4 ms·

In QC, a gate is a logical building block, akin to the logic gates in classical computing such as AND, OR, XOR, NOT. Multi-qubit gates like CZ (present in the

by vitus 2y ago

In QC, a gate is a logical building block, akin to the logic gates in classical computing such as AND, OR, XOR, NOT. Multi-qubit gates like CZ (present in the article) and CNOT are crucial for manipulating quantum states and achieving the wavefunction interference patterns that yield interesting results (quantum teleportation, Shor's algorithm, and basically any nontrivial quantum result you can think of). One of the biggest issues keeping quantum computation from taking over the world is maintaining a given quantum state for sufficiently long (coherence time), so you can do meaningful computations on it. Of course, that's meaningless if your actual operations are noisy, hence the emphasis on high fidelity as well. Classical computation would not be very useful if, say, one-fifth of the time when you multiply two numbers together you get the all-ones bit sequence. The same can be said about quantum computing. Of course, with classical computation, we are greedy and will raise a stink about a CPU that gives us errors in, say, one in 9 billion divisions. But QC isn't quite there yet. What I would be interested in (and didn't see from a preliminary skim) is how big of an incremental improvement this is from the previous record.

vitus 2y ago
To elaborate on why CZ / CNOT are so important: CNOT is controlled-NOT (CZ is controlled-Z, but NOT has a clearer classical analogue, and phases are harder to explain). Consider two inputs A and B, where A is a mixture of the zero-state (|0>; ground) and the one-state (|1>; excited) a|0> + b|1> (such that |a|^2 + |b|^2 = 1), and B is the pure zero-state |0>. In this case, A behaves somewhat like a selector: when it is |1>, B's state will be flipped from |0> to |1> (and vice versa), and when it is |0>, B's state will remain the same. But crucially, this applies for the entire superposition of states! So the portion of the wavefunction where A is in the zero-state will do nothing, but the portion where A is in the one-state, it'll flip B from its zero-state. In other words, CNOT takes the state from a|00> + b|10> to a|00> + b|11>, i.e. it entangles a qubit that was previously in the ground state with an arbitrary other qubit. Now if you measure A, then B (or the other way around), you'll always get the same result for both (provided that the quantum state is still coherent). (Normally, if you took two particles that had the same state but were not correlated, you'd have to consider all the possibilities of being in |00>, |01>, |10>, and |11>.)
- MobiusHorizons 2y ago
  This was very helpful, but I just can’t decipher this central paragraph > Consider two inputs A and B, where A is a mixture of the zero-state (|0>; ground) and the one-state (|1>; excited) a|0> + b|1> (such that |a|^2 + |b|^2 = 1), and B is the pure zero-state |0>. I’m guessing the notation is tripping me up, but I’m having trouble parsing this. A: mix of |0> and |1> B: apparently |0>, but elsewhere you refer to it being unconstrained. CNOT(A,B) = a|0> + b|1> (such that |a|^2 + |b|^2 = 1)??? The relationship between the parenthetical and the a + b bit is impenetrable, but I can’t fathom what absolute value or squaring would do in the context of single qubits.
  - itishappy 2y ago
