How entanglement is useful in a quantum computer? Wouldn't a 32 qubit quantum computer with two entangled qubits be only capable of producing an answer with precision and range of 31 bits since two of the qubits are locked together? - by Viktor T. Toth - Quora Question Review

This document contains a review of the answer by Steve Baker on the question in Quora: "How entanglement is useful in a quantum computer? Wouldn't a 32 qubit quantum computer with two entangled qubits be only capable of producing an answer with precision and range of 31 bits since two of the qubits are locked together? "
To order to read all the answers select: https://www.quora.com/How-entanglement-is-useful-in-a-quantum-computer-Wouldnt-a-32-qubit-quantum-computer-with-two-entangled-qubits-be-only-capable-of-producing-an-answ

• The text in italics is copied from the article.
  
• Immediate followed by some comments
  

Contents

• 1. Answer Review

Reflection

• Reflection 1 - Question Review

1. Answer Review

Here is the main thing to remember about quantum computers: Fundamentally, they are analog computers.

Digital computers are basically also analog computers. There main difference is they use a clock to synchronize all the calculations. The result is that you can repeat the same program and always get the same result.
For an analog computer and specific a quantum computer that is not guaranteed.

What? I hear you asking, what is this stupid idiot blabbering about? Qubits are bits, 0 or 1, nothing analog about them!

True. But these values of 0 and 1 matter only when you set up the quantum computer or when you read the result. In between…

In between, the qubits are indeed “entangled”. Nothing mysterious about this, it simply means that while the state of each individual qubit may be up to chance (if you were to measure it; by doing so, of course you’d ruin the ongoing calculation) the collective state of the system precisely obeys conservation laws.
That is probably true, but that is not important.
The problem is that if you start two analog computers, which have the same configurations and parameter settings, simultations, there is no garantee that after a certain time both are in the same state.
The same quantum computers.

And what it actually means is that when the qubits interact with one another, it’s not 0’s and 1’s that interact with each other. It’s the phase of the qubits’ wavefunctions, a continuous, i.e., analog quantity, that gives a quantum computer its anticipated capabilities.

But how accurate can they do that?

So then, you might wonder, if that is that case why not just use an analog computer in the first place? Why go through all the trouble with quantum computers, all that costly hardware, the cooling, the whole nine yards?

Well, analog computers are notoriously inaccurate. In principle they can be infinitely accurate but in practice… Try reading micron-length differences in distance using a yardstick. Try building a voltmeter that is accurate, say, to 10, 15, 30 digits. Can you? I don’t think so. In practice, most analog computers are very inaccurate, so not suitable for calculations such as, say, factorizing 100-digit (or larger) numbers.

You can may be do that for a number like 77 = 7 * 11, but this can also be a problem for very large numbers.

Quantum computers are no different. I used the words above, “precisely obeys conservation laws”.

Such a statement means nothing. The physical problems of quantum computers are much more complex.

But that is true only if the quantum system is completely isolated from its environment. If it isn’t, it can exchange small, random amounts of energy, momentum, angular momentum, maybe other conserved quantities with its environment. So it no longer obeys conservation laws precisely. Once again, we have an inaccurate analog computer.

But that fact, is a critical issue.

But in the case of quantum computers there’s something else: The famous threshold theorem. A mathematical statement that tells us that if we can keep this “environmental decoherence” below a certain threshold, there exist error correction algorithms that, at the expense of adding (perhaps lots of) extra qubits, allow the imperfect quantum computer to emulate a perfect quantum computer.

All these extra qubits are also are a source of errors.

If we can build this, we have the key to a scalable quantum computer architecture, no limit in principle as to how complex the system can be, how many useful qubits it has, and how long it can maintain coherence.

When coherence is not maintained you have a serious problem.

This threshold has not yet been reached. For all the news about successes, quantum computers with more qubits, even “quantum supremacy”, until and unless this threshold is achieved, there’s no scalable general purpose quantum computing. (I personally suspect that the threshold theorem will one day be shown to be a “no go” result, but chances are that I am wrong; in fact, billions of dollars have been invested into quantum computing betting on the opposite.)

So yes, your 32-qubit quantum computer will only produce a 32-bit result. What matters, however, is what happens in between you inputing your 32 bits and reading out that result. During that calculation, it’s not the bits but the phases of the respective qubits’ wavefunctions that interact and do the computing bit.

3.

The question: "How entanglement is useful in a quantum computer? Wouldn't a 32 qubit quantum computer with two entangled qubits be only capable of producing an answer with precision and range of 31 bits since two of the qubits are locked together? " is a very technical question. The question consists of 2 parts. The problem is, both parts can be wrong.

For an article to read, related to part 2, select this link: https://en.wikipedia.org/wiki/Trapped-ion_quantum_computer
The problem is that I think there exists no (simple) relation with the accuracy of a classical computer consisting of a word length of 32 bits versus the accuracy of two entangled qubits.

An article (paragraph) to read, related to the first part, select this link: https://en.wikipedia.org/wiki/Qubit#Quantum_entanglement
IN the first part, the Bell state is discussed, which consists of 2 entangled Qubits A and B. Both have the same probility of measuring a 0 or a 1 (Slightly different wording). The result is that when Alice measures here qubit A as a "0", she knows that Bob will measure his Qubit B as a "1". and vice versa. See also https://en.wikipedia.org/wiki/Qubit#Quantum_entanglement

The real question to answer is how accurate can you bring two qubits, each in a state, that when they are both measured, always one qubit is "0" and the other qubit is a "1". That means never both as a"0" or never both as a "1".

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