cubits in the plus state and three cubits or three cubits in the minus state. So in general the state of course is anything it can be a superposition of uh the two code words.
05:45
Okay. So in this case uh the stabilizers that you measure the stabilizers of the code are x operators.
So stabilizers one stabilizer is xx identity the other stabilizer is i xx
06:04
um so you see that um and the logical operators um let's say the x logical operator is x x and the z logical operator is just um
06:19
sorry um because I'm calling this zero and this one. So z logical is actually x
06:37
let me call this as plus logical and minus logical so as to not confuse everybody. So this is x one and this is uh z
06:54
good. All right.
So in this case you see that if that if there's a single z error a single z error with anti-commute with one of the both or one of the two stabilizers and I can use that for error
07:10
correction. Do you want me to go into this code a little bit deeper?
>> Yes. >> Good.
Fantastic. Thank you.
So, uh, if you have a Zer for example, Z u Z error on the first cubit. So, the code word becomes Z I applied on S.
This becomes a
07:29
um uh minus plus plus B plus - Z flips plus to minus and minus to plus. Okay.
So now if I measure the stabilizers, if I measure if I measure
07:46
XX I, you'll see that X here X on the first cubit and the second cubit are different. Right?
So X X on the first and the second cubit is minus one. Right?
If I just apply xxi
08:03
to this state then this is a * - - - - - - - - - - - - - - - - - - - - - plus + plus - b *
08:19
uh + - right. I apply x to the first cubid.
I pick up a minus1 sign. I apply x to the second cubid.
I pick up a plus one sign. Minus1* + one is minus one.
And similarly, you get a minus sign over here. So this is just minus of the
08:37
original state you started with. So this state s uh so sorry after the error after one after Z error on the first cubit, the state is just a minus
08:54
one value of this operator. Um and it will you can work you you can work it out.
So zi error means the state becomes minus1
09:12
minus one value of x x i but + one value of iix xx you can just work this out and then I zi error implies it's a minus one
09:29
value of xx i and minus one igon value of iixx and then for i z error it's a + one igon value of x x i but
09:48
minus one value of i xx okay did I do something >> I think I'm not following that so the logical x is x i but measuring xx.
10:06
>> Yeah. So these are the stabilizers.
So these are the logical operators. You don't want to measure them because they are the >> Yeah.
Uh so if maybe I should put bar to be consistent with the notation. Yeah.
So I should never measure this. I'm just giving this information to you that if
10:22
you wanted the logical information, these are the operators that you will need to measure. Um you see X logical should put put a minus sign between the plus and the minus states and the Z logical should take you from the plus to the minus state and you can confirm that
10:37
these two operators do that. Okay.
And these are the stabilizers. So these should commute the logical operators which they do.
Uh and you should measure them to determine if there's an error or not. So in here I've shown just worked out an example that if there is a zi
10:55
zero on the first cubit then this is a state that you get and you can confirm that if you measure so this state is a minus one this state becomes a minus one value of xx i. So when you measure this
11:10
stabilizer you'll get a minus one outcome but if you measure the second stabilizer you will get a plus one outcome. You can work this out.
Just apply ixxx to the state and so on. And then again you can work out that if the error was on the second cubit you would
11:27
um both s1 and s_ub_2 will be measured in the minus1 state um and if the error was z on the third cubit that only s2 is in the minus one state and s1 is in the plus one state. Good.
Okay.
11:43
So with this you can you know decode you can basically this is your table and you can just say you can just go measure your stabilizers. If you measure both stabilizers to be minus one then the error was on the second cubit.
If one stabiliz the first if s1 is minus1 s2 is
12:00
+ one this was the error and if s1 is + one and s1 s2 is minus1 then this is the error. All good.
Now what is the problem here? The problem is if I have two errors, right?
12:15
Suppose I have two errors. So suppose I have ZI.
If I have ZZI for example, if I can apply it on the state S and I will see that so this is ZZI applied to say a ++
12:34
b - uh a sorry a - plus + b + plus -
12:50
Okay. And this state is so this is the let's call this the error.
Yeah, it's called the U error state.
13:05
And this state is if you measure the stabilizers, if you measure the first stabilizer, this will still be + one. But if you measure the second stabilizer,
13:21
you will get a minus one because it's a minus one value of the second stabilizer. Just work through the math.
Simple. This state is the plus one state of the first stabilizer but the minus one state of the second stabilizer.
So now if you go and measure the
13:37
stabilizers, you will still get plus or minus one. But you would have thought that there was a error one error on the last cubit whereas what actually happened was uh uh two errors um on the first two
13:52
cubits. So you you get confused basically right.
So if if all errors are happening independently so that the probability of one error is more likely than the probability of two errors then you
14:08
should you should guess this error but there's a chance that actually two errors happen and you missed it. And so the probability of a of a logical error
14:23
is when two Z errors happen and this is P square right I should write it as order P square
14:39
because there's a cominatoric factor on how you can select those two errors from happening and of course there is a probability that there's three Z errors 3 Z errors is basically you're doing a logical Z on your state of course That's an
14:55
uncorrectable error. Um but if P is small then P square is smaller than P cube.
So this is why I write it like this. Yeah.
15:13
[Music] decoded this way then the probability. So suppose you had said that whenever I see + one and minus one I will assume that there was two Z errors then the you will fail if there was actually one Z
15:29
error happening here. So then the probability of logical error will be P and you can make the code that way but the logical error rate will be higher than if you had done it this way.
Right?
15:46
For some reason, if your system if your uh uh some of your cubits are such that it's more likely you'll only get correlated errors. You never get any single cubit errors for some reason, then yeah, this is probably a better code, right?
