Think of a qbit as a 2D vector. You can express it as a(1,0)+b(0,1), where (1,0) is the "1" state and (0,1) is the "0" state. In quantum mechanics a and b are complex numbers. When you measure the qbit the quantum state by magic collapses to (1,0) with probability a^2 or to (0,1) with probability b^2. The power of quantum computers come from the fact that transformations or your qbits act on the superpositioned state, not the collapsed state.
So in principle you operate on all possible classical states of your qbits simultaneously when operating on the superposition. This magic doesn't grant you unlimited powers however, since you need to look at the state to get an answer and there you only measure one definite state. The trick to quantum algorithms is then to make the "right" answer as probable as possible.
This explanation is missing a little bit of background. A classical computer's state is represented by a single number - the concatenation of all of the classical bits in its memory. The transition function in a classical computer takes a state and moves it into another state - a simple mapping.
A quantum computer's state is a 2D vector of unit length. Transitions are actually rotations on this vector. Observing the quantum state collapses it into a classical state - a single number - by projecting it onto a random nearby axis. If your vector is halfway between two axes, the computer has an equal chance of collapsing into either classical state when observed. As it turns out, this also means that acting on this vector is equivalent to acting on both of those classical states simultaneously. If you have the right kind of problem (http://en.wikipedia.org/wiki/BQP, for example), this lets you do a lot of computation in a hurry.
A k-qbit computer's state is actually a unit vector in 2^k dimensional complex space. This exponential blowup makes it difficult to simulate quantum computers on classical systems.
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Think of a qbit as a 2D vector. You can express it as a(1,0)+b(0,1), where (1,0) is the "1" state and (0,1) is the "0" state. In quantum mechanics a and b are complex numbers. When you measure the qbit the quantum state by magic collapses to (1,0) with probability a^2 or to (0,1) with probability b^2. The power of quantum computers come from the fact that transformations or your qbits act on the superpositioned state, not the collapsed state.
So in principle you operate on all possible classical states of your qbits simultaneously when operating on the superposition. This magic doesn't grant you unlimited powers however, since you need to look at the state to get an answer and there you only measure one definite state. The trick to quantum algorithms is then to make the "right" answer as probable as possible.
This explanation is missing a little bit of background. A classical computer's state is represented by a single number - the concatenation of all of the classical bits in its memory. The transition function in a classical computer takes a state and moves it into another state - a simple mapping.
A quantum computer's state is a 2D vector of unit length. Transitions are actually rotations on this vector. Observing the quantum state collapses it into a classical state - a single number - by projecting it onto a random nearby axis. If your vector is halfway between two axes, the computer has an equal chance of collapsing into either classical state when observed. As it turns out, this also means that acting on this vector is equivalent to acting on both of those classical states simultaneously. If you have the right kind of problem (http://en.wikipedia.org/wiki/BQP, for example), this lets you do a lot of computation in a hurry.
A k-qbit computer's state is actually a unit vector in 2^k dimensional complex space. This exponential blowup makes it difficult to simulate quantum computers on classical systems.
Yeah, my bad. I think I crosswired "computer" and "bit" when I was typing "2D vector".