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This is probably just a misunderstanding on my part, but everything I've seen on what quantum computers do thus far seems to suggest that the actual process of reading the entangled qubits would be equivalent to reading the value of a plate opposing a subdivided plate in a plate capacitor while the setting of initial qubits would be the equivalent of assigning a voltage to each subdivided plate. E.g. in this image:

enter image description here

You would be able to read the voltage on the red plate after setting independent voltages from a known range representing 0 at the low and 1 at the high on the 4 separate subdivisions of the opposing plate, then rounding off at some particular voltage to get a zero or one out of it for those 4 bits.

Is this wrong? If so, how does it differ from actual quantum computing?

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    $\begingroup$ Can you tell us more about where you have read about quantum computation? I don't really recognise the analogy you are giving. $\endgroup$
    – Niel de Beaudrap
    Jun 4, 2018 at 22:07

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Your capacitors cannot be in the state $\frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right)$, but qubits can.

Let's say $|0\rangle$ is 0$\,$V and $|1\rangle$ is 1$\,$V.
If you have 2 bits we can have $|00\rangle$,$|01\rangle$,$|10\rangle$,$|11\rangle$.

But the state: $\frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right)$, is in a superposition of two of these cases. The bit values can be (0,0) or (1,1). Either case is equally possible, until a measurement is made (think Schrödinger's cat: you don't know if it's alive or dead until you open the box).

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  • $\begingroup$ Wouldn't this just be using 1V out of a 0-2V range? What significance does this have in terms of the result of the calculation? $\endgroup$
    – CoryG
    Jun 4, 2018 at 13:25
  • $\begingroup$ Right, but in practice (not the theory) the superposition is irrelevant because you only get the one reading in the end. This brings me back to asking how does the result differ? $\endgroup$
    – CoryG
    Jun 4, 2018 at 14:53
  • $\begingroup$ The answer is on Pg 30 of the book in my answer, under the heading "1.4.2 Quantum parallelism". Yes you get the same reading in the end, but before doing the reading you can do computations on two states at the same time. So if you want to know f(0,0) + f(1,1) where f(x,y) is some function, you do not have to evaluate the function twice, for two different inputs. You evaluate it once, with one input. $\endgroup$
    – user1271772
    Jun 4, 2018 at 15:06
  • $\begingroup$ You can employ the same processing mechanisms in capacitors (e.g. a linear NOT gate is just another plate sandwiched between the input and output.) You can't get the superposition out of it regardless. I have voted your answer up because I appreciate the link, but still don't feel this answers the question. $\endgroup$
    – CoryG
    Jun 4, 2018 at 16:39
  • $\begingroup$ @CoryG: Did you read the "Deutsch's algorithm" section? Did you notice that with qubits you can determine whether or not the function is constant with only one function evaluation, whereas with classical bits you need to evaluate the function more than once? $\endgroup$
    – user1271772
    Jun 4, 2018 at 16:44
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It looks like you are asking about the possibility of encoding the mathematics of quantum states & measurement into some kind of analog device. And yes, I think this is possible. This reminds me of how people study analog models of gravity. The one problem I can see with this is that it will not scale very well as you increase the number of entangled quantum systems. Eg. for every extra qubit added to a system you double the number of dimensions. So for 10 qubits you would need a capacitor plate with 1024 subdivisions, and so on.

In summary, what you are proposing is to simulate a quantum system with an analog computer. We already do this with digital computers, but it just doesn't scale.

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