But this is under the
16:03
assumption that single cubit errors are more likely than two cubit errors. In fact, if if there's only two cubit Z errors that ever are going to happen, then you can just use the repetition
16:20
code and you'll always be correct. Okay.
Um so this is just the repetition code for policy errors. Now let's assume
16:36
that uh the errors you're having are uh eraser errors. So quantum eraser errors
16:58
I'm going to start with this even the simpler case of quantum biased erasure error. So in this case the error is basically
17:14
say uh if you uh remember the krauss operator formulism from this morning you know one of the crusser is like with some probability P uh uh the state zero goes to some state some other state E um
17:35
and um yes nothing happens to the one one state is always goes through but the zero state can go to some leak state E and so you can complete this um so the full cross operator would be
18:18
Let's forget this cross second cross operator for for a minute. Let's just look at this guy for a second.
Um so this is the error and we're saying that zero can go to E and whenever Z goes to E um I can measure you know I
18:36
can check that the cubit did go to E um and I can know that this error has happened okay so let's start with the repetition code so I have my state of the repetition code was some a plus plus
18:52
plus plus b - - Now let's let's act the error uh on the first cubit. So the first cubit um uh
19:08
nothing happens to the other two cubits just the first cubit goes to the E state and I add this upon a + plus b - - - - - - - - - - - - - - - - - - - - - so if you open this you you you know do the math you'll have this state the plus
19:26
state is 0 + 1 so this will be a e plus plus p e - right I just if you open this this guy
19:42
is 0 + 1 over square<unk> of two there's a there's a square root of two factor here which I'm kind of neglecting to write but um yeah and this is 0 - 1 over
20:01
So when I apply this operator to this state I just get 0 goes to e and I get 1 /<unk> two factor here. Okay.
Okay. So now this is let me just simplify
20:16
this. This is just e over two uh tensor a + b minus minus.
just saying the first cubit is an E and then the remaining two cubits are still entangled.
20:32
Okay. So now what I do is I'm going to replace whenever I see the cubid is an E, I'm going to replace it either with a zero or one with 50% probability.
You'll ask me why am I replacing it with 50%
20:48
probability. It's just to simplify the math.
It doesn't matter. I can always just replace it back in zero or I can always replace in replace it back in one.
It doesn't ma matter
21:18
actually. So let me do this first just by replacing E with zero.
So whenever I see a E, I take this state and replace it back with zero. So replace the first cubit
21:34
in zero. If I do that, the state now becomes zero.
Answer. I'm going to forget writing the square root of two.
It's just annoying. Uh assume that it's there.
21:50
Okay, now I go back and measure the stabilizers. Okay, remember the stabilizers were the stabilizers were XXI and IXX.
22:06
So if I go back and measure the stabilizers, um I can either get the stabilizer value to be + one or minus one. So if I if I start from this state a 0 + plus plus b 0 -
22:26
then um s_ub_2 applied to this state is always this state back. So s_ub_2 applied to s is always s back.
But S1 applied to SE
22:45
will be either plus one or minus one with 50/50 probability. So I'll have uh so I'll have a plus plus plus plus b
23:01
um - - 50% probability or I'll have a
23:16
so measure s1 one on the state. This is a minus plus plus plus b + - with 50% probably
23:38
is is this step okay or should I explicitly work out? I'm happy to work out, but I'm not sure how familiar you are with the math.
So, I'm not sure if I'm going to just bore everybody if I work it out or how many how many of you think I should
23:54
work it out? >> Thank you.
Okay.
24:17
Okay. So when I measure S S1 I can get two measurement outcomes either plus one or minus one.
To know what is the state corresponding to the plus one measurement outcome I can just apply the projector onto this state.
24:33
Right? Is this okay?
So if I uh okay, maybe I should do it slightly even one step lower. Okay, let's just apply S1 onto
24:52
S. If I apply S1 onto S, S1 is X uh X on the first two cubits.
So I'll have a uh so a x x i on 0 +
25:12
b x x i on uh 0 - minus. So what is this?
x acting on zero is one. So I'll have a 1 x acting on plus is plus plus
25:28
b x acting on um uh yeah this is fine plus b x acting on zero is is one uh x acting on minus is minus one so I'll just get a minus one sign here
25:44
minus uh minus okay simple Now I want to I I don't want to write I don't want to keep zeros and ones here because my code states are in you know
26:00
the plus and minus basis. So I know that um plus is just 0 plus or I know that I can write zero as plus minus over<unk> two.
I can write 1 as plus
26:18
- -<unk> 2 right forget about square root of twos I'm just going to put this back over here so here I'll have this one becomes plus minus and then I have the remaining plus+ minus p I'll have plus min - minus
26:38
and I have - okay so here I have a plus plus plus uh a - plus plus - b + - - + plus b - -
26:56
okay I'm going to put this term and this term together and then keep these two terms together. So this is just a plus plus plus plus b -
27:13
plus a - plus plus - b + - - if you remember from the first part uh of the discussion this was a plus one igon state of x i and this was a minus
27:31
one igon state of xxi. We did we did that in the first part of the exercise.
So that means that when I measure the state I will either collapse it into this plus one state or the plus one minus one state and all of the and these two are happening in at equal
27:46
probability which is why I said with 50% probability I'm going to be having either this state or I'm going to have this state. Okay, good.
So um I will measure uh S1 uh if I get a I
28:03
will get a plus one measurement outcome with 50% probability and I'll know that this is the state of the system and I'll get minus one uh measurement outcome and I know that this is the state of the system. Okay.
