Quantum physics - time dependant changes

Click For Summary
SUMMARY

The discussion focuses on a quantum mechanics problem involving a particle in a one-dimensional box that expands from width 'a' to '2a'. The key calculations involve determining the probability of the particle being in the new ground state immediately after the expansion and the required rapidity of the expansion for the calculations to hold. The energy of the ground state before the expansion is calculated as h²/8ma², while after the expansion it is h²/32ma². The probability is derived from the inner product of the original wavefunction and the new ground state wavefunction, which is identified as sqrt(1/2a) sin(πx/2a).

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically the Schrödinger equation.
  • Familiarity with wavefunctions and eigenstates in quantum systems.
  • Knowledge of inner products and probability calculations in quantum mechanics.
  • Concept of rapid expansion and its implications on wavefunction behavior.
NEXT STEPS
  • Study the implications of the Schrödinger equation on time-dependent wavefunctions.
  • Learn about the calculation of probabilities using inner products in quantum mechanics.
  • Explore the concept of eigenstates and their transformations during rapid changes in quantum systems.
  • Investigate the effects of boundary conditions on wavefunctions in quantum mechanics.
USEFUL FOR

Students and professionals in quantum mechanics, particularly those studying wavefunctions, eigenstates, and time-dependent changes in quantum systems.

david_clint
Messages
4
Reaction score
0

Homework Statement


a particle of mass m is in the groundstate of a 1D box with width a. the box then expands rapidly and symmetrically to width 2a. i). What is the probability that the particle is in the new groundstate immediately after the expansion? and ii). how rapid must this expansion be for your calculation to be valid?


Homework Equations


timescale for change must be hbar/ |E_alpha -E_ beta|
i think for probability you use that the statefunction evolves in time according to the Schrödinger equation and probability is the mod squared of the wave-function.


The Attempt at a Solution


  • pre expansion i got the energy of the groundstate as h^2/8ma^2
  • and after expansion h^2/32ma^2
  • the wavefunction is SQRT(PI/2a) cos (PI.x/a) before and after the expansion.

for part ii). i got (3/64PI) * (ma^2/h) for how rapid the expansion must be
for part i). i am stuck! [We were given the hint to use the fourth postulate to compute the probability that given the state-function (above), a measurement of energy will give the ground state energy of the expanded box (which i worked out above). ]
 
Physics news on Phys.org
david_clint said:
the wavefunction is SQRT(PI/2a) cos (PI.x/a) before and after the expansion.
But not everywhere! The wavefunction only takes that form within a certain domain of x. What is that domain, before the expansion?

For rapid expansion, you can assume that the wavefunction does not change during the expansion. But once the box expands, the eigenstates do change; even though the wavefunction was originally the ground state, it is no longer going to be the ground state. But it will have some component of the ground state in it, and you can figure out just how much by taking the inner product of the wavefunction with the new ground state. From that, you can find the probability.
 
diazona said:
But not everywhere! The wavefunction only takes that form within a certain domain of x. What is that domain, before the expansion?

For rapid expansion, you can assume that the wavefunction does not change during the expansion. But once the box expands, the eigenstates do change; even though the wavefunction was originally the ground state, it is no longer going to be the ground state. But it will have some component of the ground state in it, and you can figure out just how much by taking the inner product of the wavefunction with the new ground state. From that, you can find the probability.

The wavefunction has that form between minus and plus a/2 i think.
Is the new ground state wavefunction: sqrt(1/a) sin (PI. x/2a).
which i then multiply by the 'original' wavefunction and integrate. Also what will then be the integration limits: minus and plus a/2?
Then mod squared of the inner product would give the probability. Thanks
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K