On Mon, Nov 1, 2010 at 7:24 AM, Moritz Lenz <mor...@faui2k3.org> wrote: > On 10/22/2010 06:16 AM, Damian Conway wrote: >> That is, a C<$value> is an eigenstate of a C<$junction> if-and-only-if: >> >> $value !~~ Junction && $value ~~ $junction > > In general this definition makes it impossible to return a list of > eigenstates from the junction. Just think of junctions containing Code > objects. Or anything more complicated than the built-in value types.
[Originally sent to Moritz alone because of "Reply" not sending to the list] Is it too late in this discussion to point out that, in non-perl usage, eigenstates are associated with the operator, not with the value fed into the operator? [Added at Moritz request] In linear algebra, eigenvectors and eigenvalues are non-trivial solutions to the equation Ax=λx, where x is a vector in a vector space, A is a operator (a function from a vector space to itself) and λ is a member of the field the vector space is defined over. For a given operator A, only certain values of λ allow that equation to be solved, and those values are called the "eigenvalues" for A. Also, for a given operator A, only certain vectors x will solve the equation, and those vectors are called "eigenvectors". It should also be clear that different values of λ work with different sets of vectors x (the solutions to Ax = ax and Ax=bx are different if a != b), so it's typical to talk about the eigenvectors of A associated with a given eigenvalue λ. Since A is linear, if Ax=λx and Ay=λy, then A(ax) = a(Ax) = a(λx) = λ(ax) and A(x+y)=Ax+Ay=λx+λy=λ(x+y), so fir a given eigenvalue λ, there are typically multitudes of eigenvectors which form a vector space of their own. Eigenvectors for different eigenvalues are orthogonal, and any eigenvector can be scaled to be a unit eigenvector. If an operator has a full set of eigenvalues, one can pick a set of unit eigenvectors to act as a natural orthonormal basis for the operator. If operator A has three eigenvalues a, b, c, and three unit eigenvectors x, y, z, such that Ax=ax, Ay=by, and Az=cz, then if w = dx+ey+fz, Aw = a(dx)+b(ey)+c(fz), which is really easy to compute. In quantum mechanics, especially the Heisenberg matrix formulation (but by analogy, also every other formulation, including wave mechanics), quantum states are represented by vectors in a complex vector space, and vectors which differ by a real-valued scaling factor are generally considered equivalent. Transformations (i.e., anything which modifies the quantum state of the system, including but not limited to the passage of time) are represented by (unitary) operators on the state space. (Unitary in this case means that the norm of Ax is the same as the norm of x, for all x.) The standard notation is a bit odd, where the 'ket' |x> represents a system in state x (and therefore |x+y> a state in a superposition of x and y), The 'bra' <x| is the complement of the ket |x>, and can be multiplied by a ket to get a braket <x|y> which represents the probability that a system in state y is also in state x. |x> is, naturally, usually normalized such that <x|x> = 1. Operators act on kets and return kets, so A|x> is the braket notation way of writing the linear algebra Ax. Naturally, that means that <y|A|x> is the probability that a system that starts in state x will be in state y after the transform A. Since A is a linear operator, it has eigenvalues and eigenvectors. In the quantum mechanical world, where vectors represent states, the eigenvectors are called "eigenstates". Eigenstates |i>, |j> of an operator A have the property that <i|A|i> = <j|A|j> = 1, but <i|A|j> = 0 (informally, if you start in an eigenstate of A, then the transform leaves you unchanged). However, A|i+j> = |ai+bj>, so A can change the nature of a superposition of states. <i|i+j> = 1/2, <j|i+j> = 1/2, but <i|A|i+j> = a/(a+b), <j|A|i+j> = b/(a+b). Schrodinger's Wave Equation, in matrix notation, is of the form Hx=Ex, where H is the "Hamiltonian operator" of the system, and E is the energy of the system, so the only allowed solutions of the wave equations are for energy levels E which are eigenvalues of H,and for quantum states which are eigenstates of H. Similar equations exist for virtually every "observable", so the only allowable momenta are the eigenvalues or eigenstates of the momentum operator, the only allowable positions are the eigenvalues or eigenstates of the position operator, etc. So asking for the eigenstates of a quantum superposition is asking the wrong object for the property.