Now if you remember
28:21
so I I have measured s_ub_2 if I if I just put the measurement values s1 comma s_ub_2 uh s_ub_1 can be + one or minus one but s_ub_2 will always be + one right
28:38
if you remember um if I measure s1 to be minus1 and s2 to be + one there was two possible things that could have happened. I either had Z error on my first cubit or Z error on the second and the third cubit
28:55
and that's where my in in the poloise case that's where the confusion was asking coming from that I didn't know whether the error was this was this measurement outcome came from a Z on the first cubit or Z errors on the second and third cubit but in case of erasers I
29:11
know the eras erasia happened only on the first cubit so there is no confusion only thing only possible Z error should have been on the first cubid. So there is no confusion.
So I've removed the the uh confusion from the decoding.
29:29
So in this case I can never confuse between one uh a Z error coming from one cubit being erased or two cubit being erased because I can always distinguish did my eraser flag said only one cubit
29:45
was erased or did my eraser flag said two cubits are erased. So I've removed that confusion.
So in this case the probability of a logical error becomes p cubed actually exactly equal
30:03
to p cubed only if all the three cubits are erased will I never be able to recover the outcome uh the recover the state
30:20
>> I still have to measure both the stabilizers >> and look at the outcome >> look at the outcome and then I know that um from the from the measurement. So I I need two bits of information to decode.
I need which cubit got erased and what
30:36
is the stabilizer measurement outcome? If my first answer is cubit one got erased and this was a measurement outcome then I know the Z happened on the first cubid.
I can recover it. If my first information said two cubit got erased and this was a measurement outcome then I know that the last the
30:54
other two cubits oh sorry this was a measurement outcome then I know that the second and the third cubit got erased right so with these two pieces of information like in poly case I only had one piece of information I only had the stabilizer measurement outcomes and
31:11
that's it in the case I have two information I not only know the stabilizer measurement outcomes but I also know exactly where the error has happened. I can exactly know whether one eraser happened or two erasers happened.
>> Yes. >> Just make sure I'm understanding.
So the
31:26
reason you can tell where the eraser >> Yeah. >> Exactly.
Yeah. Yeah.
So I need to know firstly I
31:43
need to know whether the cubit went to E. But when I do that I need to make sure I don't accidentally measure the cubit like I don't distinguish between zero and one.
I just only tell whether it was erased or not. Um secondly I need to be able to distinguish where the eraser happened.
32:01
Like this cubid got erased or this cubit got erased or this cubit got erased or these two cubit gots got erased. I need the resolution.
I need the in fact uh I need spacetime resolution on exactly where the eraser has happened.
32:17
So all the advantages of eraser errors come from the exact space you know the the space-time location of erasers. If you lose if your if you if your experiments are really really bad and you cannot resolve the ratios very well then you will start losing the advantage.
32:33
>> Um Same thing.
32:56
>> Oh, no. No.
This is just saying uh you get So, this is saying plus one measurement outcome with 50% measurement. Okay.
Plus one measurement outcome with 50% probability. And then minus one
33:12
measurement outcome with 50% probably. >> Yeah.
Right. >> Uh
33:28
wait a second. Um you this is fine.
Um yeah this is fine uh in the sense that when you when you know you got a minus on measurement outcome and you know the first cubit got erased like you can just
33:45
go and apply an x to the first cubit and just fix this. Yeah.
Yeah. This is fine.
This is this is still a u this is still a minus one measurement outcome of uh s1. Yeah.
Good point. Good question.
Yeah. I I neglected to focus on this
34:03
minus one. But this minus one is totally fine and very shortly this minus will actually become important.
34:27
>> What happens when I've not biased? Yes.
So I'm coming to that just yeah and that's where this minus one sign will become important. Um uh okay so a biased what we have just seen is that a biased measurement eraser basically looks like a zer at a known
34:45
location right that's that's what we have we have basically seen now what happens if you have um uh so okay let me let me say one more point so in this case like we can show in principle that the quantum repetition
35:00
code is sufficient to correct for a biased measurement error and Uh did did Steve go over threshold like what a threshold of a code is or did somebody mention threshold of a code? Um but basically you can show that as long as the probability of this error
35:16
is less than 50% this code will work it will suppress errors actually yeah 50%. Um and the difference from the classical case is that remember for the in
35:32
classical biased erasers, my code will succeed 100% of the time. In this case, it can fail if there's three errors because a three error will a three three all all three cubits being erased means like you've
35:48
completely destroyed the superposition, right? So, in this case, you can never recover the state back.
In the classical case, you could could. So, that's like a difference between quantum and classical.
Um, good. So now let's come to the uh unbiased eraser.
Wait, where did I put
36:05
the eraser? I don't know what I did with the I'm just going to use this.
I hope this is not somebody's
36:20
I don't know what happened to it. Uh okay.
So let's look at um unbiased erasers. Okay.
36:36
Oh, perfect. Thank you.
I'm sure I'll lose it again. Um, so in case of unbiased erasers, okay, this doesn't work or simply biased eraser or unbiased
36:52
eraser. In this case, you could have two possible uh um cross operators.
So you could potentially go to the array state from the zero state or you could go to the array state from the one
37:08
state. Um and of course there's a third cross operator to complete the map.
Um in case uh like let me just say one more thing. If uh if both the cubits are
37:24
going to the same erase state then it's a regular eras sorry this is unbiased or regular uh eraser. If suppose you have cubit zero goes to
37:42
some erase state and the cubit one goes to some other erase state this is still a bias eraser. uh biaseratia doesn't mean that only one state goes to some other state and nothing happens to the other state.
Um
37:58
bias just means that you can tell the difference between erasia coming from the zero state or from the one state. You should be able to distinguish between that.
So if if your cross operators are like this and your experiment can tell whether you went to the E1 state or the E2 state, it is
38:14
still in that case it's still a biased eraser. Unbiased erasation means there's no way for you to tell whether the eraser came from the zero state or the one state.
Okay. So um what happens in that case?
38:35
Let me erase this.
