Quantum computing

The gate just applies a unitary transform on the qubit(s). The collapse is caused by noise from the environment, not the gate.

Since all of you here to be newfags I would actually do you all a favor and post videos that actually cover the topic pretty well instead of memeing like a bunch of autistic shits.

>media.ccc.de/v/31c3_-_6261_-_en_-_saal_2_-_201412291245_-_let_s_build_a_quantum_computer_-_andreas_dewes

>media.ccc.de/v/31c3_-_6157_-_en_-_saal_6_-_201412301245_-_diamonds_are_a_quantum_computer_s_best_friend_-_nicolas_wohrl

>media.ccc.de/v/28c3-4648-en-quantum_of_science

>media.ccc.de/v/30C3_-_5536_-_en_-_saal_g_-_201312271830_-_long_distance_quantum_communication_-_c_b

I wish sometimes that you guys are good people and helpful instead of manchild fucks.

Doesn't wavefunction collapse happen whenever a measurement/interaction happens? It seems like you shouldn't be able to chain a sequence of quantum logic gates and perform complex quantum computations, because as soon as the quantum data passes through the first set of gates it will start behaving like a classical system.

pretty harsh, given that most other threads up now are worse

Thats because most of the people dont belong here and those that do only shitpost instead of talking like normal people.

The gates are not measurements. They are unitary transformations (which by its very definition means it is reversible). Measurement is typically delegated to the end of a computation because doing so projects the state into a subspace ie removes the portion of the superposition which did not satisfy the measurement outcome. The non classical nature of quantum computation hinges in the inability of a classical computer simulating it which in turn means that the superposition of qubits must be manipulated in such a way that cannot be tracked a classical computer efficiently. Removing the superposition removes that edge.

The exception to this is cluster state computation (or one way quantum computing). This relies on a state that is entangled in a certain way to be generated and performs condition measurements on the state to simulate computation. IE, the choice of measurement basis depends on previous measurement outcomes. A computer would not be capable of doing this condition measurement (which sequentially removes superposition) and would only be able to simulate the statistics of the whole state then sift out the outcomes that satisfy the condition outcome rules. This is exponentially slower than a cluster state computation.

Try looking into quantum walks

youtube.com/watch?v=OaD8F45tcUU
youtube.com/watch?v=tC9x3X4YDQU
youtube.com/watch?v=oA3bsl9oCBE

In theory I get what you mean about unitary transformations, but is it even possible to build something that works in this way? As far as I understand from introductory quantum mechanics, none of the experiments have demonstrated that something like this exists.

As for the second part of your post, it seems that interesting computations do require entanglements to happen (subsequent results depending outcomes of previous ones), but when this happens you have a loss of information, so quantum computers have very limited usefulness (i.e. when it's hyped up it almost seems like a nondeterminisic Turing machine but in reality it's a very, very limited probabilistic one).

Of course it exists. I work in optics so the qubits live in the polarization of the photon. Single qubit unitary operations are merely changing the polarization of the photons which requires birefringent materials like calcite or quartz. In practice, we are working towards integration in the form of optical circuits. There, we keep everything the same polarization and instead use two paths to logically represent the qubit. Then, all single qubit manipulation is based on mixing the two paths and changing their relative phase which can be done with beamsplitters and phase shifters. Currently, these operations can be done to 10^-4 error (your computer is something like 10^-15).

In optics, nonlinearity is hard to use so, conditional gates are not feasible. Most models therefore rely on cluster state computation. The bottleneck is then in making the cluster state which currently involves probabilistic post-selection (to sidestep the nonlinearity issue.)

I am not sure what you are saying in the second part. Measurement is not loss. Loss is when photons do not get detected - the opposite of measurement.

Quantum computation lives in BQP class which contains BPP (probabilistic turing machine) so anything a probabilistic turing machine can do is efficiently simulated on a quantum computer

>Measurement is typically delegated to the end of a computation because doing so projects the state into a subspace ie removes the portion of the superposition which did not satisfy the measurement outcome.
So we would need a classical computer to keep track of the results then.

Yeah, classical computer will be used. In cluster state model, the computer only needs to do parity operations ie it doesn't have to be strong enough to do universal computation

>I am not sure what you are saying in the second part. Measurement is not loss. Loss is when photons do not get detected - the opposite of measurement.
I'm talking about the superposition of states which is the only interesting thing about quantum computation. When a measurement or interaction happens, the wavefunction collapses and the system becomes like a classical system with one state and the information about all other states in the superposition becoming irrecoverable. That's what I mean by lost.

