Does mean velocity equal group velocity of wave packets in QM?

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SUMMARY

The discussion centers on the relationship between mean velocity and group velocity of wave packets in quantum mechanics (QM). It establishes that for a particle in free space, the group velocity aligns with classical velocity as inferred from Ehrenfest's theorem. The mean velocity is defined as the sum of velocities weighted by their probabilities, while the velocity operator in the Schrödinger picture is expressed as $$\hat{\vec{v}}=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}},\hat{H}]$$. This leads to the conclusion that group velocity equals mean velocity, which also corresponds to classical velocity under specific conditions.

PREREQUISITES
  • Understanding of Ehrenfest's theorem in quantum mechanics
  • Familiarity with the concept of wave packets
  • Knowledge of the Schrödinger equation and operators
  • Basic principles of classical and quantum velocity
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  • Study the implications of Ehrenfest's theorem in various quantum systems
  • Explore the derivation and applications of the velocity operator in quantum mechanics
  • Investigate the differences between group velocity and phase velocity in wave mechanics
  • Learn about the role of probability distributions in quantum mechanics
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fxdung
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Does mean velocity of particle equal group velocity of wave packet in QM?If they do not equal which of them is classical velocity?
 
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It mean <v>=Sum of v*probability of given v
 
fxdung said:
Sum of v*probability of given v
##m \times## the expectation value of ##k##, the Fourier transform of ##x## ?
 
As your teaching Ehrenfest theorem: mxd/dt<r>=<P>,now I can understand group velocity equal mean velocity and equal classical velocity.
 
The expectation value is given by the operator of the quantity you want to calculate the expectation value for. The velocity operator (in the Schrödinger picture) is given by
$$\hat{\vec{v}}=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}},\hat{H}].$$
For a usual simple Hamiltonian
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2+V(\hat{x})$$
you get
$$\hat{\vec{v}}=\frac{1}{m} \hat{\vec{v}}.$$
 

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