38:54
So remember when I had uh this example that I worked out was when eraser started from the zero state. Okay.
So remember this when the eraser started in zero state uh and if I tried to measure S1 I would either collapse myself to
39:09
this state or to that state. Now suppose I repeat the same exercise but the eraser started in the one state.
So I start from a ++ plus plus plus b - - and the eraser happened from the one state.
39:26
So one state went to the erase state and nothing happens to the two other cubits. And now I do the same exercise.
So here I have a uh this will be e plus
39:42
plus plus minus b e - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - you picked up a minus sign here. You did when when you when this was from
39:57
zero to e, you did not pick a minus sign here. But now when it went from one to e, you did pick up a minus sign here.
>> Directional. What do you mean by directional?
>> Yeah, it's one to e because you see the
40:15
this state is zero minus one over two, right? So when I apply this on this state, you'll pick up a minus sign because of this guy, right?
40:34
>> From the math because Yeah. Yeah.
Um right. Okay.
So now I repeat the
40:52
um I I will suppose I repeat my stabilizer measurements. Uh again if I measure S2, S2 will always be + one with 100% probability.
So measurement outcome of S2 will be +
41:10
one with 100% probability. Measurement outcome of S1 will be plus or minus one.
+ one with 50% probability and minus one with 50% probability.
41:27
In this case however when you measure + one you will the state the state s after measurement will be a ++ plus minus b -
41:48
uh when s1 is measured in + one And this will be a um minus plus plus
42:04
plus b + - with 50% probability when s1 is measured in the minus one state.
42:20
Okay. Here you see there's a minus1 difference there.
This is + one and this is minus1. But you explicitly knew that the eraser came from the zero state.
So when S1 was
42:37
measured to be the minus one state, you will just you know apply an extra gate here and fix this minus sign back to plus sign. Here you have completely lost that information because you don't know whether erasia happened from the one state or the zero state.
42:55
So in this case if you measure minus one and you apply um and you know that the eraser happened on the uh on the first cubit you don't know whether you should apply an x or not right because if it if
43:10
the eraser happened from the first cubit and it's measured in the plus one state you should not apply a flip here you should apply a flip whereas it's the opposite here. So in this case just a repetition code
43:27
is not enough. This this is actually like an X error right because the difference the the difference between this minus plus sign is whether an X happened on the first cubit or not.
So in case of a regular eraser error
43:43
when you cannot differentiate between the erasers happening from the one state or the zero state the eraser does not just look like a Z error but it looks like a Z and an X error. Okay, I confused you.
Yes,
44:00
>> you can't use the repetition code to fix it. You need a code that can correct for both X and Z errors.
So like the surface code or something. So I think Taka was mentioning the surface code this morning.
Yeah. So if you have a code that can correct for both X or Z errors uh then that can correct for regular
44:16
erasers. Um uh or like if you want to go for a smaller code like the shore code for example which also corrects for X and Z error both that will correct for regular erasers.
Um
44:33
even with regular erasers though the advantage you have is I know exactly where the eraser has happened and I can uh go and correct it. It's just difficult for me to demonstrate the advantage of regular ratio with the simple example of uh the repetition code
44:58
even with um even with reg. So suppose I have a um which part should erase?
So in general
45:14
I can show that. So if I have a uh code corrects for
45:30
um some you know n uh bit flip and face flip errors.
45:45
Then the same code
46:01
so correct for twice as many eraser errors. This was clearly demonstrated in the example of the repetition code and biased erasers.
Right? The regular repetition code will correct for one phase flip error, but it will correct for up to two eraser errors, right?
46:20
I could not demonstrate this just using the repetition code for regular erasers but believe me that this is true even for regular erasers. Basically if you know
46:37
the issue with poly errors is that different errors give rise to the same syndrome and that causes a confusion like I cannot tell whether the same that the that syndrome arose from this error or one minus of that error in erasers I
46:53
can completely eliminate that confusion right because I know exactly um where the error happened. So basically you effectively need erasers on every cubit to um kind of miss all the information.
47:17
I will just stop here for a second. Let everybody absorb it and So reg biased eraser looks like a z
47:34
error but at a known location and unbiased erasure looks like a z and an x error um at a known location. So this is why like if you if you if you think about how do you model these errors in practice like when I write a
47:51
code how do I like model this in practice uh I just tell my co in uh in my code um I will just say apply x y z uh xyz ers in different places if they
48:07
if I'm calling them as erasers just give the decoder the information that that where the x y and z happened. That's all I just give the information about the location of error, not necessarily the
48:24
type. Like if I have unbiased ratio, I don't know it was if it was an X error or Y error or Z error.
I just know that one of these three happened on cubit some cubit. In case of bias derasure, I know not only which cubit was faulty but
48:39
also that it had only a Zype error. Yeah.
>> Sorry, can I say that again?
48:55
>> Both kind of right like I I need uh the erration from um not from the code but like as a separate thing I have a information about which cubit went to the E state. So suppose I have like 10 cubits.
Before I
49:12
even do these stabilizer measurements, I will just I will have some way of telling whether uh did this cubit number one go to the east state? Did cubit number two go to the east state?
Did cubit number three go to the east state? I'll take that information with that information and
49:29
the information about the stabilizer measurement outcomes. I need both of them to actually decode the final state.
The photon >> photon period >> uh photon par uh you mean the which
49:45
information? The stabilizer information or the eraser information.
Those are two different informations. Don't worry about how the information comes from.
I I'll give you a practical example in just a second. Maybe it'll become more clear with them.
But there are these are two different uh ways the two
50:03
different orthogonal information you're getting from your system. uh let me give an example of how that may happen in like a dual rail type cubid that you may have heard about or a ritburg atom cubit that you know I should talk about too.
