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It comes down to that little-endian and big-endian are just different ways of mapping non-negative integers to strings of bits; as applied to quantum computing, that means different ways of mapping computational basis states of the form |πβ© (that is, labeled by integers) to computational basis states of the form |πβπββ¦πββββ© (that is, labeled by bit strings). The LittleEndian UDT marks that a register of qubits should be interpreted where πβ is the least-significant (little end) bit in the expansion of π. From that perspective, a big-endian paper can be converted to a little-endian one by reversing the convention used to order qubits.
Sarah Kasier
@crazy4pi314
Mridul Sarkar
@mridulsar
Ok that makes sense, thank you Chris. Sarah, that was an extremely helpful picture
Sarah Kasier
@crazy4pi314
π
Sagar Mishra
@achieveordie

Hi there! I've got another difficulty in understanding a part of the topic of Circuit-Centric-Classifiers

Here's the excerpt from the paper that I am having difficulty understanding

I don't even know where to begin reasoning the paragraph, can someone help me in justifying how this is the case?

@achieveordie Happy to try and help! The basic idea is that Hilbert space (in this context, the vector space of all quantum states) is unbelievably large, but that doesn't really reflect the reality that almost all states correspond to state preparation programs that are exponentially long to run.
So if you look only at those states reachable by programs parameterized by a small set of angles, that's a very small subset of all possible states.
the Roman Mercury
@theRomanMercury

Hello,

I am measuring in bell basis . I apply H and CNOT gate.
I measured several times and this outputs I got:

Measured CNOT Β· H |00? and observed (Zero, Zero)
Measured CNOT Β· H |00? and observed (Zero, Zero)
Measured CNOT Β· H |00? and observed (Zero, Zero)
Measured CNOT Β· H |00? and observed (One, One)
Measured CNOT Β· H |00? and observed (Zero, Zero)
Measured CNOT Β· H |00? and observed (One, One)
Measured CNOT Β· H |00? and observed (One, One)
Measured CNOT Β· H |00? and observed (Zero, Zero)

I did not get (One , Zero ) output . If input (One,One), I think output should be (One,Zero) and if input ( Zero, One ) output should be ( Zero,One) .

Why I do not get this outputs ?

Sarah Kasier
@crazy4pi314
There are 4 different bell states you can prepare, some are matching parity (One, One)/(Zero, Zero) and some are opposite as you described. It just depends on which of the bell states you prepare
the Roman Mercury
@theRomanMercury
Thnx, I will try it
Sagar Mishra
@achieveordie

@cgranade Please correct me wherever I am wrong. 1) Can't we have those cases where we don't really need to map our data into a higher dimension, i.e. we can prepare our state in a euclidean space without needing to apply any tensor product to prepare them? (This assumes that the data is simple enough that we don't really require any further mapping)

2) If the above statement is true, how is having our dataset in a limited space a bad thing? Why are we treating State Preparation step as if it is at this point that we are training our model? Isn't this analogous to zero initialization of weights in classical Machine Learning? We start at the same limited value (zero) and then learn the appropriate weights and biases.

3) If I am completely off-track here, could you please tell me some resources to get started in this part of the subject? I think that this is certainly getting too much for me.

Sarah Kasier
@crazy4pi314
Can you add some more context to your question? It's not super clear what you are trying to do π

i.e. we can prepare our state in a euclidean space without needing to apply any tensor product to prepare them

The state of a quantum system is always a vector in Hilbert space rather than Euclidian space, but that's largely a mathematical artifact of how we describe, model, and simulate quantum systems. In particular, just like with the kernel trick in traditional ML, we never write down those vectors in Hilbert space but prepare them implicitly.

