Lecture 1 - Sep 1, 2020 pure state quantum mechanics |psi> in C^2, or C^d, or C^2 otimes C^2, ... or C^2^{otimes n} mixed-states QM |psi> --> |psi>)_{i=1...m} --> rho = sum_{i=1}^m p_i |psi_i>=0} "rho >= 0" <---> rho is positive semidefinite <--> eig(rho) >= 0 quantum analogue of probability simplex {p in R^d : p_1, ..., p_d >=0, sum_i p_i = 1} measure rho in |1>, ..., |d> basis (or do POVM |1><1|, ..., |d>_AB in C^{d_A} otimes C^{d_B} psi = |psi> C id: L(C^d) -> L(C^d) L(V) = linear operators over V, i.e. matrices tr_B = id otimes tr tr_A = tr otimes id |psi> = sum_ij c_ij |i> otimes |j> psi = sum_ijkl c_ij c_kl* (|i> otimes |j>) ( = sum_ij d_ij |i> (otimes) |j> |psi> in C^{d_A}, |beta> in C^{d_B} C = |alpha> = |alpha> otimes |bar{beta}> in C^{d_Ad_B} = C_{d_A} otimes C^{d_B} i.e. |psi> is unentangled, a product state psi_A = |alpha> product <---> psi_A pure (i.e. rank 1) <--> psi_B pure (I've only shown --> direction.) example 2: d = d_A = d_B C unitary? C = U psi_A = C C^dag = U U^dag = I. tr \neq 1 instead C= U / sqrt(d) psi_A = I / d, maximally mixed state U = sum_{i=1...d} |u_i>,...,|u_d> are orthonormal |psi>_AB = 1/sqrt(d) sum_i |u_i>_A |i>_B measure B, get each i with probability 1/d, residual state is |u_i>_A psi_A = 1/d sum_i |u_i>} exercise: check sum_i |u_i> = |1> otimes |1> d_A =1, d_B =3 p_A = [1] psi_B is 3x3 matrix [1 0 0] [0 0 0] [0 0 0] 2. psi_A doesn't depend on V exercise: Show that local unitaries on A affect only U, local unitaries on B affect only V. purifications given rho_A, can we find |psi>_AB such that rho_A = psi_A ? |psi>_AB is a purification of rho_A, or of psi_A need to find factorization rho_A = C C^dag next time: always possible. factorization is not unique. will explore non-uniqueness then make robust application: bit commitment no-go thm for bit commitment, based on the robustness of purification