Symmetric Logarithmic Derivative

Symmetric Logarithmic Derivative Ask Question Asked 5 days ago Modified 4 days ago Viewed 48 times 2 I try to understand the Symmetric Logarithmic Derivative which is a central concept for Quantum Fisher Information and is the extension to the quantum world of Fisher Information. There are relatively few tutorials about the Symmetric Logarithmic Derivative : the wikipedia page the grokipedia page In the wikipedia page, an Hermitian and positive semidefinite operator ρ 𝜌 is considered, together with A 𝐴 an Hermitian operator. ρ 𝜌 heavily recalls a density matrix, but trace normalization is not required. Under these assumptions, the symmetric Logarithmic derivative L ρ (A) 𝐿 𝜌 ( 𝐴 ) of A 𝐴 is defined by : i[ρ,A]= 1 2 {ρ, L ρ (A)}(1) 𝑖 [ 𝜌 , 𝐴 ] = 1 2 { 𝜌 , 𝐿 𝜌 ( 𝐴 ) } ( 1 ) where [,] [ , ] and {,} { , } are respectively the commutator and the anticommutator. In the grokipedia page, a parameterized quantum state ρ θ 𝜌 𝜃 is defined, with θ 𝜃 a real parameter, e.g. : ρ θ = e −i θ 2 σ z ρ 0 e i θ 2 σ z 𝜌 𝜃 = 𝑒 − 𝑖 𝜃 2 𝜎 𝑧 𝜌 0 𝑒 𝑖 𝜃 2 𝜎 𝑧 for one qubit. The symmetric Logarithmic Derivative L θ 𝐿 𝜃 associated with quantum states ρ θ 𝜌 𝜃 is defined by : ∂ θ ρ θ = ∂ ρ θ ∂θ = 1 2 [ L θ ρ θ + ρ θ L θ ]= 1 2 { ρ θ , L θ }(2) ∂ 𝜃 𝜌 𝜃 = ∂ 𝜌 𝜃 ∂ 𝜃 = 1 2 [ 𝐿 𝜃 𝜌 𝜃 + 𝜌 𝜃 𝐿 𝜃 ] = 1 2 { 𝜌 𝜃 , 𝐿 𝜃 } ( 2 ) I do not understand the link between (1) and (2), are they the same ? In this case, we should have : ∂ ρ θ ∂θ =i[ρ,A], L ρ (A)= L θ (3) ∂ 𝜌 𝜃 ∂ 𝜃 = 𝑖 [ 𝜌 , 𝐴 ] , 𝐿 𝜌 ( 𝐴 ) = 𝐿 𝜃 ( 3 ) If I set : U= e −i θ 2 σ z ⇒ ∂ θ (U)=− i 2 σ z U 𝑈 = 𝑒 − 𝑖 𝜃 2 𝜎 𝑧 ⇒ ∂ 𝜃 ( 𝑈 ) = − 𝑖 2 𝜎 𝑧 𝑈 then ∂ θ ρ θ =− i 2 σ z Uρ U † + i 2 Uρ U † σ z =i[ ρ θ σ z 2 − σ z 2 ρ θ ] ∂ 𝜃 𝜌 𝜃 = − 𝑖 2 𝜎 𝑧 𝑈 𝜌 𝑈 † + 𝑖 2 𝑈 𝜌 𝑈 † 𝜎 𝑧 = 𝑖 [ 𝜌 𝜃 𝜎 𝑧 2 − 𝜎 𝑧 2 𝜌 𝜃 ] So choosing A= σ z 2 𝐴 = 𝜎 𝑧 2 makes (1) and (2) be coherent. Is my understanding correct ? Remark : in fact the first equality in (3) is the Von Neumann Equation if θ 𝜃 is time. quantum-fisher-information Share Improve this question Follow edited May 29 at 19:03 asked May 28 at 12:22 deb2014 6773 3 silver badges 8 8 bronze badges The answer to your question is yes they are the same. Forget about the commutator in EQ. (1), that's just von Neumann equation with Hamiltonian A 𝐴 . –  lcv Commented May 29 at 14:25 @lcv for unitary operations and linear parameters only –  Quantum Mechanic Commented May 29 at 16:07 Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 They are the same if the state is evolving unitarily and the parameter couples to a Hamiltonian-type generator. Specifically, if ρ(θ)=U(θ)ρ(0)U(θ ) † 𝜌 ( 𝜃 ) = 𝑈 ( 𝜃 ) 𝜌 ( 0 ) 𝑈 ( 𝜃 ) † and U(θ)=exp(−iθA) 𝑈 ( 𝜃 ) = exp ⁡ ( − 𝑖 𝜃 𝐴 ) for any Hermitian A 𝐴 then ∂ρ ∂θ =i[ρ,A] ∂ 𝜌 ∂ 𝜃 = 𝑖 [ 𝜌 , 𝐴 ] and the two formulations of the symmetric logarithmic derivative are the same. But, if ρ(θ) 𝜌 ( 𝜃 ) depends on θ 𝜃 in any other way, such as if θ 𝜃 is the temperature of a thermal state, then this equivalence will no longer hold and one has to use the more general definition (2). Share Improve this answer Follow answered May 29 at 16:07 Quantum Mechanic 4,9397 7 silver badges 26 26 bronze badges Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown 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. 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