Recent content by Alex Dingo

Thank you for you're fantastic response! You answered my questions perfectly and I have a much better grasp on the significance of the result - much appreciated.

Hi, I'm currently working through a tensor product example for a two qubit system. For the expression: $$ \rho_A = \sum_{J=0}^{1}\langle J | \Psi \rangle \langle \Psi | J \rangle $$ Which has been defined as from going to a global state to a local state. Here $$ |\Psi \rangle = |\Psi^+...

Hi, Okay this makes more sense, I forgot about this relationship. I have computed this with ##\hat{A} = p1p2## I find that ## \mathrm{Tr}(p1p2)=\Sigma\langle n | p_n|n\rangle \langle n | p_n q_n |n \rangle \langle n| q_n | n \rangle ## which reduces to ## \Sigma p_n^2 q_n^2 ## as desired ?

Hi DrCloud, So the result for calculating ##| \langle \psi_1 | \psi_2 \rangle |^2## I found it equat to ##\Sigma p^2_n q^2_n ##. As for the calculation with the trace. I find ## Tr(\rho_1 \rho_2) = Tr(\langle n | \Sigma p^2_n |n\rangle \langle n | \Sigma q^2_n |n\rangle ## From this point I...

Hi Bill, If I express Ψ1=∑√(p(x))|x> and Ψ2=∑√(q(x))|x> such that they are both expressed in the basis of x and compute |⟨ψ1|ψ2⟩|^2 I find Is equal to ∑[√p(x)√q(x)]^2 <x|x> such that I am left with Σp(x)q(x) which is a summation of their probabilities over x. Can I then relate the relation...

Hi, I'm currently working on showing the relation of quantum fidelity: The quantum “fidelity” between two pure states ρ1 = |ψ1⟩⟨ψ1| and ρ2 = |ψ2⟩⟨ψ2| is given by |⟨ψ1|ψ2⟩|^2. Show that this quantity may be written as Tr(ρ1ρ2). I've been following the wikipedia page on fidelity but can't...