Symmetric and Asymmetric Electron Wave Functions in a Proton-Electron System

Click For Summary
SUMMARY

The discussion focuses on the derivation of normalization constants and energy expressions for symmetric and asymmetric electron wave functions in a proton-electron system. The symmetric wave function is defined as ψs = Ns(ψ1 + ψ2) and the asymmetric wave function as ψa = Na(ψ1 - ψ2). The normalization constants are derived as Ns = (1/√2)[(∫|ψ1|^2dτ + ∫|ψ2|^2dτ)^(-1/2)] and Na = (1/√2)[(∫|ψ1|^2dτ - ∫|ψ2|^2dτ)^(-1/2)]. Energy expressions are calculated using the Hamiltonian, resulting in Es and Ea for symmetric and asymmetric states, respectively.

PREREQUISITES
  • Quantum Mechanics fundamentals
  • Understanding of wave functions and normalization
  • Familiarity with Hamiltonian operators
  • Knowledge of integrals in quantum mechanics
NEXT STEPS
  • Study the derivation of Hamiltonian operators in quantum systems
  • Explore normalization techniques for multi-particle wave functions
  • Learn about the implications of symmetric and asymmetric states in quantum mechanics
  • Investigate energy calculations using the variational principle
USEFUL FOR

Students and researchers in quantum mechanics, particularly those studying multi-particle systems and wave function behavior in atomic physics.

helpcometk
Messages
71
Reaction score
0
1. Homework Statement [/b]

1. Consider two protons, separated by a distance R, and one electron. Assume
that the electron wave function can be symmetric, s, or asymmetric, a, with:
ψs = Ns( ψ1 + ψ2);
ψa = Na( ψ1 - ψ2):
Here 1 and 2 are the electron wave functions for large R when bound with
proton 1 or proton 2, respectively
a) Derive a expression for the normalisation constants Ns and Na. [6 Marks]
b) Assuming the Hamiltonian is given by:

look image attachment

where r1 and r2 are the distances to proton 1 and proton 2, respectively.
Derive an expression for the energies Es and Ea.

Homework Equations



for a) i found a solution here : http://farside.ph.utexas.edu/teaching/qmech/lectures/node129.html#e14.57

is this correct ?

im still confused with energues Ea and Eb

The Attempt at a Solution

 

Attachments

Physics news on Phys.org
a) Ns = (1/√2)[(∫|ψ1|^2dτ + ∫|ψ2|^2dτ)^(-1/2)] Na = (1/√2)[(∫|ψ1|^2dτ - ∫|ψ2|^2dτ)^(-1/2)] b) Es = <ψs|H|ψs> = =N^2s[∫|ψ1|^2H1dτ + ∫|ψ2|^2H2dτ + 2∫ψ1*H12*ψ2dτ]Ea = <ψa|H|ψa> = =N^2a[∫|ψ1|^2H1dτ + ∫|ψ2|^2H2dτ - 2∫ψ1*H12*ψ2dτ]
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 19 ·
Replies
19
Views
19K
Replies
9
Views
4K
  • · Replies 48 ·
2
Replies
48
Views
10K