Representing Operators as Matrices and Differential Operators

aaaa202
Messages
1,144
Reaction score
2
An operator A defined by a matrix can be written as something like:

A = Ʃi,jlei><ejl <eilAlej>

How does this representation translate to a continuous basis, e.g. position basis, where operators are not matrices but rather differential operators etc. Can we still write for e.g. the kinetic energy operator T:

T = ∫∫dr dr' lr'><rl <r'lTlr>

? Or how would T be represented.
 
Physics news on Phys.org
I know there are some subtleties to shifting to a continuous basis, but basically yes. You change \Sigma's into \int's, and then all the same logic goes through. T becomes a "matrix" with an infinite number of rows and columns, which you could think of as a function T(q_1, q_2).
 
The energy eigenstates form a countable basis. Would it be possible to do the above expansion in those rather than position eigenstates?
 
^ The eigenstates of the Hamiltonian form a countable basis only if all the states are bound states, like in a harmonic oscillator. For example, the bound states of hydrogen atom form a discrete set, but the system also has a continuous spectrum of scattering states which have positive total energy. The continuous spectrum is not a countable set.
 
but either way you can expand any operator in energy just as well as position eigenstates right? or any other observables eigenstates..
 
^ Yes, the eigenstates of any hermitian operator form a complete basis.
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
I asked a question related to a table levitating but I am going to try to be specific about my question after one of the forum mentors stated I should make my question more specific (although I'm still not sure why one couldn't have asked if a table levitating is possible according to physics). Specifically, I am interested in knowing how much justification we have for an extreme low probability thermal fluctuation that results in a "miraculous" event compared to, say, a dice roll. Does a...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Back
Top