>In optics, nonlinearity is hard to use so, conditional gates are not feasible. Most models therefore rely on cluster state computation. The bottleneck is then in making the cluster state which currently involves probabilistic post-selection (to sidestep the nonlinearity issue.)
>Quantum computation lives in BQP class which contains BPP (probabilistic turing machine) so anything a probabilistic turing machine can do is efficiently simulated on a quantum computer
You are being silly. BPP is only contained in BQP because of the way quantum Turing machines/quantum circuits are defined. A quantum TM can be "collapsed" into a classical system and this is why you get that BPP is contained in BQP. It is not because a fully quantum computation can solve the problems solvable by classical probabilistic computation. The problem, by your own admission and the point I'm trying to raise, is that quantum computations cannot have conditional branches or irreversible transformations without "collapsing" into classical computations, and this is why I'm saying quantum computers have very limited applicability in terms of the problems they can solve efficiently that classical computers can't.

I'm not sure you're fully grasping what a quantum gate is, there's no measurement involved in a gate process.

idk how much QM you know but I'll start here. Basically a quantum state can be thought of as an n-dimensional unit vector, where n is the number of possible states it can be measured in. For a qubit n is two, i.e. it can be measured to be 0 or 1. Due to the quantum nature of things, it can exist in a superposition of these two states and when written in vector form that is:[eqn]|\psi\rangle=\begin{pmatrix}
\alpha\\
\beta
\end{pmatrix}=\alpha|0\rangle+\beta|1\rangle[/eqn]where [math]\alpha[/math] and [math]\beta[/math] are the amplitudes of each state in the superposition. All a unitary transform or a gate does is rotate this vector through some angle in some direction, this can be written as a matrix that acts on the vector. The single qubit quantum NOT gate is:[eqn]\hat{X}=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} = |0\rangle\langle1|+|1\rangle\langle0|[/eqn]Now it's pretty clear that when you apply this to a qubit in state [math]|0\rangle[/math] you'll end up with it in state [math]|1\rangle[/math] and vice versa but when you apply it to a general superposition state it will swap the values of [math]\alpha[/math] and [math]\beta[/math]. No information is lost after a transform like this, it's just "rearranged", as it were.

(1/2)

2/2

As for the physical implementation of this operation, you want to make your qubit evolve through time in such a way that the operation is applied after some period. The time evolution of a quantum system is defined by its hamiltonian so the task is to give your system the right hamiltonian for the desired evolution. For a single qubit, say a pair of energy levels in an atom (assume the other levels are far off and can be ignored), you can fire a laser at it and give it a hamiltonian of this form:[eqn]\hat{H}=\frac{\hbar}{2}\begin{pmatrix}
0 & \Omega\\
\Omega & 0
\end{pmatrix}[/eqn]where [math]\Omega[/math] is proportional to the intensity of the laser. According to the Schrodinger equation the system will then evolve like this:[eqn]|\psi(t)\rangle=e^{-i\hat{H}t}|\psi(0)\rangle[/eqn]If we evaluate the evolution operator we get:[eqn]e^{-i\hat{H}t}=\begin{pmatrix}
\cos(\frac{\hbar\Omega}{2}t) & -i\sin(\frac{\hbar\Omega}{2}t)\\
-i\sin(\frac{\hbar\Omega}{2}t) & \cos(\frac{\hbar\Omega}{2}t)
\end{pmatrix}[/eqn]If we run that for [math]t=\frac{\pi}{\hbar\Omega}[/math] then we get our NOTmatrix, only with a global phase of [math]-i[/math] (which doesn't matter). Obviously in real life there's noise and stuff to consider but the point is that we can do and have done these kinds of operations on real quantum systems.

Fixed TeX gore

***If we run that for [math]t=\frac{\pi}{\hbar\Omega}[/math] then we get our NOT matrix, only with a global phase of [math]-i[/math] (which doesn't matter). Obviously in real life there's noise and stuff to consider but the point is that we can do and have done these kinds of operations on real quantum systems.

Well I don't know what's happening with that but I guess you can still read it

shut up
noone aint got time to read that

>Veeky Forums - Science & Math

Measurement doesn't imply total wavefunction collapse. It is a conditional: what does state look like conditioned on particular observation. You don't even have to use the conditional. If I have a Bell state and one photon goes through a polarizer and I measure the other photon. After taking statistics on the other photon, it will appear to have a completely random polarization. It doesn't matter what you do to the first photon, the result would be the same. Why? Because you are not measuring the first photon.