    In the statement |a> + |b>, both a and b are probability amplitudes (complex numbers). You obtain the probability density by squaring the probability amplitude (or rather, multiplying it by it's complex conjugate aa*), which gives the same result as squaring the magnitude, so it can be written |a|^2. Since there's two possible states in our system, their probabilities have to add to unity, so |a|^2 + |b|^2 = 1. https://en.wikipedia.org/wiki/Probability_amplitude https://en.wikipedia.org/wiki/Probability_amplitude
  - vlmutolo 2y ago
    This is quantum mechanics notation and basic principles. |x> is pronounced “ket x” (the second half of the intentionally-misspelled word braket), and refers to a quantum state x. The quantum states |0> and |1> refer to the classical states we’re used to. But as written in the GP comment, a general qubit state is given by a|0> + b|1>, where a and b are complex numbers. This is a “superposition”: the qubit occupies both states simultaneously and will resolve to one of them when observed/measured. For any quantum state written in this way, we know from quantum mechanics that observing the state will yield a measurement of the state |0> with probability |a|^2 (the norm of the complex number a), and will yield state |1> with probability |b|^2. The qubit measurement must yield exactly one of a or b, so the probabilities must add up to 1, hence |a|^2 + |b|^2 = 1.
    - 2y ago
Yizahi 2y ago
I have a newbie question about quantum computing. Are there any quantum computers (regardless of the name or tech involved) in the wild now, where you can pick a problem with unknown answer, compute the result on the quantum computer, and later verify that such result was correct? Like, I don't know, calculate sum of 123+987, or something else which can be later reproduced on the "normal" computer. Sorry for the layman terms.
- vitus 2y ago
  > Sorry for the layman terms. Not at all! What you're describing is whether there are problems with classical solutions that have also been solved via quantum computers in practice. It depends how exactly you define a quantum computer, but for a long time D-Wave was making a number of splashy claims about how it had some large amount of entangled qubits in its "adiabatic quantum computer" and was using it to perform quantum annealing to solve various optimization problems. (Confusingly, despite the name, quantum annealing is a classical algorithm.) As part of the showcase, this system was used to solve a sudoku puzzle among other things. There is no clear evidence that D-Wave's devices qualify as quantum computers (and if so, certainly not general-purpose ones capable of implementing Shor's algorithm), and that they achieved meaningful quantum speedup. But, they certainly did sell a handful of these devices for millions of dollars. The general challenge with a "real" (more precisely: universal) quantum computer is that in order to get any useful computation out of them, you need a fair amount of entangled particles with high coherence times, and a fair amount of gates with high fidelity. The wording of your question suggests that you're aware of the various implementations of Shor's algorithm and how the circuitry was designed with the answer in mind. So instead let me point you at https://en.wikipedia.org/wiki/HHL_algorithm https://en.wikipedia.org/wiki/HHL_algorithm which solves a certain family of systems of linear equations, which we can verify using classical algorithms like Gaussian elimination or LU factorization.
black_puppydog 2y ago
nice dig at intel there :D
- vitus 2y ago
  In all seriousness, this was intended to illustrate the enormous gulf between even this result (one error in a thousand computations) and the infamous Pentium FDIV bug. (I would describe it as lightly poking fun at Intel... I'd hope there aren't still hard feelings over a 30-year-old hardware defect.)