Um so so say if
50:21
the concept of uh erasers is fine and why they're really good I can go to like how do you practically think about implementing it
50:45
>> so what I've described so far the concept is platform term agnostic how do you actually make such a cubid is of course that yeah so in I can tell you what we do in rubric atoms so so the main the main idea is that you
51:00
need like a three-level system at least right um and you want to make sure that there's no like direct or the uh probability of errors in the computational subspace is small compared to probability of errors that take you out of the subspace and you need to be
51:16
able to measure whether you're in that extra subspace or not. So just as a very high level thing the several different species of redber atoms most of the time what people do is people encode information in the ground state
51:31
subspace. So I'm going to I can give you some chemical structure 3 P1 whatever 4 P 0 not important what's important is you take two states and these are the uh you know the stable states they have really really uh uh
51:48
they have long lifetime like several seconds can can happen um and that's why people work with this and um you do gates the uh in this within this subspace the problem though is that uh there can be decay. It's
52:03
rare, but they can decay can still take you from one uh to the zero state and um um that's bad. Um in some cases the you you use the ground state subspace
52:20
and I'm going to put this another um state the redberg state. So in rinberg atoms what happens is that uh when you're sitting in the ground state subspace the atoms don't really talk to each other.
Um they don't have sufficient wave function overlap that
52:36
they can talk to each other. If you put them in these rit states these are highly excited states they have like big kind of sphere where they can you know the wave functions are large.
That's when the two atoms can uh talk to each other by something called as a red bird blockade.
52:52
Again, not super important, but they can talk to each other when they're in the rhythmic state. So, what you do is suppose you have um so to do gates, what you do is you excite um one of the states.
So, let's
53:09
say I have um this is like just a cartoon um cartoon kind of picture. Uh don't like there's a lot of details involved which I'm not giving but not important.
Um So this is atom number one. This is atom
53:24
number two. So what you do is you u apply apply selective laser pulse so that you only take the zero state say to the red brick state.
Okay. So you couple the zero state to the red brick state.
Zero state to the red brick state. And
53:39
so now this is effectively your cubid subspace. The red brick state and the one state.
The red brick state and the one state. Now when the atoms are in the red brick state they will talk to each other and they will repel each other.
Otherwise they will not. So in this ca case the redberg states get a phase the
53:57
otherwise you don't. So this actually does and and once you're done with the gate you bring the rberg states back down to your computational subspace and those are the states with long lifetime and you just sit there.
So basic principle when you're in the
54:13
when you want to do gates when you're not not doing gates you sit in the ground state subspace lifetime is long. When you want to do gates, you excite yourself to the rberg state.
The selectively the zero state to the rberg state. The atoms interact.
They pick up a phase and then you bring them back to
54:29
your computational subspace where they can live longer again. The lifetime of the red brick states are shorter than the lifetime in the ground state subspace.
So during the gate you can happen it can happen that the red brick state decays. And where will the red rick state decay?
It will decay to the
54:45
ground state. Right.
So decay back to the zero state. So this is bad because now some errors happen because whenever you come back to the zero state, you will stop doing the gate, right?
So all of a sudden you would you were accumulating phases when
55:02
you were in the rubric state. But now all of a sudden you fell back to the zero state and now you've stopped accumulating accumulating any phases.
So that's almost like a Z error. And you don't know this has happened because you know you didn't you're back in the computational subspace.
So you were supposed to be back in the computational subspace. So this is like a poly error.
55:20
So until recently this was kind of the uh uh neutral atom cubits that people were thinking about mostly. Uh Ku the company for example still uses this kind of setup.
What you could do is uh in
55:37
um uh in contrast is that you work in a metastable subspace rather than the ground state subspace. So you know the atomic spectrum is rich uh and you work with a species where you
55:54
have um something like this. So these are some metastable states.
I'm going to call that as member states. These are still long lived.
Maybe not as longived as the ground state subspace,
56:11
but decently long. This is still like a few seconds.
This can be tens of seconds. This can be like a few seconds or a bit smaller.
Okay. Now again, this is a rberg I'm going to call this ground.
Ground.
56:26
Okay. Okay.
So now what you do is now if you want to do the gate you will say excite the zero state to the red brick state again. Do the gate the same way and bring the red brick state back down whenever you're done with the gate.
Now when the red brick state decays what
56:42
happens it will mostly decay to the ground state will mostly decay here. There'll be some decay to these states, but they are much less likely.
So, I'm
56:58
going to draw this like thicker because this is much more likely than transition zero. So now what you do is you can now measure you can check whether my my red atom
57:15
ended up in the G state or not. If it's in the G state, I know something bad happened.
I know the gate was bad. Um, so it's like an eraser.
Of course, there is some probability that the rubric state goes to these states, but uh, you know, our calculations show that this should be
57:31
like 50 to 100 times smaller than the probability that it goes to the G state. So we can define an eraser fraction.
So eraser fraction is probability of
57:46
eraser divided by total probability of error. So this includes computational error plus the ratio errors.
This includes the top is just the ratio. So you can achieve in principle like 98%
58:04
um or 0.98 as fraction uh ratio fraction which is pretty good and you can you know this makes air correction much easier or this is enough to see air correction becoming easier. Um again this is a simplified picture.
58:22
Um in fact it's it can happen that there's some type of errors where the population actually doesn't come back to either the ground state or these two states. It might get stuck in the red brick state.
So for truly converting all errors to erasia errors you not only not only have to check the population in the
58:38
G state you have to also check the population that was you know left back in the R state or maybe dec. You basically need to like check the population on every possible scenario where your atom could have gone.
If that happens, if you're able to do
58:54
that, then the erasia fraction is 98%. Uh more recently in the experiment which was done in Jeff Thompson's group, the erasia fraction was like 50 to 60%.