Sarah Kasier
@crazy4pi314
Other general question for folks here, do you use discord? Would a Q# community discord make it easier to interact/chat than this single thread?
Sagar Mishra
@achieveordie
@cgranade thank-you, is it alright for you if I contact you via your e-mail that you've provided here? @crazy4pi314 I used discord back when it wasn't very popular in '16. I don't really use it anymore though. I look forward to it if you decide on making it though
Harshit Garg
@githg22_gitlab
Please use slack. It makes everything simpler.
:-)
Sarah Kasier
@crazy4pi314
@achieveordie The example of what I have seen for making discord really work for tech communities is the Python one: https://pythondiscord.com/
@githg22_gitlab I regularly use both slack and discord but it seems to me that discord is really keeping up with demand and adding features that basically make it a strictly nicer slack. Slack is also not bad though, I initially set this up because I didn't want people to have to create a new account to chat (assuming they had a GitHub account). I know this single thread is kinda limiting, so I am open to what would work best for the folks here
Harshit Garg
@githg22_gitlab
The best thing about slack is you can create a thread for any message.
It really comes useful in this kind of huge projects
Just my 2 cents
I'll check out the python discord channel though
Mridul Sarkar
@mridulsar
is there a way to 'transpose' a gate in q#? i noticed there was a discussion above as to how gates in Q# are not recognized as a matrix bur rather an operation. Would love some insight as to what sort of approach is taken when observing a circuit
specifically in Q# because in the posts above mentioned how the approach taken in Q# is different from qiskit
2 replies
Sarah Kasier
@crazy4pi314
Apparently you can do threading here too, but its not super obvious
Sarah Kasier
@crazy4pi314
How was everyone's weekend? I can't believe it's already Tuesday π­
the Roman Mercury
@theRomanMercury
I am reading Microsoft Quantum Documentation again
I think I missed the broadcast today
Amir Ebrahimi
@amirebrahimi

Other general question for folks here, do you use discord? Would a Q# community discord make it easier to interact/chat than this single thread?

Coming to this late, but I would prefer Slack or Discord (in that order) if you are considering alternatives.

ShaunJW360
@ShaunJW360
Slack or Discord good either way π
Sarah Kasier
@crazy4pi314

Sweet, I'll get something moving soon for the chat platform stuff!

In the meantime verson 0.12 is out of the QDK, I wrote up some thought/context on the release notes here: https://twitter.com/crazy4pi314/status/1280174137973981201?s=20

Amir Ebrahimi
@amirebrahimi
Congrats on the release!
Sarah Kasier
@crazy4pi314
@/all Q# Community update!
We now have a slack setup, check out the badge on the homepage to join! https://qsharp.community/
We will continue to make use of the gitter for more transient questions/ convos but after polling at least the active folks here, there was interest in a Slack so I got that up and going!
Sarah Kasier
@crazy4pi314
the Roman Mercury
@theRomanMercury
Hello, what do you think about Randonautica ( Δ±t is an app, uses QNRG )
?
Sarah Kasier
@crazy4pi314
If there are any folks dropping by from the Azure Quantum Developer Workshop, we tend to answer quick questions here on the gitter, and if you want to work on some of our projects/learn/collaborate hop on over to our slack here: bit.ly/qsharp-slack !
the Roman Mercury
@theRomanMercury
using (register = Qubit[8])
does it give 8 qubits in 0 state or 8 qubits 0 and 1 randomly ?
@jesusdeepkai
Hello. I think you must to initialize the Qbits with Hadamard function H(register[idxQubit]);
to stablish the superposition 50/50
Rolf Huisman
@RolfHuisman

using (register = Qubit[8])

You get 8 qubits in |0> state. So its the |00000000> state.

the Roman Mercury
@theRomanMercury
I see, now I understand the it, thnx
[true, true, false, false, true, false, false, true]
this is the bitstring which we wanted right ?
Rolf Huisman
@RolfHuisman
In the example that is used indeed to only X gate the specific qubits for which the bool was true. Afterward the register has the |11001001βͺ state.
the Roman Mercury
@theRomanMercury
I had not been able to look into quantum programming for a long time because of professional exams. Now I have time for https://github.com/microsoft/MLADS2018-QuantumML and QuantumComputingas a High School Module