Dear research colleagues of RG, please watch a YouTube video titled, “A beginner’s guide to quantum computing (TED – Ideas worth spreading)” in the link given below:
https://youtu.be/QuR969uMICM
It’s a 10 minutes video. You may skip the last 4 minutes.
I got surprised after watching this video. I got a serious doubt, which I posted in the comment section of that video. It’s already three years old video and YouTube may not be a good platform for the clarification of scientific doubts; hence I deleted my comment. The same I am posting here in the below. Please go through it and provide your answers to my doubt, valuable comments etc.
Pl. Note: My comment is only with respect to the scientific content of the video, but in no way a personal criticism of the person presented that video, whom I don’t know at all.
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Below is my comment:
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The Q-computer prepares the initial state as |Head> and plays first by choosing either to flip the coin or not, but the outcome is not revealed to the opponent – not even to itself! The opponent plays in the second place akin to the Q-computer. Finally, the Q-computer plays, which is also a measurement revealing the outcome … it’s |Head> and hence, the Q-computer wins, not once, but in every round of the game.
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If the opponent plays first choosing the |Tail>, then obviously, the Q-computer loses every round of the game … which is not all mentioned in this video presentation!!!
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|Head> is like the |y; spin-up> state, prepared initially by filtering it out from the Stern-Gerlach (SG) apparatus with magnetic field direction along Y-axis, by blocking the |y; spin-down> component.
The |y; spin-up> state then passes through two sequential SG apparatuses [1], but without undergoing any measurement. The magnetic field directions in these two SG apparatuses can be arbitrary.
The output from the 2nd SG apparatus is subjected to a measurement by the Q-computer using the final SG apparatus with magnetic field direction along Y-axis, akin to the initial SG apparatus.
It’s very clear from the quantum formalism that the two sequential SG apparatuses in between the initial and final SG apparatuses, simply play the roles of identity operators [1], i.e., every time the |y; spin-up> state passes-out of any one of the two sequential SG apparatuses, it’s in the same |y; spin-up> state and hence, the final measurement obviously results in the |y; spin-up> state. Actually, the presence or absence of the in-between two SG apparatuses doesn't matter, because, they don’t perform any measurement.
Therefore, the play of Q-computer and the play of opponent are just dubious. In fact, in the case discussed in this video, the Q-computer never performed any quantum computation for winning every round of the game. In other words, it's a game played by the Q-computer with itself = A self-goal (OR) the opponent must be an ignorant to loose every round of the game, because, when the Q-computer shows the initial state, then that's the state to be bet for a win.
Reference
[1] J.J. Sakurai, Modern Quantum Mechanics, p 33 (Addison Wesley, 1994).
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Anticipating your answers, valuable comments etc.
Thank you and best regards,
N.G
The short answer to the question is, No, a quantum computer doesn’t work as claimed in the comment.
The fundamental difference between a classical computer and a quantum computer is that a classical computer describes information in terms of bits, that can take only two values, 0 or 1; a quantum computer describes information in terms of’qubits, states that are superpositions of 0 and 1 states.
What this means more explicitly: A bit can be understood as one of two antipodal points on tte surface of a sphere of unit radius ; a qubit as any point of the sphere of unit radius.
A classical computer acts on bits using gates; a quantum computer acts on qubits using transformations that map qubits to qubits. So a one qubit gate sends a point on the sphere to anither point on the sphere. A 2-qubit state is a linear combination of the states |00>,|01>,|10> and |11>, therefore it defines a space of 4 complex numbers, i.e. 8 real numbers, the sum of the squares of the modulus of which equals 1. This is the sphere in 7 dimensions (dividing out by the global phase). So the gates are the transformations, that send a point on this sphere to another point on the same sphere. Classical computing is about the classical gates; quantum computing is about these generalizations.
Furthermore, while the calculations performed using classical computing are deterministic, the calculations performed by a quantum computer are probabilistic-the result is the probability of finding the system, described by the transformations, that map states to states, Thrre’s nothing dubious about the rules, but they do require studying quantum mechanics first.
The presentation glosses over some important issues, namely what is the transformation, that realizes the flip. That transformation is known to everybody, so the probabilities can be computed in advance, for any time.
An important property of quantum systems is, however, that the eigenstates of the transformation remain as such by acting with the transformation: An eigenstate can’t become a superposition. So, if the initial state is, for sure, heads or tails, it will remain so under the transformation. That’s why the computer wins; but that’s because the initial state was chosen as a particular eigenstate.
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