If you only record data on the second photon when you measure both photons, then you will find strong polarization correlation (or anticorrelation depending on the Bell state).

If you project the bell state into the bell basis, the state that comes out still is entangled and in superposition. You can rotate one photon into a different polarization, changing the Bell state -not removing the entanglement or superposition.

The controlled phase operation (how you generate a cluster state) is a condition branch: if both photons are in vertical polarization then apply pi phase shift else apply identity. The act of cluster state computation is an example of how wavefunction collapse is not an absolute. The W state is another - a three photon state which is not maximally entangled. The list goes on... wavefunction collapse is not what you are implying.

Hierarchy collapse is no joke. You use it as a form of proof by contradiction since it is so unlikely. Whether we currently have algorithms showing BQP hardness in BPP is not relevant.

Op here:
First thanks a lot for all the sources and tips, I'm still trying to give an order to it but i will surely not run out of lecture :)

Now that I'm sure the topic ain't no meme (which was my original idea, but this site tend to corrupt your opinion (oh hi pol !)),
I'm wondering if switching my major (in engineering) from electronics to applied physics won't get me closer to the field or both ms are corrct choices ?
>or did i failed my life and should have done theorical science ??

I have some difficulties to get the concept, especially since the wikipedia article is quite unmethodical and I didn't know theclassic random walk.
Also, what are the related videos ?

Actually I had.
Speaking of which, i kind of didn't follow when Schrödinger's arrive in , how can [math]\hat{H}[\math] become a sin/cos expression ?
Also, why does Schrödinger's still the formula used when it doesn't explain the spin natively ?

How much would room temperature superconductors help quantum computers?

>how can [math]\hat{H}[/math] become a sin/cos expression ?
[eqn]e^{ix} = \cos{x} + i\sin{x}[/eqn]
Try plugging in [math]i[/math] into the Maclaurin series of [math]e^x[/math]

>Actually I had.
Appreciate it bro

>Speaking of which, i kind of didn't follow when Schrödinger's arrive in
Okay before I get into this let me just correct some errors I've noticed in my original posts. The time evolution of a state is in fact:[eqn]|\psi(t)\rangle=e^{\frac{-i\hat{H}t}{\hbar}}|\psi(0)\rangle[/eqn]the evolution operator is instead[eqn]e^{-i\hat{H}t}=\begin{pmatrix} \cos(\frac{\Omega}{2}t) & -i\sin(\frac{\Omega}{2}t)\\ -i\sin(\frac{\Omega}{2}t) & \cos(\frac{\Omega}{2}t) \end{pmatrix}[/eqn]and the time required for a NOT is in fact [math]\pi/\Omega[/math].

These errors came out of the fact that I'm used to setting [math]\hbar[/math] to 1 so some inconsistencies popped up.

Okay now to explain. You know how the Schrodinger Equation is:[eqn]i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle[/eqn]this solution satisfies the equation and gives the behaviour of a quantum system over time[eqn]|\psi(t)\rangle=e^{\frac{-i\hat{H}t}{\hbar}}|\psi(0)\rangle[/eqn]When raising [math]e[/math] to the power of an operator or matrix you need to use the series expansion like so[eqn]e^{\hat{A}} = \sum^{\infty}_{n=0}\frac{\hat{A^n}}{n!}[/eqn]for our case [math]\hat{A}=\frac{-i\hat{H}t}{\hbar}=\begin{pmatrix}
0 & \frac{\Omega}{t} t \\
\frac{\Omega}{t} t & 0
\end{pmatrix}[/math]if you work out the infinite sum with this matrix then you'll end up with the sin and cos thing.

Hopefully the math typesetting behaves this time. Does that make things clearer?

fuuuuuuuck's sake

Well for one thing they'd make superconducting qubits easier to deal with since they wouldn't need constant supercooling.

Oh, I forgot to address your last bit

>Also, why does Schrödinger's still the formula used when it doesn't explain the spin natively?
If you get the Dirac equation and do a non-relativistic approximation you get the schrodinger equation. It's like how non relativistic kinetic energy is 1/2 mv^2. It's not exactly right, but at low speeds the error is extremely small.

>1 mK starting
>1 mK
kek

Yeah and swamp any signal with blackbody radiation
>yfw coaxial cable at 40K is used as thermal microwave photon source

>click on first video
>"Hello boys and girls"

lmao
good on you man though, much smarter than me

I was using an intro from a Bloodhound Gang song there and I usually try to get my girls into programming and so I'm adressing those here