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In summary, the Gaussian Integral (eqn 2.64) involves a constant of normalization that is rationalized by changing the sign of the exponent and squaring the quantity twice. This is done in order to satisfy the requirement that the integral of the probability density function from -inf to +inf equals 1. The momentum distribution, which corresponds to a normal distribution, is related to the position distribution by their standard deviations. In order to understand the process better, someone can show the steps to transform the equation and point out where any mistakes were made.

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Hey SteamKing, thanks for stopping by to help. I'm not referring to the normalisation process here, it's the constant on the exponential I'm having trouble with.. The equation shows a rationalisation of the numerator to give (2∏(hbar)^2 / 4σ^2)^-1/4.

From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps I'm doing something wrong here?

From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps I'm doing something wrong here?

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jtbell i apologise for any confusion here but I'm not referring to the normalisation process, it's the constant on the exponential. The equation shows a rationalisation of the numerator to give 1 /(2∏(hbar)^2 / 4σ^2)^1/4.

From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps I'm doing something wrong here?

From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps I'm doing something wrong here?

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[itex] h = 2\pi \hbar[/itex]

[itex] \sqrt[4]{\frac{16\sigma^4\pi^2}{h^2 (2\pi \sigma^2)}} = \sqrt[4]{\frac{16\pi^2\sigma^2}{2\pi h^2}} = \sqrt[4]{\frac{16\sigma^2\pi^2}{2\pi\left(2\pi \hbar\right)^2}}[/itex]

[itex] = \sqrt[4]{\frac{16\sigma^2}{2\pi 4\hbar^2}} = \sqrt[4]{\frac{4\sigma^2}{2\pi \hbar^2}} = \frac{1}{\sqrt[4]{2\pi \left(\hbar^2/(4\sigma^2) \right)}}[/itex]

The momentum distribution does not describe a definite (well defined) momentum but rather corresponds to the case that both position and momentum are normal distributions, and this equation shows the relation between their standard deviations.

[itex] \sqrt[4]{\frac{16\sigma^4\pi^2}{h^2 (2\pi \sigma^2)}} = \sqrt[4]{\frac{16\pi^2\sigma^2}{2\pi h^2}} = \sqrt[4]{\frac{16\sigma^2\pi^2}{2\pi\left(2\pi \hbar\right)^2}}[/itex]

[itex] = \sqrt[4]{\frac{16\sigma^2}{2\pi 4\hbar^2}} = \sqrt[4]{\frac{4\sigma^2}{2\pi \hbar^2}} = \frac{1}{\sqrt[4]{2\pi \left(\hbar^2/(4\sigma^2) \right)}}[/itex]

The momentum distribution does not describe a definite (well defined) momentum but rather corresponds to the case that both position and momentum are normal distributions, and this equation shows the relation between their standard deviations.

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lukka said:it doesn't work out.

If you show us explicitly what you thought it should work out to be, in stepwise fashion, someone can probably tell you what your mistake is.

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(2σ √∏) / √h(2∏σ^2)^1/4 = (2∏(hbar)^2 /

here's the algebra i did..

(2σ √∏) / √h(2∏σ^2)^1/4 = √∏ / (2∏/16σ^2)^1/4 ..

..=√h (2∏/16σ^2 ^∏^2)^-1/4 = (2∏(hbar)^2 /

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i can see where i going wrong with this now.. thanks Dauto!

Gaussian Integrals for Quantum States of well Defined Momentum are mathematical techniques used in quantum mechanics to calculate the probability of a particle being in a certain state with a defined momentum. They involve integrating over a Gaussian distribution, which represents the wave function of the particle.

Gaussian Integrals are used in quantum mechanics to calculate the probability amplitudes of quantum states with well-defined momentum. They are also used to analyze the behavior of particles in quantum systems and understand the dynamics of quantum systems.

In quantum mechanics, the momentum of a particle is considered one of its fundamental properties. A well-defined momentum means that the particle has a definite momentum value and its wave function is sharply peaked at that value. This is important for understanding the behavior of particles in quantum systems.

Yes, Gaussian Integrals can be used for particles with non-zero momentum. In fact, they are commonly used to calculate the probability of a particle being in a state with a specific momentum value, regardless of whether it is zero or non-zero.

While Gaussian Integrals are a powerful tool in quantum mechanics, they do have limitations. They may not accurately describe systems with strong interactions or high energies. Additionally, they are not suitable for systems with multiple particles or entangled states.

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