Wave Function from ground to excited

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SUMMARY

The discussion focuses on the derivation of the wavefunction for the first excited state of a quantum harmonic oscillator. The initial wavefunction for the ground state is given as ground_wavelength0 = (a/pi)^(1/4) e^(-ax^2/2), where a = alpha. The transition to the first excited state is represented as excited_wavelength1 = (a/pi)^(1/4) * sqrt(2) e^(-ax^2/2). The user seeks clarification on the mathematical steps involved in this derivation, referencing resources from HyperPhysics and NYU.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the quantum harmonic oscillator.
  • Familiarity with wavefunction notation and its significance in quantum physics.
  • Basic knowledge of mathematical functions, including exponentials and square roots.
  • Access to resources such as HyperPhysics for visual aids and explanations.
NEXT STEPS
  • Study the derivation of the quantum harmonic oscillator wavefunctions in detail.
  • Learn about the mathematical techniques used in quantum mechanics, such as differential equations.
  • Explore the implications of excited states in quantum systems and their physical interpretations.
  • Review additional resources on quantum mechanics, such as the UCSD notes on harmonic oscillators.
USEFUL FOR

Students and educators in physics, particularly those focusing on quantum mechanics, as well as researchers interested in the mathematical foundations of wavefunctions in quantum systems.

terp.asessed
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I now understand how the wavefunction graphs look from the hyperphysics: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html.

However, what I fail to understand, is how the wavefunction equation was derived for the 1st excited state, from:

ground_wavelength0 = (a/pi)1/4 e(-ax2/2), where a = alpha

to

excited_wavelength1 = (a/pi)1/4 21/2e(-ax2/2), where a = alpha

If someone could explain, merci beaucoup!
 
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Awesome, thank you!
 

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