The Classical Channel

Anonymous

6:59 AM

@DaftWullie Hi!

Anonymous

So, I'm writing down my question here:

Anonymous

@DaftWullie I'm not really sure what difference measurement would have in case we are measuring a qubit in the state $e^{i\phi}|0\rangle$ and the state $|0\rangle$? Would it appear as $|0\rangle$ in both cases?

DaftWullie

@Blue yes, exactly, because the probability is the mod-square, so you never see the effect of the phase.

@Blue Or, put another way, the projector onto $|0\rangle$ is $|0\rangle\langle 0|$ while onto $e^{i\phi}|0\rangle$, it's $e^{i\phi}|0\rangle\langle 0|e^{-i\phi}=|0\rangle\langle 0|$

Anonymous

@DaftWullie Interesting. So, I can't think of any experiment to distinguish the "phase". But one thing which is still a bit of unsatisfactory to me: Why is it just that our failure to create an experiment which can distinguish the phase, dictate on which qubit that phase actually should be in? Why should physics depend on our restrictions at all?

DaftWullie

physics doesn't. quantum theory is our description of physics (as best we know it), and it doesn't contain the possibility to distinguish it. Perhaps there's a different theory that does permit it.

Anonymous

7:11 AM

@DaftWullie That does makes sense. Thanks! :) (Apparently, this one is so subtle but yet so deceivingly simple looking that I never even gave a thought to it, before :P)

Anonymous

@DaftWullie Oh, and just in case if you don't have MathJax installed for chat, you can get it here: math.ucla.edu/~robjohn/math/mathjax.html

Nelimee

@Blue quora.com/What-is-phase-kickback-and-how-does-it-occur

It helped me understand the phase kickback

(about this quantumcomputing.stackexchange.com/questions/2565/…)

Anonymous

@Nelimee Thanks, reading it

Anonymous

@Nelimee I read it. But it just mathematically shows "kick-back" without stating why it physically should occur at all

Nelimee

7:32 AM

Yes, there is no physical explanation on this link I agree

But from my experience with quantum computing so far, there are plenty of concept that are hard to understand physically but waaay easier to understand once you consider the mathematical part

Anonymous

Yeah, but the mathematics isn't satisfactory unless there's an explanation as to why the real world should obey it :P

Anonymous

Also the fact that humans can't device an experiment to distinguish the outcomes between the two possible mathematical manipulations, isn't a reason for why physically the real world should prefer one over the other.

Anonymous

10:10 AM

111

Quantum Computing – quantumcomputing.stackexchange.com

Beta Q&A site for engineers, scientists, programmers, and computing professionals interested in quantum computing.

Currently in public beta.

Anonymous

Anonymous

Wtf is up with the "visits" counter?

Anonymous

4.8 questions/day is better than before though. That's one improvement

Anonymous

Mithrandir, aagaitarino, heather and andrewO --- you guys need to cross the 2000 rep threshold soon!

Anonymous

12:23 PM

@Nelimee Your QISKit model of Cao's paper was giving the correct output for the given matrix, right?

Anonymous

How many digits of precision did it give?

Nelimee

Yup, I tested it with 2 different right-hand side and the precision was quite good

I tested it with [1,1,1,1] and [1,1,-1,-1] if I recall correctly

Anonymous

Cool :)

Anonymous

I'm starting with the QISKit part now

Anonymous

The paper is making some sense now

Nelimee

12:25 PM

You can tune github.com/nelimeee/quantum-tools/blob/master/HHL/4x4_system.py (line 114) for different values of b

Anonymous

Great, I'll check! :D

Nelimee

And tune the values line 201 & 202 for the exact solution versus the estimated one

For their example I have:
Exact solution: [-1 7 11 13]
Experimental solution: [-0.84245754+0.j 6.96035067+0.j 10.99804383+0.j 13.03406367+0.j]
Error in found solution: 0.16599956439346486

Anonymous

Looks pretty good. Thanks, I'll let you know if I get stuck anywhere :)

Nelimee

Ok! :)

Anonymous

12:46 PM

Just one theoretical point that is still not very clear to me:

