Calc group velocity and wave velocity for a wave

Click For Summary
SUMMARY

The discussion focuses on calculating group velocity and wave velocity for a wave function represented as \(\Psi=Ae^{i(px-Et)/\hbar}\), where \(p\) is momentum and \(E=p^2/2m\). The key formulas used are group velocity \(v_g = \frac{dw}{dk}\) and wave velocity \(v = \frac{w}{k}\). Participants clarify that the transformation from momentum to wave number \(k = \frac{p}{\hbar}\) and energy to angular frequency \(w = \frac{E}{\hbar}\) is derived from the general equation for momentum eigenstates and Schrödinger's time-independent equation. The discussion concludes that the formulas are established and not arbitrary.

PREREQUISITES
  • Understanding of wave functions in quantum mechanics
  • Familiarity with Schrödinger's equation
  • Knowledge of momentum and energy relationships in physics
  • Basic calculus for differentiation (dw/dk)
NEXT STEPS
  • Study the derivation of Schrödinger's time-independent equation
  • Learn about momentum eigenstates and their significance in quantum mechanics
  • Explore the relationship between wave number \(k\) and momentum \(p\)
  • Investigate the implications of group velocity in wave mechanics
USEFUL FOR

Students of quantum mechanics, physicists studying wave phenomena, and educators teaching advanced physics concepts will benefit from this discussion.

leonne
Messages
163
Reaction score
0

Homework Statement


a wave function \Psi=Aei(px-Et)/h where p is momentum and E=p2/2m calc group and wave velocity

Homework Equations


group vel=dw/dk
wave v=w/k

The Attempt at a Solution


I just have no idea what they did here to solve it.
i(px-Et)/h=i(kx-wt) than somehow got k=p/h than w=E/h than just used these to solve it but i don't get how they got it. Are these formulas? didn't seem them in the book. I kind of see how they got it. Did they just make it so the right side matches the left? Funny, as soon as i finish typing a problem here, i seem to figure it out.
 
Physics news on Phys.org
No, the formula wasn't just made up. e^(ipx/h) is the general equation for the momentum eigenstates, and since p can be any real number, it's also the general equation for a plane wave. According to Schrödinger's time-independent equation, the eigenstates' time evolution should be e^(-iEt/h), so the time-dependent wavefunction is e^(ipx/h)*e^(-iEt/h).
 
o ok thanks for the help
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 41 ·
2
Replies
41
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K