Simple Harmonic(Quantum) Oscillator

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The discussion focuses on determining the normalization constant A for the wave function of a simple harmonic oscillator, given by ψ(x) = Ae^{-ax^2}. The normalization condition requires that the integral of the probability density over the entire space equals one, leading to the equation A^2∫_{-∞}^{∞} e^{-2ax^2} dx = 1. The original poster mistakenly limited the integration to the amplitude range of -A to +A, which is incorrect as normalization should cover the entire range from -∞ to +∞. The correct approach involves recognizing that the integral can be solved easily using known results for Gaussian integrals, leading to the conclusion that A has the value (\frac{m\omega_0}{\hbar\pi})^{1/4}. The discussion highlights the importance of proper limits in normalization for quantum mechanics.
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Using the normalization condition, show that the constant A has the value (\frac{m\omega_0}{\hbar\pi})^{1/4}.

I know from text the textbook that

\psi(x)=Ae^{-ax^2}

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

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

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

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. -\infty to +\infty, the integral is then solved very easily.

A^2\int_{-\infty}^{\infty} e^{-2ax^2} dx = 1 for a = 1 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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