# What's the proper way to approximate the position uncertainty of a particle?

In this problem:

shouldn't $\Delta x\sim\lambda/\sin\theta$ be $$\Delta x\sim \frac{\lambda}{\sin\theta} - \left(\frac{-\lambda}{\sin\theta}\right) = 2\frac{\lambda}{\sin\theta}$$ instead such that the final answer is $\Delta x \Delta p_x \sim 8\pi\hbar$?

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That's kinda beside the point, since $\lambda/\sin\theta\sim2\lambda/\sin\theta$ are of similar orders of magnitude. You get a different coefficient, but the important thing is that $\Delta p_x\sim c/\lambda$ where $c$ is "some constant"... –  Alex Nelson Nov 30 '12 at 5:55
Hi, Alex.Yes, I actually do get that it's about the product of uncertainties relating to some constant but, I just wanted to make sure that, given the drawing, it should have been Δx ~ 2λ/sinθ. Was your comment a confirmation that I am correct to say that Δx ~ 2λ/sinθ? –  Deniz Nov 30 '12 at 15:15
Well, if you want to know a more precise approximation, there are note online, also available here... –  Alex Nelson Nov 30 '12 at 16:29
Hi, again. I am aware that the specific math is Δx Δp_x >= ħ/2. All I am looking for is a confirmation from someone as to whether Δx ~ λ/sinθ should be Δx ~ 2λ/sinθ for this specific problem/image above (assuming I am right). Edit: I realize that, on the surface, my question seems trivial but it's important for me for reasons that I have trouble verbalizing. –  Deniz Nov 30 '12 at 16:49

As others have remarked, Heisenberg's microscope represents nothing more than a scaling argument for the position-momentum uncertainty relation. This means the argument should be expected to lead to an expression that is accurate no better than within factors of order unity.

Having said this, the optical (diffraction) limit for determining the blurring of a point source is well-defined and given by Abbe's criterion. See this Wikipedia page: http://en.wikipedia.org/wiki/Diffraction-limited_system .

As you notice, the radius of the spot size is determined by the wavelength divided by twice the sine of the half-angle. You can determine the full width of the spot size by doubling this expression (it seems this is what you are hinting at?), if you do so, you end up at the equation used by the author you quote.

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Thanks. (Sorry for the delayed response.) –  Deniz Dec 22 '12 at 23:21