Fourier Transform of a wavefunction

children
Messages
1
Reaction score
0
Why shud one take the Fourier transform of a wavefunction and multiply the result with its conjugate to get the probability? Why can't it be Fourier transform of the probability directly?

thank you
 
Physics news on Phys.org
The wave function can't be a probability (or probability density) since it's complex. A probability must obviously be a real number between 0 and 1.

Also, you you only start by taking the Fourier transform if you're interested in the probability density of a certain value of the momentum. If you're interested in the probability density of a certain value of the position, you don't have to do a Fourier transform.
 
As Fredrick said, you don't take Fourier transform of a wave function in the process of finding the probability density. The probability density is given by (in one dimension):

P(x)=\int\psi (x)^*\psi (x) dx

which does not involve a Fourier Transform.

Instead, the Fourier transform of a wave function will give the wave function in momentum space (call it \phi). Again, as Fredrick mentioned, we can use this to find the probability density for the momentum of the particle:

P(p)=\int\phi (p)^*\phi (p) dp
 
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...
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
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...

Similar threads

Back
Top