Finding Wavefunction with just the Hamiltonian

metapuff
Messages
53
Reaction score
6
Say I have a wavefunction that's a superposition of two-particle states:
\Psi = \int dk ~f(k) c^{\dagger}_k c^{\dagger}_{-k} | 0 \rangle
Here, ##|0\rangle## is the vacuum and ##c^{\dagger}_k c^{\dagger}_{-k} | 0 \rangle## represents a pair of fermions with momenta ##k,-k##. My goal is to solve for the coefficients ##f(k)## that determine the weight of each two-particle state in the superposition. Presumably, I could do this by computing
f(k) = \langle c_{-k}c_k | \Psi \rangle
However, I don't know what ##\Psi## is. I do know the form of the Hamiltonian for this system, though. Is it possible to find ##\Psi## if I only know the Hamiltonian ##H## and the form of the basis states ##c^{\dagger}_k c^{\dagger}_{-k} | 0 \rangle##? Thanks in advance! :)
 
Physics news on Phys.org
No, it's not possible without more information. There are infinite possibilities for ##f(k)## and each possibility corresponds to a different state.
 
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
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Back
Top