Normalization of Hydrogen wavefunction

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SUMMARY

The discussion focuses on the normalization of the (1 0 0) and (2 0 0) wave functions of the hydrogen atom. The wave functions are given as (100) = (2/a^(3/2)) exp^(-r/a) and (200) = (1/((2a)^(3/2))*(2-r/a) exp^(-r/2a). It is established that normalization does not require the use of the momentum operator, as the definition of normalization is sufficient for this task. Participants are encouraged to review the concept of wave function normalization to understand the process fully.

PREREQUISITES
  • Understanding of quantum mechanics principles
  • Familiarity with hydrogen atom wave functions
  • Knowledge of spherical coordinates in quantum mechanics
  • Basic grasp of normalization in wave functions
NEXT STEPS
  • Review the definition of wave function normalization in quantum mechanics
  • Study the mathematical derivation of hydrogen atom wave functions
  • Learn about spherical coordinates and their application in quantum mechanics
  • Explore the implications of wave function normalization on physical predictions
USEFUL FOR

Students of quantum mechanics, physicists working with atomic models, and anyone studying the properties of wave functions in quantum systems will benefit from this discussion.

rini
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Homework Statement



Show that the (1 0 0) and (2 0 0) wave functions of hydrogen atom are properly normalized.

Homework Equations



I know that (n l ml):

(100) = (2/a^(3/2)) exp^ (-r/a)

(200) = (1/((2a)^(3/2))*(2-r/a) exp^(-r/2a)


The Attempt at a Solution



I started with H=-ke^2/r+p^2/2m and don't know how to convert the p operator into spherical coordinates
 
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rini said:

Homework Statement



Show that the (1 0 0) and (2 0 0) wave functions of hydrogen atom are properly normalized.

Homework Equations



I know that (n l ml):

(100) = (2/a^(3/2)) exp^ (-r/a)

(200) = (1/((2a)^(3/2))*(2-r/a) exp^(-r/2a)


The Attempt at a Solution



I started with H=-ke^2/r+p^2/2m and don't know how to convert the p operator into spherical coordinates

(100) = (2/a^(3/2)) exp^ (-r/a) is the wave function. What does it mean for it to be normalized?
 
rini said:
I started with H=-ke^2/r+p^2/2m and don't know how to convert the p operator into spherical coordinates

You don't need the p operator, or indeed any operator, in order to normalize the wave function. Look up the definition of what it means for a wave function to be "normalized."
 

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