After finding a Unitary that comes from one of your classical Cellular Automata then any power of that unitary will also be a valid unitary. So for example in classical logic their is a the “swap” gate for binary inputs, but in quantum computing the “square-root swap” gate also exists.
So you can get one of your existing unitary matrices, and (for example) take its square root. That would kind of be like a quantum system doing the classical Cellular Automata, that is interrupted halfway through the first step. (Because applying the root matrix twice is the same as applying the matrix). Similarly you can look at the 1/3rd step by applying the cube root of the matrix.
So would you consider the square root matrix a quantum elementary CA? Its not exactly equivalent to anything classical, because classically you can’t look “between the steps”.
[This is a long winded way of me saying that I don’t “get” the question. You want a unitary, U, of the form given in that equation for <y|U|x>, but you also don’t want U to be “basically equivalent” to a classical CA. How are you defining “basically equivalent”, is anything satisfying your equation automatically “basically equivalent”?]
This is a good point! I’ll send you $20 if you send me your PayPal/Venmo/ETH/??? handle. (In my flailings, I’d stumbled upon this “fractional step” business, but I don’t think I thought about it as hard as it deserved.)
How are you defining “basically equivalent”
Nyeeeh, unfortunately, sort of “I know it when I see it.” It’s kinda neat being able to take a fractional step of a classical elementary CA, but I’m dissatisfied because… ah, because the long-run behavior of the fractional-step operator is basically indistinguishable from the long-run behavior of the corresponding classical CA.
So, tentative operationalization of “basically equivalent”: U is “basically equivalent” to a classical elementary CA if the long-run behavior of U is very close to the long-run behavior of some Uclassical, i.e., uh,
∀n∈N,ϵ∈R+:∃N>n,M∈N,Ucl:∣∣UN−UMcl∣∣<ϵ
...but I can already think of at least one flaw in this operationalization, so, uh, I’m not sure. (Sorry! This being so fuzzy in my head is why I’m asking for help!)
Random thoughts. You can relatively simply get a global phase factor at each timestep if you want. I don;t think a global phase factor at each step really counts as meaningfully different though. Anyway, as an example of this:
f(xk−1,xk,xk+1)=xk−1
So that, at each (classical) timestep every single element of the CA tape just moves one step to the right. (So any patterns of 1′s and 0′s just orbit the tape in circles forever, unchanging.). Its quite a boring CA, but a simple example.
We can take the quantum CA that is exactly the same, but with some complex phase factor:
fq(xk−1,xk,xk+1,yk)=δykxk−1exp(iθ)
Where the delta function is saying “1 iff yk=xk−1 , else 0.”
This is exactly the same as the old classical one (everything moves on step to the right), but this time we also have a global phase factor applied to the total system. The total phase factor is exp(iθN), where N is the total number of cells on the tape.
Tiny bit more interesting:
fq(xk−1,xk,xk+1,yk)=δykxk−1exp(iθxk−1)
Now we only gain phase factors on values of 1, so the global phase depends on the total number of 1′s on the tape, rather than its length.
To get proper quantum stuff we need phase factors that are not global. (IE some relative phases). I feel like this equation below is a reasonable kind of place to start, but I have run out of time for now so might return to this later.
After finding a Unitary that comes from one of your classical Cellular Automata then any power of that unitary will also be a valid unitary. So for example in classical logic their is a the “swap” gate for binary inputs, but in quantum computing the “square-root swap” gate also exists.
So you can get one of your existing unitary matrices, and (for example) take its square root. That would kind of be like a quantum system doing the classical Cellular Automata, that is interrupted halfway through the first step. (Because applying the root matrix twice is the same as applying the matrix). Similarly you can look at the 1/3rd step by applying the cube root of the matrix.
So would you consider the square root matrix a quantum elementary CA? Its not exactly equivalent to anything classical, because classically you can’t look “between the steps”.
[This is a long winded way of me saying that I don’t “get” the question. You want a unitary, U, of the form given in that equation for <y|U|x>, but you also don’t want U to be “basically equivalent” to a classical CA. How are you defining “basically equivalent”, is anything satisfying your equation automatically “basically equivalent”?]
This is a good point! I’ll send you $20 if you send me your PayPal/Venmo/ETH/??? handle. (In my flailings, I’d stumbled upon this “fractional step” business, but I don’t think I thought about it as hard as it deserved.)
Nyeeeh, unfortunately, sort of “I know it when I see it.” It’s kinda neat being able to take a fractional step of a classical elementary CA, but I’m dissatisfied because… ah, because the long-run behavior of the fractional-step operator is basically indistinguishable from the long-run behavior of the corresponding classical CA.
So, tentative operationalization of “basically equivalent”: U is “basically equivalent” to a classical elementary CA if the long-run behavior of U is very close to the long-run behavior of some Uclassical, i.e., uh,
∀n∈N,ϵ∈R+:∃N>n,M∈N,Ucl:∣∣UN−UMcl∣∣<ϵ...but I can already think of at least one flaw in this operationalization, so, uh, I’m not sure. (Sorry! This being so fuzzy in my head is why I’m asking for help!)
Random thoughts. You can relatively simply get a global phase factor at each timestep if you want. I don;t think a global phase factor at each step really counts as meaningfully different though. Anyway, as an example of this:
f(xk−1,xk,xk+1)=xk−1
So that, at each (classical) timestep every single element of the CA tape just moves one step to the right. (So any patterns of 1′s and 0′s just orbit the tape in circles forever, unchanging.). Its quite a boring CA, but a simple example.
We can take the quantum CA that is exactly the same, but with some complex phase factor:
fq(xk−1,xk,xk+1,yk)=δykxk−1exp(iθ)
Where the delta function is saying “1 iff yk=xk−1 , else 0.”
This is exactly the same as the old classical one (everything moves on step to the right), but this time we also have a global phase factor applied to the total system. The total phase factor is exp(iθN), where N is the total number of cells on the tape.
Tiny bit more interesting:
fq(xk−1,xk,xk+1,yk)=δykxk−1exp(iθxk−1)
Now we only gain phase factors on values of 1, so the global phase depends on the total number of 1′s on the tape, rather than its length.
To get proper quantum stuff we need phase factors that are not global. (IE some relative phases). I feel like this equation below is a reasonable kind of place to start, but I have run out of time for now so might return to this later.
fq(xk−1,xk,xk+1,yk)=(1/√2)[δykxk−1exp(iθ)+δykxk+1exp(−iθ)]