Simple Harmonic(Quantum) Oscillator

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SUMMARY

The normalization constant A for the wave function of a simple harmonic oscillator is established as \( A = \left(\frac{m\omega_0}{\hbar\pi}\right)^{1/4} \). The wave function is given by \( \psi(x) = A e^{-ax^2} \), where \( a = \frac{\sqrt{km}}{2\hbar} \). The normalization condition requires integrating the probability density over the entire space, leading to the integral \( A^2 \int_{-\infty}^{\infty} e^{-2ax^2} dx = 1 \). The discussion highlights the importance of using the correct limits of integration and suggests that numerical methods like the trapezium rule and Simpson's Rule are insufficient for this problem.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically wave functions
  • Familiarity with normalization conditions in quantum mechanics
  • Knowledge of numerical integration techniques, including the trapezium rule and Simpson's Rule
  • Basic concepts of harmonic oscillators in physics
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  • Study the derivation of the normalization constant for quantum harmonic oscillators
  • Learn advanced numerical integration techniques suitable for quantum mechanics problems
  • Explore the properties of Gaussian integrals and their applications in quantum mechanics
  • Investigate the implications of boundary conditions in quantum systems
USEFUL FOR

Students and professionals in physics, particularly those focused on quantum mechanics, wave functions, and numerical methods for solving integrals. This discussion is beneficial for anyone looking to deepen their understanding of the simple harmonic oscillator model.

Hyperreality
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Using the normalization condition, show that the constant [tex]A[/tex] has the value [tex](\frac{m\omega_0}{\hbar\pi})^{1/4}[/tex].

I know from text the textbook that

[tex]\psi(x)=Ae^{-ax^2}[/tex]

where [tex]A[/tex] is the amplitdue and [tex]a=\frac{\sqrt{km}}{2\hbar}[/tex]

Here is my working:

Because the motion of the particle is confined to -A to +A, so the probability of finding the particle in the interval of -A to +A must be 1. Therefore the normalization condition is

[tex]\int_{-A}^{A}|\psi(x)^2| dx = 1[/tex]
[tex]A^2\int_{-A}^{A} e^{-2ax^2} dx = 1[/tex]

Here's where I'm stuck, this equation cannot be solved via integration techniques, it can only be solved using by numerical methods. I only know the "trapezium rule" and the "Simpson's Rule", I tried both of methods but nothing came up. Does this problem require some other numerical methods or is my normalization condition incorrect?
 
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Normalization is over the entire region, i.e. [tex]-\infty[/tex] to [tex]+\infty[/tex], the integral is then solved very easily.

[tex]A^2\int_{-\infty}^{\infty} e^{-2ax^2} dx = 1[/tex] for [tex]a = 1[/tex] do a change of variables to get the correct answer, hence the normalization factor.
 
Have you stated the problem correctly? It appears your limits of integration are the same as the amplitude which doesn't make sense because they have different units.
 

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