4 ms·

In a quantum walk the locations typically correspond to states, not to qubits. You're right that turning them into qubits would allow the diagram to work, but t

In a quantum walk the locations typically correspond to states, not to qubits. You're right that turning them into qubits would allow the diagram to work, but then I don't think it would match up with the "vampire fangs".

eigenket 5y ago
I don't understand this comment at all. Let me clarify my point slightly. The quantum walk is a unitary operator acting on a Hilbert space H consisting of two parts H_spin and H_space H = H_spin ⊗ H_space In this example H_spin is just a qubit and H_space is L^2(Z). In my example above the |0><0| and |1><1| are projectors acting on the spin degree of freedom (the first Hilbert space) and the sum_i |i><i+1| and sum_i |i><i-1| are shift operators acting on the space degree of freedom (the second Hilbert space). If you take the unitary I wrote and do the iteration U (H⊗I) U (H⊗I) U (H⊗I) U (H⊗I) ... U (H⊗I) where U (H⊗I) is the Hadamard operator on the spin Hilbert space and I is the identity on the space part you get exactly the quantum walk that gives the "fangs" picture.
- Strilanc 5y ago
  Sorry, I misread your comment initially. Yes, if you have an additional degree of freedom that is not being plotted (a coined walk), then the overlap is fine.
  - eigenket 5y ago
    If there is no additional degree of freedom the walk has to be trivial (assuming homogeneity).
    - Strilanc 5y ago
      In a coined walk you apply a hadamard to the least significant qubit of an integer register, then increment the register, hadamard the least significant qubit, decrement the register, and repeat. Although there is no additional degree of freedom, you are alternating between two non-commuting effects.
      - eigenket 5y ago
        This seems completely different to the sorts of walks I'm used to. When you say "qubit of an integer register" what are you actually talking about here? Do you have a qubit for each integer like a spin chain or something? The sorts of quantum walks I'm used to are exactly the ones in the article and are also mentioned in the wikipedia page here. https://en.wikipedia.org/wiki/Quantum_walk#Discrete_time https://en.wikipedia.org/wiki/Quantum_walk#Discrete_time The example given on the wiki page is the classic example that gets called a "coined walk" in the literature.
        drdeca 5y ago
        I imagine they mean there is a collection of qubits interpreted as storing an integer (would be storing an integer if they were mere bits) And so the qubit for the least significant bit, would be the one for the 1s place. I could be wrong though
        eigenket 5y ago
        Yeah thats pretty much what I thought, I would call this object a spin-chain rather than a quantum walk though. I guess it might have some "walkey" characteristics if you whack the right Hamiltonian on it though.
        drdeca 5y ago
        Hm, the way I interpreted what you described as a spin-chain seems different to me? I thought when you said "a qubit for each integer" that you meant that if dealing with integers from 1-n to n, that you would have 2n qubits, while what I thought they meant was, say you had n qubits, then the possible positions would be elements of Z/((2^n)Z), so e.g. you could start with |0000> , and then via Hadamard go to (|0000> + |0001>)/\sqrt{2} , and then by incrementing, go to (|0001> + |0010>)/\sqrt{2} and then by Hadamard, go to (|0000> - |0001> + |0010> + |0011>)/2 , and then by decrementing, go to (|1111> - |0000> + |0011> + |0010>)/2 and then repeat the above. So, next step would be another Hadamard, and go to (|1110> - |1111> - |0000> + |0001> + |0010> - |0011> + |0010> + |0011>)/sqrt(8) = (|1110> - |1111> - |0000> + |0001> + 2|0010> )/sqrt(8) Then increment, etc. I think this seems to effectively be using the least significant digit as the coin. I wouldn't say that this has a qubit for each integer, only a qubit for each digit. Would this be a spin chain? Wikipedia turns up the Heisenberg model for that, which looks to be a continuous time thing?

eigenket 5y ago

I don't understand this comment at all. Let me clarify my point slightly. The quantum walk is a unitary operator acting on a Hilbert space H consisting of two parts H_spin and H_space H = H_spin ⊗ H_space In this example H_spin is just a qubit and H_space is L^2(Z). In my example above the |0><0| and |1><1| are projectors acting on the spin degree of freedom (the first Hilbert space) and the sum_i |i><i+1| and sum_i |i><i-1| are shift operators acting on the space degree of freedom (the second Hilbert space). If you take the unitary I wrote and do the iteration U (H⊗I) U (H⊗I) U (H⊗I) U (H⊗I) ... U (H⊗I) where U (H⊗I) is the Hadamard operator on the spin Hilbert space and I is the identity on the space part you get exactly the quantum walk that gives the "fangs" picture.

Strilanc 5y ago
Sorry, I misread your comment initially. Yes, if you have an additional degree of freedom that is not being plotted (a coined walk), then the overlap is fine.
- eigenket 5y ago
  If there is no additional degree of freedom the walk has to be trivial (assuming homogeneity).
  - Strilanc 5y ago
    In a coined walk you apply a hadamard to the least significant qubit of an integer register, then increment the register, hadamard the least significant qubit, decrement the register, and repeat. Although there is no additional degree of freedom, you are alternating between two non-commuting effects.
    - eigenket 5y ago
      This seems completely different to the sorts of walks I'm used to. When you say "qubit of an integer register" what are you actually talking about here? Do you have a qubit for each integer like a spin chain or something? The sorts of quantum walks I'm used to are exactly the ones in the article and are also mentioned in the wikipedia page here. https://en.wikipedia.org/wiki/Quantum_walk#Discrete_time https://en.wikipedia.org/wiki/Quantum_walk#Discrete_time The example given on the wiki page is the classic example that gets called a "coined walk" in the literature.
      - drdeca 5y ago
        I imagine they mean there is a collection of qubits interpreted as storing an integer (would be storing an integer if they were mere bits) And so the qubit for the least significant bit, would be the one for the 1s place. I could be wrong though
        eigenket 5y ago
        Yeah thats pretty much what I thought, I would call this object a spin-chain rather than a quantum walk though. I guess it might have some "walkey" characteristics if you whack the right Hamiltonian on it though.
        drdeca 5y ago
        Hm, the way I interpreted what you described as a spin-chain seems different to me? I thought when you said "a qubit for each integer" that you meant that if dealing with integers from 1-n to n, that you would have 2n qubits, while what I thought they meant was, say you had n qubits, then the possible positions would be elements of Z/((2^n)Z), so e.g. you could start with |0000> , and then via Hadamard go to (|0000> + |0001>)/\sqrt{2} , and then by incrementing, go to (|0001> + |0010>)/\sqrt{2} and then by Hadamard, go to (|0000> - |0001> + |0010> + |0011>)/2 , and then by decrementing, go to (|1111> - |0000> + |0011> + |0010>)/2 and then repeat the above. So, next step would be another Hadamard, and go to (|1110> - |1111> - |0000> + |0001> + |0010> - |0011> + |0010> + |0011>)/sqrt(8) = (|1110> - |1111> - |0000> + |0001> + 2|0010> )/sqrt(8) Then increment, etc. I think this seems to effectively be using the least significant digit as the coin. I wouldn't say that this has a qubit for each integer, only a qubit for each digit. Would this be a spin chain? Wikipedia turns up the Heisenberg model for that, which looks to be a continuous time thing?