1) Yes, a unitary transformation like the Hadamard gate maps the state |0> to |0>+|1>, while mapping |1> to |0>-|1>. In either case, if you then just measured immediately in the {|0>,|1>} basis without doing anything else, you'd see |0> or |1> with equal probabilities, so it would have the effect of a coin flip. But of course, in other cases---e.g., if you measured in a different basis, or if you applied the Hadamard to a state that wasn't just a |0> or |1> basis state---you could see that Hadamard is not just a coin-flipping transformation, because it's able to produce interference.
2) When we talk about the different "paths" that contribute to a given amplitude, it's just a fancy way of saying that we can organize the matrix multiplications in such a way that the amplitude we want is a giant sum. So for example, suppose we apply Hadamard twice in sequence to the initial state |0>. The first Hadamard maps |0> to (|0>+|1>)/sqrt(2). The second Hadamard maps |0> to (|0>+|1>)/sqrt(2) and |1> to (|0>-|1>)/sqrt(2). So by linearity, it maps (|0>+|1>)/sqrt(2) to
So in this case, we could say that there are "two paths leading to |0>," both of which contribute 1/2 to its final amplitude (so that the amplitude is 1). There are also "two paths leading to |1>," but one contributes 1/2 to its amplitude and the other contributes -1/2, so the two contributions interfere destructively and the final amplitude of |1> is 0.
This is sometimes called the "Feynman" or "sum-over-paths" picture of quantum mechanics. As you can see, though, it's just a different way of looking at exactly the same math, namely multiplication of matrices and vectors.
So then why use the sum-over-paths picture at all? Well, a few reasons: physicists like it because it often gives them more insight into what's going on, into what are the more and less important contributions to a given process, and it can also make calculations easier. Meanwhile, computer scientists like the picture because it lets us simulate a quantum computer by a classical computer, still using exponential time but now using only a linear amount of memory, rather than the exponential amount of memory we'd need if we tried to store all 2^n amplitudes at once.
The cleanest explanation I read of information / state linearity properties of quantum operators. Thanks! (Working through Brian Hall's Quantum Theory for Mathematicians and Frederic Schuller's course at the moment, both highly recommended).
2) When we talk about the different "paths" that contribute to a given amplitude, it's just a fancy way of saying that we can organize the matrix multiplications in such a way that the amplitude we want is a giant sum. So for example, suppose we apply Hadamard twice in sequence to the initial state |0>. The first Hadamard maps |0> to (|0>+|1>)/sqrt(2). The second Hadamard maps |0> to (|0>+|1>)/sqrt(2) and |1> to (|0>-|1>)/sqrt(2). So by linearity, it maps (|0>+|1>)/sqrt(2) to
((|0>+|1>)/sqrt(2) + (|0>-|1>)/sqrt(2))/sqrt(2) = (1/2+1/2)|0> + (1/2-1/2)|1>.
So in this case, we could say that there are "two paths leading to |0>," both of which contribute 1/2 to its final amplitude (so that the amplitude is 1). There are also "two paths leading to |1>," but one contributes 1/2 to its amplitude and the other contributes -1/2, so the two contributions interfere destructively and the final amplitude of |1> is 0.
This is sometimes called the "Feynman" or "sum-over-paths" picture of quantum mechanics. As you can see, though, it's just a different way of looking at exactly the same math, namely multiplication of matrices and vectors.
So then why use the sum-over-paths picture at all? Well, a few reasons: physicists like it because it often gives them more insight into what's going on, into what are the more and less important contributions to a given process, and it can also make calculations easier. Meanwhile, computer scientists like the picture because it lets us simulate a quantum computer by a classical computer, still using exponential time but now using only a linear amount of memory, rather than the exponential amount of memory we'd need if we tried to store all 2^n amplitudes at once.