Quantum Lattice-Gas Automata

Quantum Lattice-Gas Automata Ask Question Asked 16 days ago Modified 16 days ago Viewed 32 times 0 I study the Quantum Lattice-Gas Automata (Q-LGA) in the context of Quantum Computing for Partial Differential Equations. There has been a significant development which began with : Meyer, David A. (1996) From quantum cellular automata to quantum lattice gases, Journal of Statistical Physics. If you are not familiar with LGA, here is the necessary notions for my question (taken from previous article). A Cellular Automata (CA) is a lattice of L 𝐿 cells with a field ϕ 𝜙 : ϕ: N×L (t,x) ⟼ ⟼ S ϕ t (x) 𝜙 : 𝑁 × 𝐿 ⟼ 𝑆 ( 𝑡 , 𝑥 ) ⟼ 𝜙 𝑡 ( 𝑥 ) where t 𝑡 represents the discrete time, S 𝑆 is the set of states that the field can take on each cell x 𝑥 . The field ϕ 𝜙 across the lattice evolves according to local interactions between cells : ϕ t+1 (x)= ∑ e∈E(t,x) ω(t,x+e) ϕ t (x+e)(2) 𝜙 𝑡 + 1 ( 𝑥 ) = ∑ 𝑒 ∈ 𝐸 ( 𝑡 , 𝑥 ) 𝜔 ( 𝑡 , 𝑥 + 𝑒 ) 𝜙 𝑡 ( 𝑥 + 𝑒 ) ( 2 ) where E(t,x) 𝐸 ( 𝑡 , 𝑥 ) is a set of vectors characterizing the neighborhood of x 𝑥 at time t 𝑡 . If the CA is homogeneous then both E(t,x) 𝐸 ( 𝑡 , 𝑥 ) and ω(t,x+e) 𝜔 ( 𝑡 , 𝑥 + 𝑒 ) are independent of time t 𝑡 and cell x 𝑥 , it depends then only of the direction e 𝑒 : ϕ t+1 (x)= ∑ e∈E ω e ϕ t (x+e)(3) 𝜙 𝑡 + 1 ( 𝑥 ) = ∑ 𝑒 ∈ 𝐸 𝜔 𝑒 𝜙 𝑡 ( 𝑥 + 𝑒 ) ( 3 ) Assume E={−r,…,0,…,r} 𝐸 = { − 𝑟 , … , 0 , … , 𝑟 } , this lets (3) be put in matrix form for a lattice of dimension 1 1 : ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⋮ ϕ t+1 (−1) ϕ t+1 (0) ϕ t+1 (+1) ⋮ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⋱ … ω −r … ⋱ … ω −r … ω −1 … ω −r ω 0 ω −1 … ⋱ ω 1 ω 0 ω −1 ⋱ … ω 1 ω 0 ⋱ ω r … ω 1 … ω r … ⋱ … ω r … ⋱ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⋮ ϕ t (−1) ϕ t (0) ϕ t (+1) ⋮ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ [ ⋮ 𝜙 𝑡 + 1 ( − 1 ) 𝜙 𝑡 + 1 ( 0 ) 𝜙 𝑡 + 1 ( + 1 ) ⋮ ] = [ ⋱ ⋱ ⋱ … 𝜔 − 𝑟 … 𝜔 − 1 𝜔 0 𝜔 1 … 𝜔 𝑟 … … 𝜔 − 𝑟 … 𝜔 − 1 𝜔 0 𝜔 1 … 𝜔 𝑟 … … 𝜔 − 𝑟 … 𝜔 − 1 𝜔 0 𝜔 1 … 𝜔 𝑟 … ⋱ ⋱ ⋱ ⋱ ] [ ⋮ 𝜙 𝑡 ( − 1 ) 𝜙 𝑡 ( 0 ) 𝜙 𝑡 ( + 1 ) ⋮ ] so that ϕ t+1 =U ϕ t 𝜙 𝑡 + 1 = 𝑈 𝜙 𝑡 As an example for r=2 𝑟 = 2 and a field ϕ t 𝜙 𝑡 : ϕ t = [ ϕ −2 ϕ −1 ϕ 0 ϕ 1 ϕ 2 ] T 𝜙 𝑡 = [ 𝜙 − 2 𝜙 − 1 𝜙 0 𝜙 1 𝜙 2 ] 𝑇 we have a tri-banded matrix : U= ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 𝑈 = [ 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 ] in above mentioned article, is is demonstrated a No-Go Lemma stating that : "In one dimension there exists no nontrivial, homogeneous, local, scalar QCA. More explicitly, every band c-diagonal unitary matrix (collision operator) which commutes with the 1-step translation matrix is also a translation matrix times a phase." This means that the total evolution operator W 𝑊 composed by a collision operator U 𝑈 and translation operator T 𝑇 would result in a translation matrix