Velocity of qm particle in a box

fk08
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Can I draw conclusions from the curvature of a wavefunction on the kin. energy of the particle? For instance, the lowest bound solution of a particle in the box is a half of sinus:
Psi(x) = sin(pi*x/L)

Since the second deriv. of a wavefunction is proportional to the kin. energy, this would imply that the highest kinetic energy of the particle is exactly in the middle of the box, and zero at the nodes:

d^2Psi/dx^2 = -sin(pi*x/L)
e.g for x = L---> d^2Psi/dx^2 = 0 = Ekin = p^2
 
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fk08 said:
Can I draw conclusions from the curvature of a wavefunction on the kin. energy of the particle? For instance, the lowest bound solution of a particle in the box is a half of sinus:
Psi(x) = sin(pi*x/L)

Since the second deriv. of a wavefunction is proportional to the kin. energy, this would imply that the highest kinetic energy of the particle is exactly in the middle of the box, and zero at the nodes:

d^2Psi/dx^2 = -sin(pi*x/L)
e.g for x = L---> d^2Psi/dx^2 = 0 = Ekin = p^2

In the infinitely deep square well (a box) the total energy is the kinetic energy and, therefore, \psi (x) = \sin \frac{{\pi x}}{L} is an eigenfunction of the kinetic energy operator corresponding to the kinetic energy eigenvalue E_1 = (KE)_1 = \frac{{\hbar ^2 \pi ^2 }}{{2mL^2 }}. In the ground state, then, a measurement of the kinetic energy always yields the value (KE)_1 = \frac{{\hbar ^2 \pi ^2 }}{{2mL^2 }}. The ground state energy is also the ground state kinetic energy, in this case. You assumed the kinetic energy to be proportional to the eigenfunction which is not correct.
 
In order to talk about a velocity one has to define it via an operator. It seems natural to define

{\bf v}=\frac{\bf p}{m}

or

{\bf v^2}=\frac{2{\bf E}_\text{kin}}{m}

Then one can calculate the velocity for an arbitrary state via

\langle v \rangle = \langle\psi|{\bf v}|\psi\rangle

or

\langle v^2 \rangle = \langle\psi|{\bf v^2}|\psi\rangle

respectively.

This shows that one cannot associate one piece of the wave function with velocity ore something like that. Classical entities are encoded in the whole state, not only in a single "piece" of its wave function.
 
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!
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