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From what I understand, a qubit exists in a superposition of states and once it has been measured, it must fall into one of the two possible states. Now, I have been told that once a qubit is measured, it is no longer proper to call it a qubit but a bit since it no longer exists in a superposition of states. Is this correct?

Along the same lines, if a photon with unknown polarization (the polarization state can be our qubit) hits a polarizing beam splitter, then its no longer exists in a superposition of states but must be either horizontally or vertically polarized. So would this mean that the polarization no longer is a qubit, but a bit, since it can only exist in one of two states? This would not make sense because many regimes for experimentally realizing quantum logic gates involve polarizing beam splitters. So if my reasoning is correct, that would mean that in the gate itself the qubit actually is no longer a qubit, but a bit.

One final thing, since measuring a qubit is inherent to a functional quantum computer, does this mean that quantum computers actually use bits as well as qubits?

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  • $\begingroup$ Every state is (obviously) a superposition of some other states (except in the most trivial case where the state space is the projectivization of ${\mathbb C}^1$). What do you imagine to be an example of a state that is not a superposition? $\endgroup$ – WillO Jul 19 '16 at 21:23
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One final thing, since measuring a qubit is inherent to a functional quantum computer, does this mean that quantum computers actually use bits as well as qubits?

The entire advantage of a quantum computer lies in the use of qubits. I've implemented a quantum computing algorithm(Grover's Search) on a classical computer before, and it was incredibly slow compared to classical alternatives.

It isn't accurate to think of a qubit as a bit while it is in a classical state. Even in a classical state, the physical implementation of a qubit is far more complex than that of a bit, and it can be impelled into a superposition at any time using a Hadamard gate. So no, quantum computers do not use bits. They may interact with a classical computer that uses bits, but that is a different story.

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    $\begingroup$ The entire advantage of a quantum computer lies in the use of qubits. This is not accurate. Since a computation system is described by a map $S\times G\rightarrow S$, where $S$ is the state space and $G$ is the operation space. Here for QC, both the state space (Hilbert space) and the operation space(Unitary operation) are 'bigger' than those in classical computers. So the power of QC comes from both the qubits and unitary operations. $\endgroup$ – XXDD Jul 21 '16 at 4:19
  • $\begingroup$ Unitary operations can be performed on a classical computer, and are used for some transformations. Though quantum computing would be impossible without them, unitary operations are not unique to the field. $\endgroup$ – B. Bush Jul 21 '16 at 20:54
  • $\begingroup$ The difference between QC and CC is that there are some operations in QC that can not be efficiently simulated by CC. From this point of view, a unitary operation on a Hilbert space (with an exponential dimension w.r.t. the quibt number) can not be efficiently simulated by CC. Of course, finally this is due to the exponential dimension of the Hilbert space. $\endgroup$ – XXDD Jul 23 '16 at 3:05
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There is a role for classical bits. Quantum computing requires quantum teleportation. The total set of Bell states in a particular basis are $$ |+,+\rangle~+~|-,-\rangle $$ $$ |+,+\rangle~-~|-,-\rangle $$ $$ |+,-\rangle~+~|-,+\rangle $$ $$ |+,-\rangle~-~|-,+\rangle $$ which are four qubits in a Hilbert space ${\cal H}^4$. Let me label these with $00$, $01$, $10$ and $11$, which is two bits of classical information. If I have a part of an EPR pair I can perform a measurement to determine which of these Bell states it is in. I convey to my partner which of these it is by a classical signal, which are bits. My partner can take her part of the EPR pair and run it through a Hadamard gate to rotate the state $\pi$ around the $x$ $y$ $z$ axes. My partner then reconstructs the state, but it requires this classical information.

Classical information is necessary in order to do quantum computing. Without local classical information it is not possible to "read" the output of a quantum computer or to teleport states within it.

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The qubit can be in a superposition of states thanks to the superposition principle, when you measure the spin of your photon it collapse in the vertical or horizontal state (eigenstates of the spin operator) and during the measure you extract the "bit" intended as the information about the outcome of this measure. But the photon (and then the qubit) is still there, in a collapsed state. If you perturb the photons it will evolve and it will no longer be in the collapsed state: it will be again in a complex state described by the superposition principle.

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  • $\begingroup$ Saying that a quantum "has not a defined state" is very misleading. A state $\left \lvert \uparrow \right \rangle + \left \lvert \downarrow \right \rangle$ is a perfectly well defined state. $\endgroup$ – DanielSank Jul 20 '16 at 2:29

protected by Qmechanic Jul 20 '16 at 7:41

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