I can't remember the exact number. Reason being is that there were some um populations that was read back left back
59:10
in the redber state or you know some others. there's some population and states that were not being detected.
Basically, in principle, it's possible to do those detections, but for that first experimental apparatus, they did not do it for technical reasons, and
59:27
they got a lower irration fraction. Um, yeah.
Um,
59:46
so did this answer your So this this is how you would do the this is the first bit of information about which atom got erased. Um, after you do this, you say you're
00:02
building the repetition code. After you do this, you have to again go and check the parity or the stabilizers of the repetition code.
That's a separate thing you have to do. Um, in addition to this,
00:20
what else did I want to make sure I said everything here? Um, so, oh yeah, in this case, in fact, the eraser is not just standard eraser, it is biased eraser because the eraser only
00:35
happens from the Rber state. very it doesn't really h these states have long lifetime so it's basically zero from these states it only happens from the red brick state and you go to the red brick state only if you started in the zero state
00:52
right so basically it's the first kind like when the erasation happened you know that the atom must have started in the state effectively so this is a biased eraser so in this case truly technically a repetition code is is sufficient
01:07
and the bias of Asia was also checked in the experiment and that was also very high. I can't remember the exact number but it was like triple 9 or something like that.
So it's truly truly truly really biased.
01:23
Um so also let me say this it's very very important like I I must emphasize it's important to know to get a good
01:39
space-time resolution of erasers because with the repetition good example we saw the reason erasers are helping me is because it can help me decode. can help me differentiate between two confusing configuration of errors.
01:55
Of course, in experiments because of experimental limitations, it might happen that you lose this resolution. Um, so there have been studies on how much res losing resolution is okay or what what is the cost of losing this
02:10
resolution. For example, this case like if you have a code that corrects for n bit flip and face flip errors, then it'll correct for two n eraser errors.
This is true for perfect eraser resolution. If you have imperfect
02:27
resolution, this can completely go away. Um yeah so in my group uh elsewhere as well we have lot of work going on on how much uh
02:45
liberty you can take in the experiment to kind of like you know maybe make the experiment a bit faster but at the cost of making the experiment faster you're kind of losing the resolution on u where the eras happen for example Um
03:01
or maybe you don't want to do this eraser check after every gate operation. You want to do this after five gate operations.
You know, you're going to optimize on the number of times you have to do this check. Um so that you're kind of losing the time resolution of erasers.
That can also
03:18
affect kind of how many erasers you can correct. Um again so how much how much leeway do we have in experiments to do this
03:34
>> in different platforms >> is different >> um so there there is one eraser cubid
03:50
proposal for trap time systems I I don't remember the details of that okay and this number is also a theoretical number not an experimental number right um in principle yes it's possible to do this in experiment but it has not been done yet so it's hard for
04:07
me to like compare numbers because even in superconducting cubits in kind of practical but you know experimentally optimistic numbers I can get a refraction like this But has admit has that been demonstrated
04:25
maybe I'm not sure right um and then there is also question of what is the probability of missed erasers like what is the probability erasia happened and you missed it that's also different in different experiments so it's hard to like me to compare apples to apples
04:42
superconducting platform versus redberg atom cubits I will I will give a comparison in a different sense though um even assuming you can reach the same uh eraser fraction what is the hardware
04:57
cost to build an eraser cubit right in the atomic cubit it's not that I'm like it's it's the same atom I'm not you know it's it's one atom whether I use it in the ground state subspace or some metastable sub subspace it's still one atom the cost is I have to use this
05:13
extra laser to detect whether I went to G or not or maybe two extra lasers if I want to detect some remaining population so the losses in number of lasers I have to use. But it's like if if I'm building a five cub say just for example if I'm
05:28
building five cubit code with an atom whether I use you know use them as eraser cubits or other cubits it's the same thing cost is going into laser power for example in superconducting cubits what is a cubit you have learned so far dual rail
05:44
you must have seen dual rail cubit people talk about it what is a dual rail cubit you use two cavities and then you use a transmon to check for erasers. So one you you could have just used the transpond as a cubit but now
06:00
you're using transmon plus two cavities to make that erasia cubit work. Is that better?
Which which which one's better in hardware cost to me? The atom platform is better right because in the in the in the dual rail
06:15
cubit you have you think you're reducing the overhead count. you you think you're getting this advantage of now I can double my recction capability but you're also doubling the hardware actually so how did you win truly anything
06:32
not sure it depends on how you count things um so we have a paper where we actually do some a little bit more in-depth calculation of um how many like couplers you need how many um like uh readout lines do you need for dual rails
06:48
and of course it's It's a theory calculation and I wanted the calculation to go some way. So you know um we have chosen parameters where we see that dual rail can also win over just using transmons but I still think like if I were to put my money on it I will not put my money
07:05
on dual rails. I will put it on uh reprod at least in my group the the dream is to just use the transmon as a cubit.
transmon is not a two-level system. It's a multi-level system just
07:21
like the atom. Um so can I use multi levels of the transmon instead and and make an eraser cubit within that.
Um and we've been there there are some limit of course it's not as easy otherwise we would have done it already. Uh there there are some
07:37
challenges there and so my group is working on um overcoming like trying to design a good erasia cubit in the transmon. Um if you have ideas or if you want to talk more about that I'm happy to uh we haven't solved the problems yet.
So yeah I think if we have that
07:55
then we truly have a good superconducting erasation cubid. Yeah >> don't have a question.
So how could a transmon be a multi-level system? The cartoon in my head was two level.
>> Yeah. So transmon is is a anharmonic oscillator, right?
Like you have
08:10
>> um so it is um just as um like again super high level thing, right? Um so basic trans is like a cosine potential and there levels inside.
