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Starting from $$(Δx) (Δp) \geq h/2$$ How does one derive $$a^2 (Δx)^2 + (Δp)^2 \geq a h~? $$

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    $\begingroup$ Never mind I just figured it out. Start with $$ [a (Δx) - (Δp)]^2 \geq 0 $$ Then just expand: $$ a^2(Δx)^2 + (Δp)^2 - 2 a (Δx) (Δp) \geq 0 $$ $$ a^2(Δx)^2 + (Δp)^2 \geq 2 a (Δx) (Δp) \geq 2 a h/2 = ah $$ $\endgroup$ – Tehol Beddict Jun 21 at 15:26
  • $\begingroup$ It is perfectly acceptable to answer your own question! Perhaps you should write one up! $\endgroup$ – march Jun 21 at 15:53

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