Susskind points out that the wave funcion is just a eigenfunction of the projection density matrix and that said matrix has only one eigenvalue of value – when the state is pure. All other vectors that are orthogonal to that one have value 0…
Mixed states have more than value (less than 1).
His point is that both pure and mixed’ness each have a detector.
Turing pure state was expressed in terms of the indepedence, rather than entanglement of two variables. It is detected by the closure property (of H1.U1**m) since 1 is the mean of the f function, achieved only when the variables commute.
Remember that inner product is the sum of the products of each vectors’ components, divixed by the order of the group. If one of those vectors has value 1 for each component, the inner product is just the classical definition of the mean.
Remember also that the “inner product” length is less than the product of the lengths, by cauchy schwatz, which means that our special “mean” inner product is less than the product of thd “distinguishing” norms.
So mean 1 is the accumulation point of the compact operator, which is the 1d’ness of being closed in (coset) group theory, which is a single eigenvalue of value 1 in entanglement theory of pure states.
So in general, he is Citing l2 convergence in the mean.
Before doing so, however, it is useful to introduce some more language, and one more fact about the density operator. First, the language. A quantum system whose state |ψ〉 is known exactly is said to be in a pure state. In this case the density operator is simply ρ = |ψ〉 〈ψ|. Otherwise, ρ is in a mixed state; it is said to be a mixture of the different pure states in the ensemble for ρ. In the exercises you will be asked to demonstrate a simple criterion for determining whether a state is pure or mixed: a pure state satisfies tr(ρ2) = 1, while a mixed state satisfies tr(ρ2) < 1. A few words of warning about the nomenclature: sometimes people use the term ‘mixed state’ as a catch-all to include both pure and mixed quantum states. The origin for this usage seems to be that it implies that the writer is not necessarily assuming that a state is pure. Second, the term ‘pure state’ is often used in reference to a state vector |ψ〉, to distinguish it from a density operator ρ.