Quantum particle in a 2 dimensional box

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SUMMARY

The discussion centers on the quantum mechanics of a particle in a two-dimensional box, specifically addressing the solutions for wave functions based on the parity of quantum numbers. The solutions for odd and even quantum numbers arise from the symmetry properties of the wave functions around the origin, with the ground state exhibiting zero nodes and the first excited state exhibiting one node. The wave function can be expressed as ψnm = C sin(nπ(x - (-A))/(A - (-A))) sin(mπ(y - 0)/(B - 0)), highlighting the symmetry in both x and y directions.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically wave functions
  • Familiarity with the concept of parity in quantum systems
  • Knowledge of trigonometric identities and their application in physics
  • Basic grasp of two-dimensional potential wells in quantum mechanics
NEXT STEPS
  • Study the implications of wave function symmetry in quantum mechanics
  • Explore the concept of quantum numbers and their significance in particle states
  • Learn about the mathematical derivation of wave functions in two-dimensional boxes
  • Investigate the role of nodes in quantum states and their physical interpretations
USEFUL FOR

Physics students, quantum mechanics enthusiasts, and educators looking to deepen their understanding of wave functions and quantum states in two-dimensional systems.

Unskilled
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Homework Statement


I need some help :cry:

http://www.fysik.uu.se/kurser/1tt306/filer/TentKF06+short-answers.pdf

On task 4 a) i don't understand why they have two solutions, one for odd n and the other for even n.
 
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Unskilled said:

Homework Statement


I need some help :cry:

http://www.fysik.uu.se/kurser/1tt306/filer/TentKF06+short-answers.pdf

On task 4 a) i don't understand why they have two solutions, one for odd n and the other for even n.

Hi there, I also study @ Uppsala ;) physics..

It has to do with the parity of the wave functions. Look on the symmetries around x=0. Ground state has zero nodes, 1st excited has one etc. Skiss the potetial and search for symmetry points.
 
Last edited by a moderator:
Note there's no qualititative difference between the x and y directions. You could write the wavefunction more symmetrically as:

[tex]\psi_{nm} = C \sin \left( n \pi \frac{x-(-A)}{A-(-A)} \right) \sin \left( m\pi \frac{y-0}{B-0}\right)[/tex]

They just used some trig identities to rewrite this.
 
Last edited:

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