Anonymous

$$U=\sum_{k\in\{0,1\}^t}\sum_{s\in\{0,1\}^l}\sum_{p\in\{0,1\}^m}|k\rangle\langle k|_C\otimes|s\rangle\langle s|_L\otimes|p\rangle\langle p|_Me^{-i2\pi p k \frac{f(s)}{2^{t+m+l}}}$$

Anonymous

This is the controlled-controlled-$e^{iH_0t}$ gate ^

Anonymous

And our state just before that gate is applied is $\sum_s\sum_p |p\rangle e^{i p/2^m |s\rangle}\otimes (\sum_j \beta_j |\tilde\lambda_j\rangle |u_j\rangle)$

Anonymous

After the action of the controlled-controlled-$e^{iHt}$ tensor product(ed) with the identity gate (since the operation on register B is the identity operation):

Anonymous

We're supposed to get to $$|0\rangle \otimes \sum_{j=1}^{n}\sum_{p=0}^{2^m-1}\sum_{s=0}^{2^t-1}\beta_j \exp[i \frac{p}{2^{m+l}} t_0 (2^l-\lambda_js)]|s\rangle |p\rangle |\lambda_j\rangle |u_j\rangle$$

Anonymous

12:51 PM

I'm just not sure how the $(2^l-\lambda_j s)$ term arises!

Anonymous

That looks weird!

Anonymous

@Nelimee Any idea? ^

Nelimee

Let me read and understand the first part before answering :p

Anonymous

2

A: How are two different registers being used as "control"?

You can see from the circuit diagram that in the third-last slice, both registers $L$ and $C$ are being used as controls. There's no problem with two registers being controls, after all, that's exactly what a Toffoli (controlled-controlled-NOT) gate does. It probably helps to explicitly write dow...

Nelimee

(I'll tell you, even if I don't the answer)

Anonymous

12:52 PM

I got the form of the controlled-controlled-gate from DaftWullie's answer

Anonymous

@Nelimee Okaies

Anonymous

If we can understand that part I think we'll be able to understand the whole paper. That's the only shady part :P

Anonymous

Aah, wait, that part $(2^l-\lambda_j s)$ is inside the exponent

Nelimee

Why do they need this $e^{iH_0t}$ in the first place?

Anonymous

@Nelimee Eh, that's the Hamiltonian simulation in HHL :P

Nelimee

12:57 PM

Nope, the Hamiltonian simulation I was talking about is the $e^{iAt}$. Here $H_0$ is fixed and does not depend on the input matrix (it only depends on the input size)

I mean, this $e^{iH_0t}$ is not mentioned in the original HHL paper, so it's something added by Cao and all

Anonymous

@Nelimee The $e^{-iAt}$ gate in Figure 1 is simply part of the Quantum phase estimation algorithm

Nelimee

Re-writing the expression like:
$$
|0\rangle \otimes \sum_{j=1}^{n}\sum_{p=0}^{2^m-1}\sum_{s=0}^{2^t-1}\beta_j e^{i \frac{p}{2^{m}} t_0} e^{-i\frac{p}{2^{m+l}}t_0\lambda_js}|s\rangle |p\rangle |\lambda_j\rangle |u_j\rangle
$$
may help you

@Blue @Blue I know, the $e^{iAt}$ is not a mystery :) The problem is the $e^{iH_0t}$, for the moment I don't understand it

Anonymous

Okay, I see the problem. This controlled-controlled-Hamiltonian simulation wasn't given in the original HHL

Anonymous

They modified the algorithm

Anonymous

For example they bypassed the uncomputing part

Anonymous

1:04 PM

The last step is rotation of ancilla

Anonymous

Followed by measurement

Nelimee

They added the $R_{zz}$ and $e^{iH_0t}$ parts, probably for the eigenvalue inversion. But I don't understand why, and I'm not sure it's for $\lambda^{-1}$

$U^\dagger$ is the uncomputing part

Anonymous

@Nelimee Ah, right

Nelimee

Anonymous

@Nelimee See last para on Page 2

Anonymous

1:06 PM

Anonymous

Indeed it was for eigenvalue inversion

Anonymous

The values get concentrated on $2^l/\lambda_j$

Anonymous

@Nelimee Interesting, but still can't spot the $\exp[i \frac{p}{2^{m+l}} t_0 (2^l-\lambda_js)]$ from there

Anonymous

Can you?

Transcript for

The Classical Channel