4 ms·

It can be written as a spin chain, there is a Hamiltonian that generates the unitary you wrote above. Maybe it isn't very natural to do so but I also don't real

It can be written as a spin chain, there is a Hamiltonian that generates the unitary you wrote above. Maybe it isn't very natural to do so but I also don't really that its natural to call this thing a "qantum walk" (like how is this thing walking?). The difference between integers / digits is basically just a difference in numbering, I think they're completely equivalent if you replace n with 2^n or log_2(n) or something.

drdeca 6y ago
Ah, alright. Thanks. As for how much this resembles walking, if we label the states with integers rather than their binary expansion (uh, maybe assume infinitely many bits, and represent the negative numbers with the ones starting with all 1s?), Well, the state corresponding to a given integer is sent to a linear combination of itself and the states corresponding to integers either one or 2 higher or lower. If this were a transition function for a classical random process (with non-negative coefficients which sum to 1 instead of having absolute value squared-s which sum to 1), that seems to me like it could be called a random walk. hadamard on last digit does, If n is even, |n> -> (|n> + |(n+1)>)/sqrt(2) If n is odd, |n> -> (|(n-1)> - |n>)/sqrt(2) So, combining this with the incrementing, For n even, |n> -> (|(n+1)> + |(n+2)>)/sqrt(2) If n is odd, |n> -> (|(n)> - |(n+1)>)/sqrt(2) Composing with the hadamard again For n even, |n> -> ((|n> -|(n+1)>)/sqrt(2) + (|(n+2)> + |(n+3)>)/sqrt(2) )/sqrt(2) = ( |n> - |(n+1)> + |(n+2)> + |(n+3)>)/2 For n odd, |n> -> (|(n-1)> - |n> - |(n+1)> - |(n+2)>)/2 And so, including the final decrement, the whole operation does For n even, |n> -> ( |(n-1)> - |n> + |(n+1)> + |(n+2)>)/2 For n odd, |n> -> (|(n-2)> - |(n-1)> - |n> - |(n+1)>)/2 Certainly if you measured the result after each application of this, you would observe behavior like a classical random walk of sorts (it would have a bias towards increasing when n is even and a bias towards decreasing when n is odd, but otherwise looks balanced.) Though, I don’t really see much advantage of this over simply taking what is initially being used as the least significant bit of an integer, and instead using it as your coin qubit, and using the rest of the bits for the integer, instead of using this weird mashup of the two (which seems to maybe give something almost but not quite equivalent to it.) Edit: oh! They clarified what they meant, good

drdeca 6y ago

Ah, alright. Thanks. As for how much this resembles walking, if we label the states with integers rather than their binary expansion (uh, maybe assume infinitely many bits, and represent the negative numbers with the ones starting with all 1s?), Well, the state corresponding to a given integer is sent to a linear combination of itself and the states corresponding to integers either one or 2 higher or lower. If this were a transition function for a classical random process (with non-negative coefficients which sum to 1 instead of having absolute value squared-s which sum to 1), that seems to me like it could be called a random walk. hadamard on last digit does, If n is even, |n> -> (|n> + |(n+1)>)/sqrt(2) If n is odd, |n> -> (|(n-1)> - |n>)/sqrt(2) So, combining this with the incrementing, For n even, |n> -> (|(n+1)> + |(n+2)>)/sqrt(2) If n is odd, |n> -> (|(n)> - |(n+1)>)/sqrt(2) Composing with the hadamard again For n even, |n> -> ((|n> -|(n+1)>)/sqrt(2) + (|(n+2)> + |(n+3)>)/sqrt(2) )/sqrt(2) = ( |n> - |(n+1)> + |(n+2)> + |(n+3)>)/2 For n odd, |n> -> (|(n-1)> - |n> - |(n+1)> - |(n+2)>)/2 And so, including the final decrement, the whole operation does For n even, |n> -> ( |(n-1)> - |n> + |(n+1)> + |(n+2)>)/2 For n odd, |n> -> (|(n-2)> - |(n-1)> - |n> - |(n+1)>)/2 Certainly if you measured the result after each application of this, you would observe behavior like a classical random walk of sorts (it would have a bias towards increasing when n is even and a bias towards decreasing when n is odd, but otherwise looks balanced.) Though, I don’t really see much advantage of this over simply taking what is initially being used as the least significant bit of an integer, and instead using it as your coin qubit, and using the rest of the bits for the integer, instead of using this weird mashup of the two (which seems to maybe give something almost but not quite equivalent to it.) Edit: oh! They clarified what they meant, good