Gaussian Integrals for Quantum States of well Defined Momentum

• lukka
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.

lukka

Consider the Gaussian Integral (eqn 2.64).. is anyone able to explain how the constant of normalization is rationalised?

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If by 'rationalized' you mean 'normalized' so that the integral of the pdf from -inf to + inf = 1, I would think in the usual fashion.

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?

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It comes from normalizing ##\langle p | \psi \rangle##, that is, requiring that ##| \langle p | \psi \rangle |^2 = 1##.

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?

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$h = 2\pi \hbar$

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

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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1 person
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.

Equation 2.64 implies that the normalisation constants are equivalent. that is..

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

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 / 16σ^2)^-1/4.

$\frac{2\sigma \pi^{1/2}}{h^{1/2}(2\pi \sigma^2)^{1/4}}=\left [\frac{16\sigma^4\pi^2}{2\pi h^2\sigma^2}\right ]^{1/4}=\left [\frac{16\sigma^2\pi^2}{2\pi (2\pi\hbar)^2}\right ]^{1/4}=\left [\frac{16\sigma^2\pi^2}{8\pi^3 \hbar^2}\right ]^{1/4}=\left[\frac{4\sigma^2}{2\pi \hbar^2}\right ]^{1/4}$

1 person
Hey that's great! i never thought of transforming it that way.. i guess what i was attempting to do at first before was nonsensical although I'm not sure why, it seems they should have revealed the same? I'm sure I'm on the right track with this now. Thanks for going to the time to show me where i was going wrong..

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