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- Thread starter lukka
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SteamKing

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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

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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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jtbell

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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!

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