by vitus 2y ago

vitus 2y ago

To elaborate on why CZ / CNOT are so important: CNOT is controlled-NOT (CZ is controlled-Z, but NOT has a clearer classical analogue, and phases are harder to explain). Consider two inputs A and B, where A is a mixture of the zero-state (|0>; ground) and the one-state (|1>; excited) a|0> + b|1> (such that |a|^2 + |b|^2 = 1), and B is the pure zero-state |0>. In this case, A behaves somewhat like a selector: when it is |1>, B's state will be flipped from |0> to |1> (and vice versa), and when it is |0>, B's state will remain the same. But crucially, this applies for the entire superposition of states! So the portion of the wavefunction where A is in the zero-state will do nothing, but the portion where A is in the one-state, it'll flip B from its zero-state. In other words, CNOT takes the state from a|00> + b|10> to a|00> + b|11>, i.e. it entangles a qubit that was previously in the ground state with an arbitrary other qubit. Now if you measure A, then B (or the other way around), you'll always get the same result for both (provided that the quantum state is still coherent). (Normally, if you took two particles that had the same state but were not correlated, you'd have to consider all the possibilities of being in |00>, |01>, |10>, and |11>.)

MobiusHorizons 2y ago
This was very helpful, but I just can’t decipher this central paragraph > Consider two inputs A and B, where A is a mixture of the zero-state (|0>; ground) and the one-state (|1>; excited) a|0> + b|1> (such that |a|^2 + |b|^2 = 1), and B is the pure zero-state |0>. I’m guessing the notation is tripping me up, but I’m having trouble parsing this. A: mix of |0> and |1> B: apparently |0>, but elsewhere you refer to it being unconstrained. CNOT(A,B) = a|0> + b|1> (such that |a|^2 + |b|^2 = 1)??? The relationship between the parenthetical and the a + b bit is impenetrable, but I can’t fathom what absolute value or squaring would do in the context of single qubits.
- itishappy 2y ago
  In the statement |a> + |b>, both a and b are probability amplitudes (complex numbers). You obtain the probability density by squaring the probability amplitude (or rather, multiplying it by it's complex conjugate aa*), which gives the same result as squaring the magnitude, so it can be written |a|^2. Since there's two possible states in our system, their probabilities have to add to unity, so |a|^2 + |b|^2 = 1. https://en.wikipedia.org/wiki/Probability_amplitude https://en.wikipedia.org/wiki/Probability_amplitude
- vlmutolo 2y ago
  This is quantum mechanics notation and basic principles. |x> is pronounced “ket x” (the second half of the intentionally-misspelled word braket), and refers to a quantum state x. The quantum states |0> and |1> refer to the classical states we’re used to. But as written in the GP comment, a general qubit state is given by a|0> + b|1>, where a and b are complex numbers. This is a “superposition”: the qubit occupies both states simultaneously and will resolve to one of them when observed/measured. For any quantum state written in this way, we know from quantum mechanics that observing the state will yield a measurement of the state |0> with probability |a|^2 (the norm of the complex number a), and will yield state |1> with probability |b|^2. The qubit measurement must yield exactly one of a or b, so the probabilities must add up to 1, hence |a|^2 + |b|^2 = 1.
  - 2y ago

Yizahi 2y ago

I have a newbie question about quantum computing. Are there any quantum computers (regardless of the name or tech involved) in the wild now, where you can pick a problem with unknown answer, compute the result on the quantum computer, and later verify that such result was correct? Like, I don't know, calculate sum of 123+987, or something else which can be later reproduced on the "normal" computer. Sorry for the layman terms.

vitus 2y ago
> Sorry for the layman terms. Not at all! What you're describing is whether there are problems with classical solutions that have also been solved via quantum computers in practice. It depends how exactly you define a quantum computer, but for a long time D-Wave was making a number of splashy claims about how it had some large amount of entangled qubits in its "adiabatic quantum computer" and was using it to perform quantum annealing to solve various optimization problems. (Confusingly, despite the name, quantum annealing is a classical algorithm.) As part of the showcase, this system was used to solve a sudoku puzzle among other things. There is no clear evidence that D-Wave's devices qualify as quantum computers (and if so, certainly not general-purpose ones capable of implementing Shor's algorithm), and that they achieved meaningful quantum speedup. But, they certainly did sell a handful of these devices for millions of dollars. The general challenge with a "real" (more precisely: universal) quantum computer is that in order to get any useful computation out of them, you need a fair amount of entangled particles with high coherence times, and a fair amount of gates with high fidelity. The wording of your question suggests that you're aware of the various implementations of Shor's algorithm and how the circuitry was designed with the answer in mind. So instead let me point you at https://en.wikipedia.org/wiki/HHL_algorithm https://en.wikipedia.org/wiki/HHL_algorithm which solves a certain family of systems of linear equations, which we can verify using classical algorithms like Gaussian elimination or LU factorization.

black_puppydog 2y ago

nice dig at intel there :D

vitus 2y ago
In all seriousness, this was intended to illustrate the enormous gulf between even this result (one error in a thousand computations) and the infamous Pentium FDIV bug. (I would describe it as lightly poking fun at Intel... I'd hope there aren't still hard feelings over a 30-year-old hardware defect.)