We basically work in like one kind of cosine and uh um you know there's
08:30
Okay, I'm going to kind of draw like exaggerated image here, but there are several levels energy levels here and you have the G state, you have the E state and maybe you have some other states here, right? But you choose to just operate in in these two like you're able to control these two
08:48
without ever ever meaning with low probability going outside this two subspace but it is still a multi-level system. So what if like we use say instead of these two states I use this as cubit zero this as cubit one and this
09:04
is my eraser state because the dominant error is like jumps going down right so dominant error will be one going to this state you need two jumps go from here to here and then from here to here to actually have a computational error
09:19
um so this is like the simplest we we call this the GF cubit because um or the GF cubit uh in trans language this is the F level and this is the G level so that's what we call GF cubit rather than the GE
09:35
cubit this is the E level um so the hope is at the end of the day is that we are able to use the GF cubit as a eraser cubit rather than um the GE cubit
09:50
several challenges the most important being you want to measure whether where you want to measure if you're in E or not E. But whatever measurement schemes you have come up with or whatever the simplest measurement scheme or most obvious measurement schemes are will not only tell you whether you are in E or
10:06
not E, but it'll also tell you sometimes it'll tell you the information that whether you're in G or F. And so I'm not only getting information about E, but I'm kind of measuring the cubit itself in the computational subspace and that's not good.
So how do we minimize that kind of computational subspace
10:23
measurement? That's the hard part that we have to come up with more
10:49
complicated than that. Yeah.
Yeah. basically um you know you will you'll have some readout thing which will be coupled with your transmon.
coupling is such that I mean this the if if it couples with E it
11:05
also couples with G and F and you can't kind of minimize that coupling um but while also keeping this coupling with E large yeah
11:24
so to answer your question maybe fractions you can get as large in superconducting circuits as you can and the mid brick atoms but I think the better question is how are you actually making the eraser cubit and
11:39
you're already increasing the hardware count in superconducting circuit so do you actually win anything I don't know it's a hard hard question
11:58
>> scale what breaks Yeah. So that is a problem.
But um I
12:14
mean people have like thousands and thousands of atoms uh in the labs currently being controlled by one laser. Um, so I think it's not a technologically impo impossible problem to have many lasers
12:29
to, you know, control a million whatever cubits atoms. Um, I'm not saying it's easy, but I I I know very clever people believe that it's possible and there's a lot of money in the field.
So yeah,
12:44
I mean it's the same thing, right? Like with superconducting circuits, you need like millions of them on a single chip and you need all of these buyers going in and so Yeah, I mean clearly you have people have been able to control more number of neutral atoms than they have been able to control superconducting cubits.
13:00
Yeah.
13:19
So more recently I can tell you that um there's work uh from the rubric atom group um so from Princeton for example where they have implemented the 422 code um
13:40
I always forget about whether double bracket is for quantum or single bracket is for quantum but the quantum 422 code whatever double or single bracket. Um so this code uh can um it can uh it can correct for uh
14:01
two erasers. Wait a second.
Am I saying this right? Yes, it can correct for it can detect uh this code can detect
14:18
one poly error and correct for one eraser error. It can only detect poly errors not correct them but it can correct for eraser error.
So the 42 code uh did taka talk about this at all? Okay.
Um
14:34
so the 422 code for example is four cubit code. Uh it encodes two um two logical cubits and the stabilizers are let me call this cubit 1 2 3 4.
The stabilizers are xxx and stabilizer one.
14:52
Stabilizer 2 is C C. Uh so the stabilizers are four body operators.
All X's on four cubits or all
15:08
Z is on four cubits. Uh there are two logical cubits here.
One logical is like this X X. When I when I do this that means that X is on this cubit and X is on this cubit.
Um and the let me call this X logical one. The Z logical one is
15:27
uh Z on this cubit and Z on this cubit. So this is Z logical one the uh
15:46
X. So Z logical 2 is Z on this cubid, Z on this cubid and X logical two is X on this cub X on this cubid.
16:05
Is this does this diagram make sense? This just the square just indicates the stabilizers are four body operators.
um these other indicates what are the two logical um cubits you can encode in them. So that means that Hbert space of
16:21
four is protected. The X logical operator of the first logical cubit is X on cubit number one X on cubit number two and it's Z logical is Z on cubit number one and Z on cubit number two.
Um and X logical for the
16:37
second cubit is X on one X on three and it's Z is Z on three Z on four. You can make sure all of this is correct because all of these operators commute with the stabilizers and this anti-commutes with this this x anti-commutes with this zlogical as it
16:53
should happen because they're poly x and polyz. Uh similarly this x this x logical anti-commutes with this zlogical but this xlogical commutes with this logical and this zological commutes with this logical.
So everything is proper
17:10
just make sure accommodation relations are well behaved to check that you have the right encoding. Anyway, so why is what does this code do?
Suppose I have a poly zer on cubit
17:27
number one. So there's a z error on cubit number one.
Z error commutes with this stabilizer. So it will not change the value of this stabilizer.
But Z anti-commutes with this XXX stabilizer. So if you measure X XXX X and there was
17:43
you find it to be minus one then you know that there is a Z error on one of the cubits. You don't know which cubit yet because a Z error here will also anti-commute with this guy.
A Z error here will also anticute with this. Zer here will also anticute with this.
Z here will also
18:00
anticute with that. So it's an error detecting code, not an error correcting code.
But um if you now have information about the which cubit got erased then it becomes an error correcting code right?
18:16
If you know this operator was measured minus one but you know that only one this first cubit got erased you know that Z error happened exactly here right? So this becomes an error correcting code for eraser errors.
This is the smallest error correcting code for eraser errors
18:33
possible only detects one poly errors but corrects one eraser errors. Showing you again the power of Asia errors.
