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How might one show that $1\over 2m$$[\sigma_p^2+\mathbb E(p)^2]$ +$m\omega^2\over2$$[\sigma_x^2+\mathbb E(x)^2]$ $\geq \omega \sigma_x\sigma_p$?

The immediate thing I think would help is $(\sigma_p-\sigma_x)^2\geq0 \implies\sigma_p^2+\sigma_x^2\geq 2\sigma_p\sigma_x$. But then...?

Added: $m,\omega$ are positive constants. $\sigma_x,\sigma_p$ are standard deviations of $x,p$ respectively.

Thanks.

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What are $m$ and $\omega$? –  leonbloy Sep 2 '11 at 14:54
    
Symbols defined. :) –  H. V. Sep 2 '11 at 15:04
    
@H.V. I think the connection to standard deviation is not that relevant as far as this question is concerned. So I removed the [standard-deviation] tag and added the [algebra-precalculus] tag. Hope that's ok. –  Srivatsan Sep 2 '11 at 15:13
    
@Srivatsan: Sure! –  H. V. Sep 2 '11 at 17:01
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1 Answer

up vote 1 down vote accepted

I assume this is taken from physics textbook or so. In particular, I assume $m$ (for mass), $\omega$ (angular frequency?), as well as $\sigma_x$ and $\sigma_p$ (standard deviation terms) to be positive. The symbols are now defined.

Then the The given inequality follows by applying the AM-GM inequality to the nonnegative quantities $\frac{\sigma_p^2}{m}$ and $m \omega^2 \sigma_x^2$: $$ \frac{1}{2} \left( \frac{\sigma_p^2}{m} + m \omega^2 \sigma_x^2 \right) \geq \sqrt{\frac{\sigma_p^2}{m} \times m \omega^2 \sigma_x^2} = \omega \sigma_p \sigma_x. $$

Alternatively, you can also deduce this by expanding: $$ \frac{1}{2} \left( \frac{\sigma_p}{\sqrt{m}} - \sqrt{m} \cdot \omega \sigma_x \right)^2 \geq 0. $$

The two other terms involving $\mathrm{E}(x)^2$ and $\mathrm{E}(p)^2$ are nonnegative, so they do can be safely added to left hand side of the inequality.

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You are very right, this indeed has a Physics context. :) Thanks for the enlightenment. –  H. V. Sep 2 '11 at 15:07
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