times a phase φ 𝜑 , which does not effectively provide an interaction between particles : W=TU= e iφ T 𝑊 = 𝑇 𝑈 = 𝑒 𝑖 𝜑 𝑇 I guess the translation operator is this matrix (following the example) : T= ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 𝑇 = [ 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 ] I do not understand the lemma as it is clear for me that no r>0 𝑟 > 0 -banded-matrix like U 𝑈 above can ever be unitary. One can check this by taking the norm2 of each column of U 𝑈 , which must be equal to one, so that the only solution is : U= ω 0 I 𝑈 = 𝜔 0 𝐼 which removes all interactions. Furthermore, one condition of the lemma is that U 𝑈 and T 𝑇 commutes. It is true that if U 𝑈 was a unitary, the effect of translation matrix T 𝑇 would just be to switch either columns ( UT 𝑈 𝑇 ) or rows (TU) ( 𝑇 𝑈 ) so that it would still be a unitary, but there are no reason for UT=TU 𝑈 𝑇 = 𝑇 𝑈 and in our case : UT= ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω 0 ω −1 0 0 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 𝑈 𝑇 = [ 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 ] [ 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 ] = [ 𝜔 1 0 0 0 𝜔 0 𝜔 0 𝜔 1 0 0 𝜔 − 1 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 0 ] and TU= ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ω −1 0 0 0 ω 0 ω 0 ω −1 0 0 ω 1 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 ω −1 0 0 0 ω 1 ω 0 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 𝑇 𝑈 = [ 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 ] [ 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 ] = [ 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 1 0 0 0 𝜔 − 1 𝜔 0 𝜔 0 𝜔 1 0 0 0 ] Clearly : UT≠TU 𝑈 𝑇 ≠ 𝑇 𝑈 , but if one wants W=TU 𝑊 = 𝑇 𝑈 to be unitary, then the sole solution is ω −1 = ω 1 =0 𝜔 − 1 = 𝜔 1 = 0 . so that indeed, I agree with the lemma : W=TU= ω 0 T 𝑊 = 𝑇 𝑈 = 𝜔 0 𝑇 T 𝑇 is unitary, then for W 𝑊 to be unitary, we must have ω 0 = e iφ 𝜔 0 = 𝑒 𝑖 𝜑 . and in this case : W † = ω 0 ¯ ¯ ¯ ¯ ¯ T † = T † U † 𝑊 † = 𝜔 0 ¯ 𝑇 † = 𝑇 † 𝑈 † with U= ω 0 I 𝑈 = 𝜔 0 𝐼 . Is my understanding correct ? quantum-algorithmslinear-algebra Share Improve this question Follow asked May 19 at 9:17 deb2014 6773 3 silver badges 8 8 bronze badges I read again the article, if you consider an infinite one dimension lattice, then indeed you have TU=UT 𝑇 𝑈 = 𝑈 𝑇 . –  deb2014 Commented May 19 at 10:32 Add a comment Know someone who can answer? Share a link to this question via email, Twitter, or Facebook. Your Answer Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions quantum-algorithmslinear-algebra See similar questions with these tags. The Overflow Blog What it takes to be a player in the international AI... Making the OWASP top ten in the vibe code... 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