Um and so so in Jeff's lab for example they have implemented this code um and uh
18:51
they they show that if you did take take into account the eraser information versus not how does my performance of the code improve. Um so that's an example where the eraser errors was used in practice to improve uh error
19:07
correction. So this was kind of the rough sketch of what I was thinking of.
I will stop and take errors. Uh take questions.
Uh
19:23
yes, I can I can stop and take questions. Um I can go into this code in more details.
I can go into the stabilizer formalism in more details. whatever you want me to talk about.
19:42
>> So, are both these stabilizers are the ones we talked about earlier with the uh each one's connected and >> yeah, are you talking about the surface code? >> Yeah.
>> Yeah. So, this is this is kind of like a single patch of the surface code.
19:58
>> Specifically, this is a single patch of the Doric code, but yeah. It's related.
Yeah. Was that your question if they're related or not?
>> Yeah,
20:18
>> it's the smallest surface code or code you can make.
20:35
Okay, I'm went faster than I was hoping, I guess, but uh >> physics applied physics
20:52
>> physics. So students >> no it's mixed uh I have a computer science student I have students from physics department um >> one of my students who's not here right
21:07
now but she came from like biology background yeah >> but no we we work a lot with I And this
21:24
is kind of computer architecture, right? Uh that we're thinking of.
Um so it's like computer architecture, but we also think of the actual uh physics of how you're building the cubid. So
21:40
yeah, but like there's a lot of there's a lot of work there's a lot of interest in in making eraser cubits and lot of like I know a lot of people are trying to build like eraser cubits with fluxoniums. there is more um uh instead of using transmons you use fuxoniums but
21:56
the idea there again all of them underlying idea is the same we want to just use that one element to make the cubit and not make you know bring two cavities and then more jazz on top uh yeah so right now the the game like everybody
22:13
is uh so I I think our this Adam paper was kind of like started the whole thing and so now basically after that everybody's trying to make their own keywords.
22:29
>> Oh jeez. Uh, I'm not going to be good at explaining that, but it's um um I'm not going to be good at it's a it's
22:45
um my student Tom is going to be better at explaining that, but basically the idea is that like the I I think of the transmon as like this. It's a rotor.
So, it's like a cosign potential like kind of goes away and wraps onto itself. the fluxonium uh these wells are kind of
23:01
transformed. So it kind of looks like like this and you can kind of change these wells and so your your cubid states are like these two >> different >> different yeah you just b you still use a join junction but you like shunt it a
23:18
different way so it's potential changes u so it has different like um positives and negatives compared to uh this guy um but yeah they're they're also trying to make an eraser cubid in in flexia
23:33
But those two levels are not >> uh they don't so they usually do gates with these higher levels up there. But again like you see this is kind of looking like this lambda system that the Rberg atoms have.
Um that's uh
23:51
yeah but yeah so like my student Tom for example is is working on making cubits in this kind of setup. Um some of the other students are thinking about this guy.
24:10
>> Yeah, toric code is just a surface code but wrapped around a Taurus. Um so you see I could let me just give a simple example from here, right?
Like this.
24:25
I could have a code that looks like this. So my stabilizers are XXX and my uh there are two more stabilizers which is ZZ and ZZ here.
So if I number the cubits 1 2 3 4 then one stabilizer is x1 x2 x3 x4 and another stabilizer is
24:45
z1 z3 and another stabilizer is z2 z4 one two three four yeah z2 z4 right these are the three stabilizers here for this code I have one logical
25:00
cubit the logical cubit is such that Um the Z logical is Z here and the X logical is X X like this. So Z logical is Z1 Z2 X logical is
25:20
uh X2 X4 like that. Okay.
Now assume so this this is this is one code this is encoding one piece of information one logical information this
25:37
is this still corrects this still detects for one u poly error and corrects for one eraser now what you can have what you can do is you can take this and um kind of take this think of this as a
25:54
page and kind of wrap it so that these two cubits are kind of like sitting on top of each And so that these stabilizers come together. So yeah.
Yeah. So these two body stabilizers become a four body stabilizer.
So in that case
26:10
um I actually get this code. So what I've done is I've taken the two two open Z stabilizers here and wrapped them around and made one fourbody Zstabilizer.
Um so
26:27
and then I get that then I get this code um with these stabilizers and but I've now uh I've reduced one stabilizer right like I here I had three stabilizers I've now reduced it to two stabilizers
26:43
and in doing so I've actually increased the number of logical cubits here I had one logical cubit here I had two logical cubit so the surface code is just doing that surface code is taking your surface and like closing it like um
27:00
closing it on a Taurus. So both these Z edges come together and the X edges like come together like that.
So uh as an example if I have to draw 3x3 surface code um this is a kind of plot you must have seen in Taka's talk.
27:19
Uh no one second. Yeah, this stabilizer stabilizer and the stabilizer.
So this is the surface code. Now you
27:35
just take this and kind of like bring them bring these two like this and this like that. So you're like closing this paper on a Taurus basically.
So here you will get the toy code. These two body stabilizers will become four
27:51
body stabilizers. And instead of encoding one logical cubit, you'll actually end up encoding two logical cubits.
>> Yeah.
28:07
>> Higher GS. I don't understand what that means, but there is like 3D code.
So this is like the I want to say yes to that question but like um there are um to codes which are like I can't like
28:23
draw it. It's not like like a simple take a 2D surface and wrap it on itself.
It's like more complicated 3D things that you have to kind of close on. Yeah.
High dimension thing. Yeah.
Yeah. If if that's what you were asking.
Yeah.
28:42
Okay, I guess I will stop here and everybody's free to move. But if there are more questions, please feel free to ask me.
[Applause]